Power Electronics Pune University MCQs
Power Electronics Pune University MCQs
This set of Power Electronics Multiple Choice Questions & Answers focuses on “The P-N Junction”.
1. The p-region has a greater concentration of __________ as compared to the n-region in a P-N junction.
a) holes
b) electrons
c) both holes & electrons
d) phonons
Answer: a
Explanation: Holes are the majority charge carriers in p-type material.
2. A p-type semiconductor material is doped with ____________ impurities whereas a n-type semiconductor material is doped with __________ impurities
a) acceptor, donor
b) acceptor, acceptor
c) donor, donor
d) donor, acceptor
Answer: a
Explanation: Donor impurities denote an electron to the n-type material making it a electron majority carrier & vice-versa.
3. In the p & n regions of the p-n junction the _________ & the ___________ are the majority charge carriers respectively.
a) holes, holes
b) electrons, electrons
c) holes, electrons
d) electrons, holes
Answer: c
Explanation: Holes are the majority charge carriers in p-type material & vice-versa.
4. The n-region has a greater concentration of _________ as compared to the p-region in a P-N junction diode.
a) holes
b) electrons
c) both holes & electrons
d) phonons
Answer: b
Explanation: Electrons are the majority charge carriers in n-type material.
5. Which of the below mentioned statements is false regarding a p-n junction diode?
a) Diode are uncontrolled devices
b) Diodes are rectifying devices
c) Diodes are unidirectional devices
d) Diodes have three terminals
Answer: d
Explanation: Diode is a two terminal device, anode & cathode are the two terminals.
6. In the p & n regions of the p-n junction the _________ & the ___________ are the minority charge carriers respectively.
a) holes, holes
b) electrons, electrons
c) holes, electrons
d) electrons, holes
Answer: d
Explanation: Holes are the minority charge carriers in n-type material & vice-versa.
7. Lets assume that the doping density in the p-region is 10 -9 cm -3 & in the n-region is 10 -17 cm -3 , as such the p-n junction so formed would be termed as a
a) p – n –
b) p + n –
c) p – n +
d) p + n +
Answer: b
Explanation: Doping density is greater in the p-region compared to the n-region.
8. When a physical contact between a p-region & n-region is established which of the following is most likely to take place?
a) Electrons from N-region diffuse to P-region
b) Holes from P-region diffuse to N-region
c) Both of the above mentioned statements are true
d) Nothing will happen
Answer: c
Explanation: When p & n region come together diffusion takes places & a depletion region is established with opposite charges on both the sides of the junction.
9. Which of the following is true in case of an unbiased p-n junction diode?
a) Diffusion does not take place
b) Diffusion of electrons & holes goes on infinitely
c) There is zero electrical potential across the junctions
d) Charges establish an electric field across the junctions
Answer: d
Explanation: A potential difference is established across the junctions due to recombination of holes & electrons. This growing filed stops the further diffusion.
10. Which of the following is true in case of a forward biased p-n junction diode?
a) The positive terminal of the battery sucks electrons from the p-region
b) The positive terminal of the battery injects electrons into the p-region
c) The negative terminal of the battery sucks electrons from the p-region
d) None of the above mentioned statements are true
Answer: a
Explanation: The diode is forward biased, positive is connected to p & vice-versa, as such batter provides EMF to drive electrons from n-region to p-region.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Power Diode – 1”.
1. An ideal power diode must have
a) low forward current carrying capacity
b) large reverse breakdown voltage
c) high ohmic junction resistance
d) high reverse recovery time
Answer: b
Explanation: Large reverse breakdown voltage is desirable whereas others will increases the losses.
2. To make a signal diode suitable for high current & high voltage carrying applications with minimum losses, ________
a) a lightly doped n layer is grown between the two p & n layers
b) a heavily doped n layer is grown between the two p & n layers
c) a lightly doped p layer is grown between the two p & n layers
d) a heavily doped p layer is grown between the two p & n layers
Answer: a
Explanation: The above process simply the one used to manufacture power diodes.
3. Power diode is __________
a) a three terminal semiconductor device
b) a two terminal semiconductor device
c) a four terminal semiconductor device
d) a three terminal analog device
Answer: b
Explanation: It has two terminals anode and cathode same as that of a ordinary diode. In fact, a power diode is nothing but a signal diode with a extra layer.
4. The V-I Characteristics of the diode lie in the
a) 1st & 2nd quadrant
b) 1st & 3rd quadrant
c) 1st & 4th quadrant
d) Only in the 1st quadrant
Answer: b
Explanation: First in the forward region & Third in the reverse biased mode.
5. Which of the following is true in case of a power diode with R load?
a) I grows almost linearly with V
b) I decays almost linearly with V
c) I is independent of V
d) I initial grows than decays
Answer: a
Explanation: R load therefore V and I are linear and in phase.
6. A diode is said to be reversed biased when the
a) cathode is positive with respect to the anode
b) anode is positive with respect to the cathode
c) cathode is negative with respect to the anode
d) both cathode & anode are negative
Answer: a
Explanation: K is positive w.r.t the A when the device is reversed biased.
7. A diode is said to be forward biased when the
a) cathode is positive with respect to the anode
b) anode is positive with respect to the cathode
c) anode is negative with respect to the anode
d) both cathode & anode are positive
Answer: b
Explanation: A is positive w.r.t the K when the device is forward biased.
8. In case of an ideal power diode, the leakage current flows from
a) anode to cathode
b) cathode to anode
c) in both the directions
d) leakage current does not flow
Answer: d
Explanation: Leakage current does not flow in IDEAL diode.
9. The peak inverse current I P for a power diode is given by the expression
a) I P =t + di/dt
b) I P =t * log i
c) I P =t * di/dt
d) I P =t * ∫ t*i dt
Answer: c
Explanation: The leakage current is the reveres recovery time into the rate of change of current.
10. A power diode with small softness factor has
a) small oscillatory over voltages
b) large oscillatory over voltages
c) large peak reverse current
d) small peak reverse current
Answer: b
Explanation: Peak reverse current is independent of S-factor smaller the value of S-factor larger the oscillatory over voltage.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Power Diode-2”.
1. If V & I are the forward voltage & current respectively, then the power loss across the diode would be
a) V/I
b) V 2 I 2
c) I 2 V
d) VI
Answer: d
Explanation: Simply power is voltage into current i.e. VI
2. The power loss in which of the following cases would be the maximum?
a) When both V & I are minimum
b) When both V & I are maximum
c) When only V is maximum
d) When only I is maximum
Answer: b
Explanation: P=VI Hence, it would be maximum when both V and I are maximum.
3. Even after the forward current reduces to zero value, a practical diode continues to conduct in the reverse direction for a while due to the
a) resistance of the diode
b) high junction temperature
c) stored charges in the depletion region
d) none of the mentioned
Answer: c
Explanation: Due to the stored charges during the earlier current flow, even when the current reduces to zero due to some structural properties of the device, the device takes time to sweep out the stored charges.
4. For a p-n junction diode, the peak inverse current & the reverse recovery time are dependent on
a) inverse voltage
b) forward Voltage
c) di/dt
d) all of the above mentioned
Answer: c
Explanation: It only depends upon the number stored charges which depends upon the rate of change of current.
5. In an AC-DC converter, a diode might be used as a
a) voltage source
b) phase angle controller
c) freewheeling Diode
d) filter
Answer: c
Explanation: In converters diodes are used to feed the energy back to the load in case of an inductive load.
6. When the p-n junction diode is forward biased, the width of the depletion region __________
a) increases
b) decreases
c) remains Constant
d) increases than Decreases
Answer: b
Explanation: When forward biased depletion layer decreases & finally it collapses to allow the current flow.
7. When the p-n junction diode is reversed biased, the width of the depletion region __________
a) increases
b) decreases
c) remains Constant
d) none of the above mentioned
Answer: a
Explanation: When reverse biased depletion layer increases until the breakdown value is reached.
8. In case of a practical p-n junction diode, the rise in the junction temperature ___________
a) decreases the width of the depletion region
b) increases the barrier potential
c) increases the width of the depletion region
d) width of the depletion region increases but the barrier potential remains constant
Answer: a
Explanation: The rise in temperature excites the charges, which go & recombine with the charges in the depletion region decreasing its width. Higher the temperature, lesser is the E.M.F required to turn on the device.
9. In the equilibrium state, the barrier potential across a unbiased silicon diode is _________
a) 0.3 V
b) 0.7 V
c) 1.3 V
d) 0 V
Answer: b
Explanation: Barrier potential is due to the charges that establish an electric field across the two junctions.
10. In the equilibrium state the barrier, potential across a unbiased germanium diode is __________
a) 0.3 V
b) 0.7 V
c) 1.7 V
d) 0 V
Answer: a
Explanation: Barrier potential is due to the charges that establish an electric field across the two junctions.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Types of Power Diodes-1”.
1. Shown below is the diagram of an ideal super diode. When the input voltage Vin is negative, then the output voltage Vout = ?
power-electronics-mcqs-types-power-diodes-1-q1
a) Vout/Vin
b) Vin
c) 0
d) Vin * Rl
Answer: c
Explanation: When the input voltage is negative, there would be a negative voltage on the diode so it works like an open circuit. Hence no current flows through the load and Vout is zero.
2. In order to reduce the reverse recovery time of the diodes, __________ is carried out.
a) shortening of the length of the device
b) platinum & gold doping
c) antimony doping
d) adding an extra silicon layer
Answer: b
Explanation: Platinum & gold doping improves the performance of the devices.
3. Which of the following diodes uses a metal-semiconductor junction?
a) General purpose diodes
b) Fast recovery diodes
c) Schottky diode
d) None of the mentioned
Answer: c
Explanation: Schottky diode uses a Al-Semiconductor junction.
4. Which of the below mentioned statements is false regarding Schottky diodes?
a) Schottky diodes have a Al-Silicon junction
b) There is no storage of charges in a Schottky diode
c) The majority charge carriers in a Schottky diode are holes
d) Schottky diodes can be switched off faster than a p-n junction diode of the same rating
Answer: c
Explanation: The majority charge carriers in a Schottky diode are electrons not holes.
5. A Schottky diode _____
a) has current flow due to holes only
b) has no reverse recovery time
c) has large amount of storage charges
d) has zero cut-in voltage
Answer: b
Explanation: Due to the metal-silicon junction there are no stored charges hence, no reverse recovery time, due to which the switching is faster.
6. Which of the following are/is the majority charge carriers in a Schottky diode?
a) Holes
b) Electrons
c) Both holes & Electrons carry equal current
d) None of the mentioned
Answer: b
Explanation: The metal has no holes hence major current flows due to the electrons only.
7. In a Schottky diode, the silcon is usually
a) N-type
b) P-type
c) un-doped semiconductor
d) silicon is not used
Answer: a
Explanation: Usually only n-type silicon is used because the p-type has certain limitations.
8. As compared to a p-n junction diode, a Schottky diode has ___________
a) higher cut-in voltage
b) lower reverse leakage current
c) higher operating frequency
d) higher switching time
Answer: c
Explanation: Due to the metal-silicon junction there are no stored charges, hence no reverse recovery time due to which the switching is faster.
9. A Schottky diode has __________
a) a gate terminal
b) aluminum-silicon junction
c) platinum gold junction
d) germanium-Arsenide junction
Answer: b
Explanation: Schottky diode uses a Al-Semiconductor junction.
10. A Schottky diode can be switchd off much faster than an equivalent p-n junction diode due to its
a) higher operating frequency
b) no recombination of charges
c) more compact structure
d) None of the mentioned
Answer: b
Explanation: Due to the metal-silicon junction there are no stored charges, hence no reverse recovery time due to which the switching is faster.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Types of Power Diodes-2”.
1. Ideally the voltage drop across a conducting diode must be
a) ∞
b) 0
c) higher than the forward biased voltage
d) equal to the forward biased voltage
Answer: b
Explanation: Ideal conduction = No losses.
2. When reverse breakdown occurs in a diode
a) voltage increases & current is constant
b) voltage increases & current also increases
c) both are constant
d) voltage is constant & current increases
Answer: d
Explanation: Recall the I-V curve of a diode in the 3rd quadrant.
3. Schottky diodes are also called as
a) metal diode
b) hot carrier diode
c) signaling diode
d) easy turn on diode
Answer: b
Explanation: Due to the metal used to carry the current, it is also called as a hot carrier diode.
4. In a Schottky diode, the aluminum metal acts as a __________
a) anode
b) cathode
c) gate
d) common terminal
Answer: a
Explanation: The Al always acts as the anode.
5. If the doping levels of the semiconductor is increased, then the width of the depletion layer
a) increases
b) decreases
c) is unchanged
d) keeps oscillating
Answer: b
Explanation: Higher the doping, more the number of charge carrier available to neutralize the opposite charges on the junction.
6. As compared to a p-n junction device of equal rating, the Schottky diode has
a) lower reverse voltage rating
b) lower reverse leakage current
c) higher Switching time
d) higher cut-in voltage
Answer: a
Explanation: Low reverse voltage rating is the only drawback against a p-n junction diode.
7. In a Schottky diode, the silicon layer acts as a _____________
a) anode
b) cathode
c) gate
d) common terminal
Answer: b
Explanation: The metal acts as the anode and the semiconductor as a cathode.
8. In a certain power electronics application, it is required that the voltage at the load terminals is to be kept within a certain range of voltages only. Among the device listed below, which would be the most ideal choice for this application?
a) P-n junction diode
b) Schottky diode
c) Zener diode
d) Fast recover diode
Answer: c
Explanation: Zener diode is used as a voltage regulating device.
9. Shown below is the diagram of an ideal super diode. When the input voltage Vin is positive, then the output voltage Vout = ?
power-electronics-mcqs-types-power-diodes-2-q9
a) Vout/Vin
b) Vin
c) 0
d) Vin * Rl
Answer: b
Explanation: Input is positive, hence it is amplified by the operational amplifier which switches the diode on. Current flows through the load & because of the feedback, the output voltage is equal to the input voltage.
10. Zener diodes allow a current to flow in the reverse direction, when the
a) voltage reaches above a certain value
b) temperature reaches above a certain value
c) current always flows in the reverse direction only
d) current cannot flow in the reverse direction
Answer: a
Explanation: Zener diode has voltage regulating property. When voltage reaches above a certain value , current starts to flow in the reverse direction.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Power Transistors – 1”
1. Which of the following devices does not belong to the transistor family?
a) IGBT
b) MOSFET
c) GTO
d) BJT
Answer: c
Explanation: GTO is gate turn off transistor, it belongs to the Thyristor family. All the other devices belong to the transistor family.
2. A power transistor is a
a) three layer, three junction device
b) three layer, two junction device
c) two layer, one junction device
d) four layer, three junction device
Answer: b
Explanation: It has three layers p-n-p or n-p-n forming two p-n junctions.
3. In a power transistor, ____ is the controlled parameter.
a) V BE
b) V CE
c) I B
d) I C
Answer: d
Explanation: The collector current is the controlled parameter.
4. A power transistor is a _________ device.
a) two terminal, bipolar, voltage controlled
b) two terminal, unipolar, current controlled
c) three terminal, unipolar, voltage controlled
d) three terminal, bipolar, current controlled
Answer: d
Explanation: Power transistor is simply many BJT’s connected in series parallel on a single silicon chip for power applications. It is a three terminal, bipolar, current controlled device.
5. In a power transistor, _________ is the controlling parameter.
a) V BE
b) V CE
c) I B
d) I C
Answer: c
Explanation: The base current controls the collector current. Hence, the base current Ib is the controlling parameter.
6. In a power transistor, the I B vs V BE curve is
a) a parabolic curve
b) an exponentially decaying curve
c) resembling the diode curve
d) a straight line Y = I B
Answer: c
Explanation: The B-E junction of a BJT resembles a p-n junction diode, hence the curve.
7. For a power transistor, if the base current I B is increased keeping V CE constant, then
a) I C increases
b) I C decreases
c) I C remains constant
d) none of the mentioned
Answer: a
Explanation: Ic is directly proportional to Ic.
8. The forward current gain α is given by
a) I C /I B
b) I C /I E
c) I E /I C
d) I E /I B
Answer: b
Explanation: Collector current by emitter current is the current gain, its value is close to one but never greater than.
9. The value of β is given by the expression
a) I C /I B
b) I C /I E
c) I E /I C
d) I E /I B
Answer: a
Explanation: Collector current by the base current is beta, its value is in the range 50 to 300.
10. A power BJT is used as a power control switch by biasing it in the cut off region or in the saturation region . In the on state
a) both the base-emitter & base-collector junctions are forward biased
b) the base-emitter junction is reverse biased, and the base collector junction is forward biased
c) the base-emitter junction is forward biased, and the base collector junction is reversed biased
d) both the base-collector & the base-emitter junctions are reversed biased
Answer: a
Explanation: When base-emitter & base-collector junctions are forward biased only than both the p-n junctions are forward biased and the device is on.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Power Transistors-2”.
1. For a power transistor, if the forward current gain α = 0.97, then β = ?
a) 0.03
b) 2.03
c) 49.24
d) 32.33
Answer: d
Explanation: Use the relation α = β/.
2. The power electronics devices have a very high efficiency because
a) cooling is very efficient
b) the devices traverse active region at high speed & stays at the two states, on and off
c) the devices never operate in active region
d) the devices always operate in the active region
Answer: b
Explanation: They are efficient due to their higher transition speeds.
3. For a power transistor, which of the following relations is true?
a) Ie>Ic>Ib
b) Ib>Ic>Ie
c) Ic>Ie>Ib
d) Ie=Ib
Answer: a
Explanation: Practically speaking Ie = Ib+Ic. Ie is the highest as it is the sum of the collector and base currents. The base current is the smallest.
4. High frequency operation of any device is limited by the
a) forward voltage rating
b) switching losses
c) thermal conductivity
d) heat Sink arrangements
Answer: b
Explanation: Lower the switching losses higher the frequency of operation of the device.
5. The instantaneous power loss during the delay time of a transistor is given by
a) Ic Vce
b) Ib Vbe
c) Ic Vbe
d) Ib Vce
Answer: a
Explanation: During the delay time only the collector current flows & base to emitter voltage is zero.
6. For a power transistor, the average power loss during the delay time can be given by the equation
a) Ic * Vc
b) 1/T * ∫ Td dt
c) Ic * dVc/dt * T
d) 1/T * ∫ dt
Answer: b
Explanation: During the delay time only, the collector current flows & base to emitter voltage is zero. Hence the average power can be found, simply by integrating it over the total delay time & dividing by the base time period.
7. A 1mv of i/p gives an output of 1V, the voltage gain as such would be
a) 0.001
b) 0.0001
c) 1000
d) 100
Answer: c
Explanation: 1V/1mv = 1000.
8. Which of the following relations is true for a BJT?
a) Ic ≈ Ie
b) Ib ≈ Ic
c) Ie ≈ Ib
d) Ib ≈ Ie ≈ Ic
Answer: a
Explanation: The collector & emitter current differ only by the base current, which is very very small.
9. Choose the correct statement
a) A transistor will remain on as long the the base current is applied
b) A transistor remains on after a high to low pulse is applied at the base
c) A transistor will remain on as long the the collector current is applied
d) A transistor remains on after a high to low pulse is applied at the collector
Answer: a
Explanation: Unlike the thyristor devices, all the transistor family devices remain in the conducting state as long as the firing pulses are applied. This is a very important property of the transistor devices.
10. Let’s say that a transistor is operating at the middle of the load line, then a decrease in the current gain would
a) move the Q point up
b) move the Q point down
c) result in to & fro motion of the Q point
d) not change the Q point
Answer: b
Explanation:The current gain would decreases the collector current, shifting the Q point below.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “MOSFETs-1”.
1. The MOSFET combines the areas of _______ & _________
a) field effect & MOS technology
b) semiconductor & TTL
c) mos technology & CMOS technology
d) none of the mentioned
Answer: a
Explanation: It is an enhancement of the FET devices using MOS technology.
2. Which of the following terminals does not belong to the MOSFET?
a) Drain
b) Gate
c) Base
d) Source
Answer: c
Explanation: MOSFET is a three terminal device D, G & S.
3. Choose the correct statement
a) MOSFET is a uncontrolled device
b) MOSFET is a voltage controlled device
c) MOSFET is a current controlled device
d) MOSFET is a temperature controlled device
Answer: b
Explanation: It is a voltage controlled device.
4. Choose the correct statement
i) The gate circuit impedance of MOSFET is higher than that of a BJT
ii) The gate circuit impedance of MOSFET is lower than that of a BJT
iii) The MOSFET has higher switching losses than that of a BJT
iv) The MOSFET has lower switching losses than that of a BJT
a) Both i & ii
b) Both ii & iv
c) Both i & iv
d) Only ii
Answer: c
Explanation: MOSFET requires gate signals with lower amplitude as compared to BJTs & has lower switching losses.
5. Choose the correct statement
a) MOSFET is a unipolar, voltage controlled, two terminal device
b) MOSFET is a bipolar, current controlled, three terminal device
c) MOSFET is a unipolar, voltage controlled, three terminal device
d) MOSFET is a bipolar, current controlled, two terminal device
Answer: c
Explanation: MOSFET is a three terminal device, Gate, source & drain. It is voltage controlled unlike the BJT & only electron current flows.
6. The arrow on the symbol of MOSFET indicates
a) that it is a N-channel MOSFET
b) the direction of electrons
c) the direction of conventional current flow
d) that it is a P-channel MOSFET
Answer: b
Explanation: The arrow is to indicate the direction of electrons .
7. The controlling parameter in MOSFET is
a) Vds
b) Ig
c) Vgs
d) Is
Answer: b
Explanation: The gate to source voltage is the controlling parameter in a MOSFET.
8. In the internal structure of a MOSFET, a parasitic BJT exists between the
a) source & gate terminals
b) source & drain terminals
c) drain & gate terminals
d) there is no parasitic BJT in MOSFET
Answer: b
Explanation: Examine the internal structure of a MOSFET, notice the n-p-n structure between the drain & source. A p-channel MOSFET will have a p-n-p structure.
9. In the transfer characteristics of a MOSFET, the threshold voltage is the measure of the
a) minimum voltage to induce a n-channel/p-channel for conduction
b) minimum voltage till which temperature is constant
c) minimum voltage to turn off the device
d) none of the above mentioned is true
Answer: a
Explanation: It is the minimum voltage to induce a n-channel/p-channel which will allow the device to conduct electrically through its length.
10.The output characteristics of a MOSFET, is a plot of
a) Id as a function of Vgs with Vds as a parameter
b) Id as a function of Vds with Vgs as a parameter
c) Ig as a function of Vgs with Vds as a parameter
d) Ig as a function of Vds with Vgs as a parameter
Answer: b
Explanation: It is Id vs Vds which are plotted for different values of Vgs .
This set of Power Electronics Multiple Choice Questions & Answers focuses on “MOSFETs-2”.
1. In the output characteristics of a MOSFET with low values of Vds, the value of the on-state resistance is
a) Vds/Ig
b) Vds/Id
c) 0
d) ∞
Answer: b
Explanation: The o/p characteristics Is a plot of Id verses Vds, which for low values of Vds is almost constant. Hence, the on-state resistance is constant & the slop is its constant value.
2. At turn-on the initial delay or turn on delay is the time required for the
a) input inductance to charge to the threshold value
b) input capacitance to charge to the threshold value
c) input inductance to discharge to the threshold value
d) input capacitance to discharge to the threshold value
Answer: b
Explanation: It is the time required for the input capacitance to charge to the threshold value, which depends on the device configuration. The device can start conducting only after this time.
3. Choose the correct statement
a) MOSFET suffers from secondary breakdown problems
b) MOSFET has lower switching losses as compared to other devices
c) MOSFET has high value of on-state resistance as compared to other devices
d) All of the mentioned
Answer: b
Explanation: MOSFET has lower switching losses due to its unipolar nature & less turn off time. All of the other statements are false.
4. Which among the following devices is the most suited for high frequency applications?
a) BJT
b) IGBT
c) MOSFET
d) SCR
Answer: c
Explanation: MOSFET has the least switching losses among the rest of the devices.
5. Choose the correct statement
a) MOSFET has a positive temperature co-efficient
b) MOSFET has a high gate circuit impedance
c) MOSFET is a voltage controlled device
d) All of the mentioned
Answer: d
Explanation: MOSFETs are voltage controlled devices. They have high gate circuit impedance and are PTC devices.
6. Consider an ideal MOSFET. If Vgs = 0V, then Id = ?
a) Zero
b) Maximum
c) Id
d) Idd
Answer: a
Explanation: Gate current = 0 so device is off .
7. For a MOSFET Vgs=3V, Idss=5A, and Id=2A. Find the pinch of voltage Vp
a) 4.08
b) 8.16
c) 16.32
d) 0V
Answer: b
Explanation: Use Id = Idd x [1-Vgs/Vp] 2 .
8. How does the MOSFET differ from the JFET?
a) JFET has a p-n junction
b) They are both the same
c) JFET is small in size
d) MOSFET has a base terminal
Answer: a
Explanation: None.
9. The basic advantage of the CMOS technology is that
a) It is easily available
b) It has small size
c) It has lower power consumption
d) It has better switching capabilities
Answer: c
Explanation: Complementary MOS consumes very less power as compared to all the earlier devices.
10. The N-channel MOSFET is considered better than the P-channel MOSFET due to its
a) low noise levels
b) TTL compatibility
c) lower input impedance
d) faster operation
Answer: d
Explanation: The N-channel are faster than the P-channel type.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “IGBTs-1”.
1. IGBT possess
a) low input impedance
b) high input impedance
c) high on-state resistance
d) second breakdown problems
Answer: b
Explanation: Like MOSFET IGBT possess high input impedance.
2. IGBT & BJT both posses ___
a) low on-state power losses
b) high on-state power losses
c) low switching losses
d) high input impedance
Answer: a
Explanation: Low on state power loss is one of the best parameters of both BJT & the IGBT.
3. The three terminals of the IGBT are
a) base, emitter & collector
b) gate, source & drain
c) gate, emitter & collector
d) base, source & drain
Answer: c
Explanation: IGBT is a three terminal device. It has a gate, a emitter & a collector.
4. In IGBT, the p + layer connected to the collector terminal is called as the
a) drift layer
b) injection layer
c) body layer
d) collector Layer
Answer: b
Explanation: It is called as a injection layer, because it injects holes into the n – layer.
5. The controlling parameter in IGBT is the
a) I G
b) V GE
c) I C
d) V CE
Answer: b
Explanation: The controlling parameter is the gate to emitter voltage, as the device is a voltage controlled device.
6. In IGBT, the n – layer above the p + layer is called as the
a) drift layer
b) injection layer
c) body layer
d) collector Layer
Answer: a
Explanation: It is called as the drift layer because its thickness determines the voltage blocking capabilities of the device.
7. The voltage blocking capability of the IGBT is determined by the
a) injection layer
b) body layer
c) metal used for the contacts
d) drift layer
Answer: d
Explanation: The drift layer which is a n – layer determines the voltage blocking capabilities.
8. The controlled parameter in IGBT is the
a) I G
b) V GE
c) I C
d) V CE
Answer: c
Explanation: The controlling parameter is the gate to collector current.
9. The structure of the IGBT is a
a) P-N-P structure connected by a MOS gate
b) N-N-P-P structure connected by a MOS gate
c) P-N-P-N structure connected by a MOS gate
d) N-P-N-P structure connected by a MOS gate
Answer: c
Explanation: The IGBT is a semiconductor device with four alternating layers that are controlled by a metal-oxide-semiconductor gate structure without regenerative action.
10. The major drawback of the first generation IGBTs was that, they had
a) latch-up problems
b) noise & secondary breakdown problems
c) sluggish operation
d) latch-up & secondary breakdown problems
Answer: d
Explanation: The earlier IGBT’s had latch-up problems , and secondary breakdown problems .
This set of Power Electronics Multiple Choice Questions & Answers focuses on “IGBTs-2”.
1. When latch-up occurs in an IGBT
a) Ig is no longer controllable
b) Ic is no longer controllable
c) the device turns off
d) Ic increases to a very high value
Answer: b
Explanation: After latch-up the collector emitter current is no longer in control of the gate terminal.
2. A latched up IGBT can be turned off by
a) forced commutation of current
b) forced commutation of voltage
c) use of a snubber circuit
d) none of the mentioned
Answer: a
Explanation: Forced commutation of current is the only way to turn off a latched up IGBT.
3. The static V-I curve of an IGBT is plotted with
a) Vce as the parameter
b) Ic as the parameter
c) Vge as the parameter
d) Ig as the parameter
Answer: c
Explanation: V-I curves are plotted for Ic vs Vce with the controlling parameter as a parameter.
4. Latch-up occurs in an IGBT when
a) Vce reaches a certain value
b) Ic reaches a certain value
c) Ig reaches a certain value
d) the device temperature reaches a certain value
Answer: b
Explanation: Latch up occurs when the current through the device collector current increases beyond a certain value.
5. In an IGBT, during the turn-on time
a) Vge decreases
b) Ic decreases
c) Vce decreases
d) none of the mentioned
Answer: c
Explanation: Vce decreases from 0.9 to 0.1 of the initial value whereas others increase.
6. Choose the correct statement
a) IGBTs have higher switching losses as compared to BJTs
b) IGBTs have secondary breakdown problems
c) IGBTs have lower gate drive requirements
d) IGBTs are current controlled devices
Answer: c
Explanation: Due to its high gate impedance, IGBTs require less gate drive current.
7. The approximate equivalent circuit of an IGBT consists of
a) a BJT & a MOSFET
b) a MOSFET & a MCT
c) two BJTs
d) two MOSFETs
Answer: a
Explanation: Gate of the MOSFET forms the gate terminal of the IGBT, the source of MOSFET is connected to the base of the BJT and drain to the collector.
8. An IGBT is also know as
a) MOIGT
b) COMFET
c) GEMFET
d) all of the mentioned
Answer: d
Explanation: All of the above mentioned are alternate names of IGBTs.
9. The body of an IGBT consists of a
a) p-layer
b) n-layer
c) p-n layer
d) metal
Answer: a
Explanation: IGBT has a p-n-p structure with fingers of n + layers into the p layer. The p layer has the largest cross section and forms the body of the IGBT.
10. At present, the state-of-the-art semiconductor devices are begin manufactured using
a) Semiconducting Diamond
b) Gallium-Arsenide
c) Germanium
d) Silicon-Carbide
Answer: d
Explanation: All of the above mentioned can be used but Si-Ca has certain advantages over the other materials.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “SOAs”.
1. For a transistor, the safe operating area is a plot of
a) Ib versus Vce
b) Ib versus Ic
c) Ic versus Vce
d) Ic versus time
Answer: c
Explanation: For reliable operation the collector current & voltage must remain within the SOA curves.
2. The forward safe operating area pertains to the operation when
a) the device is fired at a 50% Duty cycle
b) the device is forward-biased
c) the device is operated on AC
d) the device is operated on DC
Answer: b
Explanation: The FSOA is for forward biased operations. The FSOA is plotted for AC as well as DC for different duty cycles. Hence, option is the most appropriate choice.
3. The SOAs are plotted always on a _________ scale
a) time
b) frequency
c) logarithmic
d) polynomial
Answer: c
Explanation: The scale is always logarithmic, irrespective of the type of device.
4. As the FSOA increases, the pulse width
a) decreases
b) increases
c) remains constant
d) vanishes
Answer: b
Explanation: On reduced pulse width values, the devices can operated on higher voltages & currents.
5. The SOAs provided by the manufacturers are for
a) single pulse operation & a particular temperature
b) multi pulse operation & all the temperature
c) all the conditions
d) a particular duty cycle operation
Answer: a
Explanation: The manufacturer specifies the SOAs only for single pulse DC operation & a particular temperature . For actual operations, The SOA’s have to be modified using the thermal impedance charts.
6. A device is operating at Ic = 4A & Vce = 50V. For the device to operate at Ic = 20A ,
a) voltage should be increased
b) voltage should be reduced
c) voltage can be kept constant
d) current has to increased further
Answer: b
Explanation: For safe operation, the values should be within the limits. P = V.I – with increase in one of the values, another value should decrease.
7. For a BJT, find the maximum power dissipation when the device is safely operated at Vce = 90V and Ic = 0.5A
a) 40 Watts
b) 35 Watts
c) 45 Watts
d) 30 Watts
Answer: c
Explanation: P=90*0.5=45Watts.
8. The SOA for a MOSFET is plotted for
a) Id versus Vds
b) Ig versus Id
c) Ig versus Vds
d) Id versus Vgs
Answer: a
Explanation: It is a plot of drain current vs drain to source voltage.
9. The SOA for an IGBT is plotted for
a) Ic versus Vge
b) Ig versus Ic
c) Ig versus Vce
d) Ic versus Vce
Answer: d
Explanation: It is a plot of collector current vs collector to emitter voltage.
10. For MOSFET’s SOA, as the pulse width goes on increasing, the maximum voltage rating ____ & current rating ____
a) is constant, increases
b) increases, decreases
c) decreases, is constant
d) constant, decreases
Answer: c
Explanation: Refer MOSFET’s SOA.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Thyristors-1”.
1. A thyristor is a
a) P-N-P device
b) N-P-N device
c) P-N-P-N device
d) P-N device
Answer: c
Explanation: An SCR is a four layer p-n-p-n type device.
2. Which terminal does not belong to the SCR?
a) Anode
b) Gate
c) Base
d) Cathode
Answer: c
Explanation: The SCR is having three terminals viz. anode, cathode and the gate.
3. An SCR is a
a) four layer, four junction device
b) four layer, three junction device
c) four layer, two junction device
d) three layer, single junction device
Answer: b
Explanation: SCR is a four layer p-n-p-n device which forms three p-n junctions.
4. Choose the false statement.
a) SCR is a bidirectional device
b) SCR is a controlled device
c) In SCR the gate is the controlling terminal
d) SCR are used for high-power applications
Answer: a
Explanation: It is a unidirectional device, current only flows from anode to cathode.
5. In the SCR structure the gate terminal is located
a) near the anode terminal
b) near the cathode terminal
c) in between the anode & cathode terminal
d) none of the mentioned
Answer: b
Explanation: The gate is located near the cathode, because it allows fast turning on of the device when the gate signal is applied by forward basing the second junction.
6. The static V-I curve for the SCR is plotted for
a) Ia vs Ig , Va as a parameter
b) Ia vs Va with Ig as a parameter
c) Va vs Ig with Ia as a parameter
d) Ig vs Vg with Ia as a parameter
Answer: b
Explanation: The curve is plotted for Ia vs Va for different values of gate current Ig.
7. If the cathode of an SCR is made positive with respect to the anode & no gate current is applied then
a) all the junctions are reversed biased
b) all the junctions are forward biased
c) only the middle junction is forward biased
d) only the middle junction is reversed biased
Answer: c
Explanation: The device is in the reverse blocking state & only the middle junction is forward biased whereas other two are reversed biased.
8. For an SCR in the reverse blocking mode,
a) leakage current does not flow
b) leakage current flows from anode to cathode
c) leakage current flows from cathode to anode
d) leakage current flows from gate to anode
Answer: c
Explanation: In the reverse blocking mode, the gate current is zero & a reverse voltage is applied at the cathode-anode.
9. With the anode positive with respect to the cathode & the gate circuit open, the SCR is said to be in the
a) reverse blocking mode
b) reverse conduction mode
c) forward blocking mode
d) forward conduction mode
Answer: c
Explanation: The SCR is in the forward blocking mode with its top and bottom junctions forward biased and the middle junction reversed biased.
10. For an SCR in the forward blocking mode
a) leakage current does not flow
b) leakage current flows from anode to cathode
c) leakage current flows from cathode to anode
d) leakage current flows from gate to anode
Answer: b
Explanation: In the forward blocking mode, the gate current is zero & only the middle J2 junction is reversed biased.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Thyristors – 2”.
1. The forward break over voltage is the
a) anode-cathode voltage at which conduction starts with gate signal applied
b) anode-cathode voltage at which conduction starts with no gate signal applied
c) gate voltage at which conduction starts with no anode-cathode voltage
d) gate voltage at which conduction starts with anode-cathode voltage applied
Answer: b
Explanation: It is the forward voltage at which the middle junction breaks down without any gate signal and pushes the device into the conducting state.
2. For a forward conducting SCR device, as the forward anode to cathode voltage is increased
a) the device turns on at higher values of gate current
b) the device turns on at lower values of gate current
c) the forward impedance of the device goes on increasing
d) the forward impedance of the device goes on decreasing
Answer: b
Explanation: Higher the value of anode-cathode forward voltage, lower the gate requirements of the device. Also, the forward resistance of the device is always constant as long as the junction temperature is constant.
3. A thyristor can be bought from the forward conduction mode to forward blocking mode by
a) the dv/dt triggering method
b) applying a negative gate signal
c) applying a positive gate signal
d) applying a reverse voltage across anode-cathode terminals
Answer: d
Explanation: a) & c) are used to turn on the device, b) will damage the SCR.
4. Usually the forward voltage triggering method is not used to turn-on the SCR because
a) it increases losses
b) it causes noise production
c) it may damage the junction & destroy the device
d) relatively it’s an inefficient method
Answer: c
Explanation: In forward voltage triggering the middle junction breaks down without any gate signal and pushes the device into the conducting state. This method can permanently damage the J2 junction and make the device useless.
5. Among the following, the most suitable method to turn on the SCR device is the
a) gate triggering method
b) dv/dt triggering method
c) forward voltage triggering method
d) temperature triggering method
Answer: a
Explanation: d) & b) are unreliable methods, c) can permanently damage the SCR
Gate triggering is simple, reliable & most efficient.
6. The forward break over voltage is maximum when
a) Gate current = ∞
b) Gate current = 0
c) Gate current = -∞
d) It is independent of gate current
Answer: b
Explanation: Higher the value of anode-cathode forward voltage, lower the gate requirements of the device.
7. For the SCR to remain in the ON state
a) gate signal is continuously required
b) no continuous gate signal is required
c) no forward anode-cathode voltage is required
d) negative gate signal is continuously required
Answer: b
Explanation: Unlike the transistor devices, once the SCR is turned on by the gate terminal, the gate terminal losses its control over the device.
8. The value of anode current required to maintain the conduction of an SCR even though the gate signal is removed is called as the
a) holding current
b) latching current
c) switching current
d) peak anode current
Answer: b
Explanation: It is the minimum anode current value required to maintain the conduction of an SCR even though the gate signal is removed. It is a very important parameter when employing an SCR in any circuit.
9. In the reverse blocking mode the middle junction (J 2 ) has the characteristics of that of a
a) transistor
b) capacitor
c) inductor
d) none of the mentioned
Answer: b
Explanation: It is like a capacitor, as the dv/dt voltage triggering turns on the device. The charging current is given by,
I C = C j dV a /dt.
10. ________ are semiconductor thyristor devices which can be turned-on by light of appropriate wavelengths.
a) LGTOs
b) LASERs
c) MASERs
d) LASCRs
Answer: d
Explanation: LASCR stands for light activated SCRs, which can be turned on in made to conduct by firing appropriate light pulses at its gate region.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Thyristors-3”.
1. During the transition time or turn-on time
a) The forward anode voltage decreases from 90 % to 10 % & the anode current also decreases from 90 to 10 % of the initial value
b) The forward anode voltage increases from 10 % to 90 % & the anode current also increases from 10 % to 90 % of the initial value
c) The forward anode voltage decreases from 90 % to 10 % & the anode current increases from 10 % to 90 % of the initial value
d) The forward anode voltage increases from 10 % to 90 % & the anode current decreases from 90% to 10% of the initial value
Answer: c
Explanation: During the turn on time, the voltage across the SCR is going down and the current through it is slowly rising as it is going into the conduction mode.
2. For an SCR the total turn-on time consists of
i) Delay time
ii) Rise time and
iii) Spread time
During the delay time the
a) anode current flows only near the gate
b) anode current rises from zero to very high value
c) losses are maximum
d) anode to cathode voltage is zero
Answer: a
Explanation: Initially for a fraction of a microsecond after the gate signal is applied the anode current only flows near the gate terminal where the gate current density is maximum, as the gate current takes some time to spread all over the cross section of the device.
3. The minimum value of anode current below which it must fall to completely turn-off the device is called as the
a) holding current value
b) latching current value
c) switching current value
d) peak anode current value
Answer: a
Explanation: The device will remain in the conducting state unless the anode current falls below the holding current value.
4. For an SCR the total turn-on time consists of
i) Delay time
ii) Rise time and
iii) Spread time
During the rise time the
a) anode current flows only near the gate
b) anode current rises from zero to very high value
c) losses are maximum
d) anode to cathode voltage is zero
Answer: c
Explanation: The losses are maximum during the rise time because both Ia & Va are high.
5. The latching current is _________ than the holding current
a) lower
b) higher
c) same as
d) negative of
Answer: b
Explanation: The latching current is the value of current on which the device will remain in the on state even after removal of the gate signal. Whereas, the holding current is the threshold above which the device will work.
6. For an SCR the total turn-on time consists of
i) Delay time
ii) Rise time and the
iii) Spread time
The spread time interval depends upon
a) the value of gate current
b) junction temperature
c) area of the cathode
d) area of the anode
Answer: c
Explanation: During the spread time the conduction starts spreading all over the SCR cathode cross-section structure, which depends upon the structure of the gate & cathode. Higher the cathode area more is the time required for the charges to spread all over.
7. For effective turning off of the SCR after the anode current has reached zero value, ______________
a) chargers are injected by applying reverse anode-cathode voltage
b) chargers are removed by applying reverse anode-cathode voltage
c) chargers are injected by applying gate signal
d) chargers are removed by applying gate signal
Answer: b
Explanation: To enable the device to regain its reverse blocking capabilities, the stored charges in the junctions of the SCR must be removed.
8. To avoid commutation failure
a) circuit turn-off time must be greater than the thyristor turn-off time
b) circuit turn-off time must be lesser than the thyristor turn-off time
c) circuit turn-off time must be equal to the thyristor turn-off time
d) none of the above mentioned
Answer: a
Explanation: If the thyristor turn off time is more than the circuit turn off time, the circuit will be turned off and the thyristor will keep conducting, which is not at all desirable.
9. The gate characteristics of thyristor is a plot of
a) V g on the X-axis & I g on the Y-axis
b) I g on the X-axis & V g on the Y-axis
c) V a on the X-axis & I g on the Y-axis
d) I g on the X-axis & V a on the Y-axis
Answer: b
Explanation: It is the gate current versus the gate voltage plot and gives the minimum and maximum values of gate parameters.
10. The area under the curve of the gate characteristics of thyristor gives the
a) total average gate current
b) total average gate voltage
c) total average gate impedance
d) total average gate power dissipation
Answer: d
Explanation: As the gate characteristics is a plot of Ig vs Vg consisting of two curves one for the maximum values & other for the minimum the area between them gives the total average gate power dissipation. .
This set of Power Electronics online test focuses on “Thyristors-4”.
1. A tangent drawn from the Y-axis to the Pavg on the gate characteristics gives the value of the
a) maximum value of gate-source resistance
b) minimum value of gate-source resistance
c) maximum value of gate-source power
d) minimum value of gate-source power
Answer: b
Explanation: It gives the min gate to source resistance.
2. Higher the magnitude of the gate pulse
a) lesser is the time required to inject the charges
b) greater is the time required to inject the charges
c) greater is the value of anode current
d) lesser is the value of anode current
Answer: a
Explanation: Lesser time is required to inject the charges & turn on the device with higher gate pulse magnitude.
3.The average gate power dissipation for an SCR is 0.5 Watts the voltage applied to the gate is Vg = 10 V. What is the maximum value of current Ig for safe operation?
a) 0.25 A
b) 10 A
c) 0.05 A
d) 0.1 A
Answer: c
Explanation: Vg.Ig = 0.5 W, the power dissipation mustn’t exceed the average power dissipation.
4. For an SCR, the gate-cathode characteristic has a slop of 130. The gate power dissipation is 0.5 watts. Find Ig
a) 0.62 A
b) 620 mA
c) 62 mA
d) 6.2 mA
Answer: c
Explanation: Vg/Ig = 130 ..
Vg.Ig = 0.5 watts ..
use both the given data & find the gate current.
5. The two transistor model of the SCR can obtained by
a) bisecting the SCR vertically
b) bisecting the SCR horizontally
c) bisecting the SCRs top two & bottom two layers
d) bisecting the SCRs middle two layers
Answer: d
Explanation: The two transistor model consists of p-n-p and n-p-n transistors, of which the middle n-p layer is common in both the transistors.
6. Latching current for an SCR is 100 mA, DC source of 200 V is also connected from the SCR to the L load. Compute the minimum width of the gate pulse required to turn on the device. Take L = 0.2 H.
a) 50 μsec
b) 100 μsec
c) 150 μsec
d) 200 μsec
Answer: b
Explanation: For L load, E = L di/dt
I = E/L t
Therefore, 0.100 = 200t/0.2
T = 100 μsec.
7. The gate-source voltage is Es = 16 V and the load line has a slope of 128 V/A. Calculate the gate current for an average gate power dissipation of 0.5 W.
power-electronics-online-test-q7
a) 62.5 mA
b) 100.25 mA
c) 56.4 mA
d) 80.65 mA
Answer: a
Explanation: Load line is nothing but Rs
Es = 16V
Vg.Ig = 0.5
Rs = 128
We have Es = Ig x Rs + Vg.
8. From the two transistor Misplaced & analogy of SCR, the total anode current of SCR is ___________ in the equivalent circuit.
a) the sum of both the base currents
b) the sum of both the collector current
c) the sum of base current of T1 & collector current of T2
d) the sum of base current of T2 & collector current of T1
Answer: b
Explanation: The sum of both the collector currents of T1 and T2 forms the total anode current of SCR. Refer the model.
9. Consider the two transistor analogy of SCR, if α 1 & if α 2 are the common-base current gains of both the transistors then to turn-on the device
a) α 1 + α 2 should approach zero
b) α 1 x α 2 should approach unity
c) α 1 – α 2 should approach zero
d) α 1 + α 2 should approach unity
Answer: d
Explanation: To turn on the device sum of both the current gains should approach unity value.
10. Latching current for an SCR is 100 mA, a dc source of 200 V is also connected to the SCR which is supplying an R-L load. Compute the minimum width of the gate pulse required to turn on the device. Take L = 0.2 H & R = 20 ohm both in series.
a) 62.7 μsec
b) 100.5 μsec
c) 56.9 μsec
d) 81 μsec
Answer: b
Explanation;
E = Ri + L di/dt
Solve the above D.E for I & substitute the above values.
t = 100.503 μsec.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Thyristor Ratings”.
1. The voltage safety factor (V SF ) for an SCR is the ratio of
a) peak working voltage & peak reverse repetitive voltage
b) dv/dt & di/dt
c) peak repetitive reverse voltage & maximum value of input voltage
d) peak repetitive reverse voltage & rms value of input voltage
Answer: c
Explanation: It is peak repetitive reverse voltage/ the maximum value of input voltage.
2. The forward dv/dt rating of an SCR
a) increases with increase in the junction temperature
b) decreases with increase in the junction temperature
c) increases with decrease in the rms value of forward anode-cathode voltage
d) decreases with decrease in the rms value of forward anode-cathode voltage
Answer: a
Explanation: If the temperature is high, lesser dv/dt is required to turn on the device as the higher temperature has already excited few of the holes & electrons.
3. The finger voltage of an SCR is
a) minimum value of Vak to turn on the device with gate triggering
b) maximum value of Vak to turn on the device with gate triggering
c) minimum value of Vak to turn on the device without gate triggering
d) maximum value of Vak to turn on the device without gate triggering
Answer: a
Explanation: Finger voltage is the minimum value of Vak to turn on the device with gate triggering, it is to be avoided for accidental turn-on of the device.
4. Which among the following anode current waveforms will have the minimum junction temperature?
a) 100 % DC
b) 25 % DC
c) 50 % DC
d) AC
Answer: b
Explanation: N % DC is nothing but a wave with n % duty cycle. Lower the Duty cycle lesser is the current flowing & lesser is the temperature dissipation.
5. An SCR has half cycle surge current rating of 3000 A for 50 Hz. Calculate its one-cycle surge current rating
a) 3121.32 A
b) 2121.32 A
c) 3131.32 A
d) 2131.32 A
Answer: b
Explanation: By equating the energies involved in one cycle & subcycle
I 2 .T = Isb 2 .t
Isb = 3000 A
T = 1/50
t = T/2.
6. For a SCR the maximum rms on-state current is 35 A. If the SCR is used in a resistive circuit for a rectangular wave with conduction angle of 90°. Calculate the average & rms currents respectively.
a) I/4, I/2
b) I/2, I/√2
c) I/4, I 2 /2
d) I/4, I/√2
Answer: a
Explanation: First convert the conduction angle into N for rectangular wave.
N = 360°/Conduction angle = 4
Therefore, Iavg = I/4
Irms = I/√4 = I/2.
7. For an SCR the average & rms values of current are I/4 & I/2 respectively. Calculate the average on-state current rating (I TAV ). Take maximum RMS on-state current = 35 A.
a) 8.78 A
b) 10.10 A
c) 17.5 A
d) 24.74 A
Answer: c
Explanation: Form factor = RMS/Average vaules of current
Therefore, FF = 2
I TAV = 35/2 = 17.5.
8. The amp 2 -sec rating of the SCR specifies
a) The power dissipated by the device when fault occurs
b) The energy dissipated by the device when fault occurs
c) The energy that the device can absorb before the fault is cleared
d) The energy that the device can absorb while operating in the forward blocking mode.
Answer: c
Explanation: The amp 2 -sec is the energy that the device can handle before the fault is cleared. Its value decides how fast the fault has to be cleared to avoid damage to the device.
9. The maximum rms current of an SCR is 50 A. For a 120° sine wave conduction the form factor = 1.878
Find the average on-state current rating (I TAV ).
a) 93.9 A
b) 174 A
c) 26.62 A
d) 68.52 A
Answer: c
Explanation: I TAV = 50/FF.
10. The thermal resistance between junction & the SCR (θ jc ) has the unit
a) Ω/°C
b) W/Ω
c) °C/W
d) ΩW/°C
Answer: c
Explanation: Thermal resistance always has the unit degree temperature per watt.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Thyristor Protection”.
1. di/dt protection is provided to the thryistor by
a) connecting an inductor in parallel across the load
b) connecting an inductor in series with the load
c) connecting an inductor in parallel across the gate terminal
d) connecting an inductor in series with the gate
Answer: b
Explanation: By placing the di/dt inductor in series with the load, the change in the anode current can be limited to a small value.
2. The local hot spot formation in the cross-section of the SCR is avoided by
a) reducing the junction temperature
b) applying gate current nearer to the maximum gate current
c) using only R loads
d) proper mounting of the SCR on heat sink
Answer: b
Explanation: Applying the higher gate current spreads the ions quickly and avoids hotspot formation.
3. The dv/dt protection is provided in order to
a) limit the power loss
b) reduce the junction temperature
c) avoid accidental turn-on of the device
d) avoiding sudden large voltage across the load
Answer: c
Explanation: Accidentally some voltage spike or noise may occur in the vicinity of the device, if the magnitude is large enough it may turn on the SCR.
4. dv/dt protection is provided to the SCR by
a) connecting a capacitor in parallel with the load
b) connecting an inductor in series with the load
c) connecting a capacitor & resister in parallel with the device
d) connecting an inductor & resister in parallel with the device
Answer: c
Explanation: Snubber circuit R-C in parallel with SCR is connected for dv/dt protection.
5. Figure below shows SCR having dv/dt and di/dt protection, when the switch is closed the current through Rl =
power-electronics-questions-answers-thyristor-protection-q5
a) I(1 – e -t/τ )
b) I(1 + e -t/τ )
c) I(1 – e t/τ )
d) I(1 + e -t/τ )
Answer: a
Explanation: As soon as the switch is closed, C acts like a S.C & the voltage equation gives
Vs = I + L di/dt
Solve the above D.E.
6. The effect of over-voltages on SCR are minimized by using
a) RL circuits
b) Circuit breakers
c) Varistors
d) di/dt inductor
Answer: c
Explanation: Varistors are non-linear voltage clamping devices, RC circuits across the loads can also be used.
7. Over-current protection in SCRs is achieved through the use of
a) Varistors
b) Snubber Circuits
c) F.A.C.L.F & C.B.
d) Zener diodes
Answer: c
Explanation: FACLF stands for Fast Acting Current Limiting Fuse.
8. False triggering of the SCRs by varying flux & noise is avoided by using
a) F.A.C.L.F & C.B
b) Shielded cables & twisted gate leads
c) Snubber circuits
d) di/dt inductor in series with the gate terminal
Answer: b
Explanation: Shielded cables provide isolation from outside noise.
9. The thyristor has the following specifications
Vs = 400 V
max = 25 A/μsec
Find the value of L
power-electronics-questions-answers-thyristor-protection-q5
a) 8 μH
b) 80 μH
c) 16 μH
d) 160 μH
Answer: c
Explanation: As soon as the switch is closed, C acts like a S.C & the voltage equation gives,
Vs = I + L di/dt
Solve the above D.E.
di/dt = Vs/L e -t/τ
di/dt is maximum at t = 0, substitute the above given values & find L
10. Thyristors are used in electronic crowbar protection circuits because it possesses
a) high surge current capabilities
b) high amp 2 -sec rating
c) less switching losses
d) voltage clamping properties
Answer: a
Explanation: Crowbar protection circuits have high surge current capabilities.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Thyristor Mounting”.
1. The usual way to accomplish higher gate current for improved di/dt rating is by using
a) varistors
b) pilot thyristors
c) twisted cables
d) op-amps
Answer: b
Explanation: Pilot SCR is an SCR which is fired which activates the firing circuit and fires the main SCR.
2. Inter-digitating of gate-cathode regions in SCR devices improves the
a) I 2 t rating
b) di/dt rating
c) dv/dt rating
d) thermal resistance
Answer: b
Explanation: Inter-digitating is the inter-mixing of the gate-cathode area to improve the di/dt ratings. di/dt rating is improved by providing more cathode conduction area during the delay and rise time.
3. The dv/dt rating of SCR can be improved by using
a) cathode-short structure
b) anode-short structure
c) gate-short structure
d) centre gate thyristor
Answer: a
Explanation: Cathode shorts are realized by overlapping metal on cathode n+ layers with a narrow p-region in between.
4. The total thermal resistance between junction and ambient θ jA is 10°C/W. θ jc is 2°C/W. θ cs is 4°C/W. θ sA = ?
a) 4°C/W
b) 2°C/W
c) 10°C/W
d) 16°C/W
Answer: a
Explanation: θ jA = θ jc + θ cs + θ sA .
5. Which of the following thermal resistance parameters depends on the size of the device and the clamping pressure?
a) θ sA
b) θ cs
c) θ jc
d) None
Answer: b
Explanation: The case-to-sink thermal resistance depends on the size of the device, pressure, grease between the interface, etc.
6. The sink to ambient thermal resistance of SCR θ sA
a) depends on the flatness of the device
b) depends on the length of the device
c) depends on the current carrying capabilities
d) is independent on thyristor configuration
Answer: d
Explanation: It does not depend on any of the device configurations.
7. P av x (θ jc + θ cs + θ sA ) =
a) Maximum specified temperature
b) Energy lost
c) Difference in temperature between junction & ambient
d) Sum of junction & ambient temperature
Answer: c
Explanation: P av = (T j – T a )/θ jA
θ jA = (θ jc + θ cs + θ sA ).
8. Heat dissipation from heat sink mainly takes place by
a) radiation
b) convection
c) absorption
d) none of the mentioned
Answer: b
Explanation: Heat DISSIPATION takes place through convention only.
9. For low power SCRs _____________ type of mounting is used
a) lead
b) stud
c) bolt-down
d) press-fit
Answer: a
Explanation: Lead mounting is a very simple time of mounting used for low power devices.
10. In the ___________ type of mounting the SCR is pressed between two heat sinks
a) bolt-down mounting
b) stud-mounting
c) press-pak mounting
d) cross-fit mounting
Answer: c
Explanation: In the press-pak type the device is pressed or clamped between two heat-sinks & external pressure is applied from both the sides.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Interconnecting Thyristors”.
1. SCRs are connected in parallel to fulfill the ___________ demand
a) high voltage
b) high current
c) size
d) efficiency
Answer: b
Explanation: Number of devices connected in parallel can carry huge amounts of current.
2. The term used to measure the degree of utilization of SCRs connected in series & parallel is
a) tuf
b) string efficiency
c) voltage/current utilization ratio
d) rectification efficiency
Answer: b
Explanation: String Efficiency = Rating of the whole string/
3. To have maximum possible string efficiency
a) SCRs of same rating must be used
b) SCRs with similar V-I characteristics must be used
c) SCRs with the same dimensions must be used
d) SCRs with similar thermal characteristics must be used
Answer: b
Explanation: Having similar ratings does not mean they have similar charc.
4. For a string voltage of 3300 V, let there be six series connected SCRs each of voltage 600V. Then the string efficiency is
a) 99.36 %
b) 91.7 %
c) 98.54 %
d) 96 %
Answer: c
Explanation: String efficiency = 3300/ = 98.54.
5. The measure of reliability of string is given by the factor
a) DRF = 1 – String efficiency
b) DRF = 1 + String efficiency
c) DRF = String efficiency – 1
d) DRF = String efficiency x 2
Answer: a
Explanation: DRF is de-rating factor given by the above expression.
6. When an extra SCR is connected in series with a string
a) DRF decreases
b) DRF increases
c) DRF remains constant
d) None of the mentioned
Answer: b
Explanation:
DRF = 1 – String efficiency
String Efficiency = Rating of the whole string/
Extra SCR will reduce the string efficiency which in turn increase the DRF.
7. The most practical way of obtaining a uniform distribution of series connected SCRs is to
a) connect a resistor of value R in series with each of the series connected SCRs
b) connect a resistor of value R in parallel with each of the series connected SCRs
c) connect a resistor of value R in series with one of the series connected SCRs
d) connect a resistor of value R in parallel with one of the series connected SCRs
Answer: b
Explanation: For uniform distribution of voltage across series connected SCRs, a resistor of value R in parallel with each series connected SCR.
8. 3 SCRs are connected in series. The string efficiency is 91%. SCRs 1, 2 & 3 have leakage currents 4 mA, 8 mA & 12 mA. Which SCR will block more voltage?
a) SCR-1
b) SCR-2
c) SCR-3
d) All the three will block equal voltage
Answer: a
Explanation: The SCR with lower leakage current block more voltage.
9. Two parallel connect SCRs have same voltage drop having rated current = 2I 1 . SCR-1 carries a current of I 1 =2.6 A whereas SCR-2 carries a current of I 2 =1.4 A. Find the string efficiency.
a) 45 %
b) 77 %
c) 92 %
d) 84 %
Answer: b
Explanation: The total current would be I1+I2 & rated current is 2I 1
String efficiency = (I 1 +I 2 )/2I 1 .
10. SCRs with a rating of 1000 V & 200 A are available to be used in a string to handle 6 KV & 1 KV. Calculate the number of series & parallel units required in case the de-rating factor is 0.1. Misplaced &
a) Series = 7, Parallel = 6
b) Series = 6, Parallel = 7
c) Series = 6, Parallel = 6
d) Series = 7, Parallel = 7
Answer: a
Explanation: DRF = 1-S.E
Therefore
0.1 = =
Ns = 6.6 = 7
Np = 5.5 = 6
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Other Thyristor Members-1”.
1. _________ device from the thyristor family has its gate terminal connected to the n-type material near the anode.
a) SCR
b) RCT
c) PUT
d) SUT
Answer: c
Explanation: PUT is Programmable Unijunction Transistor which is a p-n-p-n device just like the SCR with its gate connected to the n-type material.
2. The Programmable Unijunction Transistor turns on & starts conducting when the
a) gate voltage exceeds anode voltage by a certain value
b) anode voltage exceeds gate voltage by a certain value
c) gate voltage equals the anode voltage
d) gate is given negative pulse w.r.t to cathode
Answer: b
Explanation: The device only starts to conduct when the forward anode to cathode voltage exceeds the applied gate to cathode voltage.
3. The equivalent circuit of SUS consists of
a) a diode in series with a PUT
b) a diode in parallel with a PUT
c) a diode in anti-parallel with a PUT
d) two diodes
Answer: c
Explanation: It is a diode connected in anti-parallel with a PUT.
4. From the following list of devices, choose the device that only turns-on for a fixed-value of anode-cathode voltage
a) PUT
b) SCR
c) SUS
d) BJT
Answer: c
Explanation: Unlike the other devices the SUS only turns-on for a fixed value of anode to cathode voltage.
5. The SCS is a
a) two terminal device
b) three terminal device
c) four terminal device
d) five terminal device
Answer: c
Explanation: The SCS is a four terminal device A,K,KG & AG.
6. The SCS is a four layer, four terminal thyristor. Can be turned on by
a) the anode gate
b) the cathode gate
c) either of the gates
d) gating both the gates together
Answer: c
Explanation: The SCS has two gates, anode-gate and cathode-gate. Either of the gates could be used to turn on the device.
7. The SCS can be turned on by two methods, by applying __________ and __________
a) positive pulse to the anode gate, positive pulse to the cathode gate
b) positive pulse to the anode gate, negative pulse to the cathode gate
c) negative pulse to the anode gate, positive pulse to the cathode gate
d) negative pulse to the anode gate, negative pulse to the cathode gate
Answer: c
Explanation: Either of the gates could be used to turn on the device.
8. Which of the following devices provide complete isolation between triggering circuit and power circuit?
a) PUT
b) LASCR
c) SUS
d) DIAC
Answer: b
Explanation: Complete Isolation between triggering circuit & power circuit is the major advantage of using LASCR as they are light activated or light trigged.
9. The DIAC can be represented by
a) two SCRs in anti-parallel
b) two SCRs in parallel
c) two diodes in anti-parallel
d) two diodes in parallel
Answer: c
Explanation: The DIAC is nothing but a bi-directional diode.
10. The TRIAC can be represented by
a) two SCRs in anti-parallel
b) two SCRs in parallel
c) two diodes in anti-parallel
d) two diodes in parallel
Answer: a
Explanation: The TRIAC is a bidirectional SCR.
This set of Power Electronics online quiz focuses on “Other Thyristor Members-2”.
1. The SITH is a
a) 4 terminal, self-controlled device
b) 3 terminal, self-controlled device
c) 4 terminal, un-controllable device
d) 3 terminal, un-controllable device
Answer: b
Explanation: It is just like a GTO but it is p-n-n device.
2. The SITH has
a) gate, anode, cathode
b) base, collector, emitter
c) base, anode, cathode
d) gate, emitter, collector
Answer: a
Explanation: The SITH has 3 terminals viz. gate, anode and cathode.
3. The SITH is a
a) p + n + diode with p + electrodes
b) n + p + n – diode with p + electrodes
c) p + n n + diode with p + electrodes
d) p n + p + n with p + electrodes
Answer: c
Explanation: The Static Induction Thyristor is a p + n + diode with p + electrodes.
4. The SITH is a
a) normally-off device
b) normally-on device
c) uncontrollable device
d) none of the mentioned
Answer: b
Explanation: The SITH unlike other devices is a normally on-device which can be switched off as and when required.
5. The SITH will act as a diode when the
a) anode is forward biased with zero gate-cathode voltage
b) anode is reversed biased with zero gate-cathode voltage
c) anode is forward biased and a positive gate voltage is applied
d) anode is forward biased and a negative gate voltage is applied
Answer: a
Explanation: Load current flows from anode to cathode as the p + junction is forward biased.
6. In SITH, the magnitude of the anode current can be controlled by
a) controlling the anode-cathode voltage
b) controlling the negative gate bias
c) controlling the positive gate bias
d) it cannot be controlled
Answer: b
Explanation: Controlling the negative gate bias controls the main anode current.
7. The SITH
a) has high reverse blocking capabilities as compared to CTs
b) has low reverse blocking capabilities as compared to CTs
c) has no reverse blocking capabilities
d) none of the mentioned
Answer: c
Explanation: It has no reverse blocking capabilities due to emitter-shorting.
8. The TRIAC’s terminals are
a) gate, anode, cathode
b) MT1, MT2, gate
c) gate1, gate2, anode, cathode
d) MT1, MT2, gate1, gate2
Answer: b
Explanation: TRAIC has MT1, MT2 and a gate.
9. The RCT has
a) a diode in series with the SCR
b) a diode in anti-parallel with the SCR
c) two SCR’s in anti-parallel
d) none of the mentioned
Answer: b
Explanation: A diode in anti-parallel with the SCR is called as a RCT.
10. The TRIAC is most sensitive in the _________ quadrants
a) 1st & 3rd with positive gate current
b) 1st with positive gate current & 3rd with negative gate current
c) 3st with positive gate current & 1rd with negative gate current
d) 1st & 3rd with negative gate current
Answer: b
Explanation: It is the most sensitive when all the voltage,current and gate signal have the same polarities.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “GTOs”.
1. The GTO is a
a) p-n-p-n device
b) p-n-p device
c) p-metal-n device
d) p-n single junction device
Answer: a
Explanation: Just like a SCR, the GTO is a four layer p-n-p-n device.
2. The GTO can be turned off
a) by a positive gate pulse
b) by a negative gate pulse
c) by a negative anode-cathode voltage
d) by removing the gate pulse
Answer: b
Explanation: The GTO can be turned off by applying a negative gate pulse to the gate terminal.
3. The anode current is ideally limited by the
a) gate pulse amplitude
b) internal impedance of the device
c) load Impedance
d) gate circuit impedance
Answer: c
Explanation: The SCR or any device is connected through the load, hence the magnitude of the anode current will depend on the supply voltage and load impedance.
4. In a GTO the n + layer forms the
a) anode & gate
b) cathode & gate
c) cathode
d) gate
Answer: c
Explanation: The bottom n + layer forms the cathode.
5. The turn-off gain β off of the GTO is given by
a) I g /I a
b) I a /I g
c) V g /V a
d) V g /V a
Answer: b
Explanation: β off = .
6. A GTO can be represented by two transistors T1 & T2. The current gain of both transistors are α1 and α2 respectively. A low value of gate current requires
a) low value of α1 and α2
b) low value of α1 and high value of α2
c) high value of α1 and low value of α2
d) high values of α1 and α2
Answer: b
Explanation: In order that the gate current for turning-off the device is low, α2 should be made as nearer to unity as possible whereas α1 should be small.
7. Gold doped GTOs have _____________ as compared to the conventional GTOs
a) high turn-off time
b) low negative gate current requirement
c) low reverse voltage blocking capabilities
d) lower positive gate current requirement
Answer: b
Explanation: Gold doping reduces the negative gate current requirements, different kinds of dopings have different advantages over the others.
8. Latching current for the GTOs is ________ as compared to CTs .
a) more
b) less
c) constant
d) cannot be said
Answer: a
Explanation: Latching current of GTOs is 2-4A as compared to 200 to 400 mA in case of CT’s.
9. In case of the two-transistor model Misplaced & of GTO with anode-short, the anode-short is placed between the
a) emitter of T1 & T2
b) emitter of T1 & base of T2
c) emitter of T1 & base of T1
d) emitter of T1 & collector of T2
Answer: c
Explanation: Draw the model. The anode-short resistor is connected between emitter with base of T1 transistor.
10. Choose the correct statement:
GTOs have _________ as compared to the CTs.
a) less on-state voltage drop
b) less gate drive losses
c) higher reverse blocking capabilities
d) faster switching speed
Answer: d
Explanation: GTOs have less turn-on and turn-off time, making it efficient for high frequency applications.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Diodes Circuits- 1”.
1. An ideal diode has _________ &__________
a) some forward voltage drop, some reverse recovery time
b) high switching losses, high reverse voltage drop
c) no forward voltage drop, negligible reverse recovery time
d) no reverse recovery time, high leakage current
Answer: c
Explanation: An ideal diode has no losses and negligible reverse recovery time.
2. A diode circuit is so arranged that when the switch is open it’s KVL gives
Ri+ 1/C ∫i dt = 0
When the switch is closed,
Ri+ 1/C ∫i dt = Vs
Vs is the dc supply voltage.
The diode is so connected that it is forward biased when switch is closed
The circuit is mostly likely be a
a) diode in parallel with Vs, switch, R & C
b) diode in series with R, than parallel with Vs & C
c) diode in series with the switch, R, C & Vs
d) diode in series with R,C & Vs with the switch connected in parallel across Vs
Answer: c
Explanation: Examine the equation, the same current flows through R & C. Also when switch is open, the equation R.H.S is 0. Hence, all the elements are in series.
3. A circuit is so formed such that the source-R-C-diode-switch are in series. Consider the initial voltage across the C to be zero. The diode is so connected that it is forward biased when the switch is closed. When the switch is closed,
a) the current will decay exponentially & the voltage will increase exponentially
b) the current will increase exponentially & the voltage will increase exponentially
c) the current will fall to zero & the voltage both will decay exponentially
d) the voltage and current both remain constant
Answer: a
Explanation: Instant switch is closed, the current is maximum Vs/R than starts to reduce, whereas voltage starts to increase from 0 to Vs .
4. The time constant of a series RC circuit is given by
a) R/C
b) C/R
c) RC
d) 1/RC
Answer: c
Explanation: For a series RC circuit, τ = RC.
5. A circuit is so formed such that source-R-L-diode-switch are all in series. Consider the initial current in L to be zero. The diode is so connected that it is forward biased when switch is closed.
When the switch is closed,
a) the current will decay exponentially & the voltage will increase exponentially
b) the current will increase exponentially & the voltage will decay exponentially
c) the current will fall to zero & the voltage both will decay exponentially
d) the voltage and the current both remain constant
Answer: b
Explanation: Instant switch is closed, the current is minimum zero than starts to increase till it reaches a constant value Vs/R, whereas voltage starts to reduce from Vs to 0 .
6. In the figure shown below,
power-electronics-mcqs-diode-circuits-1-q6
As the switch is pressed, the voltage across the diode
a) increases to Vs/R
b) increases to Vs
c) decreases to zero
d) remains Constant
Answer: c
Explanation: As the switch is pressed, current starts to flow & the whole supply voltage appears across the load R & voltage across the diode is zero.
7. For a diode circuit the voltage across the capacitor is given by
Vc= Vs(1-e )
Then the initial rate of change of capacitor voltage is given by
a) 0
b) ∞
c) Vs x RC
d) Vs/RC
Answer: d
Explanation: Find d/dt and put t = 0.
8. In the circuit show below,
power-electronics-mcqs-diode-circuits-1-q8
The initial current through the inductor is zero. When the switch is closed, then the current through the inductor
a) decreases from Vs/R to 0
b) increases from zero to Vs/R
c) decreases from Vs/L to 0
d) increases from zero to Vs/L
Answer: b
Explanation: Current increases from zero to maximum value gradually due to the L nature. The KVL when switch is closed gives,
Ri + Ldi/dt = Vs
Solve for i. Maximum value comes out to be Vs/R.
9. In the figure shown below,
power-electronics-mcqs-diode-circuits-1-q6
When the switch is open, the voltage across the diode
a) is Vs/R
b) is Vs
c) is zero
d) none of the mentioned
Answer: b
Explanation: When the switch is open, the diode experiences all the supply voltage.
10. The time constant of a series RL circuit is given by
a) R/L
b) L/R
c) RC
d) 1/RL
Answer: a
Explanation: The time constant τ for a series RL circuit is R/L.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Diodes Circuits-2”.
1. For the initially relaxed circuit shown below, KVL with switch in the closed position gives a certain equation. The Laplace of this equation will have the right hand side as
power-electronics-mcqs-diode-circuits-2-q1
a) Vs
b) /RC
c) Vs/s
d) RC/s
Answer: c
Explanation: When switch is closed,
Ldi/dt + 1/C ∫ idt = Vs
Laplace of the above gives,
L[sI] + 1/C [I/s] = Vs/s.
2. For the initially relaxed circuit shown below, the Laplace transform of the KVL when the switch is closed is
power-electronics-mcqs-diode-circuits-2-q1
I [ X ] = Vs/s
The value of X is
a) sL + 1/C
b) Cs + 1/sL
c) sL + 1/sC
d) Vs/CLs
Answer: c
Explanation: When switch is closed,
Ldi/dt + 1/C ∫ idt = Vs
Laplace of the above gives,
L[sI] + 1/C [I/s] = Vs/s.
3. For the initially relaxed circuit shown below, if Ω=1/√LC. Then the current is a function of ___
power-electronics-mcqs-diode-circuits-2-q1
a) cos Ωt
b) sin Ωt
c) tan Ωt
d) cos Ωt.sin Ωt
Answer: b
Explanation: When switch is closed,
Ldi/dt + 1/C ∫ idt = Vs
Laplace of the above gives,
L[sI] + 1/C [I/s] = Vs/s
I = Vs/ * (Ω/Ω 2 + s 2 )
Taking the inverse lapace gives,
I = Vs * √ * sin Ωt.
4. For the circuit shown below, the capacitor is initially charged to a voltage of Vo with the upper plate positive. After the switch is closed, the current through the load
power-electronics-mcqs-diode-circuits-2-q4
a) increases from zero to Vo/R
b) decreases from Vo/R to zero
c) increases from zero to Vo/C
d) decreases from Vo/C to zero
Answer: b
Explanation: The capacitor acts as a source. At instant switch is closed the current is maximum and than discharges till zero value through the load R.
5. For the circuit shown below, the capacitor is initially charged to a voltage of Vo with the upper plate positive. Switch is closed at t=0. The peak value of the current through the diode is
power-electronics-mcqs-diode-circuits-2-q4
a) Vo/C
b) Vo/R
c) Vo
d) Vo/
Answer: b
Explanation: When switch is closed, the equation is
Ri + 1/C ∫idt = 0
Solution of the above equation gives, Vo/R at t= 0.
6. When the switch is closed, the average current through the diode in the positive cycle is
power-electronics-mcqs-diode-circuits-2-q6
a) 0
b) Vs/R
c) Vs/
d) none of the mentioned
Answer: a
Explanation: The switch S.C’s the source.
7. When the switch is closed, the steady state current through the diode is
power-electronics-mcqs-diode-circuits-2-q7
a) Vo/C
b) Vo/R
c) Vo
d) Vo/
Answer: b
Explanation: I = Vo/R
8. When the switch is open, the current through the diode in the positive cycle is
power-electronics-mcqs-diode-circuits-2-q6
a) zero
b) Vs/R
c) Vs/
d) none of the mentioned
Answer: c
Explanation: When the switch is open, the diode is forward biased and I = Vs/. Where, Rd is the diode resistance.
9. For the circuit shown in the figure below, consider the diode as an ideal diode & R.M.S value of source voltage as Vs.
power-electronics-mcqs-diode-circuits-2-q9
The output voltage waveform at R is most likely to have
a) zero value in the positive half cycle and a peak value of 1.414Vs in the negative half cycle
b) sine-wave nature with a peak value 1.414Vs
c) zero value in the negative half cycle and a peak value of 1.414Vs in the positive half cycle
d) sine-wave nature with a peak value Vs
Answer: a
Explanation: The diode S.C’s the load in the positive half cycle.
10. For the circuit shown in the figure below, V8 is AC voltage source with peak value Vm. The waveform of the load voltage at the resistor is
power-electronics-mcqs-diode-circuits-2-q10
a) zero in the positive half & peak value of – negative half
b) zero in the negative half & peak value of – in the positive half
c) zero in the positive half & peak value of – negative half
d) zero in the positive half & peak value of – negative half
Answer: d
Explanation: Diode is reversed biased in the positive half cycle. In the negative half cycle, apply KVL to get the value of peak voltage at the load.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Diode Circuits-3”.
1. For the circuit shown in the figure below, consider the diode as an ideal diode & rms value of source voltage as Vs.
power-electronics-mcqs-diode-circuits-3-q1
The output voltage waveform at R will have
a) zero value in the positive half cycle and a peak value of 1.414Vs in the negative half cycle
b) sine-wave nature with a peak value 1.414Vs
c) zero value in the negative half cycle and a peak value of 1.414Vs in the positive half cycle
d) sine-wave nature with a peak value Vs
Answer: c
Explanation: The diode short circuits the load in the negative half cycle. The peak value in the positive half is 1.414 x Vs.
2. For the circuit shown in the figure below, Vs is the ac voltage source with peak value Vm. The waveform of the load voltage at the resistor will have
power-electronics-mcqs-diode-circuits-3-q2
a) zero value in the positive half cycle and a peak value of – in the negative half cycle
b) zero value in the negative half cycle and a peak value of – in the positive half cycle
c) zero value in the positive half cycle and a peak value of – in the negative half cycle
d) zero value in the positive half cycle and a peak value of – in the negative half
Answer: c
Explanation: Diode is reversed biased in the positive half cycle. In the negative half cycle, apply KVL to get the value of peak voltage at the load.
3. For the circuit shown below,
power-electronics-mcqs-diode-circuits-3-q3
Vs=230V, Voltage drop across the diode = 2V
The peak value of voltage at R in the positive and the negative half cycles are ___ & ___ respectively.
a) 323V,0V
b) 0V, 323V
c) 327V, 0V
d) 0V, 327V
Answer: b
Explanation: Load is S.C in the positive half cycle hence, voltage is zero. In negative half the peak value is – 2 = 323 Volts.
4. For the circuit shown in the figure below, V9 is the AC voltage source with peak value Vm. The waveform of the load voltage at the resistor has a
power-electronics-mcqs-diode-circuits-3-q4
a) peak value of in the negative half cycle
b) peak value of in the positive half cycle
c) peak value of in the positive half cycle
d) peak value of in the negative half cycle
Answer: c
Explanation: Diode is reversed biased in the negative half cycle. In the positive half cycle, apply KVL to get the value of peak voltage at the load. Peak value = Vm + V1, as V1 is aiding the positive anode of the diode.
5. The circuit shown below has the following parameters:
power-electronics-mcqs-diode-circuits-3-q5
V1 = 8 Volts
V2 = 6 Volts
Vs = 10V/√2
Voltage drop across D1 & D2 = 0.7 Volts
At the load , the peak value in the positive half cycle will be
a) 8.7 V
b) 6.7 V
c) 8 V
d) 10V/√2
Answer: a
Explanation: In the positive half, only D1 is active. Hence use KVL, Vo = 8 + 0.7 Volts.
6. The circuit shown below has the following parameters:
power-electronics-mcqs-diode-circuits-3-q5
Voltage drop across D1 & D2 = 0.7 Volts
V1 = 8 Volts
V2 = 6 Volts
Vs = 10V/√2
At the load , the peak value in the negative half cycle is
a) 8.7 V
b) 6.7 V
c) 8 V
d) 10V/√2
Answer: b
Explanation: In the negative half, only D2 is active. Hence use KVL, Vo = 6 + 0.7 Volts.
7. Consider the diode to be an ideal one and Vo = Vr + Vdc during positive half cycle. Thus, during the negative half cycle
power-electronics-mcqs-diode-circuits-3-q7
a) Vo = Vr
b) Vo = 0
c) Vo = Vdc+Vr
d) Vo = Vdc
Answer: d
Explanation: During the negative half diode is not conducting hence Vr = 0. Vo = Vdc.
8. When a diode is connected in series with an AC source & R load, the conduction time per cycle is
a) 0
b) 2π
c) π
d) π/2
Answer: c
Explanation: The diode is forward biased for half cycle and reversed biased in the another 180 degrees. Hence, per cycle it only conducts for 180 degrees or π radians.
9. When diode is connected in series to an AC source & RL load, the conduction time for the diode
a) is always less than π
b) is 0
c) is π
d) can be greater than π
Answer: d
Explanation: For an R – L load, the inductively load can make the diode to force conducted hence, the conduction time can be greater than 180 degrees.
10. For the circuit shown below,
power-electronics-mcqs-diode-circuits-3-q10
Vdc = 50 V
Cut-in voltage for D1 = 0.2 V
Cut-in voltage for D2 = 0.6 V
R = 5KΩ
Current through D1 & D2 would be,
a) 5mA, 5mA
b) 10mA, 0
c) 0, 10mA
d) 9.98mA, 9.94mA
Answer: b
Explanation: As the cut-in voltage for D1 is lesser than D2, D1 would start conducting first & S.C D2. Hence, only D1 conducts with I = 50/5000 A.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “1-Phase-Diode Rectifiers HW-1”.
1. In the process of diode based rectification, the alternating input voltage is converted into
a) an uncontrolled alternating output voltage
b) an uncontrolled direct output voltage
c) a controlled alternating output voltage
d) a controlled direct output voltage
Answer: b
Explanation: Rectification is AC to DC. In DIODE biased rectification, control is not possible.
2. In a half-wave rectifier, the
a) current & voltage both are bi-directional
b) current & voltage both are uni-directional
c) current is always uni-directional but the voltage can be bi-directional or uni-directional
d) current can be bi-directional or uni-directional but the voltage is always uni-directional
Answer: c
Explanation: Current is always in one direction only, but voltage can be bi-directional in case of an L load.
3. For a certain diode based rectifier, the output voltage is given by the equation
1/2π [ ∫Vm sin ωt d ] Where the integral runs from 0 to π
The rectifier configuration must be that of a
a) single phase full wave with R load
b) single phase full wave with RL load
c) single phase half wave with R load
d) single phase half wave with RL load
Answer: c
Explanation: Integration is 0 to π from base period of 1/2π so it is a half wave R load.
4. For a single phase half wave rectifier, with R load, the diode is reversed biased from ωt =
a) 0 to π, 2π to 2π/3
b) π to 2π, 2π/3 to 3π
c) π to 2π, 2π to 2π/3
d) 0 to π, π to 2π
Answer: b
Explanation: Diode will be reversed biased in the negative half cycles.
5. For the circuit shown below,
power-electronics-mcqs-1p-diode-rectifiers-hw-1-q5
The secondary transformer voltage Vs is given by the expression
Vs = Vm sin ωt
Find the PIV of the diode.
a) √2
b) Vs
c) Vm
d) √2 Vm
Answer: c
Explanation: PIV = √2 Vs = Vm.
6. For the circuit shown below,
power-electronics-mcqs-1p-diode-rectifiers-hw-1-q6
The peak value of the load current occurs at ωt = ?
a) 0
b) π
c) 2π
d) Data is insufficient
Answer: b
Explanation: Due to the L nature, load current is maximum when the diode will be com-mutated i.e at π.
7. Find the rms value of the output voltage for the circuit shown below.
power-electronics-mcqs-1p-diode-rectifiers-hw-1-q7
Voltage across the secondary is given by Vm sinωt.
a) Vm
b) 2Vm
c) Vm/2
d) Vm 2 /2
Answer: c
Explanation: The above is a HW diode rectifier, the RMS o/p voltage equation is given by
Vor = √ [ ∫ π Vm 2 sin 2 ωt. d ] Solving above equation we get, Vor = Vm/2.
8. In a 1-Phase HW diode rectifier with R load, the average value of load current is given by
Take Input = Vm sinωt
a) Vm/R
b) Vm/2R
c) Vm/πR
d) Zero
Answer: c
Explanation: Vo = √ [ ∫ π Vm sinωt. d] Vo = Vm/π
I = Vo/R = Vm/πR.
9. In the circuit shown below,
power-electronics-mcqs-1p-diode-rectifiers-hw-1-q9
The switch is closed at ωt = 0. The load current or capacitor current has the maximum value at ωt =
a) 0
b) π
c) 2π
d) none of the mentioned
Answer: a
Explanation: The instant switch is closed the load current will be zero due to the nature of the capacitor.
10. Find the average value of output current for a 1-phase HW diode rectifier with R load, having RMS output current = 100A.
a) 200R A
b) 100/R√2 A
c) 200/R√2 A
d) 200/Rπ A
Answer: d
Explanation: I = Vm/2R
Therfore, Vm = 200R
I = Vm/πR = 200R/πR.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “1-Phase-Diode Rectifiers HW-2”.
1. A 1-phase 230V, 1KW heater is connected across a 1-phase HW rectifier . The power delivered to the heater is
a) 300 W
b) 400 W
c) 500 W
d) 600 W
Answer: c
Explanation: R = /1000
V = /2
P = V 2 /R = 500W.
2. A 1-phase half wave diode rectifier with R load, has input voltage of 240 V. The input power factor is
a) Unity
b) 0.707 lag
c) 0.56 lag
d) 0.865 lag
Answer: c
Explanation: Input p.f = V/Vs
Vrms is the RMS value of output voltage. Vrms = /2
Vs = 230
pf = 0.707.
3. A 1-phase half wave diode rectifier with R = 1 KΩ, has input voltage of 240 V. The diode peak current is
a) Zero
b) 240mA
c) 24mA
d) 0.24mA
Answer: b
Explanation: Diode peak current = peak current through the load = Vo/R = Vm/2R.
4. For the below given circuit, after the switch is closed the voltage across the load remains constant.
power-electronics-mcqs-1p-diode-rectifiers-hw-2-q4
Assuming that all initial conditions are zero. The element across the load would be a/an
a) resistor
b) capacitor
c) inductor
d) data not sufficient
Answer: b
Explanation: As the voltage remains constant as soon as the switch is closed, the element is most likely to be a capacitor.
5. For the below given circuit,
power-electronics-mcqs-1p-diode-rectifiers-hw-2-q5
After the supply voltage is given the
a) diode starts conducting
b) diode starts conducting only when Vs exceeds Vdc
c) diode never conducts
d) diode stops conducting only when Vs exceeds Vdc
Answer: b
Explanation: The diode will be forward biased only when Vs will be greater than Vdc.
6. For the below given circuit,
power-electronics-mcqs-1p-diode-rectifiers-hw-2-q6
With Vs = Vm sin ωt . The expression for the average voltage is
a) Vm
b) Vm/2π
c) Vm/π
d) Vm/2
Answer: c
Explanation: Due to the FD, we get 1st quadrant operation.
Where, output voltage = 1/2π [ ∫Vm sin ωt d ], integration runs from 0 to 180 degrees.
7. For the below given circuit, the
power-electronics-mcqs-1p-diode-rectifiers-hw-2-q7
a) output voltage is never positive
b) output current is never positive
c) output current is never zero
d) output voltage is never zero
Answer: d
Explanation: The output voltage is the voltage across the resister and Vdc. Even if the current falls to zero, the output voltage will be equal to Vdc.
8. For the below given circuit,
power-electronics-mcqs-1p-diode-rectifiers-hw-2-q8
Vs = 325 sin ωt
The ripple voltage is
a) 125.32 V
b) 255.65 V
c) 325 V
d) 459.12 V
Answer: a
Explanation: Ripple voltage = √(Vrms 2 + Vavg 2 )
Vrms = Vm/2
Vavg = Vm/π
9. For a single phase half wave rectifier, the rectifier efficiency is always constant & it is
a) 4/π 2
b) 8/π 2
c) 100
d) 2/π 2
Answer: a
Explanation:
Rectifier efficiency = Pdc/Pac
Pdc = /π 2
Pac = 4/.
10. For the below given circuit,
power-electronics-mcqs-1p-diode-rectifiers-hw-2-q7
The power delivered to the load in Watts is
a) I.Vdc
b) I.Vdc + I 2 .R
c) I.Vdc – I 2 .R
d) I.Vdc + I 2 .R
Answer: b
Explanation: P = power delivered to the resister + power delivered to the emf source.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “1-Phase-Diode Rectifiers FW-1”.
1. A single-phase full wave mid-point type diode rectifier requires __________ number of diodes whereas bridge type requires _________
a) 1,2
b) 2,4
c) 4,8
d) 3,2
Answer: b
Explanation: A bridge type requires 4 diodes which are connected in a bridge, and the mid-point has 2 diodes.
2. A single-phase full wave rectifier is a
a) single pulse rectifier
b) multiple pulse rectifier
c) two pulse rectifier
d) three pulse rectifier
Answer: c
Explanation: It is a two-pulse rectifier as it generates 2 pulses per cycle.
3. The below shown circuit is that of a
power-electronics-mcqs-1p-diode-rectifiers-fw-1-q3
a) full wave B-2 type connection
b) full wave M-2 type connection
c) half wave B-2 type connection
d) half wave M-2 type connection
Answer: b
Explanation: Full wave M-2 type employs a transformer and two diodes.
4. In a 1-phase full wave bridge rectifier with M-2 type of connection has secondary side voltage Vs = Vm sin ωt, with R load & ideal diodes.
The expression for the average value of the output voltage can be given by
a) 2Vm/π
b) Vm/π
c) Vm/√2
d) 2Vm/√2
Answer: a
Explanation: The voltage waveform is a pulsating voltage with peak value Vm & symmetrical about π.
Vo = ∫ π Vm sinωt d
5. The below shown circuit is that of a
power-electronics-mcqs-1p-diode-rectifiers-fw-1-q5
a) full wave B-2 type connection
b) full wave M-2 type connection
c) half wave B-2 type connection
d) half wave B-2 type connection
Answer: a
Explanation: Full wave B-2 type uses 4 diodes in a bridge connection.
6. In a 1-phase full wave bridge rectifier with M-2 type of connection has secondary side voltage Vs = Vm sin ωt,
with R load & ideal diodes.
The expression for the rms value of the output voltage can be given by
a) Vm/π
b) Vm/√2
c) Vm
d) Vm 2
Answer: b
Explanation: The voltage waveform is a pulsating voltage with peak value Vm & symmetrical about π.
Vo = ∫ π Vm 2 sin 2 ωt d = Vm/√2.
7. For the circuit shown below, find the power delivered to the R load
power-electronics-mcqs-1p-diode-rectifiers-fw-1-q3
Where,
Vs = 230V
Vs is the secondary side single winding rms voltage.
R = 1KΩ
a) 46 W
b) 52.9 W
c) 67.2 W
d) 69 W
Answer: b
Explanation: Power delivered to the load is V.I = V 2 /R
Where, for the given circuit, V = Vs.
8. The PIV experienced by the diodes in the mid-point type configuration is
a) Vm
b) 2Vm
c) 4Vm
d) Vm/2
Answer: b
Explanation: In the m-2 type connection, each diode experiences a reverse voltage of 2Vm.
9. For the circuit shown below, find the value of the average output current.
power-electronics-mcqs-1p-diode-rectifiers-fw-1-q3
Where,
Vs = 230V
R = 1KΩ
Vs is the secondary side single winding rms voltage.
a) 100mA
b) 107mA
c) 200mA
d) 207mA
Answer: d
Explanation: I = Vo/R
Vo = 2Vm/π.
10. In the circuit, let Im be the peak value of the sinusoidal source current. The average value of the diode current for the below given configuration is
power-electronics-mcqs-1p-diode-rectifiers-fw-1-q5
a) Im
b) Im/2
c) Im/π
d) Im/√2
Answer: b
Explanation: The peak current through the diodes is same as that passing from the load. Each diode pair conducts for 180 degrees. Hence, 1/2 of a cycle. Average current = Im/2.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “1-Phase-Diode Rectifiers FW-2”.
1. The PIV experienced by each of the diodes for the below shown rectifier configuration is
power-electronics-mcqs-1p-diode-rectifiers-fw-2-q1
a) Vm
b) 2Vm
c) 3Vm
d) Vm/2
Answer: a
Explanation: When any diode is reversed biased due to the negative half cycle, the maximum peak value through it will be Vm.
2. For the circuit shown in the figure below,
power-electronics-mcqs-1p-diode-rectifiers-fw-2-q1
Vs = 230 V
R = 10Ω
Find the average value of output current.
a) 207.04 A
b) 20.704 A
c) 2.0704 A
d) 207.04 mA
Answer: b
Explanation: I = Vo/R
Vo = 2Vm/π.
3. Choose the correct statement regarding the below given circuit.
power-electronics-mcqs-1p-diode-rectifiers-fw-2-q3
a) The load current is never negative
b) The load current is never zero
c) The load current is never positive
d) The load voltage is never negative
Answer: a
Explanation: Due to the inductive nature of the load, the Diodes are force conducted & voltage goes negative but the current can never fall below zero.
4. For a single phase, full bridge, diode rectifier excited from a 230 V, 50 Hz source. With R = 10 Ω & the inductance large enough to maintain continues conduction, the average and rms values of diode currents will be
a) 7.85 A, 8 A
b) 10.35 A, 7.85 A
c) 10.35 A, 14.6 A
d) 8 A, 8 A
Answer: c
Explanation:
Id = Io/2 = Vo/2R
Id = Io/√2 = Vo/R√2.
5. The circuit shown below, will have the output voltage waveform similar to that of a
power-electronics-mcqs-1p-diode-rectifiers-fw-2-q5
a) half wave rectifier with RL load
b) full wave bridge rectifier with RL load
c) full wave bridge rectifier with R load
d) full wave bridge rectifier with RC load
Answer: c
Explanation: The FD short circuits the load & voltage waveform is similar to that of a Full wave bridge rectifier with R load.
6. For the circuit shown below, the load current attains the maximum value at ωt =
power-electronics-mcqs-1p-diode-rectifiers-fw-2-q3
a) 0
b) π
c) 2π
d) none of the mentioned
Answer: b
Explanation: As the load is RL, the load current will be maximum when the output voltage waveform falls to zero i.e. at π. At π the inductor is charged to its maximum value and starts delivering power to the source.
7. For a single phase, full bridge, diode rectifier excited from a 230 V, 50 Hz source. With R = 10 Ω & the inductance large enough to maintain continuous conduction, the value of the supply power factor will be
a) 0.707 lag
b) 0.9 lag
c) 0.86 lag
d) Unity
Answer: b
Explanation:
Pf = Vs.Is.cosθ/Vo.Io
Io = Vo/R A
Vo = 2Vm/π Volts.
8. The rectification efficiency for B-2 type & M-2 type full wave diode rectifiers are ___ & ___ respectively.
a) 8/π & 4/π
b) 4/π & 8/π
c) 8/π & 8/π
d) 4/π & 4/π
Answer: c
Explanation: B-2 type has efficiency 8/π. M-2 type has efficiency half of that of a B-2 type.
9. A load of R = 60 Ω is fed from 1phase, 230 V, 50 Hz supply through a step-up transformer & than a diode. The transformer turns ratio = 2. The power delivered to the load is
a) 614 Watts
b) 714 Watts
c) 814 Watts
d) 914 Watts
Answer: b
Explanation: P = Vo 2 /R
Vo = Vm/π
AC supplied to the rectifier is 2 x 230 = 460 V
Therefore, Vo = √2 x 460 / π = 207.04
P = 714.43 W.
10. For the circuit shown below, D11 & D14 conduct from?
power-electronics-mcqs-1p-diode-rectifiers-fw-2-q5
Assume that anode of D12 is positive at ωt = 0 and likewise.
a) 0 to π
b) π to 2π
c) 2π to 3π
d) 0 to π/2
Answer: b
Explanation: In the first cycle i.e. 0 to π, D12 and D13 conduct. In the next cycle i.e. π to 2π, D11 and D14 conduct.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “3-Phase Diode Rectifiers-1”.
1. In a three-phase half wave rectifier usually, the primary side of the transformer is delta connected because
a) it has no neutral connection
b) we can get greater output voltage
c) it provides a path for the triplen harmonics
d) it provides better temperature stability
Answer: c
Explanation: The delta connected winding help circulating and eliminating the triplen harmonics.
2. The diode rectifier circuit given below is that of a
power-electronics-mcqs-3phase-rectifiers-1-q2
a) three-phase half wave common cathode arrangement
b) three-phase half wave common anode arrangement
c) three-phase full wave common cathode arrangement
d) three-phase full wave common anode arrangement
Answer: b
Explanation: Common anode arrangement because all the anodes are connected to the load side.
3. In a three-phase half wave diode rectifier using 3 diodes, each diode conducts for
a) 90 degrees
b) 120 degrees
c) 180 degrees
d) 360 degrees
Answer: b
Explanation: Each diode conducts for 120 degrees, starting from ωt = 30 degrees.
4. In the below shown diode rectifier circuit,
power-electronics-mcqs-3phase-rectifiers-1-q4
The diodes D1, D2 & D3 are connected to phases R,Y and B respectively as shown
The phase sequence is R-Y-B.
The diode D1 would conduct from
a) 0 to 90 degrees
b) 30 to 150 degrees
c) 0 to 180 degrees
d) 30 to 180 degrees
Answer: b
Explanation: It conducts from 30 to 150, for 90 degrees. D1 starts conducting first as it will be the most positive as it is connected to the R phase.
5. In the below shown diode rectifier circuit,
power-electronics-mcqs-3phase-rectifiers-1-q4
The diodes D1, D2 & D3 are connected to phases R,Y and B receptively as shown
The phase sequence is R-Y-B.
The diode D3 conducts from
a) 0 to 270 degrees
b) 270 to 390 degrees
c) 270 to 450 degrees
d) 270 to 360 degrees
Answer: c
Explanation: It conducts from 270 to 450, for 120 degrees. D1 starts conducting first as it will be the most positive as it is connected to the R phase & likewise.
6. In a three-phase half wave diode rectifier using 3 diodes,
a) All diodes conduct together
b) Only two diodes conduct at a time
c) Only one diode conducts at a time
d) None of the above mentioned
Answer: c
Explanation: 3 diodes, each conduct for 120 degree at a time.
7. In a three-phase half wave diode rectifier, if Vmp is the maximum phase voltage, then the output voltage on a R load varies from
a) 0 to Vmp
b) 0.5 Vmp to Vmp
c) Vmp to 3Vmp
d) –Vmp to Vmp
Answer: b
Explanation: The voltage value is positive and varies from Vmp to Vmp.
8. The average value of the output voltage, in a 3-phase half wave diode rectifier with Vml as the maximum line voltage value, is given by the expression
a) Vml/3π
b) 2Vml/3π
c) 3Vml/2π
d) 3Vml
Answer: c
Expression: The average value can be obtained by
3 x [ 1/2π x Vml sin ωt d ] The integration runs from π/6 to 5π/6 as the diode is conducting for 120 degrees each.
9. In a three-phase half wave 6-pulse mid-point type diode rectifier, each diode conducts for
a) 120°
b) 60°
c) 90°
d) 180°
Answer: b
Explanation: In a six-pulse rectifier, each diode conducts once every one cycle, 60° x
6 diodes = 360°.
10. A step-down delta-star transformer, with per-phase turns ration of 5, is fed from a 3-phase, 1100 V, 50 Hz source. The secondary of this transformer through a 3-pulse type rectifier feeds a R load of 10 Ω. Find the maximum value of the load current .
a) √2 x 22 A
b) 1 x 11 A
c) √2 x 11 A
d) 1 x 22 A
Answer: a
Explanation:
Vph = 1100/5 = 220 V
Vmp = √2 x 220 V
Imp = Vmp/R.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Three-Phase Rectifiers-2”.
1. A step-down delta-star transformer, with per-phase turns ratio of 5 is fed from a 3-phase 1100 V, 50 Hz source. The secondary of this transformer is connected through a 3-pulse type rectifier, which is feeding feeding an R load. Find the average value of output voltage.
a) 220 V
b) 257 V
c) 1100/√3 V
d) 206 V
Answer: b
Explanation:
Vph = 1100/5 = 220 V
Vmp = √2 x 220 V
Vo = 3√3/2π x Vmp = /.
2. The circuit shown below is that of a
power-electronics-mcqs-q2
a) 3-phase, 6-pulse, diode rectifier
b) 3-phase, 6-pulse, diode inverter
c) 3-phase, 3-pulse, diode rectifier
d) 3-phase, 3-pulse, diode inverter
Answer: a
Explanation: A 3-phase, 6-pulse rectifier consists of 6 diodes connected in 3 legs. Two diodes conduct at a time.
3. A step-down delta-star transformer, with per-phase turns ratio of 5 is fed from a 3-phase 1100 V source. The secondary of this transformer is connected through a 3-pulse type rectifier, which is feeding an R load.
The power delivered to the load is 6839.3 Watts.
The maximum value of the load current is √2 x 22 A.
Fin, the rms value of output voltage Vo
a) 257.3 V
b) 220 V
c) 261.52 V
d) 248.32 V
Answer: c
Explanation: Power delivered to the load = Vo 2 /R
Imp = Vmp/R
Therefore, R = Vmp/Imp = / = 10 Ω
Put R in equation & find the required R.M.S voltage.
4. From the diode rectifier circuit shown below, with phase sequence R-Y-B, diodes D3 & D5 conduct when
power-electronics-mcqs-q2
a) R is the most positive & B is the most negative
b) R is the most positive & Y is the most negative
c) R is the most negative & B is the most positive
d) R is the most negative & Y is the most positive
Answer: b
Explanation: Which diode will conduct depends on where is it in connected? as in in which phase?. D3’s anode is connected to the R phase, hence it will turn on when R is the most positive.
5. From the diode rectifier circuit shown below, with phase sequence R-Y-B, from ωt = 150° to 270°
power-electronics-mcqs-q2
a) D1
b) D2
c) D3
d) None of the diodes conduct
Answer: b
Explanation: Construct the phase voltage waveforms on a graph. At 150 degree, D2 is forward biased while the other positive group diodes i.e. D2 and D3 remain reserved biased.
6. A 3-phase 6-pulse diode rectifier is shown below with phase sequence R-Y-B. The negative group of diodes conduct in sequence
power-electronics-mcqs-q2
a) D4-D5-D6
b) D5-D6-D4
c) D6-D5-D4
d) D6-D4-D5
Answer: b
Explanation: The conduction sequence always depends on the phase sequence, which diode is conducting will depend upon which phase voltage is active at that moment.
7. For a 3-phase 6-pulse diode rectifier, has Vml as the maximum line voltage value on R load. The peak current through each diode is
a) Vml/2R
b) 2Vml/R
c) Vml/R
d) Insufficient Data
Answer: c
Explanation: Two diodes conduct at a time, constructing the equivalent circuit with supply, R & replacing the conducting diodes by S.C & non-conducting as O.C, the required value can be found out.
8. A 3-phase bridge rectifier, has the average output voltage as 286.48 V. Find the maximum value of line voltage
a) 100 V
b) 200 V
c) 300 V
d) 400 V
Answer: c
Explanation: Vo = 3Vml/π
Vml = /3 = 300 V.
9. A 3-phase bridge rectifier charges a 240 V battery. The rectifier is given a 3-phase, 230 V supply. The current limiting resistance in series with the battery is of 8 Ω.
Find the average value of battery charging current.
a) 12.56 A
b) 8.82 A
c) 9.69 A
d) 6.54 A
Answer: b
Explanation: Vo = /π = 310.56 V
Draw the battery charging circuit,
Vo = E +
Io = /R = /8.
10. A 3-phase bridge rectifier charges a 240-V battery. The rectifier is given a 3-phase 230 V supply. The current limiting resistance in series with the battery is 8 Ω.
Find the power delivered to the battery .
a) Pdc = 2000 W
b) Pdc = 1226 W
c) Pdc = 2356 W
d) Pdc = 2116 W
Answer: d
Explanation: Vo = /π = 310.56 V
Draw the battery charging circuit,
Vo = E +
Io = /R = /8 = 8.82A
Pdc = 240 x 8.82 = 2116 W.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Filters-1”.
1. In practice the output from the diode rectifier has
a) AC component only
b) DC component only
c) AC + DC component
d) None of the mentioned
Answer: c
Explanation: The output contents along with the DC components the AC harmonics which does no useful work & reduces the efficiency.
2. Choose the correct statement
a) The AC component in the output of rectifier does the useful work
b) The AC component in the output of rectifier increases the efficiency of the system
c) The AC component in the output of rectifier causes ohmic losses
d) The AC component in the output of rectifier does not affect the operation
Answer: c
Explanation: A rectifier is used to convert AC to DC. Lower the AC components in the output lower are the ohmic losses.
3. An L filter is connected ________
a) in series
b) in parallel
c) in both series and parallel
d) none of the mentioned
Answer: a
Explanation: Inductor has a very important property that the current through it cannot change rapidly. We can make use of this property by connecting it in series.
4. In case of an L filter connected with a rectifier in series with the load, it offers ________ impedance to ac whereas _______ resistance to dc respectively.
a) high, high
b) high, low
c) low, high
d) low, low
Answer: b
Explanation: It offers high impedance to AC such as the AC ripples do not pass through the load.
5. In case of a C filter, the AC is not allowed to pass to the load by
a) offering it high impedance
b) offering it low impedance
c) short circuiting the AC component
d) open circuiting the AC component
Answer: c
Explanation: AC ripples are not allowed to pass, by S.C the AC ripples as the C is always connected in parallel with the load.
6. A capacitor filter or C filter can be used in a rectifier by connecting it
a) in parallel with the load
b) in series with the load
c) in parallel with the supply
d) in series with the supply
Answer: a
Explanation: AC ripples are not allowed to pass, by S.C the AC ripples as the C is always connected in parallel with the load.
7. Examine the below shown circuit.
power-electronics-questions-answers-filters-1-q7
While D7 & D10 are reversed biased the
a) capacitor is charging through Vs
b) capacitor is discharging through Vs
c) capacitor is charging through R
d) capacitor is discharging through R
Answer: d
Explanation: When the two are reversed biased and the other two SCR’s are not yet gated, the stored energy in the capacitor is supply the load R.
8. When a certain type of filter is connected across the R load of a full wave bridge type diode biased rectifier, the following output voltage waveform is obtained
power-electronics-questions-answers-filters-1-q8
The filter connected is most likely to be a/an
a) L filter
b) C filter
c) LC filter
d) None of the above mentioned
Answer: b
Explanation: It is to be noted that the waveform is that of a voltage parameter. The voltage goes up for a very small time and than goes down for longer time. For every half wave the peak of the waveform is to the left, if it were to be on the right it would be more likely to be a L filter.
9. For the below given circuit,
power-electronics-questions-answers-filters-1-q7
When the output current rises, the capacitor current
a) is increasing
b) is decreasing
c) is constant
d) is zero
Answer: b
Explanation: When the load current is increasing, the capacitor is charging through the source voltage & its current keeps on falling & voltage rises.
10. A single-phase diode B-2 rectifier is fed from a 250 V, 50 Hz source & connected to a load of R = 400 Ω.
Design a capacitor filter such that the ripple factor of the output voltage is less than 5 %. Find the value of the capacitance of the C filter.
a) 156.4 μF
b) 500 μF
c) 189 μF
d) 246 μF
Answer: c
Explanation: Use,
C = K * [1 + 1/√2RF] Where, K = 1/4fR.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Filters-2”.
1. A diode rectifier with a C filter will have average output voltage _______ than that obtained by a rectifier without the C filter.
a) lower than
b) higher than
c) same as
d) none of the mentioned
Answer: b
Explanation: C filter improves the efficiency and eliminates ripples by keeping the output voltage constant.
2. An inductor filter connected in series with a resistive load provides a
a) smoothing of the output voltage waveform
b) smoothing of the input voltage waveform
c) smoothing of the output current waveform
d) smoothing of the input current waveform
Answer: c
Explanation: Filter is always connected to the load side . Inductor has the property of keeping the current smooth and constant.
3. In the below given configuration, L is connected as a filter across the R load.
power-electronics-multiple-choice-questions-answers-q3
The average ammeter current is
a) 2Vm/R
b) 2Vm/πR
c) Vm/πR
d) Vm/R
Answer: b
Explanation: I = Vo/R
Vo = 2Vm/π.
4. The current ripple factor is the ratio of
a) Average value/RMS value
b) RMS value/Average value
c) Average value/Maximum value
d) Maximum value/RMS value
Answer: b
Explanation: CRF = Ir/Io.
5. In case of an L filter, the ripple current increases with
a) increase in Load
b) decrease in Load
c) increase in the value of L
d) ripple current never increases
Answer: b
Explanation: When the load is reduced the resistance R is increased & ripple current increases.
6. In case of a C filter, if R is increased
a) ripple factor is reduced
b) ripple factor is increased
c) ripple factor is not affected
d) increases noise in the circuit
Answer: a
Explanation: The time constant RC is increased therefore the ripple factor is reduced.
7. When a certain type of filter is connected across the R load, of a full wave bridge diode biased rectifier the following output current waveform is obtained.
power-electronics-multiple-choice-questions-answers-q7
The filter connected is most likely a
a) L filter
b) C filter
c) LC filter
d) None of the above mentioned
Answer: a
Explanation: The above waveform is that of the output current. The peak per half cycle is on the right, hence it is more likely to be an L filter.
8. C filters are suitable for ___________ load resistances and L filters are suitable for _____________ load resistances.
a) low, low
b) high, high
c) high, low
d) low, high
Answer: c
Explanation: In case of L filter, if R is lowered, the time constant increases, therefore ripple factor reduces. In case of a C filter, the time constant RC is increased, therefore, ripple factor is reduced.
9. An LC filter will have ripple factor value___________ Misplaced &
a) lower than that obtained by L filter but higher than that obtained by C filter
b) lower than that obtained by C filter but higher than that obtained by L filter
c) lower than that obtained by either L or C filter
d) higher than that obtained by either L or C filter
Answer: c
Explanation: Simply, an LC filter combines both the advantages of L and C filters.
10. In a single-phase full wave rectifier ___________ order harmonics are the most dominant
a) first
b) second
c) third
d) fourth
Answer: b
Explanation: The output voltage is given by,
Vo = – cos 2ωt – cos 4ωt . . . .
Hence the other values after the second factor become very very small hence they can be eliminated.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Firing Circuits-1”.
1. In a single pulse semi-converter using two SCRs, the triggering circuit must produce
a) two firing pulses in each half cycle
b) one firing pulse in each half cycle
c) three firing pulses in each cycle
d) one firing pulse in each cycle
Answer: b
Explanation: A single phase semi-converter has only two SCRs & two diodes. Hence, only two pulses are required in each cycle, one in each half.
2. In a 3-phase full converter using six SCRs, gating circuit must provide
a) one firing pulse every 30°
b) one firing pulse every 90°
c) one firing pulse every 60°
d) three firing pulses per cycle
Answer: c
Explanation: 60° x 6 = 360°.
3. In the complete firing circuit, the driver circuit consists of
a) pulse generator & power supply
b) gate leads & power supply
c) pulse amplifier & pulse transformer
d) pulse detector & pulse amplifier
Answer: c
Explanation: The driver circuit consists of a pulse amplifier to increase the magnitude of the gate pulse to a sufficient value. The pulse transformer then provides pulses to individual SCRs.
4. Find the average gate power dissipation (P gav ) when the maximum allowable gate power dissipation (P gm ) = 10 kW, with a duty cycle = 50 %.
a) 10 KW
b) 5 KW
c) 2.5 KW
d) 7 KW
Answer: b
Explanation: (P gm ) = (P gav )/Duty Cycle.
5. The magnitude of gate voltage and gate current for triggering an SCR is
a) inversely proportional to the temperature
b) directly proportional to the temperature
c) inversely proportional to the anode current requirement
d) directly proportional to the anode current requirement
Answer: a
Explanation: Higher the temperature lesser will be the gate current required as the temperature must have already excited some of the atoms.
6. Find the amplitude of the gate current pulse, when the gate-cathode curve is given by the relation Vg = [ x Ig] The peak gate drive power is 5 Watts.
a) 359mA
b) 659mA
c) 1.359 A
d) 1.659 A
Answer: b
Explanation:
.Ig = 5 Watts
Ig = 0.59 A.
7. The gate-cathode curve for an SCR is given by the relation Vg = Ig. The gate voltage source is a rectangular pulse of peak value 15 V and current = 0.659 A. Find the source resistance.
power-electronics-questions-answers-firing-circuits-1-q7
a) 90.2 Ω
b) 11.24 Ω
c) 46.2 Ω
d) 39 Ω
Answer: b
Explanation: Es = Rs.Ig + Vg
Vg = 1+10 Ig
Therefore Rs = /0.659.
8. Find the triggering frequency when the average gate power dissipation = 0.3 W and the peak gate drive power is 5 Watts. The gate source has a pulse width of 20 μsec duration.
a) 3 kHz
b) 0.3 kHz
c) 30 kHz
d) 0.03 mHz
Answer: a
Explanation: (P gm ) = (P gav )/Duty Cycle
Duty Cycle = f x T = (P gav )/(P gm )
Duty Cycle = 0.3/5
T = 20 μsec.
0.3/5 = f x T
f = /(5 x 20 x 10 -6 ) = 3000 Hz.
9. The duty cycle can be written as
a) f x T
b) f/T
c) T/f
d) f
Answer: a
Explanation: The duty cycle is defined as the ratio of pulse-on period to periodic time of pulse.
The pulse on period is T, and the periodic time is 1/f.
It is to be noted that T = pulse width whereas f = = frequency of firing or pulse repetition rate.
10. The major function of the pulse transformer is to
a) increase the voltage amplitude
b) reduce harmonics
c) isolate low & high power circuit
d) create periodic pulses
Answer: c
Explanation: Isolation of the two circuit is done by the transformer, as the transformer is a magnetically coupled device and any mishap at the load side will not damage the other side of the circuitry.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Firing Circuits – 2”.
1. In a resistance firing circuit the firing angle
a) cannot be greater than 120°
b) cannot be greater than 90°
c) cannot be greater than 180°
d) cannot be greater than 160°
Answer: b
Explanation: The R firing circuits cannot be used for alpha greater than 90 degrees.
2. For a R firing circuit, the maximum value of source voltage is 100 V. Find the resistance to be inserted to limit the gate current to 2 A.
a) 5 Ω
b) 50 Ω
c) 500 Ω
d) 0.5 Ω
Answer: b
Explanation: R = 100/2 = 50 Ohm.
3. The diode in the R firing circuit
a) ensures that the gate voltage is a half wave DC pulse
b) ensures that the gate voltage is a full wave DC pulse
c) ensures that the gate voltage is a half wave AC pulse
d) ensures that the gate voltage is a full wave AC pulse
Answer: b
Explanation: The diode is placed between the resistances and gate which ensures that the current flows in one direction only.
4. In case of an RC half wave triggering circuit, the firing angle can be ideally varied between
a) 0 to 180
b) 0 to 90
c) 0 to 120
d) 0 to 360
Answer: a
Explanation: Unlike the R firing circuit, the RC firing circuits can be used to obtain firing angle greater than 180. Although practically 0 and 180 degree is improbable.
5. In the figure given below, the resistance R1 is used to
power-electronics-questions-answers-firing-circuits-2-q5
a) keep the gate circuit voltage drop minimum
b) limit the gate current to a safe value when R2 = 0
c) limit the gate current to a safe value when R2 is very large
d) allow the gate power dissipation
Answer: b
Explanation: As R2 is the variable, R1 makes sure that the current does not exceed the maximum value when R2 is kept at zero position.
6. In case of a R firing with R2 as the variable resistance, V gp and V gt the value of R2 is so adjusted such that
a) V gp = V gt
b) V gp > V gt
c) V gp < V gt
d) V gp = V gt = 0
Answer: a
Explanation: For turning on the device, the peak of gate voltage must be equal to the gate triggering voltage.
7. In case of a R firing circuit with V gp > V gt
a) α = 90°
b) α > 90°
c) α < 90°
d) α = 0°
Answer: c
Explanation: For the values of Vgp great than the gate triggering voltage the firing angle is less than 90°. And for V gp = V gt the firing angle is equal to 90°. Α cannot go beyond 90° in case of a R firing circuit.
8. The figure shown below is that of a
power-electronics-questions-answers-firing-circuits-2-q8
a) R firing circuit
b) RC half-wave firing circuit
c) RC full-wave firing circuit
d) UJT triggering circuit
Answer: b
Explanation: The given circuit is a RC half-wave firing circuit.
9. The figure shown below is that of an RC firing circuit.
power-electronics-questions-answers-firing-circuits-2-q8
In case of negative cycle at Vs, the capacitor C
a) charges through D2 with lower plate negative
b) charges through D1 with lower plate negative
c) charges through D2 with lower plate positive
d) charges through D1 with lower plate positive
Answer: c
Explanation: The current flows through Vs+ – C – D2 – Load – Vs.
10. Find the value of R in case of an RC firing circuit which is to be turned on with a source voltage of 150 V and the following parameters.
Igt = 2A
Vd = 1.5V
Vgt = 125 V
a) 11.75 Ω
b) 54.25 Ω
c) 96 Ω
d) 5 Ω
Answer: a
Explanation:
R = /Igt.
This set of Power Electronics question bank focuses on “Firing Circuits-3”.
1. For the following RC triggering circuit with R load and a firing angle of α, the voltage across the R load is zero for
power-electronics-question-bank-q1
a) ωt = 0 to α and ωt = π to 2π+α
b) ωt = 0 to α
c) ωt = π to 2π+α
d) ωt = α to 2π
Answer: a
Explanation: The SCR is triggered at α therefore voltage appears across the load from ωt = α to π only as in case of a half wave firing circuit the SCR is naturally commuted at π.
2. For the following circuit, the SCR is
power-electronics-question-bank-q1
a) forced commutated at ωt = π
b) forced commutated at ωt = 2π
c) naturally commutated at ωt = π
d) naturally commutated at ωt = 2π
Answer: c
Explanation: Circuit is that of a half-wave RC firing circuit where natural commutation takes place at π due to the sinusoidal voltage source.
3. For the following RC triggering circuit with R load and a firing angle of α, the voltage across the SCR will be zero for
power-electronics-question-bank-q1
a) ωt = 0 to α & ωt = π to 2π+α
b) ωt = 0 to π
c) ωt = α to 2π
d) ωt = α to π
Answer: d
Explanation: The SCR is triggered at α therefore voltage appears across the load from ωt = α to π only as in case of a half wave firing circuit the SCR is naturally commuted at π.
4. In case of an RC full wave firing circuit with R load, the voltage across the load is zero for____________
a) ωt = 0 to α and ωt = π to 2π+α
b) ωt = 0 to α
c) ωt = π to 2π+α
d) ωt = α to 2π
Answer: b
Explanation: The SCR is not triggered until α, hence no current flows until α.
5. For an RC full wave firing circuit the empirical formula for calculating the value of RC is
a) RC = 157/ω
b) RC = 157 x ω
c) RC = ω/157
d) RC = 157 x ω 2
Answer: a
Explanation: RC = 157/ω, Where ω is the angular frequency.
6. Pulse triggering can be only used by the _____________ type of triggering circuit
a) R
b) RC
c) UJT
d) RLC
Answer: c
Explanation: R & RC produce prolonged pulses which increases the gate power dissipation. There is no RLC firing circuit.
7. The UJT terminals are
a) E, B1 and B2
b) E1, E2 and B
c) E, G and C
d) G, S and D
Answer: a
Explanation: The terminals are Emitter, Base1 & Base 2.
8. In case of the UJT firing circuit, when the UJT turns on
a) the capacitor starts to charge
b) the capacitor starts to discharge
c) the capacitor remains unaffected
d) there is no capacitor in a UJT firing circuit
Answer: b
Explanation: When the UJT is fired, C starts to discharge.
9. In the UJT firing circuit, the pulses are generated while the
a) capacitor charges
b) capacitor discharges
c) capacitor voltage is zero
d) there is no capacitor in a UJT firing circuit
Answer: b
Answer: When the C starts to discharge through the resistance with time constant RC the UJT turns on & hence pulses are generated.
10. Find the value of the charging resistor in case of a UJT firing circuit with firing frequency of 2 kHz, C = 0.04 μF, η = 0.72
a) 5.62 kΩ
b) 37 kΩ
c) 4.23 kΩ
d) 9.82 kΩ
Answer: d
Explanation: T = 1/f = 1/2 kHz
R = T/C ln.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Gating Circuits”.
1. If the RC firing circuit used for firing an SCR is to be used to fire a TRIAC then
a) the capacitor should be removed
b) the diode should be replaced by a diac
c) the diode should be replaced by a bjt
d) the diode should be shorted using a resistor
Answer: b
Explanation: The TRIAC is a bidirectional SCR, hence it will need gating in both the directions. This can be achieved by replacing the diode by a DIAC .
2. In the thyristor gating circuit, the supply to the pulse amplifier is provided by the
a) zcd
b) isolation transformer
c) synchronizing transformer
d) control signal generator
Answer: b
Explanation: Isolation transformer provides the supply to the amplifier and also provides the necessary isolation for the load and triggering circuit.
3. In the thyristor gating circuit, the ZCD is used to
a) amplify the voltage
b) produce a train of pulses
c) convert AC input the ramp voltage
d) used to step-down the voltage
Answer: c
Explanation: It is used to convert the AC synchronizing input voltage into ramp voltage & synchronizes it with the zero crossing of the AC supply.
4. The firing-angle delay is
a) inversely proportional to the synchronizing transformer voltage
b) inversely proportional to the control signal voltage
c) directly proportional to the synchronizing transformer voltage
d) directly proportional to the control signal voltage
Answer: d
Explanation: If Ec is lowered the firing angle decreases & vice-verse.
5. The pulse gating is not suitable of
a) R loads
b) RC loads
c) RL loads
d) It is suitable of every type of load
Answer: c
Explanation: It is not suitable of RL load because initiation of SCR conduction is not well defined in these types of loads.
6. In case of a cosine firing scheme, __________ is used to get a cosine wave
a) ic 555
b) a comparator
c) an integrator circuit
d) a differentiator circuit
Answer: c
Explanation: The Sync. Transformer is connected to a integrator to obtain a cosine-wave.
7. If the gating circuits generator negative pulses, then those can be removed by using
a) schmit triggers
b) clippers
c) clampers
d) zener diodes
Answer: b
Explanation: The clippers can be used to clip the negative part.
8. The improved version of the UJT oscillator triggering circuit is the
a) ramp & pedal triggering
b) rc triggering
c) cosine-pulse triggering
d) ramp triggering
Answer: a
Explanation: The ramp & pedal triggering is the improved version of the UJT oscillator triggering circuit.
9. R B1 = 3 kΩ & R B2 = 6 kΩ. Find the intrinsic stand-off ratio of the UJT.
a) 9
b) 1/3
c) 2/3
d) 3
Answer: b
Explanation:
η = R B1 /(R B1 +R B1 ).
10. The decaying factor in the wave shape of the output pulses from the pulse transformer is its
a) transformer ratio
b) inductance
c) capacitance
d) resistance
Answer: b
L is the decaying factor in the waveform which emerge from the PT.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “SCR Commutation Techniques-1”.
1. The thyristor turn-off requires that the anode current
a) falls below the holding current
b) falls below the latching current
c) rises above the holding current
d) rises above the latching current
Answer: a
Explanation: For effective turn-off of the SCR the anode current must fall below the holding current value.
2. In case of class A type commutation or load commutation with low value of R load the
a) L is connected across R
b) L-C is connected across R
c) L is connected in series with R
d) L-C is connected in series with R
Answer: d
Explanation: In case of Class A commutation the requirement is that the circuit should be an under-damped RLC circuit.
3. The class A commutation or load commutation is possible in case of
a) dc circuits only
b) ac circuits only
c) both DC and AC circuits
d) none of the above mentioned
Answer: a
Explanation: The nature of the circuit should be such that when energized from the source, current must a a natural tendency to decay to zero for load commutation to occur in a SCR circuit.
4. In case of class B commutation or resonant-pulse commutation with L = 5 μH and C = 20 μC with initial voltage across the capacitor = 230 V. Find the peak value of resonant current.
a) 560 A
b) 460 A
c) 360 A
d) 260 A
Answer: b
Explanation: Ip = Vs x √C/L.
5. In case of class B commutation or resonant-pulse commutation with L = 5 μH and C = 20 μC with the initial voltage across the capacitor = 230 V. Find the conduction time for auxiliary thyristor.
a) 0.23 μs
b) 6.57 μ
c) 31.41 μs
d) 56 μs
Answer: c
Explanation: Ip = Vs x √C/L
ω = 1/√LC
t = π/ω.
6. An SCR is connected in series with L = 5 mH and C = 20 μF. Find the resonant frequency of the circuit.
a) 2569 rad/s
b) 3162 rad/s
c) 2400 rad/s
d) 7889 rad/s
Answer: b
Explanation: ω = 1/√LC.
7. The type of commutation when the load is commutated by transferring its load current to another incoming thyristor is
a) class A or load commutation
b) class B or resonant commutation
c) class C or complementary commutation
d) class D or impulse commutation
Answer: c
Explanation: In the Class C type commutation also called as complementary commutation the load is commutated by transferring the current th another device.
8. The type of commutation in which the pulse to turn off the SCR is obtained by separate voltage source is
a) class B commutation
b) class C commutation
c) class D commutation
d) class E commutation
Answer: d
Explanation: In class E commutation, another voltage source is used. It is also called as external pulse commutation.
9. The natural reversal of ac supply voltage commutates the SCR in case of
a) forced commutation
b) only line commutation
c) only natural commutation
d) both line & natural commutation
Answer: d
Explanation: Both line and natural commuataion are used in converters.
10. ___________ commutation technique is commonly employed in series inverters.
a) line
b) load
c) forced
d) external-pulse
Answer: b
Explanation: Load commuation is used in inverter in which L and C are connected in series with the load or C in parallel with the load such that overall load circuit is under damped.
This set of Power Electronics Questions & Answers for entrance exams focuses on “SCR Commutation Techniques-2”.
1. For the circuit shown in the figure below,
power-electronics-questions-answers-entrance-exams-q1
C = 4 pF
L = 16 μH
Vs = 200V
The capacitor voltage after the SCR is self commutated is
a) -100 V
b) -200 V
c) -400 V
d) 0 V
Answer: b
Explanation: It is simply the negative of Vs as it is a series circuit & after turn off the whole voltage appears across the thyristor.
2. For the circuit shown below,
power-electronics-questions-answers-entrance-exams-q2
The maximum value of current through thyristors T1 & TA can be given by
a) Vs/R, Vs√C/L
b) Vs √C/L, Vs/R
c) Vs/R + LC, Vs
d) Vs/LC, Vs/R
Answer: a
Explanation: I1 =
T2 = .
3. For the circuit shown in the figure below,
power-electronics-questions-answers-entrance-exams-q1
C = 4 pF
L = 16 μH
Vs = 200 V
The peak thyristor current is
a) 800 A
b) 400 A
c) 100 A
d) 50 A
Answer: c
Explanation: The peak SCR current is Vs/2.
4. For the circuit shown in the figure below, ___ type of commutation would take place for the SCR
power-electronics-questions-answers-entrance-exams-q2
a) line
b) load
c) forced
d) external-pulse
Answer: c
Explanation: The given figure is that of a class B commutation or resonant-pulse commutation, which is a type of forced commutation technique.
5. For the circuit shown in the figure below,
power-electronics-questions-answers-entrance-exams-q5
R1 = 50 Ω, R2 = 100 Ω, Vs = 100 V. At the time of turn-on of the SCR, initial thyristor current is
a) 0 A
b) 10 A
c) 5 A
d) 3.5 A
Answer: c
Explanation: Io = Vs x [+].
6. Natural commutation of an SCR takes place when
a) voltage across the device becomes negative
b) voltage across the device becomes positive
c) gate current becomes zero
d) anode current becomes zero
Answer: d
Explanation: Anode current becomes zero and turns off the device, hence the name line commutation.
7. ___________ commutation is usually used in phase-controlled rectifiers
a) line
b) load
c) forced
d) external-pulse
Answer: a
Explanation: Line commuataion is used in converters.
8. Parallel-capacitor commutation is
a) line commutation
b) load commutation
c) forced commutation
d) external-pulse commutation
Answer: c
Explanation: Parallel capacitor is another name for forced commutation.
9. Class E commutation is a/an
a) line commutation technique
b) load commutation technique
c) forced commutation technique
d) external-pulse commutation technique
Answer: d
Explanation: As an external source is used it is a external-pulse commutation technique.
10. Below is the circuit of a resonant-pulse commutation, the maximum voltage to which the capacitor C charges is
power-electronics-questions-answers-entrance-exams-q2
a) Vs
b) Vs/2
c) 2
d) √2Vs
Answer: a
Explanation: The capacitor charges from 0 to Vs with the right side plate positive.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Single Phase HW AC-DC-1”.
1. A voltage source Vs = Vm sinωt is connected in series with a resistance R and an SCR. At some firing angle delay of α a positive gate pulse is applied to the SCR which turns on the SCR. Considering ideal conditions, at the instant α the voltage at the resistor terminals Vo
a) falls to zero
b) falls to –Vm sin α
c) rises to Vm sin α
d) rises to Vm sin ωt
Answer: c
Explanation: At the firing angle, the SCR turns on and thus the resistor or load R is connected to the source.
2. The average output voltage is maximum when SCR is triggered at ωt =
a) π
b) 0
c) π/2
d) π/4
Answer: b
Explanation: The sooner the conduction starts higher the average power. Though practically a device cannot be triggered at exactly zero degrees.
3. In the below given circuit, the SCR is triggered at a certain firing angle α.
power-electronics-questions-answers-1phase-hw-acdc-1-q3
The load current falls to zero at ωt =
a) α
b) α/2
c) π
d) π/2
Answer: c
Explanation: The SCR is naturally commutated at 90°.
4. In the method of phase control, the phase relationship between ___ & ___ is controlled by varying the firing angle
a) supply current, supply voltage
b) end of the load current, end of the load voltage
c) start of the load current, start of the load voltage
d) load current, load voltage
Answer: c
Explanation: By varying the firing angle, the load current and load voltage can be controlled.
5. In a single phase half-wave thyristor circuit with R load & Vs=Vm sinωt, the maximum value of the load current can be given by
a) 2Vm/R
b) Vs/R
c) Vm/2
d) Vs/2
Answer: c
Explanation: Vm is the peak value of the load as well as supply voltage. I = Vm/R.
6. Below shown ideal circuit is triggered at a firing angle of α. The voltage across the SCR is zero from ωt =
power-electronics-questions-answers-1phase-hw-acdc-1-q3
a) 0 to α
b) 0 to π
c) α to 2π
d) α to π
Answer: d
Explanation: Before the SCR is triggered all of the supply voltage appears across it that is from 0 to α, at α the device is gated and it starts to conduct acting like a S.C . Hence voltage across it is zero from α to π. At π the SCR is again naturally commutated due to the reversal of the AC supply & a negative voltage appears across it.
7. Below shown ideal circuit is triggered at a firing angle of α. From ωt = π to 2π+α
power-electronics-questions-answers-1phase-hw-acdc-1-q3
a) SCR voltage is zero
b) Load voltage is maximum
c) SCR voltage is –Vmsinα
d) SCR acts like a short circuit
Answer: c
Explanation: From ωt = π to 2π+α, negative cycle is active and the SCR experiences revered biased voltage by Vs.
8. For a single phase thyristor circuit with R load & firing angle α, the conduction angle can be given by
a) π+α
b) 2π+α
c) π-α
d) α
Answer: c
Explanation: SCR is fired at α, hence it conducts from α till it is commutated by the reversal of the AC supply i.e at π radians.
9. For the below given circuit with firing angle α & angular frequency of ω, the circuit turn-off time in secs can be given by
power-electronics-questions-answers-1phase-hw-acdc-1-q3
a) α/π
b) π/α
c) ω/π
d) π/ω
Answer: d
Explanation: The circuit turns of at ωt = π as it is a half wave circuit.
Hence t off = π/ω.
10. A single-phase half-wave thyristor circuit with R load is triggered at an angle of α = 0°. As such, the maximum value of the average output voltage would be given by
Consider Vs = Vm sinωt.
a) Vm
b) 2Vm/π
c) Vm/π
d) Vm/α
Answer: c
Explanation: For the whole cycle of 2π, the R load would be active only for half of the cycle. Hence, Vo = Vm/π.
This set of Power Electronics Interview Questions and Answers focuses on “Single-Phase HW AC-DC-2”.
1. For a certain SCR configuration, the below shown waveform is obtained. Find the value of the average output voltage with α = 30° & Vs = 240 V.
power-electronics-interview-questions-answers-q1
a) 50.27 V
b) 100.8 V
c) 140 V
d) 120 V
Answer: b
Explanation: Vm = √2 Vs = 339.41 V
V avg = ∫ Vm Sin ωt d where the integration runs from α to π
V avg = Vm/2π .
2. A single-phase half wave circuit has Vs = 230 V with a R load of 100 Ω. Find the average load current at α = 30°.
a) 1.45 A
b) 0.57 A
c) 0.96 A
d) 2.3 A
Answer: c
Explanation: Io = Vo/R
Vo = x .
3. A single phase 230 V, 1 kW heater is connected across a 1-phase 230 V, 50 Hz supply through an SCR. The firing angle of the SCR is adjusted to give a rms voltage of 155 V. Find the power absorbed by the heater element.
a) 250 watts
b) 1 kW
c) 454 watts
d) 378 watts
Answer: c
Explanation: R = 230 2 /1kW = 52.9 Ω
Now, P = Vrms 2 /R = 155 2 /52.9.
4. For a single phase half-wave thyristor circuit with R load, the power delivered to the resistive load is
a) x
b) 2 /R
c) 2 /R
d) /R
Answer: c
Explanation: P = I 2 .R
V = IR
I = V/R
Hence, P = V 2 /R.
5. For a single phase half-wave thyristor circuit with R load, the input power factor is given by
a) rms source voltage/total rms line current
b) rms input power/power delivered to the load
c) cos α
d) power delivered to load/input VA
Answer: d
Explanation: input = power delivered to load/input VA.
6. For the below shown circuit,
power-electronics-interview-questions-answers-q6
The load voltage waveform from ωt = π to ωt = π+α = β is
a) gradually rising
b) going from positive to negative
c) gradually decaying
d) zero
Answer: c
Explanation: In this case, the Inductor L is force conducting the SCR & supply current to the source the current is gradually decaying as the inductor is discharging.
7.For the below shown circuit,
power-electronics-interview-questions-answers-q6
The SCR is gated at α. At ωt = π
a) load voltage & load current are both zero
b) only load voltage is zero
c) only load current is zero
d) both are non zero & non negative
Answer: b
Explanation: At π the L force conducts the SCR & current starts gradually decaying whereas at π the voltage is zero as it goes from positive to negative at π.
8. In case of a single-phase half-wave circuit with RL load, with firing angle α and extinction angle β, the conduction angle γ can be written as
a) γ = β+α
b) γ = β-α
c) γ = β/α
d) γ = α/β
Answer: b
Explanation: Extinction angle is some angle > π at which load current reduces to zero. It depends upon the inductor value.
9. For a half-wave single phase thyristor rectifier circuit with any type of load the _______ + _________ waves must ideally give the supply voltage waveform.
power-electronics-interview-questions-answers-q6
a) load voltage, load current
b) load voltage, thyristor voltage
c) supply current, load voltage
d) supply current, thyristor voltage
Answer: b
Explanation: The supply voltage either appears across at the thyristor or the load.
10. In case of a single-phase half-wave circuit with RL load, with firing angle α and extinction angle β, the thyristor is reversed biased from
a) β to α
b) β to 2π+α
c) β to 2π
d) β to 2β
Answer: c
Explanation: As the load current goes to zero from β, the SCR is reversed biased or in the OFF state till it is turned on at 2π again by the positive AC cycle.
This set of Power Electronics Multiple Choice Questions and Answers for freshers focuses on “Single-Phase HW AC-DC-3”.
1. Choose the correct statement with respect to the below given circuit.
power-electronics-questions-answers-freshers-q1
a) The load current can be negative
b) The load voltage can never be negative
c) The load voltage can never be zero
d) The load voltage can never be positive
Answer: b
Explanation: The voltage cannot be negative due to the FD .
2. In the below given circuit, the FD is forward biased at ωt =
power-electronics-questions-answers-freshers-q1
a) 0
b) α
c) π
d) 2π
Answer: c
Explanation: It is forward biased at π by the conducting SCR & the current starts to through the FD & Load.
3. In the below given circuit, when the commutating diode or FD is conducting than the
power-electronics-questions-answers-freshers-q1
a) SCR has reverse bias voltage and the load current is zero
b) SCR has reverse bias voltage and the load current is positive
c) SCR has forward bias voltage and the load current is zero
d) SCR has forward bias voltage and the load current is positive
Answer: b
Explanation: When the FD is forward biased at π by the conducting SCR & the current starts to through the FD & load.
4. The output voltage waveform of the below given circuit would be the same that obtained from a
power-electronics-questions-answers-freshers-q1
a) full-wave R load circuit
b) half-wave R load circuit
c) half-wave RL load circuit
d) full-wave RL load with freewheeling diode
Answer: b
Explanation: The wave from will be like a half wave diode rectifier circuit. Which is the same as that obtained from a half-wave R load circuit and the above given circuit.
5. In a single-phase half-wave circuit with RL load and a freewheeling diode, the load voltage during the freewheeling period will be
a) zero
b) positive
c) negative
d) positive than negative
Answer: a
Explanation: The FD short circuits the load and voltage across a short circuit would be = 0.
6. In a single-phase half-wave circuit with RL load and a freewheeling diode, the freewheeling period is
a) 0 to π
b) α to π+α
c) π to 2π+α
d) π/2 to 2π-α
Answer: c
Explanation: Freewheeling period is the one in which the FD diode conducts.
7. A single-phase half wave rectifier with a FD is supplied by Vs = 240 V, AC with a load R = 10 Ω, L = 0.5 mH and a firing angle α = 30°. Find the average value of the load voltage.
a) 50 V
b) 100 V
c) 150 V
d) 200 V
Answer: b
Explanation: Vo = x
Vm = √2Vs.
8. A single-phase HW rectifier with a FD is supplied by Vs = 240 V, 50 Hz, AC with a load R = 10 Ω, L = 0.5 mH and a firing angle α = 30°. Find the average value of the load current.
a) 10 A
b) 0.063 A
c) 6.3 A
d) 0.1 A
Answer: a
Explanation: Vo = x
Vm = √2Vs
Io = Vm/R .
9. A single phase half-wave controlled rectifier has 400 sin314t as the input voltage and R as the load. For a firing angle of 60°,the average output voltage is
a) 200/π
b) 300/π
c) 100/π
d) 400/π
Answer: b
Explanation: Vo = x = 400/2π x = 300/π.
10. Choose the incorrect statement with respect to the use of FD in half-wave circuits.
a) Input pf is improved
b) Load current waveform is improved
c) It prevents the load voltage from becoming negative
d) Reduces the reverse voltage faced by the SCR
Answer: d
Explanation: PIV is unaffected with the use of FD .
This set of Power Electronics Inteview Questions and Answers for freshers focuses on “Single-Phase HW AC-DC-4”.
1. A single-phase half-wave rectifier with a FD is supplied by Vs = 240 V, 50 Hz, AC source with a load R = 10 Ω, L = 0.8 mH. The firing angle is so adjusted such that the output voltage obtained is 100 V. Find the firing angle.
a) 62°
b) 102°
c) 31°
d) 47°
Answer: c
Explanation: Vo = Vm/2π cos.
2. For the below shown circuit, a motor load is connected to the SCR through supply Vs. The minimum value of the firing angle to turn on the SCR would be
power-electronics-interview-questions-answers-freshers-q2
a) Sin -1
b) Sin -1
c) Cos -1
d) Cos -1
Answer: a
Explanation: The firing angle should be such that the value of V exceeds E. So SCR turns on at excatlly Vm sinωt = E
Therefore ωt = min. angle = Sin -1 .
3. Choose the correct statement with respect to the below given circuit.
power-electronics-interview-questions-answers-freshers-q2
a) The load current can never be zero
b) The source current can never be zero
c) The load voltage can never be zero
d) All the given statements are false
Answer: c
Explanation: Even if no current is flowing through the load when the SCR is off, voltage = E will exists at the load terminals at all times.
4. By using a freewheeling diode in a rectifier with RL load, the power consumed by the load
a) increases
b) decreases
c) is not affected
d) decreases to zero
Answer: a
Explanation: The FD feeds inductor current again to the load.
5. A 230 V, 50 Hz, one-pulse SCR controlled converter has extinction angle β = 210°. Find the circuit turn-off time
a) 10 m-sec
b) 0
c) 8.3 m-sec
d) 5.4 m-sec
Answer: c
t = 2π-β/ω
Where, π = 90°
ω = 2xπx50
β = 210°.
6. A 230V, 50Hz, single-pulse SCR is feeding a RL load with α = 40° and β = 210°. Find the value of average output voltage
a) 54 V
b) 106 V
c) 84 V
d) 32 V
Answer: c
Explanation:
Vo = Vm/2π x
Where Vm = √2Vs.
7. A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60°, find the average current in the bulb
a) 2.5 A
b) 7.7 A
c) 9.6 A
d) 3 A
Answer: b
Explaantion: First find the average voltage, than Io = Vo/R
Vo = x .
8. A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60° and output AC power of 2127 Watts, find the rectification efficiency
a) 98.6 %
b) 42 %
c) 28 %
d) 19 %
Answer: c
Explanation:
Vo = x = 77.64 V
Pdc = Vo 2 xR = 602.8 W
Rectification efficiency = Pdc/Pac = 28.32 %.
9. Find the PIV of the SCR
power-electronics-interview-questions-answers-freshers-q9
a) Vs
b) 2Vs
c) √2Vs
d) Vs/2
Answer: c
Explanation: The maximum negative voltage faced by the SCR is –Vm which is nothing but √2Vs.
10. An SCR rectifier circuit is designed such that the average output voltage is 77.64 V & RMS value of output voltage is 145.873 V. Find the voltage ripple factor.
a) 1.879
b) 0.532
c) 1.590
d) 0.629
Answer: c
Explanation: FF = 145.873/77.64 = 1.879
VRF = √(FF 2 -1) = 1.5908.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Single Phase FW AC-DC-1”.
1. The below shown single phase controlled rectifier configuration is a
power-electronics-questions-answers-1phase-fw-acdc-1-q1
a) two pulse M-2 Connection
b) two pulse B-2 Connection
c) single pulse M-2 Connection
d) single pulse B-2 Connection
Answer: a
Explanation: It is a two pulse i.e full wave M-2 connection.
2. In the below shown circuit, considering the load to be purely R load, the thyristor T2 would conduct from ωt =
power-electronics-questions-answers-1phase-fw-acdc-1-q1
a) 0 to π
b) α to π
c) π+α to 2π
d) π to 2π+α
Answer: c
Explanation: The bottom winding forward biases the SCR T2 at π, conduction starts after the pulse is given at α & stops when T2 is naturally commutated at 2π.
3. For a single pulse M-2 rectifier with RL load and firing angle α >0°, the value of the inductance L is such that the load current is continuous.
power-electronics-questions-answers-1phase-fw-acdc-1-q1
Choose the correct statement.
a) The load current can never be positive
b) The load current can never be negative
c) The load current can never be zero
d) The load current is a constant DC
Answer: c
Explanation: As mentioned, L is such that the load current is continuous. Hence, it never falls below a predefined positive value.
4. For the circuit below, the transformer ratio is 1:1 and Vs = Vm sinωt. When T1 is forward biased and is triggered at α than T2 is
power-electronics-questions-answers-1phase-fw-acdc-1-q1
a) subjected to a forward voltage of Vm sinα
b) subjected to a forward voltage of 2Vm sinα
c) subjected to a reverse voltage of Vm sinα
d) subjected to a reverse voltage of 2Vm sinα
Answer: d
Explanation: When T1 is gated, the other SCR T2 is reversed biased due to the windings so connected.
5. The circuit turn-off time of M-2 type connection is
power-electronics-questions-answers-1phase-fw-acdc-1-q1
a) π/ω
b) α/ω
c) π-α/ω
d) π+α/ω
Answer: c
Explanation: When T1 is gated, SCR T2 is off from ωt = α to π
Thus t = π-α/ω
Same can be said when T2 is gated and T2 is off.
6. Find the expression for the average value of the output voltage for the below given circuit. Consider the load current to be continuous, firing angle = α, transformer ration 1:1 and Vs = Vm sinωt.
a) cosα
b)
c) cosα
d)
Answer: c
Explanation: As the load current is continues, the voltage is positive from α to π and negative from π to π+α and so on. Thus,
Vavg = 1/Period ∫ Vm sinωt d
Where period = π
And the integral runs from α to α+π.
7. A fully controlled converter uses
a) diodes only
b) thyristors only
c) both diodes and thyristors
d) none of the mentioned
Answer: b
Explanation: Fully controlled implys that all the elements are “fully controlled” hence, it uses SCRs only except the FD.
8. A single phase full-converter using R load is a _________ quadrant converter and that using an RL load without FD is a __________ quadrant converter
a) one, one
b) two, one
c) one, two
d) two, two
Answer: c
Explanation: In R load both V and I are positive, in RL load the voltage can be negative but current is always positive.
9. A single phase full controlled bridge converter uses
a) 4 SCRs and 2 diodes
b) 4 SCRs
c) 6 SCRs
d) 4 SCRs and 2 diodes
Answer: b
Explanation: 4 SCR’s are connected in a bridge fashion.
10. In a B-2 type full controlled bridge converter
a) one SCR conducts at a time
b) two SCRs conduct at a time
c) three SCRs conduct at a time
d) four SCRs conduct at a time
Answer: b
Explanation: B-2 is the bridge type controller, in which 2 devices conduct at a time. One acting as the current supplying path and other acts as a return path.
This set of Power Electronics Interview Questions and Answers for experienced people focuses on “Single-Phase FW AC-DC-2”.
1. For the below given circuit,
power-electronics-interview-questions-answers-experienced-q1
a) T2 and T3 are gated together
b) T1 and T4 are gated together
c) T1 and T3 are gated together
d) T1 and T2 are gated together
Answer: d
Explanation: T1-T2 are gated together at a time and T3-T4 are gated together at a time.
2. With respect to the below given circuit, from ωt = 0 to α, ____ thyristor block the forward voltage
power-electronics-interview-questions-answers-experienced-q2
a) T1 and T2
b) T2
c) T3 and T4
d) T4
Answer: a
Explanation: T1 and T2 are not yet gated, hence they block the forward voltage.
3. With respect to the below given circuit, from ωt = 0 to α,
power-electronics-interview-questions-answers-experienced-q2
a) voltage across all the thyristors is zero
b) negative voltage appears across T1 and T2
c) load voltage is Vm/2
d) voltage across T3 and T4 is zero
Answer: d
Explanation: T1 and T2 are not yet gated, hence they block the forward voltage where as no voltage exists across T3 and T4.
4. For the below given circuit, when the devices T1 and T2 are conducting from ωt = α till they are naturally commutated at π voltage across the other two pair of devices is
power-electronics-interview-questions-answers-experienced-q1
a) zero
b) positive
c) negative
d) none of the mentioned
Answer: c
Explanation: When T1 and T2 are conducting they act like S.C’s and the other pair of devices i.e. T3 and T4 has to take care of the reverse voltage.
5. A single-phase full converter B-2 type connection has a RLE type of motor load connected. The minimum requirement to turn-on the device is
a) α > 30°
b) Vm sinα > E
c) Vm sinα < E
d) α < 30°
Answer: b
Explanation: Device will not turn on until the supply voltage exceeds the motor back emf E.
6. For the below shown circuit with continuous load current.
power-electronics-interview-questions-answers-experienced-q2
Each thyristor pair conducts for
a) π radians
b) 2π radians
c) < π radians
d) > π radians
Answer: a
Explanation: As the load current is continuous each pair conducts from π to π+α i.e π radians.
7. For the below given circuit
power-electronics-interview-questions-answers-experienced-q2
a) all the devices are gated together
b) T3 and T4 are gated together
c) T3 and T1 are gated together
d) T3 and T2 are gated together
Answer: d
Explanation: T1-T2 are gated together at a time and T3-T4 are gated together at a time.
8. For the below shown circuit has dis-continuous load current waveform.
power-electronics-interview-questions-answers-experienced-q1
Each thyristor pair conducts for
a) π radians
b) 2π radians
c) < π radians
d) > π radians
Answer: c
Explanation: Load current is discontinues than SCR pair conducts for less than 90° depending on the value of the inductor L.
9. A single-phase full converter with B-2 type of connection has a continuous load current waveform. The thyristor pairs T3, T4 is triggered at ωt =
a) 0
b) α
c) π+α
d) π-α
Answer: c
Explanation: First, at α T1-T2 conduct, then after π+α the next pair is excited i.e. T3-T4 and likewise.
10. The PIV for each diode in a single-phase, full converter with B-2 type of controlled rectifier is
Consider supply voltage as Vs = Vm sinωt.
a) 2Vm
b) Vm
c) Vm/2
d) Vm/√2
Answer: b
Explanation: The peak inverse voltage handled by each device is Vm in case of B-2 type of configuration.
This set of Power Electronics Questions and Answers for experienced people focuses on “Single-Phase FW AC-DC-3”.
1. A motor load is connected to a single-phase full converter B-2 type controlled rectifier. The net energy is transferred from ac source to the motor when
a) π+α > 90
b) π-α > α
c) π+α > α
d) π-α > 90
Answer: b
Explanation: Converter will work as a line commuted inverter when π-α > α.
2. The below shown rectifier configuration has continues load current, find the expression for the average value of output voltage when the supply Vs = Vm sinωt is connected.
power-electronics-questions-answers-experienced-q2
a) cosα
b) cosα
c)
d)
Answer: b
Explanation: Vo = 1/π x [∫ Vm sinωt d] Where the integration runs from α to π+α.
3. Find the expression of the rms value of output voltage for a single-phase M-2 type rectifier with RL load and continues load current. Transformer ratio is 1:1 with supply voltage Vs = Vm sinωt
a) Vm/√2
b) Vs
c) 2Vs
d) Vs/√2
Answer: b
Explanation: Vor 2 = 1/π x [∫ Vm 2 sin 2 ωt d], Where the integration runs from α to π+α
Vor = Vm 2 /2 = Vs.
4. A single-phase full controlled converted with RLE load will act like a line-commutated inverter when the firing angle α
a) α > 180°
b) α > 90°
c) α < 90°
d) α = 90°
Answer: b
Explanation: It will act like a inverter i.e. most of the current would flow from the battery or back emf E to the source. For α=90° it will not act like a converter nor an inverter.
5. In converter operation, with output voltage = Vo and RLE load.
a) Vo < E
b) Vo = E
c) Vo > E
d) None of the mentioned
Answer: c
Explanation: The output voltage obtained by a converter is always greater than the counter or back emf E.
Vo = IR + E.
6. In inverter operation, with output voltage = Vo and a RLE load connected
a) Vo < E
b) Vo = E
c) Vo > E
d) None of the mentioned
Answer: a
Explanation: In inverter operation Vo < E.
7. A motor load is connected to a single-phase full converter B-2 type controlled rectifier, the net energy is transferred from ac source to the motor when
a) π+α > 90
b) π-α < α
c) π+α < α
d) π-α > α
Answer: b
Explanation: It works as a converter when π-α < α.
8. The below shown rectifier configuration has continues load current, find the expression of the RMS value of output voltage when the supply Vs = Vm sinωt
a) Vs 2 /2
b) 2Vs/π
c) Vs
d) √Vs/2
Answer: c
Explanation: Vor 2 = 1/π x [∫Vm 2 sin 2 ωt d] Where the integration runs from α to π+α
Vor = Vm 2 /2 = Vs.
9. Choose the correct statement
a) M-2 type connection requires SCRs with higher PIV as compared to those in a B-2 type
b) M-2 type connection requires SCRs with lower PIV as compared to those in a B-2 type
c) The average output voltage in M-2 type is more than that obtained from a B-2 type configuration of the same rating
d) The average output voltage in M-2 type is less than that obtained from a B-2 type configuration of the same rating
Answer: a
Explanation: The PIV of diodes in M-2 is 2Vm, whereas that in B-2 type of connection is Vm.
10. An SCR has the peak forward voltage = 1000 V. Find the maximum voltage that the SCR can handle if employed in a M-2 type full controlled converter circuit. Use factor of safety = 2.5
a) 500 V
b) 400 V
c) 200 V
d) 1000 V
Answer: c
Explanation: In M-2 type configuration maximum voltage handled is 2Vm
Therefore, 1000/2×2.5 = 200 V.
This set of Power Electronics Aptitude Test focuses on “Single-Phase FW AC-DC-4”.
1. An SCR has the peak forward voltage = 1000 V. Find the maximum voltage that the SCR can handle if employed in a B-2 type full controlled converter circuit. Use factor of safety = 2.5
a) 500 V
b) 400 V
c) 200 V
d) 1000 V
Answer: b
Explanation: In M-2 type configuration maximum voltage handled is Vm
Therefore, 1000/2.5 = 400 V.
2. A single-phase full converter bridge, is connected to a RLE load. The source has a rms voltage of 230 V and the average load current is 10 A. Find the firing angle for which the power flows from AC source to the DC load. Consider E = 120 V, R = 0.4 Ω.
a) 26.4°
b) 53.2°
c) 142°
d) 39°
Answer: b
Explanation: We have,
Vo = E + IR
Vo = 2Vm/π x cosα
Substitute the given values to discover α = 53.208°.
3. A single-phase full converter bridge, is connected to a RLE load. The source has a rms voltage of 230 V and the average load current is 10 A. Find the firing angle for which the power flows from the DC load to the AC source. Consider E = 120 V, R = 0.4 Ω, L = 2 Henry.
a) 124°
b) 153°
c) 142°
d) 309°
Answer: a
Explanation: We have,
Vo = E + IR
Vo = 2Vm/π x cosα . . .
Substitute the given values to discover α = 124°
Note that as the power is flowing from DC to AC E has to be negative.
E = -120V.
Use 1 to find the firing angle.
4. For the below shown converter configuration, find the expression for the average value of voltage across the resister R with Vs = Vm sinωt and firing angle = α.
power-electronics-aptitude-test-q4
a) Vm/π cosα
b) 2Vm/π cosα
c) Vm/π
d) 2Vm/π
Answer: c
Explanation: Vo = Vo = 1/π x [∫ Vm sinωt d] where the integration would run from α to π.
5. A single phase full controlled bridge converter is connected to load having R = 0.4 Ω, E = 120, L = 0.2 mH, Vs = 230V. The RMS value of load current is 10 A for a firing angle of 53.21°. Find the input power factor.
a) 0.5043 lag
b) 0.5391 lag
c) 0.6120 lag
d) Unity
Answer: b
Explanation: Io = Irms = 10A
Vs x Irms x cosɸ = E Io + Irms 2 R.
6. A single phase full converter feeds power to a RLE load with R = 6 Ω and E = 60 V. Find the average value of the load current when the supply voltage is 230 V rms AC and a firing angle = 50°.
a) 0
b) 3.9A
c) 12.1 A
d) 6.05 A
Answer: c
Explanation: Vo = 2Vm/π cosα = 133.084 V
I = Vo – E/R = 133.084-60 / 6 = 12.181 A.
7. A single phase full wave mid-point SCR converter, uses a 230/200 V transformer with the centre tap on the secondary side. The P.I.V. per SCR is
a) 100 V
b) 214V
c) 141V
d) 282V
Answer: d
Explanation: PIV for M-2 is Vm = √2Vs = √2 x 200.
8. In a single phase full converter, for discontinuous load current and extinction angle β > π, each SCR conducts for
a) α
b) β
c) β-α
d) β+α
Answer: c
Explanation: Each device would conduct for β-α.
9. A single-phase two pulse converter feeds RL load with a sufficient smoothing such that the load current does not fall to zero. If the resistance of the load circuit is increased then the
a) ripple content is not affected
b) ripple content of current increases
c) ripple content of current decreases
d) load current may fall to zero
Answer: b
Explanation: If the resistance of the load circuit is increased then the ripple content of current increases.
10. In case of controlled rectifiers, the nature of the load current depends upon the
a) type of load and firing angle
b) only on the type of load
c) only on the firing angle
d) it is independent of all the parameters
Answer: a
Explanation: It depends on both as firing angle will decide how fast and how much current flows. The load R, Rl or RLE can also effect the current depending upon the values of L and E.
This set of Power Electronics Multiple Choice Questions & Answers for campus interviews focuses on “Single-Phase FW AC-DC-5”.
1. In the circuit shown below, find the expression of average output voltage if due to some faults the SCR T3 gets permanently open circuited.
power-electronics-questions-answers-campus-interviews-q1
a) 2Vm/π cosα
b) Vm/π cosα
c) Vm/√π
d) Zero
Answer: b
Explanation: If one of the device gets O.C the circuit would behave just like a half-wave rectifier. Only Two SCRs T1 & T2 will operate, even T4 is useless in this case.
Vo = 1/2π x [ ∫ Vm sinωt d ] where the integration would run from α to π+α.
2. A single phase full converter, feeds power to a RLE load with R = 6 Ω and E = 60 V. The supply voltage is 230 V AC and the firing angle is 50°. Find the power delivered to the battery.
a) 542 W
b) 100 W
c) 960 W
d) 730 W
Answer: d
Explanation: Vo = 2Vm/π cosα = 133.084 V
I = Vo – E/R = 133.084-60 / 6 = 12.181 A
Power delivered = E x I = 60 x 12.181 = 730.86 Watts.
3. In a single phase full converter with resistive load and firing angle α, the load current is
a) zero at α, π+α, …
b) remains zero for duration α
c) Vm/R sinα at α, π+α, …
d) remains zero for duration π-α
Answer: c
Explanation: The current starts to flow from α, π+α, …
4. In a single phase semi converter with resistive load and a firing angle α, each SCR and freewheeling diode would conduct for
a) α, 0°
b) π-α, α
c) π+α, α
d) π-α, 0°
Answer: d
Explanation: FD does not come into the picture for resistive loads.
5. A single phase full converter has average & peak voltage values of 133 V and 325 V respectively. Find the value of the firing angle.
a) 40°
b) 140°
c) 50°
d) 130°
Answer: c
Explanation: Vm = 325V
Vo = 2Vm/π cosα = 133 V.
6. A freewheeling diode placed across a RL load provides
a) fast turn-on time
b) slow turn-off time
c) poor utilization factor
d) better power factor
Answer: d
Explanation: FD improves efficiency by providing freewheeling action which increase the power delivered to the load.
7. A single phase full converter has discontinues load current. The converter is supplying a dc shunt motor load. When the current falls to zero the output voltage is
a) Zero
b) Vo
c) Vm
d) E
Answer: d
Explanation: As the load is RLE output voltage will always exists even if the current goes to zero due to E.
8. A single phase full converter has discontinues load current. The converter is supplying a DC shunt motor load. When the current falls to zero
power-electronics-questions-answers-campus-interviews-q8
a) none of the devices conduct
b) all of the devices conduct
c) only one device conducts
d) only one pair of devices conducts
Answer: a
Explanation: The current falls to zero when the L has discharged completely and the devices are not yet fired.
9. For the circuit shown below to act as a line commutated inverter
power-electronics-questions-answers-campus-interviews-q8
a) 180°>α>90° & E is removed
b) 180°>α>90° & E is reversed
c) 180°>α<90° & E is removed
d) 180°>α<90° & E is reversed
Answer: b
Explanation: For inverter operation the firing angle should be above 180 and E should be reversed.
10. If in the below given converter configuration, any 3 of the 4 SCRs are replaced by diodes
a) the average output voltage would increase
b) the average output voltage would decrease
c) the average output voltage would not change
d) the circuit would malfunction
Answer: a
Explanation: If diodes are put in place of any 3 SCRs the conduction angle would increase and hence the voltage.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Single Phase Semi-Converters-1”.
1. In the below shown semi-converter circuit, T1 and T2 are fired at an angle α, then from ωt = α to π
power-electronics-questions-answers-single-phase-semi-converters-1-q1
a) T1 and T2 conduct
b) T1 and D1 conduct
c) D1 and D2 conduct
d) FD conducts
Answer: b
Explanation: At ωt = α T1 is gated and current starts to flow from T1-R-L-E-D1-Vs-T1.
2. In the below shown semi-converter circuit T1 & T2 are fired at an angle α, then from ωt = π to α+π
power-electronics-questions-answers-single-phase-semi-converters-1-q1
a) T1 is conducting
b) T2 is conducting
c) D1 is conducting
d) FD conducting
Answer: d
Explanation: None of the SCRs are gated, FD is forward biased and starts to conduct charging the L.
3. In the below shown semi-converter circuit T1 & T2 are fired at an angle α, the output voltage is zero when
power-electronics-questions-answers-single-phase-semi-converters-1-q1
a) π<ωt<α
b) 0<ωt<α+π
c) π<ωt<π+α
d) π<ωt<2π
Answer: c
Explanation: FD is conducting as none of the SCRs are gated which S.C’s the load circuit hence zero voltage.
4. In a semi-converter with RLE load during the freewheeling period, the energy is
a) fed back to the source
b) fed to the inductor and absorbed by E
c) absorbed by the L & E and dissipated at R
d) fed to the L & E and dissipated at R
Answer: d
Explanation: The energy is fed back fed to the L & E and dissipated at R.
5. A semi-converter circuit gives the following voltage waveform. Find the expression for the average output voltage with Vs = Vm sinωt and firing angle α
power-electronics-questions-answers-single-phase-semi-converters-1-q5
a) cosα
b)
c) cosα
d)
Answer: b
Explanation: Vo = 1/π ∫ Vm sinωt d, Where the integration runs from α to π.
6. For the below given circuit configuration, the SCR is fired at angle α. For continuous current mode from ωt = π to 2π
power-electronics-questions-answers-single-phase-semi-converters-1-q6
a) FD conducts
b) SCR and D1 conduct
c) D3 and D2 conducts
d) None of the mentioned
Answer: c
Explanation: From ωt = π to 2π, the negative cycle is active and the SCR is reversed biased. D3 and D2 are ative and are supplying power to the load.
7. For the below given circuit configuration, the SCR is fired at angle α. For continuous current mode, find the expression for the average output voltage
power-electronics-questions-answers-single-phase-semi-converters-1-q6
a)
b)
c)
d)
Answer: c
Explanation: The output voltage will have two parts
V1 = [ ∫ Vm sinωt d ] where the integration would run from α to π
V2 = [ ∫ Vm sinωt d ] where the integration would run from π to 2π
Vo = 1/2π x V1+V2.
8. A semi-converter circuit gives the following voltage waveform on R load. Find the expression for the average output current with Vs = Vm sinωt and firing angle α
power-electronics-questions-answers-single-phase-semi-converters-1-q5
a) cosα
b)
c) cosα
d)
Answer: b
Explanation: Vo = 1/π ∫ Vm sinωt d Where the integration runs from α to π
Io = Vo/R.
9. A semi-converter with RLE load and a freewheeling diode has discontinuous load current with firing angle α and extinction angle β. The freewheeling period is
a) π>ωt<β
b) π>ωt<β
c) π<ωt>β
d) π>ωt>β
Answer: a
Explanation: The freewheeling period is from π to the extinction angle.
10. A semi-converter with RLE load and a freewheeling diode has discontinuous load current with firing angle α and extinction angle β. If β < π and Vm sinβ < E then
a) the converter operates as a line commutated inverter
b) the conduction period is absent
c) the freewheeling period is absent
d) none of the mentioned
Answer: c
Explanation: As the extinction angle is less than 90 degrees, the freewheeling period is absent.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Single Phase Semi-Converters-2”.
1. A single-phase symmetrical semi-converter employs
a) one SCR and one diode in each leg
b) two SCRs and two diodes in each leg
c) two SCRs in each leg
d) two diodes in each leg
Answer: a
Explanation: A symmetrical semi-converter will have one SCR and one diode in each leg. Two legs connected in parallel with each other having a symmetrical configuration.
2. A single-phase asymmetrical semi-converter employs
a) one SCR and one diode in each leg
b) two SCRs in one leg and two diodes in the other
c) two SCRs in both the legs
d) two diodes in both the legs
Answer: b
Explanation: An asymmetrical semi-converter will two SCRs in one leg and two diodes in the other.
3. In the below given circuit, from ωt = π to α+π and continuous load current configuration
power-electronics-questions-answers-single-phase-semi-converters-2-q3
a) T1 and T2 conduct
b) T2 and D1 conduct
c) T1 and D2 conduct
d) T1 and D1 conduct
Answer: c
Explanation: At ωt = π, D1 is reversed biased but T1 is still ON due to the nature of the load. Hence, the load current or inductor current flows from T1 and D2 and short circuiting the load terminals. Hence, due to the diode D2 freewheeling action takes place without even having a FD diode.
4. Find the circuit turn-off time (T c ) for the below given semi-converter configuration. Take α = 30° and Vs = 230√2 sin100t
power-electronics-questions-answers-single-phase-semi-converters-2-q3
a) 20π radians
b) 5π/6 radians
c) π/120 radians
d) 3/π radians
Answer: c
Explanation: For the given circuit, ωt = π-α. tc = π – α/ω
α = π/6 and ω = 100 …
tc = /100.
5. For the circuit shown below, T1 and T2 are both fired at an angle α. Then from ωt = π to ωt = π+α
power-electronics-questions-answers-single-phase-semi-converters-2-q5
a) None of the devices conduct
b) T1 and one diode will conduct
c) T2 and one diode will conduct
d) Both the diodes will conduct
Answer: d
Explanation: From ωt = π to ωt = π+α, None of the SCRs are fired and both the diodes are forward biased due to the nature of the load. As such, they start to conduct and freewheel all the inductor energy.
6. In any AC-DC circuit, the freewheeling action
a) improves the power handling capabilities
b) increases the THD
c) improves CDF
d) all of the mentioned
Answer: c
Explanation: Freewheeling action reduces THD , Improves CDF (current distortion factor and has no effect over the power handling capabilities as such.
7. A single-phase semi-converter is connected to a 230 V source and is feeding a load R = 10 Ω in series with a large inductance that makes the load current ripple free. Find the average output current for α = 45°.
a) 14 A
b) 17 A
c) 10 A
d) 0 A
Answer: b
Explanation: Vo =
Io = Vo/R.
8. A single-phase semi-converter is connected to a 230 V source and is feeding a load R = 10 Ω in series with a large inductance that makes the load current ripple free. For α = 45°, find the rectification efficiency. The RMS value of output voltage is 219.3 V
a) 96.54 %
b) 75.25 %
c) 89.45 %
d) 80.58 %
Answer: d
Explanation: Vo = = 176.72V
Io = Vo/R
Irms = Io = Vo/R = 17.67 A
Vrms = 219.3 V
η = / = 0.8058 = 80.58 %.
9. A single-phase semi-converter circuit is supplying power to a motor load. The average value of load voltage is 176.72 V and the rms value is 219.3 V. Find the VRF .
a) 0
b) 0.735
c) 0.569
d) 2.48
Answer: b
Explanation: First find the form factor ,
FF = Vrms/Vo = 1.241
Now, VRF = √(FF 2 -1) = 0.735.
10. For the same triggering angle and ratings
a) a semi-converter operates at lower output voltage than a full converter
b) a semi-converter operates at higher output voltage than a full converter
c) a semi-converter has lower values of input p.f as compared to a full converter
d) a semi-converter has more THD as compared to a full converter
Answer: b
Explanation: A semi-converter gives more output voltage than an equivalent full converter circuit. It also has less THD and high p.f .
This set of Power Electronics Problems focuses on “Single-Phase Semi-Converters-3”.
1. A single-phase semi-converter is having continuous conduction, as such each thyristor will conduct for an angle of
a) α
b) π
c) α+π
d) π-α
Answer: d
Explanation: From 0 to α there is no conduction. SCRs are gated at α and than naturally commutated at π. Hence, conduction angle = π-α.
2. Consider the following statements regarding phase controlled converters:
i) They do not provide smooth variation of output voltage.
ii) They inject harmonics into the power system.
iii) They draw non-unity power factor current for finite firing angles.
Which of the above statements are correct?
a) All the three
b) i and ii only
c) i and iii only
d) ii and iii only
Answer: a
Explanation: All the above given statements are correct. They all are the major drawbacks of using phase controlled converters.
3. Consider the following statement:
The overlap angle of a phase controlled converter will increase ___________
i) if the firing angle is increased
ii) if the supply frequency is increased
ii) if the supply voltage magnitude is lowered
Of these statements
a) all are correct
b) only the first and the last one are correct
c) only the first one is incorrect
d) all are incorrect
Answer: c
Explanation: The period during which both the incoming and outgoing SCRs are conducting is called as an overlap angle. This happens when inductance comes into picture either on the load or the source side.
4. All the modern AC-DC converters are using GTOs instead of SCRs because
a) Temperature is an issue in SCRs
b) Switching losses are high in SCRs
c) GTOs have reliable commutation
d) GTOs are cheaply available today
Answer: c
Explanation: GTOs unlike conventional SCRs can be turned off by providing a negative gate pulse. Hence, using GTOs the turn off as well as the turn-on time can be controlled.
5. Find the average output dc voltage of a single-phase semi-converter with Vs=230 V and firing angle of 30°. The converter is operating under continuous conduction.
a) 193 V
b) 256 V
c) 0 V
d) 230 V
Answer: a
Explanation: Vo = √2Vs /π.
6. In the below given circuit, the average value of output voltage is 191 V. Find the value of the average current through the given load if E = 100 V and R = 5 Ω. Assume that the load current is continuous.
power-electronics-problems-q6
a) 0 A
b) 10.2 A
c) 18.2 A
d) 29.2 A
Answer: c
Explanation: Vo = 191 V
Io = Vo-E/R = 191-100/5 = 18.2 A.
7. In the below given circuit, the average value of output voltage is 191 V. Find the value of the power delivered to the voltage source E. Take E = 100 V and R = 5 Ω. Assume that the load current is continuous.
power-electronics-problems-q6
a) 1.82 kW
b) 18.2 kW
c) 0.182 kW
d) 18.2 W
Answer: a
Explanation: Note that as the load current is continuous, L can be neglected.
Vo = 191V.
Io = Vo-E/R = 191-100/5 = 18.2A
P = E x Io = 100 x 18.2 = 1.82 kW.
8. A single-phase semi-converter is operated from a 240 V, 60 Hz, AC source. It is fired at an angle of 45°. Find the value of average output voltage.
a) 176 V
b) 184 V
c) 167 V
d) 148 V
Answer: b
Explanation: For a semi-converter, the output voltage Vo = √2Vs /π.
9. A single-phase semi-converter is operated from a 230 V, 60 Hz, AC source. A DMM connected at its output terminals read 219.3 V. Find the value of FF .
a) 1.24
b) 0.735
c) 1.11
d) 1
Answer: a
Explanation: A DMM always reads the RMS value. Therefore, Vrms = 219.3 V
Vo = √2Vs /π = 176.72 V
FF = Vrms/Vo = 1.241.
10. Name the circuit shown below.
power-electronics-problems-q10
a) Half controlled semi-converter symmetrical configuration
b) Half controlled semi-converter asymmetrical configuration
c) Half controlled full-converter asymmetrical configuration
d) Half controlled full-converter symmetrical configuration
Answer: b
Explanation: It is using only two SCRs, hence it is a half controlled semi-converter. It is an asymmetrical configuration as each leg has same devices.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Three Phase Converter-1”.
1. A three-phase, three-pulse, M-3 type controlled converter uses ____________ number of SCRs.
a) 1
b) 2
c) 3
d) 4
Answer: c
Explanation: It uses three SCRs with a three-phase transformer. M-3 type 3-pulse converters are not practically used.
2. A three-phase, three-pulse, M-3 type controlled converter has firing angle for one of the SCRs set as 15°. This SCR would start conducting at
a) 0°
b) 15°
c) 30°
d) 45°
Answer: d
Explanation: In a three phase controller, the actually conduction starts at 30° + α. Hence, ωt = 30+15 = 45°.
3. In a three-phase, three-pulse, M-3 type controlled converter T1 starts to conduct at 30 + n°. At what angles do T2 and T3 start to conduct? Assume that the conduction sequence is T1-T2-T3.
a) 2n° and 3n°
b) 150 + n° and 270 + n°
c) n° each
d) 30 + n° and 60 + n°
Answer: b
Explanation: In three-phase three pulse converter the conduction can start only after 30°. As each SCR conducts from 120°, T2 would conduct on 30+120+n° = 150+n° and so on.
4. A three-phase three pulse type controlled converter is constructed using 3 SCR devices. The circuit is supplying an R load with α < 30°. As such, each SCR device would conduct for
a) 60° each cycle
b) 120° each cycle
c) 180° each cycle
d) 360° each cycle
Answer: b
Explanation: Each conduct for 120° per cycle is the firing angle is less than 30°. 120 x 3 = 360°.
5. Find the expression for average output voltage at R for the below given configuration. Take firing angle as α = 15°, transformer ratio as 1:1:1 and Vmp as the maximum value of phase voltage at the supply.
power-electronics-questions-answers-three-phase-converters-1-q5
a) x cosα
b) x cosα
c) x sinα
d) x sinα
Answer: b
Explanation: The circuit is that of a three-pulse M-3 connection. The firing angle is less than 30°. Therefore, each device conducts for an angle of 120°.
Vo = 3 x [ 1/2π ∫ Vmp sinωt d ] Where, the integration runs from α+π/6 to α+5π/6.
Vo = x Vmp x cosα.
6. In the circuit shown below, SCR T1 conducts first. If T1 is fired at an angle of α > 30°, then T1 would conduct from
power-electronics-questions-answers-three-phase-converters-1-q5
a) α to 180°
b) 30 + α to 180°
c) 30 + α to 150°
d) 30 + α to 120°
Answer: b
Explanation: When firing angle is more than 30°, T1 would conduct from 30 + α to 180°. Irrespective of the firing angle, T1 will be turned on at 180° because it conducts first which means it is connected to the R phase and the phase sequence is R-Y-B. As R starts at 0° its value is 0 at 180° which reverse biases the SCR T1.
7. Find the expression for average output voltage for the given circuit if firing angle is greater than 30°. Take Vmp = secondary side maximum value of phase voltage.
power-electronics-questions-answers-three-phase-converters-1-q5
a) x cosα
b) x
c) x [1+cos].
d) x [3+cos].
Answer: c
Explanation: Vo = 3 x [ 1/2π ∫ Vmp sinωt d ] Where, the integration runs from α+π/6 to π. Because conduction takes place from 30 + α to 180° for T1 and than the waveform is symmetrical for all other SCRs.
Vo = x Vmp x [1+cos].
8. A three-phase three-pulse converter would operate as a line commutated inverter when
a) 30° < α < 60°
b) 90° < α <180°
c) 90° > α
d) it can never operate as a line commutated inverter
Answer: b
Explanation: The output voltage is proportional to cosine of the firing angle α. If α goes above 90° then the output voltage is negative, meaning that it is operating as an inverter.
9. A three-phase M-3 converter is operated from a 3-phase, 230 V, 50 Hz supply with load resistance R = 10 Ω. Find the value of firing angle if an average output voltage of 50% of the maximum possible output voltage is required.
Hint: α > 30°.
a) 92.7°
b) 67.7°
c) 45°
d) 75.7°
Answer: b
Explanation: We need , Vo = 0.5 Vom.
α>30° hence we use the equation Vo = x Vmp x [1+cos] √3Vmp = Vml = √2×230
Therefore, Vo = x √3 Vml x [1+cos] = 0.5Vom
x [1+cos] = Vo x 2π/3Vml = Vo/Vom = 1/2
α = 67.7°.
10. A three-phase half-wave controlled converter is fed from a 3-phase, 400 V source and is connected to a load which takes a constant current of 36 A. Find,the value of average output voltage and average current rating of SCR for a firing angle of 30°.
a) 234 V, 36 A
b) 234 V, 12 A
c) 135 V, 36 A
d) 135 V, 12 A
Answer: b
Explanation: Vo = x Vmp x cosα = 233.874 V.
Ia = Io/3 = 12 A.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Three Phase Converters-2”.
1. A three phase full converter will require __________ number of SCRs.
a) 3
b) 6
c) 9
d) 2
Answer: b
Explanation: Three legs having two SCRs each, six in total.
2. A three phase six pulse full converter works as a ac to dc converter for firing angles in the range
a) α > 90
b) 90 < α < 180
c) 0 < α < 90
d) 0 < α < 360
Answer: c
Explanation: When α is less than 90°, the SCRs conduct for 120° and the current and voltage are positive on an average hence, the power flows from AC source to DC load.
3. For the below given circuit, α = 60°. T2 will start conduction at ωt = __________ Assume the inductor L value to be negligible.
power-electronics-questions-answers-three-phase-converters-2-q3
a) 60°
b) 120°
c) 90°
d) 150°
Answer: d
Expansion: Assuming the phase sequence is R-Y-B. T1 would start conducting at 30+60 = 90°, T2 at 90+210/2 = 150°. This is because after T1, T3 would conduct from the upper group, as T2 belongs to the lower group it will start to conduct exactly between T1 and T3 i.e. between 90 and 210 which is 150°.
4. For a three phase full controlled converter, with 3 thyristors in the upper or positive group and 3 thyristors in the lower or negative group, at any given time
a) two thyristors are conducting from each group
b) one thyristor is conducting from each group
c) one thyristor is conducting from either of the groups
d) all 6 thyristors are conducting at a time
Answer: b
Explanation: Let’s say T1, T3 and T5 belong to the positive group and T2, T4 and T6 to the negative group. At any given time one SCR from each group conducts. e.g. T1 and T6 or T1 and T2.
5. In case of a three phase full controlled converter with 6 SCRs, commutation occurs every
a) 120°
b) 60°
c) 180°
d) 30°
Answer: b
Explanation: Every SCR conducts for 120°. This means that the SCRs from the positive group are fired 120° among themselves, same is true for SCRs from negative group. For example, if T1 starts conducting at 90° it will conduct till 90+120 = 210°. But while T1 is conducting, half of the time i.e. from 90 to 150, T6 is conducting and another half of the time T2 is conducting. Hence, commutation takes place every 60 degrees irrespective of the firing angle. Construct the firing sequence table for better understanding.
6. For the below given circuit, the conduction sequence for the negative group of SCRs is
power-electronics-questions-answers-three-phase-converters-2-q3
a) T4-T6-T2
b) T1-T2-T3
c) T2-T6-1
d) T2-T4-T6
Answer: d
Explanation: The negative group of SCRs has T2, T4 and T6. The conduct as T2-T4-T6, as T2 is connected to the B phase, T4 to the R phase and like-wise.
7. For a three-phase full controlled converter with R load, the average value of output voltage is zero for
a) α = 0°
b) α = 90°
c) α = 180°
d) It can never be zero
Answer: b
Explanation: For α = 90 degrees, the voltage waveform is equally symmetrical about the ωt axis, hence the average value is zero. This can also be found by using the formula for average output voltage,
Vo = cos α,
For α = 90°, cosα = 0, Vo = zero.
8. A three-phase full converter charges a battery from a three-phase supply of 230 V. The battery emf is 200 V and the internal resistance of the battery is 0.5 Ω. Find the value of the continuous current which is flowing through the battery if its terminal voltage is 210 V
a) 10 A
b) 20 A
c) 0.5 A
d) 25 A
Answer: b
Explanation: Vo = 210 V
Vo = E + Io x R
210 = 200 + 0.5 x Io
Io = 20 A.
9. A three-phase full converter charges a battery from a three-phase supply of 230 V. The battery emf is 200 V. Find the value of the firing angle if the battery terminal voltage is 210 V.
a) 36.54°
b) 56.7°
c) 89.3°
d) 47.45°
Answer: d
Explanation: Vo = cos α
α = cos -1 = 47.453°.
10. A three-phase full converter charges a battery from a three-phase supply of 230 V. Find the value of the power delivered to the load if a continues current of 20A is flowing through the battery of emf 200 V and internal resistance of 0.5 Ω.
a) 0 W
b) 5600 W
c) 4200 W
d) 1040 W
Answer: c
Explanation: Iavg = Irms = 20 A
P = E x Iavg + Irms 2 x R = 4200 W.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Three Phase Converters- 3”.
1. A three-phase full converter supplied from a 230 V source is working as a line commutated inverter. The load consists of RLE type with R = 5 Ω, E = 200 V and L = 1 mH. A continues current of 10 A is flowing through the load, find the value of the firing angle delay.
a) 119°
b) 127°
c) 156°
d) 143°
Answer: a
Explanation: Vo = 200 – 10×5 = 150V as the circuit is operating as an inverter Vo = -150V
Now, Vo = cos α
α = cos -1 = 118.88°.
2. A three-phase full converter is driving a DC motor. If a continues current of Im amperes is flowing through the motor load, then find the rms value of supply current drawn by the converter to drive the motor.
a) Im/√2
b) Im 2 /3
c) √2Im/√3
d) √2Im/3
Answer: c
Explanation: The RMS value of the supply current I S over π radians would be
(I S ) 2 = x 2 x = Im√2/√3.
3. Name the below given circuit.
power-electronics-questions-answers-three-phase-converters-3-q3
a) Full controlled, bridge converter
b) Full controlled, semi converter
c) Bridge type semi-converter
d) Half controlled, full converter
Answer: c
Explanation: It uses 3 SCRs and 3 diodes, hence it is a semi-converter. Option and make no sense, because there can be no full controlled semi-converter.
4. In the below given circuit, each SCR and diode conduct for
power-electronics-questions-answers-three-phase-converters-3-q3
a) 60° and 120° respectively
b) 120° and 60° respectively
c) 120°
d) 60°
Answer: c
Explanation: At any given time, one SCR and one diode is conducting, each conduct for 120° per cycle.
5. In the below given circuit, __ and __ conduct along with T2.
power-electronics-questions-answers-three-phase-converters-3-q3
a) T1, T3
b) D1, D2
c) D1, D3
d) T1, T2
Answer: c
Explanation: When one SCR conducts, a diode conducts along with it at a time to provide the path of current flow. . For example, if T2 starts conducting at 90° it will conduct till 90+120 = 210°. But while T2 is conducting, half of the time i.e. from 90 to 150 D1 is conducting and another half of the time D3 is conducting. T2 and D2 cannot conduct together as it will cause a short circuit. Hence, T2-D1 conduct for 60° and then T2-D3 conduct for another 60°.
6. In a three-phase semi-converter, at a time one SCR and one diode conduct simultaneously. With SCR T1 conducting which diode is most likely to conduct along with T1?
power-electronics-questions-answers-three-phase-converters-3-q6
a) D2 only
b) D3 only
c) D1 and D2
d) D2 and D3
Answer: d
Explanation: T1 and D1 together will cause a S.C. D2 or D3 any of these two can conduct along with T1 depending on which phase voltage is currently active RB or RY.
7. What is the value of voltage at the output terminal when the freewheeling diode is conducting?
power-electronics-questions-answers-three-phase-converters-3-q6
a) Zero
b) Maximum
c) E
d) It could be anything depending on α
Answer: a
Explanation: When FD is conducting it will short circuit the load terminal resulting in zero voltage. It won’t be E because the terminals are shorted. It can be E when none of the devices are conducting .
8. A 3-phase full converter feeds power to an R load of 10 Ω. For a firing angle delay of 30° the load takes 5 kW. An inductor of large value is also connected to the load to make the current ripple free. Find the value of per phase input voltage.
a) 133 V
b) 230/√3 V
c) 191/√3 V
d) 298/√3 V
Answer: c
Explanation: Ior = Vo/R = cos α
P = 5 kW = Ior 2 x R = [ cos α] 2 x 1/R] Therefore, Vs = √50000 x = 191.22 V
Vs = 191/√3 V.
9. A three-phase semi-converter circuit is given a supply of 400 V. It produces at the output terminals an average voltage of 381 V. Find the rectification efficiency of the converter circuit.
a) 99.65 %
b) 95.25 %
c) 91 %
d) 86.5 %
Answer: b
Explanation: Rectification efficiency = Pdc/Pac
Pdc = Vo x Io . . .
Pac = Vrms x Irms . . .
For a semi-converter Irms = Io
Therefore, Rectification efficiency = 381/400 = 95.25 %.
10. In a 3-phase semi-converter, for firing angle less than 60° the freewheeling diode conducts for
a) 30°
b) 60°
c) 120°
d) 0
Answer: d
Explanation: In case of a semi-converter operating with α < 60°, FD does not comes into play, as the voltage never falls to zero and gives no chance for the inductor to discharge.
This set of Basic Power Electronics Interview questions and answers focuses on “Three-Phase Converters-4”.
1. In a 3-phase full wave converter, if V is the maximum value of line voltage at the input, then each SCR is subjected to a peak negative voltage of
a) V
b) 3V
c) √3V
d) V/2
Answer: c
Explanation: PIV = √3 Vml in case of a semi-converter.
2. In a 3-phase semi-converter, firing angle is less than 60°, as such each SCR and diode conduct respectively for __________
a) 60, 60
b) 90, 30
c) 120, 120
d) 180, 180
Answer: c
Explanation: Each will conduct for 120° per cycle whatever the firing angle is.
3. The effect of source inductance on the performance of a 3-phase controlled converter is to
a) increase the average load voltage
b) reduce the average load voltage
c) make the load current continuous
d) remove ripples from the load current
Answer: b
Explanation: It reduces the average value of the output voltage by introducing a overlap delay μ.
4. Which of the following converter circuits would require a neutral point?
a) 3-phase semi-converter
b) 3-phase full converter
c) 3-phase full converter with freewheeling diode
d) 3-phase half wave converter
Answer: d
Explanation: Half wave converter would require delta-star transformer, the secondary winding star connection requires a neutral point.
5. The range of firing angle for a 3-phase, 3-pulse converter feeding a resistive load is __________ .
a) 0 to 180
b) 0 to 150
c) 30 to 150
d) 30 to 180
Answer: b
Explanation: Firing angle for a 3-phase, 3-pulse converter feeding a resistive load is 0 to 150 degrees.
6. A 3-phase bridge converter is given a three-phase supply in the phase sequence R-Y-B. Let the neutral to R phase voltage be Vm sinωt. The first SCR is fired at an angle of 15°. What is the maximum value at the output terminals at this instant?
a) Vm/√2
b) Vm
c) 1.5 Vm
d) 3 Vm
Answer: c
Explanation: In case of a 3 phase bridge converter, the maximum value of voltage at the output terminal is always 1.5 Vm.
7. The PIV experienced by each SCRs in M-3 converter is __________ times that in a 3-phase full converter having the same output voltage.
a) 0.5
b) 1
c) 2
d) 1.5
Answer: c
Explanation: In case of M-3 type of connection, the devices have to handle more peak inverse voltage than the 3-phase full converter which has 6 SCRs.
8. Each SCR of a 3-phase 6-pulse converter conducts for
a) 120 degrees
b) 60 degrees
c) 180 degrees
d) 360 degrees
Answer: a
Explanation: A 3-phase 6-pulse converter is nothing but the 3-pulse full controlled converter using 6 devices each conducting for 120°.
9. A 3-phase full converter has an average output voltage of 365 V for zero degree firing and resistive load. For a firing angle of 90 degree, the output voltage would be
a) 125 V
b) 569 V
c) 365 V
d) zero
Answer: d
Explanation: Cos 90 = 0.
10. Semi-converters are
a) single quadrant converters
b) double quadrant converters
c) three quadrant converters
d) none of the mentioned
Answer: a
Explanation: Semi-converters are single quadrant converters, because the voltage and current can only be both positive due to the diodes connected.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Advance Three Phase Converters”.
1. In a 3 phase M-6 controlled converter for continuous conduction mode, each SCR conducts for __________ per cycle.
a) 2π/6 radians
b) 30 degrees
c) 3π/2 radians
d) 120 degrees
Answer: a
Explanation: Each SCR conducts for 60° or 2π/6 radians. 6 x 60 = 360°.
2. In a 3 phase, 12-pulse controlled converter for continuous conduction mode, each SCR conducts for __________ per cycle.
a) π/6 radians
b) 60 degrees
c) 3π/2 radians
d) 12 degrees
Answer: a
Explanation: Each SCR conducts for 30° or π/6 radians. 12 x 30 = 360°.
3. A 3-phase full converter delivers a ripple free load current of 10 A with a firing angle delay of 45°. The input voltage is 3-phase, 400 V, 50 Hz. The source current is given by the following relation.
power-electronics-questions-answers-advance-three-phase-convters-q3
Find the fundamental component of the source current amplitude.
a) 11.03
b) 2.205
c) 11.03 sin 45
d) 46.98
Answer: a
Explanation: Put n = 1 and the rest of the given values in the above given equation.
4. A 3-phase full converter delivers a ripple free load current of 10 A with a firing angle delay of 45°. Find the DF .
a) 1.414
b) 0
c) 0.707
d) 0.569
Answer: c
Explanation: DF = cosα = cos45 = 0.707
Note that ripple free current does not mean that DF = 0. It means the current is continues in magnitude and direction, current will always contain harmonics whether it is rippled or ripple free.
5. What is the relationship between DF, CDF and PF?
a) PF = CDF = DF
b) PF = CDF/DF
c) PF = DF/CDF
d) PF = CDF x DF
Answer: d
Explanation: PF = x .
6. The commutation period when both incoming and outgoing SCRs are conducting due to source inductance is called as the
a) conduction delay
b) overlap period
c) one on one period
d) distorting angle
Answer: b
Explanation: Due to source inductance, SCRs cannot start and stop conducting immediately, hence a time occurs when both incoming and outgoing SCRs are conducting together. This delay is called as commutation angle or overlap period .
7. A 3-phase full converter delivers a ripple free load current of 10 A with a firing angle delay of 45°. The input voltage is 3-phase, 400 V, 50 Hz. The source current is given by the following relation.
power-electronics-questions-answers-advance-three-phase-convters-q3
Find the value of 2nd harmonic source current amplitutue.
a) 11.25 A
b) 0.256 A
c) 2.69 sin A
d) 0 A
Answer: d
Explanation: 2nd harmonics are absent in a 3-phase full converter, it has only odd number of harmonics i.e. 3rd, 5th etc.
8. In a p-pulse converter, each SCR conducts for
a) p radians
b) p degrees
c) p/2π radians
d) 2π/p radians
Answer: d
Explanation: In a p-pulse controlled converter, each device conducts for p radians. p-pulse converter would have p devices. Hence, 2π/p x p = 2π = 360°.
9. Which of the below mentioned converter can operate in both 3-pulse and 6-pulse modes?
a) 3-phase half wave controller
b) 3-phase full converter
c) 3-phase semi-converter
d) None of the mentioned
Answer: c
Explanation: Semi-converters can operate in 6-pulse modes gating the SCRs at appropriate intervals.
10. A M-6 controlled converter or 6-pulse half-wave controlled converter is obtained by using a transformer having
a) a double delta connected secondary winding
b) a double star connected secondary winding
c) a double delta connected primary winding
d) 6-windings on both primary and secondary sides
Answer: b
Explanation: M-6 requires a transformer having a delta connected primary and a double star connected secondary such that 6 SCRs are connected to it on the secondary side.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Dual Converters-1”.
1. Dual converters provide
a) two quadrant operation
b) three quadrant operation
c) four quadrant operation
d) none of the mentioned
Answer: c
Explanation: Dual converters provide four quadrent operation, which means voltage can be positive or negative and so can be the current. Hence, AC-DC, DC-AC any converter configuration can be used.
2. A dual converters has
a) two full converters in series
b) two half converters in series
c) two full converters in anti-parallel
d) two half converters in anti-parallel
Answer: c
Explanation: Dual converters have two full converters connected in anti-parallel which provides a four quadrant operation.
3. The major advantage of using dual converters is that
a) it is cheaply available
b) it has better pf
c) no mechanical switch is required to change the mode of operation
d) its operating frequency is very high
Answer: c
Explanation: No mechanical arrangement is required to change from inverter to converter and converter to inverter, which was required in earlier methods.
4. The four quadrant operation of dual converters can be obtained by
a) moving the mechanical lever
b) adding inductance to the circuit
c) changing the firing angle value
d) none of the mentioned
Answer: c
Explanation: The four quadrant operation can be obtained simply by adjusting appropriate values of firing angles for both the connected converters.
5. A single full converter alone can given a
a) four quadrant operation
b) three quadrant operation
c) two quadrant operation
d) none of the mentioned
Answer: c
Explanation: A single full converter alone gives two quadrent operation, hence for all four quadrant operation two full converter circuits are connected in anti-parallel.
6. Find the error in the below given dual converter circuit.
power-electronics-questions-answers-dual-converters-1-q6
i) Load is not connected in the right position
ii) Only 4 SCRs must be used
iii) Voltage source is not connected for one of the converter circuit
iv) Voltage source is not connected in the proper place
a) All 4
b) Both and
c) Both and
d) Both and
Answer: c
Explanation: The right connection for single-phase dual converter is shown below.
power-electronics-questions-answers-dual-converters-1-q7-q8
7. In the below given circuit, the right side converter C2 operates in the ___ and ___ quadrant.
power-electronics-questions-answers-dual-converters-1-q7-q8
a) second, fourth
b) first, fourth
c) second, third
d) first, third
Answer: c
Explanation: The C2 converter will supply the load current in direction opposite to that supplied by the converter C1. For converter C2, when α> 90 it operates in 2nd quadrant and if α<90 both current and voltage are negative, C2 is in inverter mode and operates in 3rd quadrant.
8. Name the below given circuit.
power-electronics-questions-answers-dual-converters-1-q7-q8
a) Single-phase dual converter circulating current type
b) Single-phase dual converter non-circulating current type
c) Three-phase dual converter non-circulating current type
d) Three-phase dual converter circulating current type
Answer: b
Explanation: The circuit is a single phase dual converter circuit. As the there is no reactor in series, it is a non-circulating type.
9. For a single-phase dual converter, with converters C1 and C2 connected in anti-parallel, which relation among the following is true to keep the average voltages from C1 and C2 equal? C1 and C2 have firing angles α1 and α2 respectively.
a) α1 = α2
b) α1 + α2 = 360°
c) α1 + α2 = 180°
d) none of the mentioned
Answer: c
Explanation: By maintaining α1 + α2 = 180°, one converter can operate as converter and another as an inverter hence, the average output voltages are equal . This can be proved as follows
Vom cos α1 = Vom cos α2
Vom = 2Vm/π . .
cos α1 = cos α2
cos α1 = cos
α1 + α2 = 180°.
10. In non-circulating current mode dual converters, the circulating current is avoided by
a) connecting a series reactor
b) maintaining α1 + α2 = 180°
c) operating only one converter
d) adding an extra SCR
Answer: c
Explanation: Reactor is added in circulating current mode not in non-circulating mode. The circulating current is avoided by using only one of the converters.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Dual Converters-2”.
1. In circulating current mode dual converters, the circulating current is avoided by
a) connecting a series reactor
b) maintaining α1 + α2 = 180°
c) operating only one converter
d) adding an extra SCR
Answer: a
Explanation: Reactor is added in circulating current mode not in non-circulating mode. The reactor limits the current to a reasonable value.
2. Choose the correct statement
a) Circulating current exists only in circulating current mode
b) Circulating current exists only in non-circulating current mode
c) Circulating current exists in both the circulating and non-circulating current mode
d) none of the mentioned
Answer: c
Explanation: The circulating current does exist in both the converter circuits, but it is avoided by using a reactor in non-circulating type and by operating only one circuit in case of circulating type.
3. What causes circulating current in dual converters?
a) Temperature issues
b) Inductance in load circuit
c) Out of phase voltages from both the converters
d) none of the mentioned
Answer: c
Explanation: In case of practical dual converters, the voltages from both the converter circuits though equal in magnitude is out of phase. This indifference in voltages causes circulating currents to flow.
4. Name the below given circuit.
power-electronics-questions-answers-dual-converters-2-q4
a) Single-phase dual converter with circulating current type
b) Single-phase dual converter with non-circulating current type
c) Three-phase dual converter with non-circulating current type
d) Three-phase dual converter with circulating current type
Answer: c
Explanation: The circuit is a three phase dual converter circuit. As the there is no reactor in series, it is a non-circulating type.
5. In case of three-phase dual converter, one of the converter circuits is fired at an angle of 60°. For both the converter circuits to have equal average output voltage, what is the value of the firing angle for the other converter circuit?
a) 60°
b) 120°
c) 100°
d) Insufficient data
Answer: b
Explanation: For equal average output voltage, α1 + α2 = 180°.
6. In case of circulating current type dual converters, the reactor is inserted between
a) supply and converter
b) across the load
c) between the converters
d) no reactor is used in case of circulating type dual converter
Answer: c
Explanation: Reactor is added in circulating current mode between both the converters. The reactor limits the current to a reasonable value.
7. Choose the correct statement.
a) Circulating current type is faster in operation
b) Non-circulating current type is faster in operation
c) Both the types have the same speed of operation
d) Circulating current improves power factor
Answer: a
Explanation: In case of non-circulating type, to shift the operation from one mode to another mode a delay of 10 to 20msec is required to let the current decay to zero value and let the outgoing SCRs safely turn off. This delay is not required in case of circulating current mode.
8. Circulating current flows
a) from load to converters
b) from one converter to another converter
c) in the whole circuit
d) none of the mentioned
Answer: b
Explanation: The circulating current flows only between the converters and not to through the load.
9. The reactor in circulating current type dual converters
a) increases losses
b) reduces power factor
c) increase the weight of the circuit
d) all of the above
Answer: d
Explanation: All of the above mentioned are the major drawbacks of using reactors to reduce circulation current.
10. If V1 and V2 are the instantaneous voltages of the two converter circuits in the dual convert, then the output voltage is
a) V1 + V2
b) /2
c) V1 – V2
d) 2
Answer: b
Explanation: The load voltage is the average value of the instantaneous converter outputs.
This set of Advanced Power Electronics Multiple Choice Questions & Answers focuses on “Dual Converters-3”.
1. Name the circuit shown in the figure below.
advanced-power-electronics-questions-answers-q1-q4-q8-q10
a) Single-phase circulating current type dual converter
b) Single-phase non-circulating current type dual converter
c) Three-phase circulating current type dual converter
d) None of the mentioned
Answer: d
Explanation: The converters are connected in parallel not in anti-parallel.
2. Dual converters handle ________ during no load.
a) very high temperature
b) no current
c) only circulating current
d) load current
Answer: c
Explanation: During no load period, both the converters handle only the circulating current as the load current is zero.
3. In a three phase dual converter, converter 1 is operating with α = 95° and converter 2 is operating with α2 = 85°. Choose the correct statement.
a) Converter 1 is operating as a rectifier and converter 2 as an inverter
b) Both the converters are operating as a rectifier
c) Converter 1 is operating as a inverter and converter 2 as an rectifier
d) Both the converters are operating as an inverter
Answer: c
Explanation: As α1 > 90, converter 1 is operating as a inverter and with α2<90, converter 2 is working as an inverter.
4. The circuit shown below is that of a
advanced-power-electronics-questions-answers-q1-q4-q8-q10
a) single-phase circulating current type dual converter
b) three-phase non-circulating current type dual converter
c) three-phase circulating current type dual converter
d) none of the mentioned
Answer: c
Explanation: It is a 3-phase as it uses 6 SCRs on either sides, it is circulating current type as there is a reactor placed on either sides of the load to compensate for the circulating current.
5. In circulating current type of dual converters, the nature of voltage across the reactor is
a) triangular
b) pulsating
c) constant
d) alternating
Answer: d
Explanation: The current is alternating in nature.
6. When a line commutated converter operates in the inverter mode then
a) it draws real and reactive power from the supply
b) it delivers both real and reactive power to the supply
c) it delivers real power to the supply
d) it delivers reactive power to the supply
Answer: c
Explanation: A converter is given an AC supply, when it is in the inverter mode, it is sending power to the AC supply and absorbing power power from the supply.
7. The reactor is required in a circulating current type dual converter to
a) improve the power factor
b) smooth-en the waveform of circulating current
c) limit the circulating current
d) increase the circulating current
Answer: c
Explanation: Reactor is introduced in the circulating current mode between both the converters. The reactor limits the current to a reasonable value.
8. In the below given figure the converter 1 has firing angle = α1, converter 2 has firing angle = α2. For the dual converter to operate in the second quadrant.
advanced-power-electronics-questions-answers-q1-q4-q8-q10
a) α1 > 90°
b) α2 > 90°
c) α1 < 90°
b) α2 < 90°
Answer: b
Explanation: Converter 1 would control the first and the fourth quadrant and converter 2 would operate in the second and the third quadrant. In the second quadrant, voltage should be positive and current should be negative, hence inverter operation hence α2 > 90°.
9. If V1 and V2 are the converter output voltages then the reactor voltage is
a) V1 + V2
b) V1 – V2
c) V1 x V2
d) none of the mentioned
Answer: b
Explanation: The reactor voltage Vr is the difference of the converter output voltages.
10. In the below given figure, the converter 1 has firing angle = α1, converter 2 has firing angle = α2. For the dual converter to operate in the fourth quadrant.
advanced-power-electronics-questions-answers-q1-q4-q8-q10
a) α1 > 90°
b) α2 > 90°
c) α1 < 90°
b) α2 < 90°
Answer: a
Explanation: Converter 1 would control the first and the forth quadrant and converter 2 would operate in the second and the third quadrant. In the fourth quadrant, voltage should be negative and current should be positive, hence C1 as an inverter is operated with α1 > 90°.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Choppers-1″.
1. In the ___________ type of chopper, two stage conversions takes place.
a) AC-DC
b) AC link
c) DC link
d) None of the mentioned
Answer: In AC link chopper, DC is converter to AC than stepped up/down than again AC to DC conversation takes place.
2. Choppers converter
a) AC to DC
b) DC to AC
c) DC to DC
d) AC to AC
Answer: c
Explanation: Choppers are used to step up or step down DC voltage/current levels. Hence, they are DC to DC converters.
3. A chopper may be thought as a
a) Inverter with DC input
b) DC equivalent of an AC transformer
c) Diode rectifier
d) DC equivalent of an induction motor
Answer: b
Explanation: It is a DC equivalent of an AC transformer because it behaves in the similar manner i.e. converting fixed DC to variable DC.
4. Which device can be used in a chopper circuit?
a) BJT
b) MOSFET
c) GTO
d) All of the mentioned
Answer: d
Explanation: All of the devices which can be used as a switch can be used in a chopper.
5. A chopper is a
a) Time ratio controller
b) AC to DC converter
c) DC transformer
d) High speed semiconductor switch
Answer: d
Explanation: It is a high speed on/off semiconductor switch. Note that it behaves like a DC transformer, does not mean it is a DC transformer. There is no DC transformer.
6. What is the duty cycle of a chopper ?
a) Ton/Toff
b) Ton/T
c) T/Ton
d) Toff x Ton
Answer: b
Explanation: It is the time during which the chopper is on relative to the whole period .
7. The load voltage of a chopper can be controlled by varying the
a) duty cycle
b) firing angle
c) reactor position
d) extinction angle
Answer: a
Explanation: The output voltage can be changed by changing the duty cycle .
8. The values of duty cycle lies between
a) 0<α<1
b) 0>α>-1
c) 0<=α<=1
d) 1<α<100
Answer: c
Explanation: The duty cycle is between 0 and 1. It can be 0 if the chopper switch is never on and it can be 1 when the chopper switch is always on.
9. If T is the time period for a chopper circuit and α is its duty cycle, then the chopping frequency is
a) Ton/α
b) Toff/α
c) α/Toff
d) α/Ton
Answer: d
Explanation: α = Ton/T
T = Ton/ α
f = 1/T = α/Ton.
10. Find the output voltage expression for a step down chopper with Vs as the input voltage and α as the duty cycle.
a) Vo = Vs/α
b) Vo = Vs x α
c) Vo = Vs 2 /α
d) Vo = 2Vs/απ
Answer: b
Explanation: The chopper output voltage is Duty cycle x the input voltage .
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Choppers-2″.
1. The below given figure is that of a _________
power-electronics-questions-answers-choppers-2-q1-q2-q3
a) step-up/step-down chopper
b) step-down chopper
c) step-up chopper
d) none of the mentioned
Answer: b
Explanation: When the switch is closed current flows through the load and also charges the inductor L, when the switch is off, the current freewheels through L-R-FD-L and voltage is zero because the FD short circuits the load.
2. In the below given circuit, when switch is on
power-electronics-questions-answers-choppers-2-q1-q2-q3
a) voltage is non-zero and current is rising
b) voltage is non-zero and current is decaying
c) voltage is zero and current is rising
d) voltage is zero and current is decaying
Answer: a
Explanation: When the switch is closed current flows through the load and also charges the inductor L, hence voltage is Vo and current is rising.
3. In the below given circuit, when switch is off
power-electronics-questions-answers-choppers-2-q1-q2-q3
a) voltage is non-zero and current is rising
b) voltage is non-zero and current is decaying
c) voltage is zero and current is rising
d) voltage is zero and current is decaying
Answer: d
Explanation: When the switch is off, the current freewheels through L-R-FD-L and voltage is zero because the FD short circuits the load.
4. In a step down chopper, if Vs = 100 V and the chopper is operated at a duty cycle of 75 %. Find the output voltage.
a) 100 V
b) 75 V
c) 25 V
d) none of the mentioned
Answer: b
Explanation: Vo = Duty cycle x Vs = 0.75 x 100 = 75 V.
5. If Vo is the output voltage and Vs is the input DC voltage, then for the below given circuit.
power-electronics-questions-answers-choppers-2-q5-q6-q7-q8
a) Vo = Vs
b) Vo < Vs
c) Vo > Vs
d) Insufficient information
Answer: c
Explanation: The given circuit is that of a step-up chopper. For a step up chopper the output voltage is always greater than the input voltage.
6. In the below given circuit, when switch is on
power-electronics-questions-answers-choppers-2-q5-q6-q7-q8
a) voltage is non-zero and current is rising
b) voltage is zero and current is decaying
c) voltage and current both are non-zero
d) voltage and current both are zero
Answer: c
Explanation: When the switch is on, the current flows from Vs-L-SW-Vs and charges the inductor , hence voltage and current both are zero.
7. In the below given circuit, when switch is first switched closed and then opened the
power-electronics-questions-answers-choppers-2-q5-q6-q7-q8
a) voltage is non-zero and current is rising
b) voltage is zero and current is decaying
c) voltage is non-zero and current is decaying
d) voltage is zero and current is rising
Answer: c
Explanation: When the switch is off, the inductor is has stored energy from the earlier chopper on operation. The current starts to flow from Vs-L-D-Load-Vs, hence current is decaying as the inductor is discharging and the voltage is Vs + Vl .
8. What is the voltage across the load when the SW is first closed and then opened?
power-electronics-questions-answers-choppers-2-q5-q6-q7-q8
a) Vs
b) duty cycle x Vs
c) Vs + L
d) zero
Answer: c
Explanation: When the switch is off, the inductor is has stored energy from the earlier chopper on operation. The current starts to flow from Vs-L-D-Load-Vs, hence current is decaying as the inductor is discharging and the load voltage is Vs + Vl
Vl = L
Vo = Vs + L.
9. Find the expression for output voltage for a step-up chopper, assume linear variation of load current and α as the duty cycle.
a) Vs
b) Vs/α
c) Vs/
d) Vs/√2
Answer: c
Explanation: During Ton, the energy stored in the inductor is
W in = Vs x T on x /2
During T off , the energy released by the inductor is
W out = x T off x /2
Considering ideal condition,
W in = W out
Vo = Vs x
α = Ton/T
Vo = T/T-Ton = Vs/1-α.
10. Find the output voltage for a step-up chopper when it is operated at a duty cycle of 50 % and Vs = 240 V.
a) 240 V
b) 480 V
c) 560 V
d) 120 V
Answer: b
Explanation: Vo = Vs/1-α.
This set of Power Electronics Test focuses on “Choppers-3”.
1. If a step up chopper’s switch is always kept off then
a) Vo = 0
b) Vo = ∞
c) Vo = Vs
d) Vo > Vs
Answer: c
Explanation; If it is said the a chopper is always kept off, that means the switch is always open. As such, Ton = 0
Duty cycle = 0
Vo = Vs/1-duty cycle . . . .
2. If a step up chopper’s switch is always kept open then
a) Vo = 0
b) Vo = ∞
c) Vo = Vs
d) Vo > Vs
Answer: b
Explanation: If it is always in then, Ton = T. Duty cycle = 1.
Vo = Vs/1-duty cycle = Vs/0
Therefore, Vo = ∞.
3. Find the average value of output voltage of a basic step-down chopper, with duty cycle = α and load = R Ω.
power-electronics-questions-answers-test-q3-q4
a) I = Vs x α
b) I = /R
c) I = 0
d) I = Vs/R
Answer: b
Explanation: Vo = Vs x α
I = Vo/R.
4. For the below given circuit, find the output current at the instant of commutation.
power-electronics-questions-answers-test-q3-q4
a) Vo/R
b) Vs/R
c) 0
d) α/2
Answer: b
Explanation: The circuit is that of a step-down chopper. The output current is commutated by the switch at the instant t = T on . Therefore, output current at the instant of commutation is Vs/R.
5. For a step-down chopper, find the rms value of output voltage. Let α be the duty cycle and Vs be the input voltage.
a) α x Vs
b) Vs/α
c) √α x Vs
d) Vs/2
Answer: c
Explanation: [(T on /T).Vs 2 ] 1/2 = √α x Vs.
6. A step down chopper is operated at 240V at duty cycle of 75%. Find the value of RMS switch current. Take R = 10 Ω.
a) 2.07 A
b) 200 mA
c) 1.58 A
d) 2.4 A
Answer: a
Explanation: The switch could be anything, a IGBT, a SCR, a GTO etc. The current through the switch is the same as the current through the load for a step down chopper.
Vo = Vs/1-α
α = 0.75
I = √α x Vs/R.
7. Find the expression for effective input resistance of a step down chopper. With R load and duty cycle = α.
a) R x α
b) R/2
c) 0
d) R/α
Answer: d
Explanation: Effective input resistance of the chopper circuit = / = / = R/α.
8. A step-up chopper has input voltage of 220 V and output voltage of 660 V. If the conducting time of the IGBT based chopper is 100 μs, compute Toff width of the output voltage pulse.
a) 100 μs
b) 150 μs
c) 50 μs
d) Insufficient data
Answer: c
Explanation: 660 = 220/1-α
α = 2/3 = Ton/T
Ton = 2T/3 = 100μs. This gives chopping period T = 100 x 3/2 = 150 μs
Toff = T – Ton = 150 – 100 = 50 μs.
9. For a step-up chopper, when the duty cycle is increased the average value of the output voltage
a) increases
b) decreases
c) remains the same
d) none of the mentioned
Answer: a
Explanation: Vo = Vs/1-duty cycle
Hence, as duty cycle increases the output voltage increases.
10. For a step-down chopper, when the duty cycle is increased the average value of the output voltage
a) increases
b) decreases
c) remains the same
d) none of the mentioned
Answer: a
Explanation; Vo = Duty cycle x Vs. Hence, output voltage is directly proportional to the duty cycle.
This set of Power Electronics Quiz focuses on “Choppers-4”.
1. For a step-down chopper shown in the figure below, dc source voltage = 230 V and the voltage drop across the chopper = 2 V. Find the value of average output voltage for a duty cycle of 0.4.
power-electronics-questions-answers-quiz-q1-q2-q3-q4
a) 230 V
b) 92 V
c) 90 V
d) 91.2 V
Answer: d
Explanation: The voltage drop across the switch occurs only when it is on.
Vo = x Duty cycle = 91.2 V.
2. For the step-down chopper shown in the figure below, dc source voltage = 230 V and the voltage drop across the chopper = 2 V. Find the value of rms output voltage for a duty cycle of 0.4.
power-electronics-questions-answers-quiz-q1-q2-q3-q4
a) 91.2 V
b) 230 V
c) 144.2 V
d) 145 V
Answer: c
Explanation: The voltage drop across the switch will occurs only when it is on.
Vo = √0.4 x = 144.2 V.
3. For the step-down chopper shown in the figure below, dc source voltage = 230 V. Find the power delivered to the load of R = 10 Ω. Duty cycle = 40 %. Take voltage drop at the switch to be 2 V.
power-electronics-questions-answers-quiz-q1-q2-q3-q4
a) 1 kW
b) 2.08 kW
c) 569 W
d) 207 W
Answer: b
Explanation:
Vo = √0.4 x = 144.2 V
P = 144.2 2 /10 = 2079.3 W = 2.08 kW.
4. For the step-down chopper shown in the figure below, dc source voltage = 230 V. Find the power input to the chopper If load of R = 10 Ω is connected. Duty cycle = 40 %.
power-electronics-questions-answers-quiz-q1-q2-q3-q4
a) 2300 W
b) 2097 W
c) 2560 W
d) 5290 W
Answer: b
Explanation: Power input to the chopper is Vs . Io
Vs = 230 V
Vo = x Duty cycle = 91.2 V
Io = Vo/R = 91.2/10
P i = Vs.Io = 2097.6 W.
5. The below given chopper circuit is that of a
power-electronics-questions-answers-quiz-q5
a) step-up chopper
b) step-down chopper
c) step-up/step-down chopper
d) none of the mentioned
Answer: c
Explanation: It is a step-up/step-down chopper. By controlling the duty cycle it can be made to behave like a step-up or a step down chopper.
6. The expression for a step-up/step-down chopper with α as the duty cycle and Vs as the dc input voltage is
a) Vs/1 – α
b) α x Vs
c) Vs
d) Vs
Answer: c
Explanation: A step-up, step-down chopper can behave as a step up chopper for α > 0.5 and step –down chopper for α < 0.5.
7. For a step-up/step-down chopper, if the duty cycle > 0.5 then
a) Vo = Vs
b) Vo < Vs
c) Vo > Vs
d) None of the mentioned
Answer: c
Explanation: Vo = Vs
For α is greater than 0.5, the chopper behaves as a step-up chopper. Hence, Vo > Vs.
8. A step down chopper has Vs = 230 V and R = 10 Ω. For a duty cycle of 0.4, the power taken by the chopper is 2097 Watts. Find the chopper efficiency. Take the voltage drop across the chopper switch as 2 V.
a) 98 %
b) 89.96 %
c) 99.14 %
d) 96.54 %
Answer: c
Explanation: Vo = √0.4 x = 144.2 V
Po = 144.2 2 /10 = 2079.3 Watts
Pi = 2097 Watts
η = Po/Pi = 99.14 %.
9. A step down chopper has input dc voltage of 220 V and R = 10 Ω in series with L = 65 mH. If the load current varies linearly between 11 A and 17 A, then find the duty cycle α.
a) 1
b) 0.4
c) 0.6
d) 0.7
Answer: d
Explanation: Average load current = 11+17 / 2 = 14A
Average load voltage = IxR = 14×10 = 140V
But, Vo = α x Vs
Therefore, α x 220 = 140, α = 0.7.
10. For a step-up/step-down chopper, if α = 0.5 then
a) Vo = Vs
b) Vo < Vs
c) Vo > Vs
d) none of the mentioned
Answer: a
Explanation:
Vo = Vs
For α = 0.5
Vo = Vs x .
This set of tricky Power Electronics questions and answers focuses on “Choppers-5”.
1. In the figure shown below, the capacitor C is used to
tricky-power-electronics-questions-answers-q1-q2-q4
a) maintain the load voltage constant
b) protect the load
c) protect the chopper switch
d) maintain the load current constant
Answer: a
Explanation: It is a step-up/step-down chopper, the C maintains the load voltage constant.
2. In the below shown step-up/step-down chopper circuit, when the switch is on
q4.png”> tricky-power-electronics-questions-answers-q1-q2-q4
a) current through inductor rises and the load voltage is zero
b) current through inductor rises and the load voltage is Vs
c) current through inductor decays and the load voltage is zero
d) current through inductor decays and the load voltage is Vs
Answer: b
Explanation: During T on , current flows from Vs-CH-L-Vs. Current through L rises from I1 to I2 and Vs appears across the inductor and load as they are in parallel. Voltage drop at the diode has to be considered as zero.
3. For a step-up/step-down chopper, if α < 0.5 then
a) Vo = Vs
b) Vo < Vs
c) Vo > Vs
d) none of the mentioned
Answer: b
Explanation: Vo = Vs
For duty cycle is less than 0.5, the circuit behaves like a step down chopper. Hence, Vo < Vs.
4. In the below shown step-up/step-down chopper circuit, when the chopper switch if first on and then off, then the current flows from path
q4.png”> tricky-power-electronics-questions-answers-q1-q2-q4
a) Vs-CH-L-Vs
b) L-Load-D-L
c) L-D-Load-L
d) no current flows
Answer: b
Explanation: When the chopper switch is off, polarity of emf induced in the inductor L gets reversed biased. Emf induced in L forward biases the diode D and L starts to discharge from inductor-load-diode-inductor.
5. For a step-up/step-down chopper, if current increases from I 1 to I 2 linearly during T on , then find the energy stored in the inductor during T on
a) zero
b) Vs x (I 1 + I 2 )
c) Vs x [ (I 1 + I 2 )/2 ] x T on
d) Vs x [ (I 1 + I 2 )/2 ] x T
Answer: c
Explanation: Iavg = [ (I 1 + I 2 )/2 ] Energy stored = W in = Vs x [ (I 1 + I 2 )/2 ] x T on .
6. A step-down chopper is also called as a
a) first-quadrant chopper
b) second-quadrant chopper
c) third-quadrant chopper
d) fourth-quadrant chopper
Answer: a
Explanation: It is called as a first quadrant chopper as the current always flows from source to load and the current and voltage both are always positive.
7. For the type-B chopper shown below
tricky-power-electronics-questions-answers-q7-q8-q9
a) during T on , Vo = 0 and L stores energy
b) during T on , Vo is not 0 and L stores energy
c) during T on , Vo = 0 and L releases energy
d) during T on , Vo is not 0 and L releases energy
Answer: a
Explanation: During T on , SW is on and vo = 0 as the switch S.C’s the load. E drives current through L and charges it.
8. For the type-B chopper shown below
tricky-power-electronics-questions-answers-q7-q8-q9
a) during T off , Vo = and D is conducting
b) during T off , Vo = and D is not-conducting
c) during T off , Vo = 0
d) during T off , Vo = Vs
Answer: a
Explanation: When switch Is off, Vo = E + Ldi/dt exceeds source voltage Vs and D is forward biased and begins to conduct.
9. The below given chopper is that of a
tricky-power-electronics-questions-answers-q7-q8-q9
a) step-up chopper
b) step-down chopper
c) step-up/step-down chopper
d) depends on the duty cycle
Answer: a
Explanation: The load voltage Vo = E + L is greater than Vs, hence it is a step-up chopper.
10. The type-C chopper or two quadrant type-A chopper has
a) type-A and type-B choppers in series
b) type-A and type-B choppers in parallel
c) two type-A choppers in series
d) two type-A choppers in parallel
Answer: b
Explanation: The type-C chopper is a two quadrant chopper, it operates in the 1st and the 2nd quadrant. It has type-A and type-B choppers connected in parallel.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Chopper Control Strategies”.
1. In case of TRC , _________ is varied
a) duty cycle
b) firing angle
c) supply frequency
d) supply voltage magnitude
Answer: a
Explanation: In case of TRC, α is varied to change the average value of output voltage.
2. In constant frequency TRC or pulse width modulation scheme, ________ is varied.
a) Vs
b) Ton
c) T
d) f
Answer: b
Explanation: In pulse width modulation scheme also known as constant frequency time ratio control, the on-time Ton is varied keeping total time T constant. Toff also varies automatically with Ton to keep Ton constant.
3. In case of variable frequency system __________ is varied
a) T
b) Ton
c) Toff
d) supply frequency
Answer: a
Explanation: In frequency modulation scheme, the frequency hence T is varied by either keeping Ton constant or keeping Toff constant.
4. In pulse width modulation scheme, _________ is kept constant.
a) Vs
b) Ton
c) T
d) Toff
Answer: b
Explanation: In pulse width modulation scheme also known as constant frequency time ratio control, the on-time Ton is varied keeping total time T constant. Toff also varies automatically with Ton to keep Ton constant.
5. In case of a constant frequency system, Ton = T. If the chopping frequency 2 kHz, find the value of Toff.
a) ms
b) ms
c) μs
d) μs
Answer: b
Explanation: Ton = T
T = 1/2kHz = 0.5 ms
Ton = T/4
Therefore, Toff = T – Ton = T – T/4 = 3T/4 = 3/8 msec.
6. In case of frequency modulation system, ________ is kept constant.
a) T
b) Ton
c) Toff
d) Either Ton or Toff
Answer: d
Explanation: In frequency modulation scheme, the frequency hence T is varied by either keeping Ton constant or keeping Toff constant.
7. The control strategy in which on and off time is guided by the pervious set of values of a certain parameter is called as
a) time ratio control
b) pulse width modulation
c) current limit control
d) constant frequency scheme
Answer: c
Explanation: In current ratio control, the on and off time is guided by the pervious set of value of load current.
8. In the current limit control method, the chopper is switched off when
a) load current reaches the lower limit
b) load current reaches the upper limit
c) load current falls to zero
d) none of the mentioned
Answer: b
Explanation: In CLC, when the current reaches the upper limit the chopper is switched off and current starts to decay. When current reaches a predefined lower limit, the chopper is again switched on and current starts to rise, the process is thus again repeated.
9. Which of the following mentioned control strategy/strategies would require a feedback loop?
a) pwm
b) constant frequency system
c) current limit control
d) none of the mentioned
Answer: c
Explanation: CLC would require a feedback loop as it is required to measure the values of current constantly before switching on or off the chopper.
10. In the current limit control method, when the load current reaches a predefined lower value, then
a) the chopper is switched off
b) the chopper is switched on
c) the source voltage is removed
d) load voltage goes to zero
Answer: b
Explanation: In CLC, when the current reaches the upper limit the chopper is switched off and current starts to decay. When current reaches a predefined lower limit, the chopper is again switched on and current starts to rise, the process is thus again repeated.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Type C Chopper”.
1. Name the below given circuit.
power-electronics-questions-answers-type-c-chopper-q1
a) Type D chopper
b) Type B chopper
c) Two-quadrant type A chopper
d) Two-quadrant type B chopper
Answer: c
Explanation: It is a type C chopper or Two quadrant type A chopper. It operates in the first and second quadrant.
2. In the below given circuit output voltage will be 0 when
power-electronics-questions-answers-type-c-chopper-q1
a) SW1 is on
b) FD is not conducting
c) D2 is conducting
d) SW2 is on
Answer: d
Explanation: Output voltage will be zero if SW 2 is on, because it will short circuit the load circuit and output voltage will be zero.
3. In the below given circuit
power-electronics-questions-answers-type-c-chopper-q1
a) current can never be zero
b) current can flow in both the directions of the load
c) voltage can never be zero
d) none of the mentioned
Answer: b
Explanation: In type C chopper, current can flow in both the directions. It would be negative if SW2 in on or D2 conducts.
4. In the below given type C chopper circuit, _______ and _________ together operate like type A chopper.
power-electronics-questions-answers-type-c-chopper-q1
a) CH1, FD
b) CH2, FD
c) CH1, D2
d) CH2, D2
Answer: a
Explanation: In the above shown type C chopper circuit, when CH1 is on and current flows VS-CH1-L-E-Vs. When CH1 is switched off, current freewheels through the FD. Hence, current and voltage both are positive giving type A chopper operation.
5. In the below given type C chopper circuit, ________ and ___________ together operate like type B chopper.
power-electronics-questions-answers-type-c-chopper-q1
a) CH1, FD
b) CH2, FD
c) CH1, D2
d) CH2, D2
Answer: d
Explanation: In the above shown type C chopper circuit, when CH2 is on current flows in the negative direction E-L-CH2-E and hence voltage is zero. When D2 operates E and L deliver power to the supply Vs hence, giving a type B chopper operation.
6. Which type of chopper is used in the regenerative braking of DC motors?
a) type A
b) type B
c) type C
d) type D
Answer: c
Explanation: In regenerative braking of DC motors, power needs to flow from load to source as well as source to load depending on whether the machine is in the motoring mode or generative braking.
7. Type C chopper consists of
a) two diodes and two switches
b) one diode and one switch
c) one diode and three switches
d) three diodes and two switches
Answer: a
Explanation: Type C consists of, two diodes and two switches. One diode switch pair is used for operation in a particular quadrant.
8. In a type C chopper, if only one switch is operated
a) only one quadrant operation will be obtained
b) two quadrant operation can be obtained
c) the chopper won’t work
d) none of the mentioned
Answer: a
Explanation: Only one quadrant operation will be operated if only one switch is used. It can be either 1st and 2nd quadrant depending on which switch is operated.
9. A type C chopper is operated with only positive load current. Choose the correct statement.
power-electronics-questions-answers-type-c-chopper-q1
a) CH1 would remain idle
b) D2 would remain idle
c) CH2 would remain idle
d) Both D2 and CH2 would remain idle
Answer: d
Explanation: It is a type C chopper or two quadrant type A chopper. It operates in the first and second quadrant. CH1 and FD in the first quadrant and CH2 and D2 in second quadrant.
10. A type C chopper is operated with only negative load current. Choose the correct statement.
power-electronics-questions-answers-type-c-chopper-q1
a) CH1 would remain idle
b) FD would remain idle
c) CH2 would remain idle
d) Both FD and CH1 would remain idle
Answer: d
Explanation: It is a type C chopper or two quadrant type A chopper. It operates in the first and second quadrant. CH1 and FD in the first quadrant and CH2 and D2 in second quadrant.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Type D Chopper”.
1. A type D chopper is a
a) two quadrant type-B chopper
b) two quadrant type-A chopper
c) two quadrant type-C chopper
d) none of the mentioned
Answer: a
Explanation: It is a two quadrant type B chopper as load current is always positive but the voltage can be positive or negative.
2. In a type-D chopper
a) current can flow in both the directions of the load
b) current cannot flow in both the directions of the load
c) voltage can only be positive
d) voltage can only be negative
Answer: b
Explanation: It is a two quadrant type B chopper. In quadrant I and IV current is only positive.
3. What is the load voltage value when both the chopper switches are on?
power-electronics-questions-answers-type-d-chopper-q3-q6-q8
a) 0
b) Vs
c) 2Vs
d) Vs/2
Answer: b
Explanation: When both the switches are on, load is directly in parallel with Vs .
4. For a type D chopper, the average value of output voltage will be positive when
a) T on = T off
b) T on < T off
c) T off = 0
d) T on > T off
Answer: d
Explanation: When the chopper on time is more than the chopper off time than the average voltage will be positive. When the chopper switches are off, both the diodes are operating and the voltage is negative.
5. For a type D chopper, if duty cycle = 0.5 then the
a) average voltage is positive
b) average voltage is negative
c) average voltage is zero
d) chopper cannot be operated with duty cycle = 0.5
Answer: c
Explanation: When α = 0.5. T on = T off hence the positive and negative value of voltages become equal. Making the average voltage = 0 V.
6. In the below given circuit, if only SW1 is operated then
power-electronics-questions-answers-type-d-chopper-q3-q6-q8
a) the power flows from load to source
b) the power flows from source to load
c) the average voltage is positive
d) none of the mentioned
Answer: d
Explanation: If only one switch is operated, circuit is not completed and the current does not flow.
7. For a type D chopper, if duty cycle α < 0.5 then the
a) average voltage is positive
b) average voltage is negative
c) average voltage is zero
d) none of the mentioned
Answer: b
Explanation: α = T on /T
If α < 0.5, T on < T off
The average value of voltage will be negative, as the switches will be off most of the times.
8. In the below given circuit, if Vo = -Vs then
power-electronics-questions-answers-type-d-chopper-q3-q6-q8
a) both the chopper switches are on
b) only SW2 and D1 is active
c) both the chopper switches are off and both the diodes D1 and D2 are active
d) both the chopper switches are on and both the diodes D1 and D2 are reversed biased
Answer: c
Explanation: When diodes are operating, power flows from load to source and the inductor in the load reverses its polarity.
9. Find the expression for average output voltage in a type D chopper on RL load.
a) Vs x Duty cycle
b) Vs x T
c) Vs x (T – T )/T
d) Vs 2 /2
Answer: c
Explanation:
During Ton = Vs x Ton/T
During Toff = – Vs x Toff/T
Vo = Vs /T.
10. Find the average output voltage for a type D chopper if it is given a dc supply of 220 V. Chopper frequency is 2 kHz and both the chopper switches are operated for 0.3 ms.
a) 220 V
b) 108 V
c) 98 V
d) 44 V
Answer: d
Explanation: T = 1/f = 0.5 ms
Ton = 0.3 ms
Toff = 0.2 ms
Vo = Vs /T = 220 x = 44 V.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Type E Chopper-I”.
1. Identify the given circuit
power-electronics-questions-answers-type-e-chopper-1-q1
a) type D chopper
b) two quadrant type C chopper
c) type E chopper
d) type F chopper
Answer: c
Explanation: It is a type E chopper. Type C consists of 4 diodes and 4 IGBT switches.
2. A type C chopper consists of __________ diodes and _________ switches in anti-parallel.
a) 2, 2
b) 3, 3
c) 4, 4
d) 3, 4
Answer: c
Explanation: Type C consists of 4 diodes and 4 switches arranged in anti-parallel. The switches and diodes are always numbered occurring to the quadrant in which they operate.
3. A type C chopper can operate in
a) Ist and IInd quadrants
b) IInd and IIIrd quadrants
c) Ist, IInd and IIIrd quadrants
d) all the four quadrants
Answer: d
Explanation: Type C can operate in all the four quadrants by controlling appropriate switches and reversing the emf in the load circuit.
4. For the first quadrant operation of type E chopper
power-electronics-questions-answers-type-e-chopper-1-q1
a) Only CH1 is on
b) CH1 and CH2 is on
c) CH1 and CH3 is on
d) CH1 and CH4 is on
Answer: d
Explanation: With CH1 and CH4 on, load voltage = Vs and load current flows. Both V and I are positive giving Ist quadrant operation.
5. For the second quadrant operation of type E chopper
power-electronics-questions-answers-type-e-chopper-1-q1
a) Only CH2 is on
b) CH1 and CH2 is on
c) CH2 and CH3 is on
d) CH2 and CH4 is on
Answer: d
Explanation: With CH2 on, negative current flows through L, CH2, CH3 and CH4. When CH2 is turned off, current is fed back to the diodes by D1 and D4. Load voltage is positive and current is negative, hence second current operation is operated.
6. For a type E chopper, when CH2 is first on and then switched off then the
power-electronics-questions-answers-type-e-chopper-1-q1
a) power flows from source to load
b) power flows from load to source
c) load is short circuited
d) none of the mentioned
Answer: b
Explanation: With CH2 on, negative current flows through L, CH2, CH3 and CH4. When CH2 is turned off, current is fed back to the diodes by D1 and D4. Load voltage is positive and current is negative, hence second current operation is operated.
7. What is the expression for load voltage when the chopper is operated in the second quadrant?
a) Vs
b) E
c) 0
d) E + Ldi/dt
Answer: d
Explanation: In the second quadrant, the chopper acts as a step up chopper and current flows from the load to source. With CH2 on, negative current flows through L, CH2, CH3 and CH4.
8. For a type E chopper operating in the first quadrant, find the expression for average output voltage.
a) Vs
b) E – Ldi/dt
c) 0
d) E + Ldi/dt
Answer: a
Explanation: With CH1 and CH4 on, load voltage = Vs and load current flows.
9. The load emf E must be reversed for
a) first and second quadrant operation
b) third quadrant operation
c) fourth quadrant operation
d) both third and fourth quadrant operation
Answer: d
Explanation: The load E must be reversed for both third and fourth quadrant operation for proper current flow.
10. For the third quadrant operation of type E chopper
power-electronics-questions-answers-type-e-chopper-1-q1
a) CH1 is on
b) CH2 and CH1 are on
c) CH3 is on
d) None of the mentioned
Answer: d
Explanation: For third quadrant operation of type E chopper, CH2 is kept on and CH3 is operated. First both CH2 and CH3 are on, and then CH3 is switched off thus the current freewheels through CH2 and D4.
This set of tough Power Electronics questions and answers focuses on “Type E Choppers-2”.
1. The below given circuit can
tough-power-electronics-questions-answers-q1
a) operate in all the four quadrants
b) operate in only the first and second quadrant
c) operate in only the fourth and third quadrant
d) operate in only the first and fourth quadrant
Answer: b
Explanation: For operation in the third and the fourth quadrant, the polarity of the load emf E should be reversed.
2. In the below shown chopper circuit, when CH3 is on
tough-power-electronics-questions-answers-q1
a) both load voltage and current are zero
b) both load voltage and current are positive
c) both load voltage and current are negative
d) none of the mentioned
Answer: c
Explanation: When CH3 is switched on, IIIrd quadrant operation is obtained in which both load current and load voltage are negative. D2-CH3-Vs.
3. In the below given type E chopper circuit, when CH3 and CH2 are switched on and then CH3 is switched off after some time, then
tough-power-electronics-questions-answers-q1
a) load current falls to zero
b) negative current freewheels through CH2, D4
c) positive current freewheels through CH2, D4
d) none of the mentioned
Answer: b
Explanation: When CH3 is first on then, load gets connected to Vs and charges the inductor L. When it CH3 switched off, current has to flows through through the same direction because of the L. Current starts to freewheel from CH2 and D4.
4. For the type E chopper to be operated in the fourth quadrant
a) only one switch is operated
b) two switches are operated
c) three switches are operated
d) all the switches are operated
Answer: a
Explanation: For IVth quadrant, only CH4 is operated.
The difference between “operated” and “on” should be noted. Operated means it can is literally operated i.e. it is switched on and off and on and off, whereas on means it is on continuously.
5. For the fourth quadrant operation, the current flows through
tough-power-electronics-questions-answers-q1
a) CH1, D2, L and D1
b) CH4, D3, L and D4
c) CH4, D2, L and E
d) CH1, D2, L and D3
Answer: c
Explanation: With CH4 on in the fourth quadrant, positive current flows through CH4, D2, L and E.
6. When CH4 is turned on and then off
tough-power-electronics-questions-answers-q1
a) load current falls to zero
b) negative current freewheels through D2, D3
c) positive current freewheels through D2, D3
d) none of the mentioned
Answer: c
Explanation: When CH4 is on, positive current flows through CH4, D2, L and E. When CH4 is opened, current is fed back to the source from D2, D3.
7. Find the expression for the ripple factor in terms of duty cycle α for a type A step down chopper
a) √
b) α 2 /2
c) √α/2
d) √
Answer: a
Explanation: RF = Vr/Vo
Vr = √α x Vs = √(αVs 2 – α 2 Vs 2 ) = Vs √(α- α 2 )
RF = Vs.√(α- α 2 )/Vs.α = √.
8. The AC ripple voltage is given by
a) Vrms/Vo
b) Vrms 2 /π
c) √(Vrms 2 – Vo 2 )
d) √
Answer: c
Explanation: The term AC ripple voltage is used for knowing the harmonic content of a waveform, without calculating its harmonic components. It is the rms difference of ac and dc voltage.
9. The expression for the thyristor current is given by
I = (α 2 .Vs – αE)/R
Find the value of firing angle α, for which the thyristor current is maximum.
a) 2Vs/π
b) 180°
c) E/2Vs
d) Vs/R
Answer: c
Explanation: In the given expression, to find the maximum value of α
dI/dα = 0
Therefore, dI/dα = /R = 0
α = E/2Vs.
10. In a type E chopper, if all the four chopper switches are closed simultaneously then
a) load is short circuited
b) supply is short circuited
c) both load and supply are shorted
d) none of the mentioned.
Answer: b
Expression: Before anything, the supply will be shorted by CH1 and CH2 first.
11. A type A step down chopper has Vs = 220 V and is connected to RLE load. With R = 1 Ω, E = 24 V and L large enough to maintain continuous conduction. Find the average value of load current for a duty cycle of 30 %.
a) 100 A
b) 22 A
c) 42 A
d) 16.5 A
Answer: c
Explanation: I = /R = 42 A.
12. A step down chopper has Vs = 220 V and is connected to RLE load. With R = 1 Ω, E = 24 V and L = 5 mH. The chopping period is 2000 μs and the on-period is 600 μs. Find the value of minimum steady state output current.
a) 33 A
b) 42 A
c) 0 A
d) 51 A
Answer: d
Explanation:
I = x [ (1 – e -m )/(1 – e -n ) ] –
m = Ton / = R.Ton/L = 0.12
n = T/ = R.T/L
T = 2000μs
n = 0.4
I = x [ (1 – e -0.12 )/(1 – e -0.4 ) ] – = 51.46 A.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Thyristor Chopper Circuits – 1”.
1. The process of commutating a SCR by applying a reverse voltage to an SCR through a previously charged capacitor is called as
a) capacitor commutation
b) forced commutation
c) voltage commutation
d) current commutation
Answer: c
Explanation: It is called as voltage commutation, which is a type of forced commutation.
2. In case of current commutation of SCR
a) a diode is connected in series with the main SCR
b) a diode is connected in parallel with the main SCR
c) a diode is connected in anti-parallel with the main SCR
d) none of the mentioned
Answer: c
Explanation: The voltage drop across the anti parallel connected SCR reverse biases the main SCR.
3. Below given circuit is a
power-electronics-questions-answers-thyristor-chopper-circuits-1-q3
a) current commutated chopper
b) voltage commutated chopper
c) load commutated chopper
d) none of the mentioned
Answer: b
Explanation: It is a voltage commutated chopper or impulse-commutated chopper. If the diode was directly in parallel with the main SCR it would be a current commutated chopper circuit.
4. In the below shown voltage commutated chopper circuit, the chopper operation can only start when
power-electronics-questions-answers-thyristor-chopper-circuits-1-q3
a) t1 is gated
b) the capacitor is charged
c) the diode is forward biased
d) the capacitor is dis-charged
Answer: b
Explanation: The chopper can only start when the C in the commutating circuit is charged first.
5. In a voltage commutated chopper circuit, the capacitor C can be charged to Vs by
power-electronics-questions-answers-thyristor-chopper-circuits-1-q3
a) gating T1
b) gating TA
c) closing the switch
d) both gating TA and closing the switch
Answer: d
Explanation: When switch is closed, the C gets charged through Vs, C, Switch. When TA is triggered C gets charged through Vs, C, TA and the load.
6. In the below given circuit, when the main thyristor T1 is triggered after charging the capacitor, the commutation current flows through the path
power-electronics-questions-answers-thyristor-chopper-circuits-1-q3
a) Vs – T1 – Load – Vs
b) S – C – T1 – Load – S
c) C – T1 – L – D – C
d) C – T1 – TA – C
Answer: c
Explanation: The commutation current is nothing but the capacitor current.
7. In the below given circuit, when the main SCR T1 is commutated at t 2 , the load voltage is
power-electronics-questions-answers-thyristor-chopper-circuits-1-q3
a) Vs
b) 0
c) 2Vs
d) Vs + Ldi/dt
Answer: c
Explanation: The main SCR T1 is commutated by the triggering of the TA which then makes the capacitor revere bias the main SCR T1. The load current then flows through Vs-C-TA-Load. Hence, Vo = Vs + .
As the capacitor is already charged to a voltage Vs. Vo = 2Vs.
8. In a voltage commutated SCR, the main SCR is commutated by
a) triggering the auxiliary SCR
b) sending a negative pulse to the main SCR
c) removing the supply Vs
d) none of the mentioned
Answer: a
Explanation: The main SCR T1 is commutated by the triggering of the TA which then makes the capacitor revere bias the main SCR T1.
9. Which type of commutation circuit does not work on no load?
a) Voltage commutation
b) Current commutation
c) Both voltage and current commutation
d) None of the mentioned
Answer: a
Explanation: This is because the capacitor won’t be able to charge on no load.
10. What is the expression for the value of the capacitor in the voltage commutation circuit? Let tc be the circuit turn off time and Io be the constant load current.
a) Io x
b) Vs/Io
c) /Vs
d) none of the mentioned
Answer: c
Explanation: Capacitor current = C dv/dt
For constant load current Io, the above expression can be
Io = C x or C = /Vs.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Thyristor Chopper Circuits-2”.
1. For the below given voltage commutation circuit, find the peak value of voltage across the freewheeling diode FD.
power-electronics-questions-answers-thyristor-chopper-circuits-2-q1
a) Vs
b) 0
c) 2Vs
d) Vs + Ldi/dt
Answer: c
Explanation: The load current then flows through Vs-C-TA-Load. Hence, Vo = Vs + . As the capacitor is already charged to a voltage Vs. Vo = Voltage across FD = 2Vs.
2. A voltage commutated chopper feeds power to an electric car . The battery voltage is 60 V. The SCR turn-off time is 20 μs. Calculate the value of capacitor required for the commutating circuit.
a) 400 μF
b) 40 nF
c) 40 μF
d) 0.4 mF
Answer: c
Explanation: Capacitor current = C dv/dt
For constant load current Io, the above expression can be
Io = C x or C = /Vs
C = ( 40 x 60 x 10 -6 )/60 = 40 x 10 -6 Farads.
3. Identify the below given circuit.
power-electronics-questions-answers-thyristor-chopper-circuits-2-q1
a) current commutated chopper
b) voltage commutated chopper
c) load commutated chopper
d) load commutated inverter
Answer: a
Explanation: Diode in anti-parallel with the main SCR. Hence .
4. A voltage commutated chopper feeds power to an electric car . The battery voltage is 60 V. Starting current is 60 A and the SCR turn-off time is 20 μs. Calculate the value of inductance required for the commutating circuit.
a) 400 μH
b) 40 nH
c) 40 μH
d) 0.4 mH
Answer: c
Explanation: L = 2 x C
C = /Vs
C = ( 40 x 60 x 10 -6 )/60 = 40 x 10 -6 F
L = 2 x 40 x 10 -6 = 40 uH.
5. In the below given circuit, at the instant T1 the main SCR is triggered, the voltage across the load terminal is
power-electronics-questions-answers-thyristor-chopper-circuits-2-q1
a) Vs
b) 0
c) 2Vs
d) Vs + Ldi/dt
Answer: a
Explanation: When the main SCR is gated, the commutating circuit remains inactive. Hence the load is directly connected to Vs through the SCR. Considering SCR as an ideal switch Vo = Vs.
6. In current commutated copper circuit
a) the capacitor cannot charge on no load
b) the auxiliary thyristor is turned on automatically
c) the auxiliary thyristor is naturally commutated
d) the commutation is not reliable
Answer: c
Explanation: It is naturally commutated as its commutating current passes through zero value in the ringing circuit formed by L and C.
7. A current commutated chopper is fed from a source of 230 V. Its commutating components are L = 20 μH and C = 50 μF. Find the value of peak commutating current.
a) 233 A
b) 59 A
c) 363 A
d) 198 A
Answer: c
Explanation: I = V √ = 230 x √ = 363.66 A.
8. A voltage commutated chopper has the following parameters:
Vs = 220 V. Commutation circuit: L = 20 μH and C = 50 μF.
Find the effective on period for a constant load current of 80 A and Ton = 800 μsecs.
a) 1.075 μs
b) 10.75 ms
c) 1075 μs
d) 1.075 sec
Answer: c
Explanation: The effective on period Ton ’ is given by
Ton + 2Vs.C/Io.
9. A voltage commutated chopper has the following parameters:
Vs = 220 V. Commutation circuit: L = 20 μH and C = 50 μF.
Find the turn-off time for the main SCR with a constant load current of 80 A and Ton = 800 μsecs.
a) 137.5 μs
b) 1075 μs
c) 49.67 μs
d) 200 μs
Answer: a
Explanation: C = /Vs
tc = C.Vs/I = 137.5 μsecs.
10. The load commutated chopper circuit consists of
a) two thyristors and one commutating capacitor
b) four thyristors and one commutating capacitor
c) two thyristors and two commutating capacitors
d) four thyristors and two commutating capacitors
Answer: b
Explanation: It consists of 4 SCRs and one capacitor. 4 SCRs forming 2 pairs of 2 SCRs conduct alternately.
This set of tough Power Electronics Multiple Choice Questions focuses on “Thyristor Chopper Circuits-3”.
1. Find the expression for peak capacitor voltage in case of a current commutated chopper circuit.
a) Io √
b) Vs + √
c) Vs + Io √
d) Zero
Answer: c
Explanation: 1/2 CVc 2 = 1/2 LIo 2
Vc = Io √
Vc = Vs + Io √.
2. Identify the below given chopper circuit.
tough-power-electronics-questions-q2
a) Type D chopper
b) Type C chopper with commutating capacitor
c) Load commutated step-down chopper
d) Voltage commutated step-up chopper
Answer: c
Explanation: Load commutated chopper uses 4 SCRs and one commuting capacitor. As there is no L in the circuit, it is a step down chopper.
3. In the below given commutated chopper circuit, __ and __ SCRs act together in pairs.
tough-power-electronics-questions-q2
a) T1-T2, T3-T4
b) T1-T3, T2-T4
c) T1-T4, T2-T4
d) T1, T2
Answer: a
Explanation: T1-T2 and T3-T4 act together as one pair for conducting the load current alternately.
4. When T1 and T2 are conducting, which components act as the commutating components?
tough-power-electronics-questions-q2
a) T3 and T4
b) All the SCRs
c) T3, T4 and C
d) T3, T4, C and FD
Answer: c
Explanation: Commutating components are the one which will help the main SCRs T1 and T2 in this case turn off safely. FD is not a commutating component, it only freewheels the current during inductive loads.
5. What is the voltage across the load when T1 and T2 are gated after charging the capacitor to Vs?
tough-power-electronics-questions-q2
a) Vs
b) 0
c) 2Vs
d) Vs + Ldi/dt
Answer: c
Explanation: The main SCRs T1 and T2 are conducting, and the path is Vs –T1 – C – T2 – Load. Hence, Vo = Vs + . As the capacitor is already charged to a voltage Vs. Vo = 2Vs.
6. What is the expression for design of commutation capacitance in case of a load commutated chopper circuit?
a) /Vs
b) Cannot be determined
c) /2Vs
d) /Vs
Answer: c
Explanation: Total change in the voltage is 2xVs in time T on .
I = C x (2Vs/ T on )
C = /2Vs.
7. Which type of commutating circuit will have highest switching losses for same rating of all the components?
a) current commutated chopper
b) voltage commutated chopper
c) load commutated chopper
d) forced commutated chopper
Answer: c
Explanation: Load commutated choppers have 4 SCRs which require frequent turning on and turning off which increases the switching losses.
8. The speed of a separately excited dc motor is controlled below base speed by using a type-A SCR based chopper. The supply voltage is 220 V dc and the armature resistance is 0.5 Ω. Calculate the minimum value of firing angle if the motor drives a constant torque load requiring an average armature current of 30 A.
a) 1/2
b) 3/44
c) 59/46
d) 7/12
Answer: b
Explanation: For a motor, Vo = E + Ir
The minimum possible speed = 0. This gives the motor counter emf E = 0.
Therefore, Vo = αVs = Ir
α x 220 = 30 x 0.5
α = 15/220 = 3/44.
9. For the step down chopper shown in the figure below, Vs = 100 V and the duty cycle of the switch is 0.8. The load is sufficiently inductive so that the load current is ripple free. The average current through the diode under steady state is
tough-power-electronics-questions-q9
a) 2 A
b) 1.6 A
c) 8 A
d) 6.4 A
Answer: b
Explanation:
The diode is active only during Toff.
If Ton = 0.8 then Toff = 0.2
Output voltage Vo = 0.8 x 100 = 80 V
When D is active during Toff, consider inductor as a short circuit thus, Id = /R = 80 x 0.2/10 = 1.6 A.
10. A chopper circuit is fed from an input voltage of 20 V delivers a power of 16 watts. If the chopper efficiency is 80 %, then find the value of input current.
a) 0.64 A
b) 1 A
c) 0.8 A
d) 1.25 A
Answer: b
Explanation: 0.8 = ouput/input
Input power = 16/0.8
16/0.8 = V x I
V = 20 V. Hence, I = 1 A.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Single Phase VSI-1”
1. Inverters converts
a) dc power to dc power
b) dc power to ac power
c) ac power to ac power
d) ac power to dc power
Answer: b
Explanation: Inverter is a dc to ac converter.
2. Line-commutated inverters have
a) AC on the supply side and DC on the load side
b) AC on both supply and load side
c) DC on both supply and load side
d) DC on the supply side and AC on the load side
Answer: b
Explanation: Line-commuted inverters are actually phase-controlled converters operated in the inverting mode. They cannot act as standalone inverters as they require a AC supply. It is to be noted that “line commutated inverter” is not the conventional inverter. The conventional inverters are forced or load commutated inverters.
3. In a VSI
a) the internal impedance of the DC source is negligible
b) the internal impedance of the DC source is very very high
c) the internal impedance of the AC source is negligible
d) the IGBTs are fired at 0 degrees.
Answer: a
Explanation: A VSI is the one in which the internal impedance of the source is negligible. It has a stiff DC source at its input.
4. VSIs using GTOs are turned off by
a) load commutation
b) line commutation
c) applying a negative gate pulse
d) removing the base signal
Answer: c
Explanation: GTOs are gate turn off SCRs in which turn-off is achieved by applying a negative gate pulse.
5. VSIs using IGBTs are turned off by
a) load commutation
b) line commutation
c) applying a negative gate pulse
d) removing the base signal
Answer: d
Explanation: IGBT is a transistor family device. It can be turned off simply by removing the gate signal. All the transistor devices operated in the same way in inverters.
6. __________ based inverters do not require self-commutation.
a) IGBT
b) GTO
c) PMOSFET
d) SCR
Answer: d
Explanation: All the devices can be turned off by their gate/base singles expect SCR. SCRs require external commutation circuits.
7. Identify the circuit given below.
power-electronics-questions-answers-single-phase-vsi-1-q7-q9
a) Half wave series inverter
b) Full wave series inverter
c) Half wave bridge inverter
d) Half wave parallel inverter
Answer: c
Explanation: It is a half-wave circuit as it has only 2 SCRs are connected. It is a half wave bridge type thyristorised inverter.
8. Single phase half bridge inverters requires
a) two wire ac supply
b) two wire dc supply
c) three wire ac supply
d) three wire dc supply
Answer: d
Explanation: They require two voltage sources Vs/2 and Vs/2.
9. What is the voltage across the R load when only T1 is conducting?
power-electronics-questions-answers-single-phase-vsi-1-q7-q9
a) Vs
b) Vs/2
c) 2Vs
d) zero
Answer: b
Explanation: Considering T1 as an ideal switch, the load is directly connected to the upper voltage source when T1 is on.
10. In a single-phase half wave inverter ________ SCR are/is gated at a time.
a) one
b) two
c) three
d) none of the mentioned
Answer: a
Explanation: Only one SCR is gated at a time, gating the both of them will short the supply.
Sanfoundry Global Education & Learning Series – Power Electronics.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Single Phase VSI-2”.
1. What is the voltage across the R load when only T2 is conducting?
power-electronics-questions-answers-single-phase-vsi-2-q1-q5-q6
a) Vs
b) Vs/2
c) 2Vs
d) Zero
Answer: b
Explanation: Considering T2 as an ideal switch, the load is directly connected to the lower voltage source when T2 is on.
2. The voltage in a single phase half wave inverter varies between
a) Vs and 0
b) Vs/2 and 0
c) Vs/2 and –Vs/2
d) Vs and –Vs
Answer: c
Explanation: It varies between –Vs/2 and Vs/2 as the upper and lower thyristors/IGBTs are fired in sequence.
3. Below given is a
power-electronics-questions-answers-single-phase-vsi-2-q3
a) SCR based inverter
b) MOSFET based inverter
c) IGBT based inverter
d) None of the mentioned
Answer: b
Explanation: It is a half wave inverter with MOSFET switches. The diodes are used to bypass the negative current.
4. The output of a single-phase half bridge inverter on R load is ideally
a) a sine wave
b) a square wave
c) a triangular wave
d) constant dc
Answer: b
Explanation: Due to rapid switching on and off of the devices, it seems to be a square wave. But practically it is a B-wave.
5. In the below given circuit, T1 is fired at 0,T and so on and T2 at T/2, 3T/2 etc. What is the frequency of the alternating voltage obtained?
power-electronics-questions-answers-single-phase-vsi-2-q1-q5-q6
a) 50 Hz
b) T Hz
c) 1/T Hz
d) T/2 Hz
Answer: c
Explanation: T1 conducts from 0 to T/2 and T2 from T/2 to T. Hence, 1 cycle is completed in T time. f = 1/T Hz.
6. In the below circuit, if Vs/2 = 50 V. Find, the rms AC voltage that would be ideally obtained.
power-electronics-questions-answers-single-phase-vsi-2-q1-q5-q6
a) 50 V
b) 100 V
c) 0.707 x 50 V
d) 0.707 x 100 V
Answer: c
Explanation: Peak value would be = 50 V
RMS = Peak/√2 = Peak x 0.707 V.
7. Find the conduction time of the diodes if the SCRs are fired at 0 and T/2 respectively in a single phase half wave inverter with R load.
a) 0
b) T/2
c) 2/T
d) insufficient data
Answer: a
Explanation: With R load, the diodes do not come into play.
8. The output current wave of a single-phase full bridge inverter on RL load is
a) a sine wave
b) a square wave
c) a triangular wave
d) constant dc
Answer: c
Explanation: On RL load, the SCRs are revised biased due to the voltage drops across the diodes and negative current flows.
9. Single-phase full bridge inverters requires
a) 4 SCRs and 2 diodes
b) 4 SCRs and 4 diodes
c) 2 SCRs and 4 diodes
d) 2 SCRs and 2 diodes
Answer: b
Explanation: Full bridge inverters require 4 SCR and 4 diodes along with a two wire dc source.
10. Identify the circuit given below.
power-electronics-questions-answers-single-phase-vsi-2-q10
a) Full wave series inverter
b) Half wave series inverter
c) Full wave bridge inverter
d) Full wave parallel inverter
Answer: c
Explanation: It is a full wave bridge inverter using 4 diodes and 4 SCRs.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Single Phase VSI-3”.
1. The output voltage from a single phase full wave bridge inverter varies from
a) Vs to –Vs
b) Vs to zero
c) Vs/2 to zero
d) –Vs/2 to Vs/2
Answer: a
Explanation: The output from a full wave bridge inverter varies from Vs to –Vs.
2. In a single phase full wave bridge inverter, when the output is Vs or –Vs
a) one SCR and one diode are conducting
b) four SCRs are conducting
c) two SCRs are conducting
d) two diodes are conducting
Answer: c
Explanation: When two SCRs are conducting, T1-T2 or T3-T4, the output voltage magnitude is Vs. When the diodes are conducting, freewheeling action is taking place and voltage is zero.
3. In the following circuit, the diodes D1 to D4 are used to
power-electronics-questions-answers-single-phase-vsi-3-q3
a) reduce the switching losses
b) send current back to the dc source when SCRs are off
c) send current back to the load when SCRs are off
d) send current back to the dc source when SCRs are conducting
Answer: b
Explanation: Diodes are connected in anti-parallel with the SCRs during inductive loads for freewheeling actions.
4. For a full wave bridge inverter, the output voltage
a) Vo = Vs/2 for 0 < t < T/2
b) Vo = Vs for 0 < t <T/2
c) Vo = Vs for T/2< t < T
d) Vo = -Vs for T/2< t < 3T/2
Answer: b
Explanation: For the first half cycle, Vo = Vs and for the second half cycle Vo = -Vs.
5. For a half wave bridge inverter, the output voltage
a) Vo = – Vs/2 for 0 < t < T/2
b) Vo = – Vs/2 for T/2< t<T
c) Vo = – Vs for 0 < t < T/2
d) Vo = Vs/2 for T/2< t < T
Answer: b
Explanation: In case of a half wave inverter, for the first half cycle, Vo = Vs/2 and for the second half cycle Vo = -Vs/2.
6. The fundamental component of output voltage for a half wave bridge inverter is given by
a) sinωt
b) sinωt
c) sinωt
d)
Answer: b
Explanation: The fourier analysis of half wave inverter gives,
power-electronics-questions-answers-single-phase-vsi-3-q6
Put n = 1 for the fundamental component.
7. The fundamental component of output voltage for a full wave bridge inverter is given by
a) sinωt
b) sinωt
c) sinωt
d)
Answer: b
Explanation: The fourier analysis of full wave inverter gives,
Vo =
power-electronics-questions-answers-single-phase-vsi-3-q7
Put n = 1 for the fundamental component.
8. A single phase half bridge inverter has a dc voltage source Vs/2 = 115 V. Find the rms value of the fundamental component of output voltage.
a) 510 V
b) 103.5 V
c) 120 V
d) 96 V
Answer: b
Explanation: The fundamental component of voltage = sinωt.
Peak value Vm = 2Vs/π
Rms value = 2Vs/π√2 = 103.552 V.
9. A single phase half bridge inverter has load R = 2 Ω and a dc voltage source Vs/2 = 115 V. Find the rms value of the fundamental load current.
a) 10.25 A
b) 51.7 A
c) 86 A
d) 24.8 A
Answer: b
Explanation: Peak value Vm = 2Vs/π
Rms value = 2Vs/π√2 = 103.552 V
Io = Vo/R = 103.552/2 = 51.776 A.
10. A single phase half bridge inverter has load R = 2 Ω and a dc voltage source Vs/2 = 115 V. Find the power delivered to the load due to the fundamental component.
a) 536 kW
b) 53.61 kW
c) 5.361 kW
d) 536 W
Answer: c
Explanation: Peak value Vm = 2Vs/π
Rms value = 2Vs/π√2 = 103.552 V
Io = Vo/R = 103.552/2 = 51.776 A
P = Io 2 x R = 5361.5 Watts.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Single Phase VSI-4”.
1. A single phase full bridge inverter has a dc voltage source Vs = 230 V. Find the rms value of the fundamental component of output voltage.
a) 90 V
b) 207 V
c) 350 V
d) 196 V
Answer: b
Explanation: The fundamental component of voltage = sinωt.
Peak value Vm = 4Vs/π
Rms value = 4Vs/π√2.
2. A single phase full bridge inverter has load R = 2 Ω, and dc voltage source Vs = 230 V. Find the rms value of the fundamental load current.
a) 96 A
b) 0 A
c) 103 A
d) none of the mentioned
Answer: c
Explanation: The fundamental component of voltage = sinωt.
Peak value Vm = 4Vs/π
Rms voltage = 4Vs/π√2 = 207 V
Current = 207/2 = 103.5 A.
3. A certain full bridge type inverter circuit has its rms value of fundamental load current component given by W. The fundamental frequency component of the load current would be given by
a) W sin ωt
b) sin ωt
c) √2 W sin ωt
d) sin ωt
Answer: c
Explanation: The fundamental frequency component of the load current is given by Im sin ωt
As W = Irms . . .
Im = √2 Irms = √2 W.
4. In a half wave bridge inverter circuit, the power delivered to the load by each source is given by
a) Vs x Is
b) /2
c) 2
d) None of the mentioned
Answer: b
Explanation: Power delivered by each source each is x Is.
5. In a half wave circuit, forced commutation is essential when the
a) load is inductive
b) load is resistive
c) source voltage is below 150 V
d) none of the mentioned
Answer: b
Explanation: When the load is resistive , the diodes do not conduct, hence they cannot help stop the conduction of the SCRs. Hence, forced commutation in such cases becomes essential.
6. A single phase full bridge inverter circuit, has load R = 2 Ω and dc source Vs = 230 V. Find the value of power delivered to the load in watts only due to the fundamental component of the load current.
a) 5361.5 W
b) 2142.5 W
c) 21424.5 W
d) 214.2 W
Answer: c
Explanation: The fundamental component of voltage = sinωt
Peak value Vm = 4Vs/π
Rms voltage = 4Vs/π√2 = 207 V
RMS Current = 207/2 = 103.5 A
P = 2 x R = 21424.5 W.
7. A single phase full bridge inverter is fed from a dc source such that the fundamental component of output voltage = 230 V. Find the rms value of SCR and diode current respectively, for a R load of 2 Ω.
a) 115 A, 80 A
b) 81.33 A, 36.2 A
c) 36.2 A, 0 A
d) 81.33 A, 0 A
Answer: d
Explanation: Fundamental component of load current = V/R = 230/2 = 115 A.
SCR current = 115/2 = 81.33 A
Diode current = 0 as the diodes do not come into picture for R loads.
8. For a full bridge inverter with the following load: R = 2 Ω, X L = 8 Ω and X C = 6 Ω.
a) The output voltage lags the current by 45°
b) The output current lags the voltage by 45°
c) The output current lags the voltage by 90°
d) The output current lags the voltage by more than 90°
Answer: b
Explanation: As the inductive effect is more than the capacitive effect, of course the current will lag the voltage by an angle P.
P = tan -1 (X L – X C )/R = tan -1 = 45°.
9. A single phase full bridge inverter has RLC load. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the expression for the load voltage up to the fifth harmonic.
a) 292 sin 314t + 97.62 sin 314t + 58.57 sin 318t + 28.31 sin 318t + 3.686 sin 318t
b) 292 sin 314t + 97.62 sin + 58.57 sin
c) 292 sin 314t + 97.62 sin + 58.57 sin + 28.31 sin + 3.686 sin
d) 292 sin 512t + 25.62 sin 249t + 6.74 sin 508t
Answer: b
Explanation: In a single phase full bridge inverter only odd harmonics are present. i.e. 1,3,5 etc.
Vo = sin ωt + sin 3ωt + sin 5ωt
= 292 V
ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.
10. A single phase full bridge inverter has RLC load with R = 4 Ω, L = 35 mH and C = 155 μF. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the rms value of the fundamental load current.
a) 28.31 A
b) 20.02 A
c) 16.69 A
d) 26.90 A
Answer: b
Explanation: Average value of fundamental output voltage = 4Vs/π = 292.85 V
X L = 2 x 3.14 x 50 x 0.035 = 10.99 Ω
X C = 1/(2 x 3.14 x 50 x 155 x 10 -6 ) = 20.54 Ω
Z = 10.345 Ω
I = V/Z = 28.31
I = V/Z√2 = / = 20.02 A.
This set of Power Electronics assessment questions focuses on “Single Phase VSI-5”.
1. A single phase full bridge inverter has RLC load with R = 4 Ω, Xl = 11 Ω and Xc = 20.54 Ω. The dc input voltage is 230 V. Find the value of fundamental load power.
a) 1633 W
b) 1603 W
c) 1576 W
d) none of the mentioned
Answer: b
Explanation: Average value of fundamental output voltage = 4Vs/π = 292.85 V
Z = [R 2 + 2 ] 1/2
I = V/Z = 28.31
I = V/Z√2 = / = 20.02 A
Fundamental Load power = 2 x R = 2 x 4 = 1603.2 Watts.
2. A single phase full bridge inverter has RLC load with R = 4 Ω, L = 35 mH and C = 155 μF. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the angle by which the third harmonic current will lead/lag the third harmonic output voltage.
a) 67.25°
b) 96.4°
c) 49.87°
d) 81.3°
Answer: d
Explanation:
X L = 2 x 3.14 x 50 x 0.035 = 10.99 Ω
X C = 1/(2 x 3.14 x 50 x 155 x 10 -6 ) = 20.54 Ω
For the third harmonic component
X L = 10.99 x 3 = 33 Ω
X C = 20.54/3 = 6.846 Ω
R = 4 Ω
P = tan -1 (X L – X C )/R = 81.3°.
3. A single phase half bridge inverter has RLC load. The dc input voltage = 115 V and the output frequency is 50 Hz. The expression for the load voltage up to the fifth harmonic will be given by
a) 146 sin 314t + 48.81 sin 314t + 58.57 sin 318t + 28.31 sin 318t + 3.686 sin 318t
b) 146 sin 314t + 48.81 sin + 29.28 sin
c) 146 sin 314t + 48.81 sin + 58.57 sin
d) none of the mentioned
Answer: b
Explanation: In a single phase HALF bridge inverter only odd harmonics are present. i.e. 1,3,5 etc.
Vo = sin ωt + sin 3ωt + sin 5ωt . . .
= 146 V
ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.
4. Let Vs be the amplitude of the output voltage and P be the output power for a single-phase half bridge inverter. Then the corresponding values for a full bridge inverter would be
a) 2Vs, 4P
b) 2Vs, 2P
c) Vs, P
d) 2Vs, P
Answer: a
Explanation: The voltage is doubled in full and in half bridge configuration. As the power is proportional to the square of the current which is proportional to the voltage it is 4 times that obtained by a half ridge configuration.
5. In VSI
a) both voltage and current depend on the load impedance
b) only voltage depends on the load impedance
c) only current depends on the load impedance
d) none of the mentioned
Answer: c
Explanation: In VSIs the voltage is independent on the load impedance Z.
6. The harmonic factor of nth harmonic is given by
a) V n
b) V 1 /V n
c) V n /V 1
d) None of the mentioned
Answer: c
Explanation: It is the ratio of rms value of the nth harmonic voltage component to the rms value of the fundamental voltage component (V 1 ).
7. ____________ is the measure of the contribution of any individual harmonic to the inverter output voltage.
a) THD
b) Distortion Factor
c) Harmonic Factor
d) TUF
Answer: c
Explanation: The HF or Harmonic factor is the ratio of the nth harmonic voltage component to the fundamental voltage component. Hence it shows how much a particular harmonic is contributing in the total output of the circuit.
8. The Total Harmonic Distortion is the ratio of
a) rms value of all the harmonic components to the rms value of the fundamental component
b) average value of all the harmonic components to the rms value of the fundamental component
c) rms value of all the third harmonic component to the rms value of the fundamental component
d) rms value of all the fundamental component to the rms value of all the harmonic components
Answer: a
Explanation: THD = Vo/V 1
V 1 = Fundamental component
Vo = value of all the harmonic components expect the fundamental.
9. The HF is the ratio of
a) an average and a rms value
b) a rms and an average value
c) two volt-ampere values
d) two rms values
Answer: d
Explanation: HF is the ratio of rms value of the nth harmonic voltage component to the rms value of the fundamental voltage component.
10. The total harmonic distortion is the measure of
a) input vs output power factor
b) temperature sensitivity
c) waveform distortion
d) contribution of each harmonic to the total output
Answer: c
Explanation: Lower the value of THD, closer is the waveform to a sine-wave.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Quality of Inverters”.
1. If V r is the rms value of the inverter output voltage and V 1 is the rms value of the fundamental component, then the total harmonic distortion is given by
a) V r /V 1
b) (V r + V 1 r
c) [ (V r /V 1 ) 2 – 1 ] 1/2
d) [ (V r /V 1 ) 1/2 + 1 ] 2
Answer: c
Explanation: THD is the ratio of rms value of all the harmonic components to the rms value of the fundamental component.
Rms value of all the harmonic components V H = (V r 2 – V 1 2 ) 1/2
THD = (V r 2 – V 1 2 ) 1/2 / V 1 .
2. The distortion factor is the ratio of
a) total rms output voltage to fundamental rms output voltage
b) fundamental rms output voltage to fundamental average output voltage
c) total rms output voltage to rms value of all the harmonic components
d) fundamental rms output voltage to total rms output voltage
Answer: d
Explanation: μ = V 1 /V r .
3. A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the fundamental power delivered to the load.
a) 1.07 W
b) 107.2 W
c) 1.07 kW
d) 107.2 kW
Answer: c
Explanation: The fundamental power is the power delivered to the load due to the fundamental components of voltage and current.
V 1 = 2Vs/√2π = 103.552 V
I 1 = V 1 /R = 10.3552 A
P = (I 1 ) 2 x R = 1072.3 W = 1.0723 kW.
4. A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the average current through each SCR inverter switch.
a) 11.5 A
b) 5.75 A
c) 23 A
d) none of the mentioned
Answer: b
Explanation: Peak current through each SCR = Vs/2R = 11.5 A
As each SCR would conduct for 180° of the total 360° cycle, average current = peak current/2 = 11.5/2 = 5.75 A.
5. Find the distortion factor , for a single phase half wave bridge inverter with dc source Vs = 1 kV.
a) 0.87
b) 1
c) 0.9
d) 0.7
Answer: c
Explanation: μ = V 1 /V r
V 1 = rms value of fundamental component = 2Vs/π√2
V r = Total rms output voltage = Vs/2
μ = x = 2√2/π = 0.9.
6. A single phase inverter gives rms value of output voltage as 115 V and the fundamental output voltage of as 103.5 V. Find the THD .
a) 0.4 %
b) 40.8 %
c) 48.3 %
d) 4.83 %
Answer: c
Explanation:
V r = 115 V and V 1 = 103.5 V
THD =[ (V r 2 – V 1 2 ) 1/2 / V 1 ] x 100 %.
7. What would be the harmonic factor of lowest order harmonic in case of a half wave bridge inverter?
a) 1/1
b) 1/3
c) 1/2
d) Insufficient data
Answer: b
Explanation: Let Vs be the input voltage.
The 3rd harmonic is the lowest order harmonic.
Vo = 2Vs/
For fundamental component n = 1 and for the lowest order harmonic n = 3
ρ 3 = V 3 /V 1 = 1/3 or 33.33 %.
8. A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the rms output voltage and the peak reverse blocking voltage of each transistor.
a) 30 V, 60 V
b) 30 V, 30 V
c) 60 V, 60 V
d) 60 V, 30 V
Answer: c
Explanation: Rms output voltage = PIV of each transistor = Vs. Vs = 60 V.
9. A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the fundamental frequency output power.
a) 1200 W
b) 856 W
c) 972 W
d) 760 W
Answer: c
Explanation: V 1 = 4Vs/π√2 = 54.02 V
P = (V 1 ) 2 /R = 972.72 W.
10. Determine the distortion factor for a full bridge inverter with supply Vs = 60 V.
a) 0.8
b) 0.7
c) 0.9
d) 1
Answer: c
Explanation: V 1 = 4Vs/π√2 = 54.02 V
rms value of output voltage V r = Vs = 60 V.
μ = V 1 /V r = 54.02/60 = 0.9.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Force-Commutated Inverters”.
1. In voltage fed thyristor inverters __________ commutation is required.
a) load
b) forced
c) self
d) any commutation technique can be used
Answer: b
Explanation: In VSI, the input voltage source Vs keeps the SCRs always forward biased. Hence, forced commutation becomes essential.
2. The McMurray circuit is a
a) commutation circuit
b) force commutated VSI
c) self commutated VSI
d) none of the mentioned
Answer: b
Explanation: McMurray is a popularyly used forced commutated VSI circuit.
3. Forced commutation requires
a) a precharged inductor
b) a precharged capacitor
c) an overdamped RLC load
d) a very high frequency ac source
Answer: b
Explanation: Forced commutation requires a pre charged capacitor of correct polarity to turn-off the earlier conducting SCR.
4. Which elements are not present in the original McMurray inverter but are present in the modified McMurray inverters?
a) Two auxiliary thyristors and the di/dt inductor
b) Two auxiliary diodes and the damping resistor
c) One auxiliary SCR and one auxiliary diode
d) None of the mentioned
Answer: b
Explanation: In the original inverter circuit given by McMurray the elements DA1, DA2 and the damping resistor R d were not present.
5. Let Im be the maximum load current and Vm be the minimum supply voltage value. Than the expression for the design of commutation circuit parameters in a single-phase modified McMurray half-bridge inverter is given by
power-electronics-questions-answers-force-commutated-inverters-q5
a) /Vm
b) Im/
c) Vm/
d) /Im
Answer: c
Explanation: The design is carried out on the basis of the worst operating conditions which consist of the minimum supply voltage Vm and the maximum load current. gives the empirical formula for calculation of the L and C commutation circuit parameters.
6. The number of diodes, SCRs and other components in full-bridge inverter McMurray inverter is ____________ of those in half-bridge McMurray inverter.
a) same
b) double
c) three times
d) none of the mentioned
Answer: b
Explanation: The modified McMurray full-bridge requires twice the components that required in the half-wave type.
7. In the single-phase modified McMurray full-bridge inverter, for commutating the main SCRs T1 and T2
a) The capacitor is charged
b) TA1 is triggered
c) The commutation circuit is switched on
d) TA1 and TA2 are triggered
Answer: d
Explanation: TA1 and TA2 are triggered together, which then turn-off the main SCRs. TA1 and TA2 are the auxiliary SCRs in the commutation circuit of the McMurray inverter.
8. How many diodes are there in total in the single-phase, modified McMurray full-bridge inverter?
a) 4
b) 6
c) 8
d) 10
Answer: c
Explanation: 4 belong to the main full-bridge circuit and 4 in the commutation circuit .
9. The McMurray-Bedford half-bridge inverter requires
a) 4 SCRs, 2 diodes, 2 capacitors and 2 inductors
b) 4 SCRs, 4 diodes, 2 capacitors and 2 inductors
c) 2 SCRs, 4 diodes, 2 capacitors and 1 inductor
d) 2 SCRs, 2 diodes, 2 capacitors and 2 inductors
Answer: d
Explanation: McMurray-Bedford type requires less number of SCRs and diodes as compared to the McMurray type, however the number of capacitors and inductors required is the same.
10. The single-phase McMurray-Bedford type bridge inverter is a/an
a) auxiliary-commutated inverter
b) complementary-commutated inverter
c) supplementary-commutated inverter
d) none of the mentioned
Answer: b
Explanation: If one SCR gets turned on, the other conducting SCR gets turned off. This type of commutation is called as complementary commutation.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “3-Phase Bridge Inverters-1”.
1. Identify the below given circuit.
power-electronics-questions-answers-3-phase-bridge-inverters-1-q1
a) Three-phase bridge regulator
b) Three-phase bridge type semi-converter circuit
c) Three-phase bridge thyristor inverter
d) Three-phase bridge IGBT inverter
Answer: c
Explanation: It is a three-phase bridge type inverter. As SCRs are used as the switching device, it is called as a thyristor inverter.
2. A three-phase bridge inverter requires minimum of _____________ switching devices.
a) 3
b) 4
c) 6
d) 8
Answer: c
Explanation: It requires a min. 6 devices, 2 in each leg. Switching devices could be anything BJT, MOSFET or an IGBT. SCRs are used when very high power ratings are required.
3. The below given inverter circuit is a __ step inverter.
power-electronics-questions-answers-3-phase-bridge-inverters-1-q1
a) 3
b) 2
c) 6
d) none of the mentioned
Answer: c
Explanation: The three-phase bridge type is a 6-step inverter. That means, the firing changes from one SCR to another 6 times per cycle.
4. In the three-phase bridge inverter, each step consists of
a) 30°
b) 60°
c) 90°
d) will depend on the value of the firing angle
Answer: b
Explanation: It is a 6 step inverter. Hence, 360/6 = 60°. This means that the SCRs are gated every 60° in proper sequence.
5. In inverters, to make the supply voltage constant
a) an inductor is placed in series with the load
b) capacitor is connected in parallel to the load side
c) capacitor is connected in parallel to the supply side
d) none of the mentioned
Answer: c
Explanation: A large C connected across the input terminal keep the supply voltage from altering.
6. In the 180° mode VSI, ___________ devices conduct at a time.
a) 5
b) 2
c) 3
d) 4
Answer: c
Explanation: Three devices conduct at a time. One from the upper pair and two from the lower pair or vice-versa.
7. In the figure given below, for 180° mode of operation if T1 is fired at 0°. Then SCRs T3 and T5 should be fired at _________ and _________ respectively.
power-electronics-questions-answers-3-phase-bridge-inverters-1-q1
a) 180°, 360°
b) 90°, 180°
c) 120°, 240°
d) none of the mentioned
Answer: c
Explanation: T1-T4 form the first pair. T3-T6 form the second pair, and like-wise. For the 180° mode, each SCR conducts for 180°, but the groups of SCRs lag the prior group by an angle of 120°. e.g. If T1 is fired at 0 then T3 must be fired at an angle of 0 + 120° and T5 at 120 + 120 = 240°.
8. For a three phase bridge inverter in the 180° mode, ___________ devices are conducting from 120° to 180°.
power-electronics-questions-answers-3-phase-bridge-inverters-1-q1
a) T1, T6, T5
b) T2, T6, T5
c) T1, T6, T5
d) T1, T2, T3
Answer: d
Explanation:
Group I T1 T1 T1 T4 T4 T4 T1 T4
Group II T6 T6 T3 T3 T3 T6 T6 T6
Group III T5 T2 T2 T2 T5 T5 T5 T2
Step No I II III IV V VI I II
Each step consists of 60°. 120° to 180° will be step III.
9. _________ SCRs conduct from 300° to 360°.
power-electronics-questions-answers-3-phase-bridge-inverters-1-q1
a) T1, T2, T3
b) T4, T5, T6
c) T4, T3, T2
d) T1, T6, T5
Answer: b
Explanation:
Group I T1 T1 T1 T4 T4 T4 T1 T4
Group II T6 T6 T3 T3 T3 T6 T6 T6
Group III T5 T2 T2 T2 T5 T5 T5 T2
Step No I II III IV V VI I II
Each step consists of 60°. 300° to 360° will be step VI.
10. The diodes D4 and D1 will conduct from
power-electronics-questions-answers-3-phase-bridge-inverters-1-q1
a) they will never conduct
b) 300° to 360°
c) 120° to 180°
d) insufficient information
Answer: d
Explanation: This will depend on the nature on the load, which is not mentioned in the above problem.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “3-Phase Bridge Inverters-2”.
1. The conducting SCRs from 180 to 240 degrees would be
power-electronics-questions-answers-3-phase-bridge-inverters-2-q1
a) 6, 1, 2
b) 2, 3, 4
c) 3, 4, 5
d) 5, 6, 1
Answer: b
Explanation: 180 to 240 degrees i.e. step IV.
2. What is the peak value of phase voltage in case of 3-phase VSI with 180° mode. The supply side consists of a constant dc voltage source of Vs.
a) Vs
b) 3Vs/2
c) 2Vs/3
d) 3Vs
Answer: c
Explanation: Apply KVL to the equivalent circuit of the inverter by closing and opening the proper switches. The phase voltage is a stepped sine-wave with peak value 2Vs/3.
3. What is the R phase voltage when T1, T6 and T5 are conducting? Consider a star connected R load.
power-electronics-questions-answers-3-phase-bridge-inverters-2-q1
a) Vs
b) 2Vs/3
c) Vs/3
d) –Vs/3
Answer: c
Explanation: They conducting SCRs can be represented as closed switches. The load terminals R and B are connected to the positive bus but the terminal Y is connected to the negative bus. V RN = V BN = Vs/3.
V YN = -2Vs/3.
4. What are the phase voltages when T6, T1 and T2 are conducting? Consider a star connected R load.
power-electronics-questions-answers-3-phase-bridge-inverters-2-q1
a) V YN = V BN = Vs/3, V RN = 0
b) V YN = V RN = – Vs/3, V BN = 2Vs/3
c) V RN = V BN = – Vs/3, V YN = 2Vs/3
d) V YN = V BN = – Vs/3, V RN = 2Vs/3
Answer: c
Explanation: When 6, 1 and 2 are conducting and the others are off. The conducting SCRs can be represented as closed switches. The load terminals Y and B are connected to the negative bus but the terminal R is connected to the positive bus. Supply voltage is Vs.
V YN = V BN = -Vs/3.
V RN = 2Vs/3.
It should be noted that at any time, the summation of these phase voltages must be zero.
5. What are the phase voltages when T1, T2 and T3 are conducting? Consider a star connected R load.
power-electronics-questions-answers-3-phase-bridge-inverters-2-q1
a) V YN = V BN = V RN = 0
b) V YN = V RN = Vs/3, V BN = – 2Vs/3
c) V RN = V BN = – Vs/3, V YN = 2Vs/3
d) V YN = V BN = Vs/3, V RN = – 2Vs/3
Answer: b
Explanation: When 6, 1 and 2 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals R and Y are connected to the positive bus and the terminal B is connected to the nrgative bus. Supply voltage is Vs.
V RN = V YN = Vs/3.
V BN = -2Vs/3.
It should be noted that at any time, the summation of these phase voltages must be zero.
6. What is the maximum line voltage value in case of a three-phase VSI in 180° mode?
a) 2Vs
b) Vs
c) 3Vs
d) 2Vs/3
Answer: b
Explanation: The line voltage waveform has a peak value of Vs. Any line voltage value can be found by just adding the two phase voltage value.
e.g. Vab = Vrn – Vbn
Vab = – = Vs.
7. Find V YB , when T6, T1 and T2 are conducting. Consider a star connected R load.
power-electronics-questions-answers-3-phase-bridge-inverters-2-q1
a) Vs
b) Vs/3
c) 0
d) 2Vs/3
Answer: c
Explanation: When 6, 1 and 2 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals Y and B are connected to the negative bus but the terminal R is connected to the positive bus. Supply voltage is Vs.
V YN = V BN = -Vs/3.
V YB = V YN – V BN = – = -Vs/3 + Vs/3 = 0.
8. Find V RY , when T6, T1 and T2 are conducting. Consider a star connected R load.
power-electronics-questions-answers-3-phase-bridge-inverters-2-q1
a) Vs
b) Vs/3
c) 0
d) 2Vs/3
Answer: a
Explanation: When 6, 1 and 2 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals Y and B are connected to the negative bus but the terminal R is connected to the positive bus. Supply voltage is Vs.
V YN = V BN = -Vs/3.
V RN = 2Vs/3
V RY = V RN – V YN = – = 3Vs/3 = Vs.
9. What is the Y phase voltage, when T2, T3 and T4 are conducting?
power-electronics-questions-answers-3-phase-bridge-inverters-2-q1
a) 0
b) Vs/3
c) 2Vs/3
d) -Vs/3
Answer: c
Explanation: Construct the equivalent circuit, considering the conducting devices as S.C and the non-conducting devices as open circuit.
When 2, 3 and 4 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals R and B are connected to the negative bus but the terminal Y is connected to the positive bus. Supply voltage is Vs.
V RN = V BN = -Vs/3.
V YN = 2Vs/3.
10. What is the voltage between the B phase and the neutral, when T2, T3 and T4 are conducting?
power-electronics-questions-answers-3-phase-bridge-inverters-2-q1
a) -2Vs/3
b) Vs/3
c) 2Vs/3
d) -Vs/3
Answer: d
Explanation: Construct the equivalent circuit, considering the conducting devices as S.C and the non-conducting devices as open circuit.
When 2, 3 and 4 are conducting and the others are off. They conducting SCRs can be represented as closed switches. The load terminals R and B are connected to the negative bus but the terminal Y is connected to the positive bus. Supply voltage is Vs.
V RN = V BN = -Vs/3.
V YN = 2Vs/3.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “3-Phase Bridge Inverters-3”.
1. The 120° mode of operation of a three phase bridge inverter requires ___________ number of steps.
a) 2
b) 4
c) 6
d) 8
Answer: c
Explanation: Like the 180 mode, the 120° mode also requires six steps, each of 60° duration.
2. In case of the 120° mode of operation, __________ devices conduct at a time.
a) 2
b) 3
c) 4
d) none of the mentioned
Answer: a
Explanation: Unlike the 180° mode, in the 120° mode one 2 devices are conducting at a time as each conduct for 120°.
3. Safe commutation can be achieved in case of the ____________ operating mode.
a) 180°
b) 120°
c) 360°
d) none of the mentioned
Answer: b
Explanation: In the 120 mode of operation, there is a 60° interval between turning off of T1 and turning on of T4 , hence SCRs are commutated safely.
4. What is the R phase voltage when only T1 and T2 are conducting from 60° to 120°. Consider star connected R load.
power-electronics-questions-answers-3-phase-bridge-inverters-3-q4
a) Vs
b) Vs/2
c) –Vs/2
d) 0
Answer: b
Explanation: As only 2 SCRs are conducting at a time, it is 120 mode.
When T1 and T2 are conducting, R phase is connected to the positive bus through the T1 SCR and B phase is connected to the negative bus through the T2 SCR.
Considering that the load is a balanced R load.
V RN = Vs/2
V YN = -Vs/2.
5. If T1 is gated at 0 °, T3 and T5 will start conducting at _______ and _________ respectively.
a) 180°, 270°
b) 120°, 240°
c) 180°, 300°
d) 240°, 360°
Answer: b
Explanation: As T1 is gated at 0, T3 will be gated after 120° and T5 after 120+120 = 240°.
Devices T1 T1 0 T4 T4 0 T1 T1 0
T6 0 T3 T3 0 T6 T6 0 T3
0 T2 T2 0 T5 T5 0 T2 T2
6. What is the B phase voltage when only T1 and T2 are conducting from 60° to 120°. Consider star connected R load.
power-electronics-questions-answers-3-phase-bridge-inverters-3-q4
a) Vs
b) Vs/2
c) –Vs/2
d) 0
Answer: c
Explanation: When T1 and T2 are conducting, R phase is connected to the positive bus through the T1 SCR and B phase is connected to the negative bus through the T2 SCR. Considering that the load is a balanced R load.
Devices T1 T1 0 T4 T4 0 T1 T1 0
T6 0 T3 T3 0 T6 T6 0 T3
0 T2 T2 0 T5 T5 0 T2 T2
V RN = Vs/2
V BN = -Vs/2.
7. What is the R phase voltage and Y phase voltage when only T3 and T4 are conducting? Consider a star connected balanced R load.
power-electronics-questions-answers-3-phase-bridge-inverters-3-q4
a) Vs, -Vs
b) Vs/2, -Vs/2
c) –Vs/2, Vs/2
d) –Vs, Vs
Answer: c
Explanation: When T3 and T4 are conducting, R phase is connected to the negative bus through the T4 SCR and the Y phase is connected to the positive bus through the T3 SCR. Considering that the load is a balanced R load.
Devices T1 T1 0 T4 T4 0 T1 T1 0
T6 0 T3 T3 0 T6 T6 0 T3
0 T2 T2 0 T5 T5 0 T2 T2
V YN = Vs/2
V RN = -Vs/2.
8. If the SCR T1 is gated at 0°, then for 120° mode of operation, from ωt = 240° to 300° __________ devices would conduct.
power-electronics-questions-answers-3-phase-bridge-inverters-3-q4
a) T3, T4
b) T4, T5
c) T1, T6
d) T5, T6
Answer: b
Explanation:
Devices T1 T1 0 T4 T4 0 T1 T1 0
T6 0 T3 T3 0 T6 T6 0 T3
0 T2 T2 0 T5 T5 0 T2 T2
ωt = 240° to 300°, would form the 5th step.
9. Find the line voltage V YR when only T3 and T4 are conducting? Consider a star connected balanced R load.
power-electronics-questions-answers-3-phase-bridge-inverters-3-q4
a) 2Vs/3
b) Vs/2
c) Vs
d) –Vs
Answer: c
Explanation: When T3 and T4 are conducting, R phase is connected to the negative bus through the T4 SCR and the Y phase is connected to the positive bus through the T3 SCR. Considering that the load is a balanced R load.
V YN = Vs/2
V RN = -Vs/2.
V YR = V YN – V RN = Vs.
10. The peak value of the line voltage in case of 120° mode of operation of a three-phase bridge inverter is
a) Vs/2
b) 3Vs/2
c) Vs/√2
d) Vs
Answer: d
Explanation: The peak value for 120° mode is Vs. The line voltage waveform is a sine wave with a peak value of Vs .
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Voltage Control in Inverters”.
1. The external control of ac output voltage can be achieved in an inverter by
a) connecting a cyclo-converter
b) connecting an ac voltage controller between the output of the inverter and the load
c) connecting an ac voltage controller between the dc source and inverter
d) connecting an ac voltage controller between the load and the dc source
Answer: b
Explanation: By connecting a AC voltage controller, the ac output from the inverter can be varied and then fed to the load.
2. The series-inverter control method is a/an
a) internal voltage control method
b) external frequency control method
c) external voltage control method
d) none of the mentioned
Answer: c
Explanation: It is a external voltage control method where the outputs of the two inverters are connected to the transformers where the secondary of the transformer sums up the two input voltages.
3. In the series-inverter control method
a) two inverters are connected back-to-back
b) the output from the inverter is taken serially
c) output voltages of two inverters are summed up with the help of a transformer
d) output voltages of two inverters are summed up with the help of a third inverter
Answer: c
Explanation: It is a external voltage control method where the outputs of the two inverters are connected to the transformers where the secondary of the transformer sums up the two input voltages.
4. In case of the series inverter control, if two inverters are connected in series through a transformer, and two secondary voltages are V1 and V2, then the resultant output is given by
a) V1 + V2
b) √(V1 2 + V2 2 )
c) [V1 2 + V2 2 + 2.V1.V2.cosθ] 1/2
d) [V1 2 + V2 2 + 2.V1.V2.sinθ] 1/2
Answer: c
Explanation: The resultant output will be the phasor sum of V1 and V2.
5. External control of dc input voltage can be obtained by the use of a
a) transformer
b) chopper
c) inverter
d) converter
Answer: b
Explanation: A chopper is used to control the dc input voltage to the inverter, which then converters the variable dc to variable ac.
6. In the external control of dc input voltage
a) a chopper is placed just after the inverter block
b) a chopper is placed just after the filter block
c) a chopper is placed before the filter and the inverter block
d) none of the mentioned
Answer: c
Explanation: Constant AC – Rectifier – Chopper – Filter – Inverter.
7. __________ method is an internal method for controlling the inverter output voltage.
a) series connection of inverters
b) chopper method
c) commutating capacitor
d) pulse width modulation
Answer: d
Explanation: The PWM method, is an internal controlling method.
8. In the PWM method
a) external commutating capacitors are required
b) more average output voltage can be obtained
c) lower order harmonics are minimized
d) higher order harmonics are minimized
Answer: c
Explanation: In all the PWM methods, only odd harmonics are present. The lower order harmonics are eliminated along with its output voltage control.
9. Which of the following is not a PWM technique?
a) Single-pulse width modulation
b) Multiple-pulse width modulation
c) Triangular-pulse width modulation
d) Sinusoidal-pulse width modulation
Answer: c
Explanation: There is no such “Triangular” PWM.
10. In pulse width modulation
a) the output voltage is modulated
b) the input voltage is modulated
c) the gating pulses are modulated
d) none of the mentioned
Answer: c
Explanation: In PWM, the gating pluses are modulated, i.e. the gating pulses or firing pulses are made to go on of rapidly which changes the output voltage values accordingly.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “PWM Inverters-1”.
1. In the single-pulse width modulation method, the output voltage waveform is symmetrical about __________
a) π
b) 2π
c) π/2
d) π/4
Answer: c
Explanation: The waveform is a positive in the first half cycle and symmetrical about π/2 in the first half.
2. In the single-pulse width modulation method, the output voltage waveform is symmetrical about ____________ in the negative half cycle.
a) 2π
b) 3π/2
c) π/2
d) 3π/4
Answer: b
Explanation: In the negative half the wave is symmetrical about 3π/2.
3. The shape of the output voltage waveform in a single PWM is
a) square wave
b) triangular wave
c) quasi-square wave
d) sine wave
Answer: c
Explanation: Positive and the negative half cycles of the output voltage are symmetrical about π/2 and 3π/2 respectively. The shape of the waveform obtained is called as quasi-square wave.
4. In the single-pulse width modulation method, the Fourier coefficient b n is given by
a) [ sin sin ].
b) 0
c) [sin sin].
d) [sin sin].
Answer: c
Explanation: The Fourier analysis is as under:
b n = ∫ Vs sin nωt .d , Where the integration would run from to
2d is the width of the pulse.
5. In the single-pulse width modulation method, the Fourier coefficient a n is given by
a) [ cos cos ].
b) 0
c) [sin sin].
d) [sin sin].
Answer: b
Explanation: As the positive and the negative half cycles are identical the coefficient a n = 0.
6. In the single-pulse width modulation method, when the pulse width of 2d is equal to its maximum value of π radians, then the fundamental component of output voltage is given by
a) Vs
b) 4Vs/π
c) 0
d) 2Vs/π
Answer: b
Explanation: The Fourier representation of the output voltage is given by
power-electronics-questions-answers-pwm-inverters-1-q6
Put 2d = π & n = 1.
7. In case of a single-pulse width modulation with the pulse width = 2d, the peak value of the fundamental component of voltage is given by the expression
a) 4Vs/π
b) Vs
c) sin 2d
d) sin d
Answer: d
Explanation: For the fundamental component put n = 1.
power-electronics-questions-answers-pwm-inverters-1-q6
Vo = sin sin
Hence the peak value is sin d.
8. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the nth harmonic from the output voltage
a) d = π
b) 2d = π
c) nd = π
d) nd = 2π
Answer: c
Explanation: To eliminate, the nth harmonic, nd is made equal to π radians, or d = π/n.
From the below expression,
power-electronics-questions-answers-pwm-inverters-1-q6
when nd = π. sin nd = 0 hence, that output voltage harmonic is eliminated.
9. Find the peak value of the fundamental component of voltage with a pulse width of 2d = 90 and Vs = 240 V for single-pulse modulation in a full wave bridge inverter.
a) 305 V
b) 216 V
c) 0 V
d) 610 V
Answer: b
Explanation: The peak value of the fundamental component of voltage is given by sin d.
10. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the 3rd harmonic from the output voltage waveform, the value of the pulse width must be
a) 0°
b) 60°
c) 120°
d) 180°
Answer: c
Explanation: To eliminate the nth harmonic, nd = π.
Therefore, d = π/n = π/3 = 60°
Hence, 2d = 120°.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “PWM Inverters – 2”.
1. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the 5th harmonic from the output voltage waveform, the value of the pulse width must be equal to
a) 72°
b) 86°
c) 91°
d) 5°
Answer: c
Explanation: To eliminate the nth harmonic, nd = π.
Therefore, d = π/n = π/5 = 36°
Hence, 2d = 72°.
2. Several equidistant pulses per half cycle are used in ___________ type of modulation technique.
a) single-pulse
b) multiple-pulse
c) sine-pulse
d) equidistant-pulse
Answer: b
Explanation: In MPM, several equidistant pulses per half cycle are used.
3. In the multiple-pulse width modulation method, the Fourier coefficient a n is
a) [ cos cos ].
b) 0
c) [sin sin].
d) [sin sin].
Answer: b
Explanation: As the positive and the negative half cycles are identical the coefficient a n = 0.
4. In the multiple-pulse width modulation method with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, has the Fourier coefficient b n =
a) [ cosɣ cos ].
b) 0
c) [sinɣ sin].
d) [sinɣ sin].
Answer: d
Explanation: The Fourier analysis is as under:
b n = ∫ Vs sin nωt .d , Where the integration would run from ɣ to ɣ
b n = [sinɣ sin].
5. The amplitude of the nth harmonic of the two-pulse MPM waveform is given by __________
Let d be the width of a single pulse and ɣ be the distance from 0 to the centre of the first pulse.
a) sinɣ sin
b) sinɣ sin
c) sin sin
d) sinɣ sin
Answer: a
Explanation: The Fourier series representation of such a wave will be given by
power-electronics-questions-answers-pwm-inverters-2-q5
b n = [sinɣ sin] a n = 0
Therefore,
Hence, the amplitude is the term excluding the sin nωt factor.
6. In case of MPM with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, for eliminating the nth harmonic from the output voltage, which of the following condition must be satisfied?
a) d = 2π
b) d = π
c) d = n°
d) d = 2π/n
Answer: d
Explanation:
power-electronics-questions-answers-pwm-inverters-2-q5
If d = 2π/n, sin = 0 and the whole term is eliminated.
7. In case of MPM with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, for eliminating the nth harmonic from the output voltage, the value of gamma ɣ must be equal to
a) 0
b) π
c) π/n
d) d/n
Answer: c
Explanation:
power-electronics-questions-answers-pwm-inverters-2-q5
If gamma = π/n , the whole nth output voltage comes becomes zero.
8. Find the peak value of the fundamental component of voltage with MPM with two pulses having pulse width = 36° and ɣ = 54°. The Fourier representation of the waveform is as follows.
power-electronics-questions-answers-pwm-inverters-2-q5
a) 0.7484 x Vs
b) 1.414 x Vs
c) 0.637 x Vs
d) 2.54 x Vs
Answer: c
Explanation: For the fundamental component put n = 1.
n = 36° and ɣ = 54° . . .
Vo1 = 8Vs/π x sin54 x sin18 = 0.637 Vs.
9. In the MPM method, the comparator is given _______ and _______ types of waveform at its input.
a) square, sine
b) square, quasi-square
c) sine, triangular
d) square, triangular
Answer: d
Explanation: To generate the modulated waves of equal width, square wave which is the reference signal is compared with the triangular wave which is the carrier signal wave.
10. In MPM, the square wave is the ________ signal whereas the triangular wave is the ________ signal.
a) reference, carrier
b) base, reference
c) carrier, reference
d) none of the mentioned
Answer: a
Explanation: To generate the modulated waves of equal width, square wave which is the reference signal is compared with the triangular wave which is the carrier signal wave.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “PWM Inverters-3”.
1. In the multiple pulse width modulation method, the firing pulses are generate during the interval when the
a) triangular wave exceeds the square modulating wave
b) square modulating wave exceeds the triangular wave
c) square wave amplitude is same as the triangular wave’s amplitude
d) none of the mentioned
Answer: a
Explanation: The firing pulses to turn on the SCRs are generated when the triangular carrier signal exceeds the square reference signal.
2. In MPM, ____________ order harmonics can be eliminated by a proper choice of __________ and _________
a) higher, d, ɣ
b) lower, d, ɣ
c) higher and lower, d, ɣ
d) none of the mentioned
Answer: d
Explanation: In multiple pulse width modulation, the lower order harmonics can be eliminated by proper choice of 2d and ɣ.
3. In ___________ type of modulation method, the pulse width is not equal for all the pulses.
a) multiple pulse width modulation
b) single pulse width modulation
c) sinusoidal pulse width modulation
d) none of the mentioned
Answer: c
Explanation: In SPWM, the pulse width is a sinusoidal function of the angular position of the pulse in a cycle.
4. In sinusoidal pulse width modulation, __________ wave is compared with a ___________ type of wave.
a) square, sinusoidal
b) sinusoidal, triangular
c) sinusoidal, quasi-square
d) none of the mentioned
Answer: b
Explanation: In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.
5. In the sinusoidal pulse width modulation, __________ is the carrier wave signal.
a) square wave
b) triangular wave
c) sinusoidal wave
d) quasi-square wave
Answer: b
Explanation: In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.
6. In the sinusoidal pulse width modulation, ____________ is the reference wave signal.
a) square wave
b) triangular wave
c) sinusoidal wave
d) quasi-square wave
Answer: c
Explanation: In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.
7. In sinusoidal pulse width modulation, the comparator output is high when the
a) triangular wave has magnitude higher than the sinusoidal wave
b) sinusoidal wave has magnitude higher than the triangular wave
c) triangular wave has magnitude equal to the sinusoidal wave
d) none of the mentioned
Answer: b
Explanation: The comparator output is high when the sinusoidal wave has magnitude higher than the triangular wave .
8. In PWM, the comparator output is further given to a ____________
a) integrator
b) scr devices
c) trigger pulse generator
d) snubber circuit
Answer: c
Explanation: The comparator output is processed in a trigger pulse generator in such a manner that the output voltage wave of the inverter has a pulse width in agreement with the comparator output pulse width.
9. The modulation index is given by
Vr = peak value of the reference wave.
Vc = peak value of the carrier wave.
a) Vr/Vc
b) Vc/Vr
c)
d) 1/
Answer: a
Explanation: MI = Vr/Vc.
10. By controlling the modulation index , __________ can be controlled.
a) gain
b) output frequency
c) harmonic content of the output voltage
d) cosine component of the output voltage
Answer: c
Explanation: MI controls the output voltage waveform.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “PWM Inverters-4”.
1. In pulse width modulated inverters, the output voltage is controlled by controlling the
a) input frequency
b) modulating index
c) amplification factor
d) none of the mentioned
Answer: b
Explanation: MI = Vr/Vc.
MI controls the output voltage waveform, as it is proportional to the fundamental component of the output voltage.
2. In case of sinusoidal pulse width modulation with MI < 1, if the number of pulses per half cycle = 5, then
a) harmonics of order 5 and 7 become significant
b) harmonics of order 5 and 7 are eliminated
c) harmonics of order 9 and 11 become significant
d) harmonics of order 9 and 11 are eliminated
Answer: c
Explanation: For MI less than one, largest harmonic amplitudes in the output voltage are associated with harmonics of order 2N 1. Thus, by increasing the number of pulses per half cycle , the order of the dominate harmonic can be raised, which can be filtered out easily.
3. In case of sinusoidal pulse width modulation with MI < 1, the order of the dominate harmonic can be raised by
a) increasing the number of pulses
b) reducing the number of pulses
c) lowering the input voltage frequency
d) raising the input voltage frequency
Answer: a
Explanation: For MI less than one, largest harmonic amplitudes in the output voltage are associated with harmonics of order 2N 1. Thus, by increasing the number of pulses per half cycle , the order of the dominate harmonic can be raised.
4. In case of sinusoidal pulse width modulation with MI < 1, if the number of pulses per half cycle = 6, then
a) harmonics of order 7 and 9 become significant
b) harmonics of order 7 and 9 are eliminated
c) harmonics of order 11 and 13 become significant
d) harmonics of order 11 and 13 are eliminated
Answer: c
Explanation: For MI less than one, largest harmonic amplitudes in the output voltage are associated with harmonics of order 2N 1. Thus, by increasing the number of pulses per half cycle , the order of the dominate harmonic can be raised, which can be filtered out easily.
5. Increasing the number of pulses , ____________
a) reduces the output voltage amplitude
b) reduces the inverter efficiency
c) improves the inverter efficiency
d) none of the mentioned
Answer: b
Explanation: Increasing N, increasing the switching frequency of the SCRs. This amounts to more switching losses and therefore efficiency lowers down.
6. In single-phase modulation of PWM inverters, the pulse width is 120°. For an input voltage of 220 V dc, the rms value of output voltage is
a) 185 V
b) 254 V
c) 127 V
d) 179 V
Answer: d
Explanation: Vo = x 1/2
Where, d = 2π/3 and Vs = 220 V.
7. In MPM the amplitudes of square wave and triangular wave are respectively 1 V and 2 V. For generating 5 pulses per half cycle, the pulse width should be ___________
a) 36°
b) 24°
c) 12°
d) 18°
Answer: d
Explanation: In multiple pulse width modulation, the pulse width is given by
[ 1 – ] x
Where, Vr = 1
Vc = 2
N = 5.
Hence, pulse width = 180/10 = 18°.
8. In an inverter, if the fundamental output frequency is 50 Hz, then the frequency of the lowest order harmonic will be
a) 50 Hz
b) 150 Hz
c) 250 Hz
d) 350 Hz
Answer: b
Explanation: The 3rd harmonic is the lowest order harmonic,
Hence, 50 x 3 = 150 Hz.
9. Calculate the pulse width in case of MPM, if the amplitudes of square wave and triangular wave are respectively 2 V and 3 V respectively. 16 pulses per cycle are generated.
a) 18°
b) 7.5°
c) 6.4°
d) 9°
Answer: b
Explanation: In multiple pulse width modulation, the pulse width is given by
[ 1 – ] x
Where, Vr = 2 V
Vc = 3 V
N = Pulses per half cycle = 16/2 = 8
Hence, pulse width = 180/ = 7.5°.
10. In a PWM inverter, if the frequency of the lowest harmonic is 180 Hz, then the frequency of the fundamental component would be ___________
a) 50 Hz
b) 60 Hz
c) 540 Hz
d) 90 Hz
Answer: b
Explanation: The 3rd harmonic is the lowest order harmonic,
Hence, 180/3 = 60 Hz.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “PWM Inverters-5”.
1. In case of multiple pulse width modulation method, if the amplitudes of the reference wave and the carrier wave are made equal then, the pulse width =
a) ∞
b) 0
c) 100 °
d) none of the mentioned
Answer: b
Explanation: In multiple pulse width modulation, the pulse width is given by
[ 1 – ] x
If Vr = Vc
Pulse width = 0.
2. In an inverter, if the fundamental output frequency is 45 Hz, then the frequency of the second lowest order harmonic will be
a) 45 Hz
b) 135 Hz
c) 225 Hz
d) 9 Hz
Answer: c
Explanation: The 5th harmonic is the second lowest order harmonic, as only odd numbers of harmonics are present.
Hence, 45 x 5 = 225 Hz.
3. In an inverter, if the fundamental output frequency is 45 Hz, then the frequency of the lowest order harmonic will be
a) 45 Hz
b) 225 Hz
c) 15 Hz
d) 135 Hz
Answer: d
Explanation: The 3rd harmonic is the lowest order harmonic, as only odd numbers of harmonics are present.
Hence, 45 x 3 = 135 Hz.
4. A VSI will have a better performance if its
a) load inductance is small and source inductance is large
b) both load inductance and source inductance are small
c) both load inductance and source inductance are large
d) none of the mentioned
Answer: b
Explanation: Higher value of source inductance will increase the overlap angle and cause commutation issues. Hence, both load inductance and source inductance should be small.
5. A single-phase bridge inverter has a square wave output voltage waveform, with odd harmonics present. What is the percentage of the fifth harmonic component to the fundamental component?
a) 50 %
b) 25 %
c) 20 %
d) 5 %
Answer: c
Explanation: For the fundamental comoponent, n = 1 and for the fifth harmonic n = 5
1/5 = 0.2 = 20 %.
6. Find, the maximum rms value of the fundamental component of the output voltage for the below given single pulse width modulated circuit. Take Vs = 68 V.
power-electronics-questions-answers-pwm-inverters-5-q6
a) 86.6 sin
b) 64 sin
c) 244 sin
d) none of the mentioned
Answer: a
Explanation: Vo1 = sin = 86.6 sin.
7. Control of frequency and control of voltage in 3-phase inverters is
a) possible only through inverter control circuit
b) possible through the control circuit of inverter and converter
c) possible through inverter control of frequency and through converter control for voltage
d) none of the mentioned
Answer: c
Explanation: Control of frequency and control of voltage in 3-phase inverters is possible through inverter control of frequency and through converter control for voltage.
8. Output voltage of a single-phase bridge inverter, fed from a fixed dc source is varied by
a) varying the switching frequency
b) pulse-width modulation
c) pulse amplitude modulation
d) all of the mentioned
Answer: b
Explanation: The output voltage is controlled by PWM techniques.
9. A single-phase bridge inverter, fed from a 230 V dc is connected to the load R = 10 Ω and L = 0.03 H. Determine the fundamental component of rms output current. Fundamental output frequency of the square wave output = 50 Hz.
a) 30 A
b) 15 A
c) 2.3 A
d) 20.7 A
Answer: b
Explanation: Vo1 = 4Vs/π√2 = 207.10 V.
Now, the load impedance at the fundamental frequency i.e. at 50 Hz will be
Z1 = [10 2 + 2 ] 1/2 = 13.7414 Ω
Hence, I = V1/Z1 = / = 15.07 A.
10. A single-phase bridge inverter, fed from a 230 V dc is connected to the load R = 10 Ω and L = 0.03 H. The output is a quasi-square wave with an on period of 0.5 of a cycle. Determine the fundamental component of rms output voltage.
a) 207.10 V
b) 146.42 V
c) 265.4 V
d) 129 V
Answer: b
Explanation: For quai-square wave or single-pulse width modulated wave pulse width 2d = 0.5 x 180° = 90° or d = 45°
Hence, Vo1 = sind = / sin45° = 146.423 V.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Harmonic Reduction”.
1. Which of the following is used as a harmonic reduction technique in inverters?
a) Amplitude modulation
b) Cycloconverter control
c) Transformer connection
d) Series connection of two inverters
Answer: c
Explanation: Two or more inverters are connected together by means of a transformer to get the net output voltage with reduced harmonic content.
2. For harmonic reduction by transformer connection, the output voltages from the two inverters must be
a) similar and in-phase with each other
b) dissimilar but in-phase with each other
c) similar but phase shifted from each other
d) dissimilar and phase shifted from each other
Answer: c
Explanation: The essential condition of this scheme is that the output voltages from the two inverters must be similar but phase shifted from each other.
3. The output voltage obtained by connecting two inverters through a transformer is a
a) square wave
b) sine wave
c) quasi-square wave
d) none of the mentioned
Answer: c
Explanation: The net output voltage is the addition of the two inverter voltages, as they are phase shifted from each other, the waveform obtained is a quasi-square wave.
4. Pulses of different widths and heights are superimposed in case of __________ harmonic reduction technique.
a) transformer connection
b) pulse width modulation
c) stepped-wave inverter
d) none of the mentioned
Answer: c
Explanation: In the steeped wave inverters pulses of different widths and heights are superimposed to produce a resultant stepped wave with reduced harmonic content.
5. In case of stepped wave inverters,
a) both the transformers have 1:1 turns ratio
b) both the transformers have 1:2 turns ratio
c) both the transformers have different transformer ratio
d) none of the mentioned
Answer: c
Explanation: The two transformers have different turns ratio, hence, different voltages levels are obtained at the secondary’s which are the added to get the net output voltage with reduced harmonic content.
6. In stepped wave inverters, one of the inverters are gated such as to obtain
a) one level modulation
b) two level modulation
c) zero level modulation
d) none of the mentioned
Answer: b
Explanation: In one-level modulation, during the first half cycle, the output voltage is either zero or positive. During the second half cycle the output voltage would be either zero or negative.
7. In three-level modulation
a) the output voltage is zero in the first half cycle.
b) the output voltage either zero or positive in the first half cycle.
c) the output voltage either zero or negative in the first half cycle.
d) the output voltage either zero, positive or negative in the first half cycle.
Answer: d
Explanation: In stepped wave inverters, one of the inverters is so gated as to obtain a three level modulation.
8. In single-phase modulation of PWM inverters, the lowest harmonic can be eliminated if the pulse width is made equal to __________
a) 30°
b) 0°
c) 120°
d) 60°
Answer: c
Explanation: In single-phase modulation of PWM inverters, the nth harmonic can be eliminated, if the pulse width is made equal to .
Lowest harmonic is for n = 3.
2d = 2π/3 = 120°.
9. In single-phase modulation of PWM inverters, the 5th order harmonic can be eliminated if the pulse width is made equal to ___________
a) 30°
b) 36°
c) 72°
d) None of the mentioned
Answer: c
Explanation: In single-phase modulation of PWM inverters, the nth harmonic can be eliminated, if the pulse width is made equal to .
10. The waveform obtained by __________ is more near to a sinusoidal wave
a) PWM inverters
b) Stepped wave inverters
c) parallel connected inverters
d) None of the mentioned
Answer: b
Explanation: The stepped wave output, is more nearer to a sine wave. More the number of steps, closer is the wave to a sinusoidal wave.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Current Source Inverters”.
1. In voltage source inverters , the amplitude of the output voltage is
a) independent of the load
b) dependent on the load
c) dependent only on L loads
d) none of the mentioned
Answer: a
Explanation: In VSIs the input voltage is maintained at a constant value and the amplitude of the output voltage does not depend on the load conditions. However, the waveform of the load current as well as its magnitude depends upon the nature of the load impedance.
2. In voltage source inverters , the output currents _____________
a) amplitude depends upon the load impedance
b) waveform depends upon the load impedance
c) amplitude as well as the nature of the waveform depends on the load
d) both amplitude and waveform are independent of the load impedance
Answer: c
Explanation: In VSIs the input voltage is maintained at a constant value and the amplitude of the output voltage does not depend on the load conditions. However, the waveform of the load current as well as its magnitude depends upon the nature of the load impedance.
3. In current source inverters
a) the amplitude of the output current is independent of the load
b) the amplitude of the output current dependents on the load
c) the amplitude of the output voltage is independent of the load
d) none of the mentioned
Answer: a
Explanation: In CSIs, the amplitude of the output current is independent on the load impedance, as the input current is kept constant.
4. In current source inverters , the output voltage’s
a) amplitude depends upon the load impedance
b) waveform depends upon the load impedance
c) amplitude as well as the nature of the waveform depends on the load
d) both amplitude and waveform are independent of the load impedance
Answer: c
Explanation: In CSIs, the amplitude of the output current is independent on the load impedance, as the input current is kept constant. However, the magnitude of output voltage and its waveform depends upon the nature of the load impedance.
5. In current source inverters
a) L filter is used after the CSI
b) L filter is used before the CSI
c) C filter is used after the CSI
d) C filter is used before the CSI
Answer: b
Explanation: In order that the current input to the CSI must remain ripple free an constant, L-filter is used before the CSI .
6. A CSI converters
a) the input dc current to an an current at output
b) the input ac current to dc current at output
c) the input dc current to amplified dc current at the output
d) the input ac current to amplified ac current at the output
Answer: a
Explanation: CSI converts the input dc current to an ac current at its output terminals.
7. In the below shown, the ideal single-phase CSI circuit has constant input source current of I amps. The output current waveform varies from
power-electronics-questions-answers-current-source-inverters-q7
a) 2I to – 2I
b) I/2 to –I/2
c) I to –I
d) none of the mentioned
Answer: c
Explanation: Considering the SCRs as ideal voltage sources, when T1, T2 are on the load current is = I. When T3, T4 conduct, the load current is negative with respect to the previous value and = -I.
8. T1, T2 are triggered at 0 and T3, T4 at T/2. If both the pairs of SCRs conduct for equal duration and the load consists of a capacitor. Then
power-electronics-questions-answers-current-source-inverters-q7
a) the load current waveform is a square wave
b) the load voltage waveform is a square wave
c) the load current is constant dc
d) the load voltage is constant dc
Answer: a
Explanation: The load current waveform is a square wave with frequency = 1/T. As the load consists of a capacitor, dv/dt must be constant over every half cycle. This slope is positive from zero to T/2 and negative from T/2 to T.
9. In a 3-phase VSI operating in square-wave mode, the output line voltage is free from
a) 3rd harmonic
b) 7th harmonic
c) 11th harmonic
d) 13th harmonic
Answer: a
Explanation: The line voltage is the difference of two phase voltages, the 3rd harmonic gets cancelled out, though it is present in the phase voltages.
10. Force-commutated CSIs need
a) capacitors for their commutation
b) inductors for their commutation
c) diodes for their commutation
d) none of the mentioned
Answer: a
Explanation: All the CSIs need capacitors for their commutation if force commutation is required. Force commutation is essential for lagging power factors.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “AC Voltage Controllers-1”.
1. AC voltage controllers convert
a) fixed ac to fixed dc
b) variable ac to variable dc
c) fixed ac to variable ac
d) variable ac to fixed ac
Answer: c
Explanation: Voltage controllers convert the fixed ac voltage to variable ac by changing the values of the firing angle.
2. In AC voltage controllers the
a) variable ac with fixed frequency is obtained
b) variable ac with variable frequency is obtained
c) variable dc with fixed frequency is obtained
d) variable dc with variable frequency is obtained
Answer: a
Explanation: Voltage controllers convert the fixed ac voltage to variable ac by changing the values of the firing angle. The available ac obtained has the same fixed frequency as the input ac.
3. Earlier then the semiconductor technology, ___________ devices were used for voltage control applications.
a) cycloconverters
b) vacuum tubes
c) tap changing transformer
d) induction machine
Answer: c
Explanation: A tap changing transformer can give variable ac from fixed ac without a change in frequency.
4. The AC voltage controllers are used in __________ applications.
a) power generation
b) electric heating
c) conveyor belt motion
d) power transmission
Answer: b
Explanation: In electric heating, variable ac supply is needed. The devices are fired appropriately to apply enough temperature.
5. In the principle of phase control
a) the load is on for some cycles and off for some cycles
b) control is achieved by adjusting the firing angle of the devices
c) control is achieved by adjusting the number of on off cycles
d) control cannot be achieved
Answer: b
Explanation: Switching devices is so operated that the load gets connected to ac source for a part of each half cycle.
6. A single-phase half wave voltage controller consists of
a) one SCR is parallel with one diode
b) one SCR is anti parallel with one diode
c) two SCRs in parallel
d) two SCRs in anti parallel
Answer: b
Explanation: As it is half wave, it consists of one SCR in anti parallel with one diode.
7. In the below given voltage controller circuit
power-electronics-questions-answers-ac-voltage-controllers-1-q7
a) the positive half cycle at the load is same as the supply Vs
b) the negative half cycle at the load is same as the supply Vs
c) the positive and negative half cycles at the load are identical to the supply
d) none of the mentioned
Answer: b
Explanation: As T1 is triggered at an angle α, the conduction in the positive have cycle will start at α. In the negative half cycle, the diode is forward biased and load is connected as it is to the supply.
8. The below shown controller circuit is a
power-electronics-questions-answers-ac-voltage-controllers-1-q7
a) half wave controller
b) full wave controller
c) none of the mentioned
d) will depend upon the firing angle
Answer: a
Explanation: As it consists one diode and one SCR only, the control is only in one cycle , hence it is a half wave controller.
9. The below given controller circuit is a
power-electronics-questions-answers-ac-voltage-controllers-1-q9
a) half wave controller
b) full wave controller
c) none of the mentioned
d) will depend upon the firing angle
Answer: b
Explanation: As it consists of two SCRs, it is a full wave controller.
10. In the below given voltage controller circuit
power-electronics-questions-answers-ac-voltage-controllers-1-q9
a) only the negative cycle can be controlled
b) only the positive cycle can be controlled
c) both the cycles can be controlled
d) none of the mentioned
Answer: c
Explanation: As it consists of two SCRs, it is a full wave controller and both the half cycles can be controlled by varying their respective firing angles.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “AC Voltage Controllers-2”.
1. In the below shown circuit, the diode conducts for
power-electronics-questions-answers-ac-voltage-controllers-2-q1
a) 90°
b) > 90°
c) < 90°
d) 0°
Answer: a
Explanation: The diode conducts from π to 2π in the negative half cycle. If it were a non resistive load, the diode would conduct for less than 90°, as the inductor would force conduct the SCR for some time.
2. The SCR T1 is fired at an angle of α, and the supply Vs = Vm sinωt. Find the average value of the output voltage.
power-electronics-questions-answers-ac-voltage-controllers-2-q1
a)
b)
c)
d) Vm
Answer: c
Explanation: Vo = ∫Vm sinωt d. Where the integration would be run from α to 2π as the conduction takes place from α to 2π.
Vo = .
3. The below given output voltage waveform can be obtained by a
power-electronics-questions-answers-ac-voltage-controllers-2-q3
a) half wave ac voltage controller
b) full wave ac voltage controller
c) half wave controller with firing angle = 0° for T1
d) full wave controller with firing angle = 0° for both T1 and T2
Answer: a
Explanation: As the positive half is chopped off due to some value of firing angle, the firing angle is not equal to zero. As the negative half is a half sine wave, either it is a full wave controller with firing angle for T2 set to zero or it is a half wave controller with a Thyristor and a diode.
4. In the below shown AC converter circuit with firing angle = α for both the devices, T2 will conduct from
power-electronics-questions-answers-ac-voltage-controllers-2-q4
a) α to π
b) π + α to 2π
c) π to 2π
d) α to 2π
Answer: b
Explanation: T2 is triggered at π + α, it conducts from π + α to 2π after which it is line commutated.
5. The below given output voltage waveform can be obtained by a
power-electronics-questions-answers-ac-voltage-controllers-2-q5
a) half wave ac voltage controller
b) full wave ac voltage controller
c) full wave inverter circuit
d) none of the mentioned
Answer: b
Explanation: As the control is in both the directions or cycles, it is a full wave ac voltage controller circuit.
6. In the integral cycle control method of ac voltage controller
a) the average power delivered to the load is controlled
b) the instantaneous power delivered to the load is controlled
c) the frequency of output voltage is controlled
d) none of the mentioned
Answer: a
Explanation: In integral cycle control, power is delivered for m cycles and not delivered for n cycles. Hence, the average value of the power delivered is controlled by manipulating m and n.
7. In the integral cycle control of ac voltage controller, is the load is on for n cycles and off for m cycles, then the periodicity is given by? Consider the output is sinusoidal.
a) m/2π
b) n/2π
c) m/π
d) n/π
Answer: b
Explanation: Over a complete cycle of 2π x the power is delivered for n cycles.
Hence, P = n/2π.
8. If k is the duty cycles of the controller, then the rms value of the output voltage in case of a integral cycle control circuit will be?
Consider the input to be sinusoidal with peak value Vm and rms value Vs.
a) Vs x k
b) Vs/k
c) Vs x √k
d) Vs
Answer: c
Explanation: Vrms = [ x ∫Vm 2 sin 2 ωt d ] 1/2
Where the integral runs from 0 to 2π.
Vrms = k = Vs x √k.
9. Find the power delivered to the load in the integral cycle control method of ac voltage control, having a sine input of Vs, R load and duty cycle = k.
power-electronics-questions-answers-ac-voltage-controllers-2-q4
a) Vs 2 /R
b) k.Vs 2 /R
c) √k .Vs 2 /R
d) 0
Answer: b
Explanation: Output voltage = Vs x √k
P = 2 /R.
10. In the integral cycle control method with duty cycle = k and maximum load current = Im. Find the value of average SCR current.
power-electronics-questions-answers-ac-voltage-controllers-2-q4
a) Im/k.π
b) Im
c) k.Im
d) k.Im/π
Answer: d
Explanation: As each SCR conducts for π radians during each cycle of n on cycles, the average value of SCR current is
I = ∫Im sin ωt d. Where the integration is from 0 to π.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “AC Voltage Controllers-3”.
1. The below given circuit has Vs = 230V and R = 20 Ω. Find the value of the average output voltage at the R load for a firing angle of 45°.
power-electronics-questions-answers-ac-voltage-controllers-3-q1
a) 224 V
b) -15.17 V
c) 15.17 V
d) –224 V
Answer: b
Explanation: Vo = [ x ]/2π = -15.17 V.
Negative value is due to the fact that the average value in the positive half cycle is less than that in the negative half cycle.
2. The below given circuit has Vs = 230V and R = 20 Ω. Find the value of the average output load current at the R load for a firing angle of 45°.
power-electronics-questions-answers-ac-voltage-controllers-3-q1
a) – 0.7585
b) 0.7585
c) -0.6396
d) -0.5
Answer: a
Explanation: Vo = [ x ]/2π = -15.17 V.
Io = /R = -0.7585
Negative value is due to the fact that the average value in the positive half cycle is less than that in the negative half cycle.
3. A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the rms output voltage.
a) 189 V
b) 260 V
c) 156 V
d) 178 V
Answer: d
Explanation: Vrms = Vo x √k
k = = 6/10 = 0.6
Vrms = √0.6 x 230 = 178.157 V.
4. A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the input pf.
a) 0.6
b) 0.7746
c) 0.855
d) 0.236
Answer: b
Explanation: k = = 6/10 = 0.6
input pf = √0.6 = 0.7746.
5. A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the power delivered to the load.
a) 2.1 W
b) 2.1 kW
c) 516 W
d) 5.16 kW
Answer: b
Explanation: Vrms = Vo x √k
k = = 6/10 = 0.6
Vrms = √0.6 x 230 = 178.157 V
P = 2 /R = 2116 W = 2.116 kW.
6. A single phase voltage controller has input of 230 V and a load of 15 Ω resistive. For 6 cycles on and 4 cycles off, determine the average value of SCR current.
a) 21.68 A
b) 200 mA
c) 4.14 A
d) 2.07 A
Answer: c
Explanation: Peak current Im = /15 = 21.681 A
k = = 6/10 = 0.6
Avg current = /π = 4.14 A.
7. Pulse gating is suitable for
a) R loads only
b) R and RL loads
c) RL loads only
d) all types of loads
Answer: a
Explanation: In RL loads with pulse gating, the incoming SCR may be fired during the interval when it is reversed biased by the outgoing SCR, thus it won’t get turn on even after the outgoing SCR is not reveres basing the incoming SCR is the pulse width is over before forward biasing the SCR.
8. In continues gating
a) overlap angle is very high
b) SCR is heated up
c) size of the pulse transformer is small
d) commutation cannot be achieved effectively
Answer: b
Explanation: As the gating is applied for a longer duration, the device is heated up.
9. High frequency gating uses a
a) train of pulses
b) continuous gating block
c) carrier signal
d) none of the above
Answer: a
Explanation: In high frequency gating a train of pulses are used to overcome the thermal problems due to continuous gating.
10. A single-phase voltage controller, using one SCR in anti parallel with a diode, feeds a load R and Vs = 230 V. For a firing angel of 90° for the SCR, the PMMC voltage connected across R would read
a) 0
b) 51.8 V
c) –51.8 V
d) –36.82 V
Answer: c
Explanation: As firing angle is 90, there is ideally be no conduction in the positive half. Hence, the average value will be zero.
Vo = /2π x = – 51.8 V.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Sequence Controller”.
1. Sequence control of ac voltage controllers is employed for the improvement of _________
a) output frequency
b) input frequency
c) commutation
d) system power factor
Answer: d
Explanation: It is used to improve the power factor at both the input and output side.
2. A two stage sequence control is
a) two SCRs in anti parallel
b) two voltage controllers in parallel
c) two voltage controllers in series
d) a voltage controller having two voltage level
Answer: b
Explanation: Sequence control of ac voltage controller means the use of two or more stages of voltage controllers in parallel for the regulation of output voltage.
3. From the below given statements regarding sequence control of ac voltage, which of them are true?
i) It improves system power factor
ii) It reduces the harmonic content at the output
iii) Wider control of output voltage is possible
a) only
b) only and
c) only and
d) all of them are true
Answer: d
Explanation: All of the above are the advantages of using two or more stages of voltage controllers in parallel. Harmonic content is reduced at two common parameters get cancelled out.
4. The below given circuit is that of a
power-electronics-questions-answers-sequence-controller-q4
a) four stage sequence controller
b) two stage sequence controller
c) full wave ac voltage controller
d) none of the mentioned
Answer: b
Explanation: Sequence control of ac voltage controller means the use of two or more stages of voltage controllers in parallel for the regulation of output voltage.
5. If T1, T2 are kept off and T3, T4 are having a firing angle of 180° each, then the output voltage is ____ if the turns ratio is 1:1
power-electronics-questions-answers-sequence-controller-q4
a) Vs
b) √2Vs
c) 0
d) 2Vs
Answer: c
Explanation: If T3 and T4 are fired at an angle of 180°, they are equivalent to an off switch, as they are fired and naturally commutated at the same time, hence they are never on.
Hence, as all the four SCRs are always off, the output voltage is zero.
6. For obtaining the output voltage control from V to 2V, the firing angle must be
power-electronics-questions-answers-sequence-controller-q4
a) always 0 for T3, T4
b) always 180 for T3, T4
c) 0 to 180 from T3, T4
d) none of the mentioned
Answer: a
Explanation: For obtaining the voltage control from V to 2V, the lower controller i.e. T3, T4 pair must always be on hence the firing angle from them must always be 0° ideally.
7. For obtaining the output voltage control from V to 2V, the firing angle must be
power-electronics-questions-answers-sequence-controller-q4
a) always 0 for T1, T2
b) always 180 for T1, T2
c) 0 to 180 from T1, T2
d) none of the mentioned
Answer: c
Explanation: For obtaining the voltage control from V to 2V, the upper controller must be controlled by varying its firing angle from 0 to 180° whereas the T3, T4 pair must always be on hence the firing angle from them must always be 0° ideally.
8. A single-phase two stage sequence controller is designed to work on 230 V supply, and upper and lower current ratings must be 20 A and 21 A respectively. Find the transformer rating.
a) 230 VA
b) 4600 VA
c) 9430 VA
d) 9200 VA
Answer: c
Explanation: Transformer rating is Vs.
9. In a N-stage sequence controller, each secondary is rated for __________
a) n x Vs
b) Vs
c) Vs/n
d) Vs x
Answer: c
Explanation: In a N-stage sequence controller, n voltage controllers are used each having to anti parallel SCR pairs and each secondary is rated for Vs/n.
10. A single-phase sinusoidal voltage controller has
a) one primary and n secondary windings
b) one primary and secondary windings
c) n primary and n secondary windings
d) primary and n secondary windings
Answer: b
Explanation: A sinusoidal voltage controller is used to obtain continuous voltage control over wide range with low harmonic content.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Cycloconverters”.
1. A cycloconverter is a _________
a) one stage power converter
b) one stage voltage converter
c) one stage frequency converter
d) none of the mentioned
Answer: c
Explanation: A cycloconverter converters input power at one frequency to output power at a different frequency with one-stage conversion.
2. Applications of cycloconverters include
a) speed control of ac drives
b) induction heating
c) static VAr compensation
d) all of the mentioned
Answer: d
Explanation: Speed of induction machines can be controlled by controlling the input frequency. In induction heating, eddy current is proportional to the square of input frequency.
3. The single phase mid-point type cycloconverter uses __________ number of SCRs.
a) 4
b) 8
c) 6
d) none of the mentioned
Answer: a
Explanation: 2 negative and 2 positive SCRs are employed in mid-point type cycloconverter.
4. The single phase bridge type cycloconverter uses __________ number of SCRs.
a) 4
b) 8
c) 6
d) none of the mentioned
Answer: b
Explanation: 4 negative and 4 positive SCRs are employed in bridge type cycloconverter.
5. In the positive half cycle from ωt = 0 to π
power-electronics-questions-answers-cycloconverters-q5
a) P1 and P2 are forward biased
b) N1 and P2 are forward biased
c) P1 and N2 are forward biased
d) None of the mentioned
Answer: c
Explanation: In the positive half cycle, upper terminal is positive of the upper secondary and the lower terminal is negative for the lower secondary. Hence, P1 and N2 are forward biased.
6. In the below given cycloconverter circuit, _________ and _________ conduct in one cycle together.
power-electronics-questions-answers-cycloconverters-q5
a) P1, P2 and N1, N2
b) P1, N2 and N1, P2
c) N1, P1 and N1, P2
d) None of the mentioned
Answer: b
Explanation: In the positive half cycle, upper terminal is positive of the upper secondary and the lower terminal is negative for the lower secondary. Hence, P1 and N2 are forward biased. And likewise, N1 and P2 will conduct in the next half cycle.
7. In the positive half cycle, _________ SCRs are forward biased.
power-electronics-questions-answers-cycloconverters-q7
a) P1, P2, N1, N2
b) P1, P2, P3, P4
c) N1, N2, N3, N4
d) P3, P4, N1, N2
Answer: a
Explanation: Pairs P1, P2 and N1, N2 are forward biased from ωt = 0 to π.
8. The principle of three phase cycloconverter is to
a) add and remove number of SCRs
b) vary progressively the firing angle of the devices
c) keep the firing angle as 0° for all the devices
d) none of the mentioned
Answer: b
Explanation: The basic principle of three phase cycloconverter is to vary progressively the firing angle of the controlling devices.
9. In three phase cycloconverters, the reduction factor is given by
a) input frequency/output frequency
b) -1
c) -1/2
d) 1/2
Answer: b
Explanation: Reduction factor = output frequency/input frequency.
10. In a three phase half-wave cycloconverter ___________
a) both inverting and converting action takes place
b) only inversion action takes place
c) only converting action takes place
d) none of the mentioned
Answer: a
Explanation: During the conversion process, the current flows in both the directions, hence both inverting and converting action takes place.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Applications of Power Electronics-1”.
1. SMPS is used for
a) obtaining controlled ac power supply
b) obtaining controlled dc power supply
c) storage of dc power
d) switch from one source to another
Answer: b
Explanation: SMPS is used for obtaining controlled dc power supply.
2. SPMS are based on the ________ principle.
a) Phase control
b) Integral control
c) Chopper
d) MOSFET
Answer: c
Explanation: SMPS are based on the chopper principle. The output dc voltage is controlled by varying the duty cycle of the chopper circuit.
3. Choose the incorrect statement.
a) SMPS is less sensitive to input voltage variations
b) SMPS is smaller as compared to rectifiers
c) SMPS has low input ripple
d) SMPS is a source of radio interference
Answer: c
Explanation: SMPS has higher output ripple and its regulation is worse.
4. _________ is used for critical loads where temporary power failure can cause a great deal of inconvenience.
a) SMPS
b) UPS
c) MPS
d) RCCB
Answer: b
Explanation: Uninterruptible Power Supply is used where loads where temporary power failure can cause a great deal of inconvenience.
5. __________ is used in the rotating type UPS system to supply the mains.
a) DC motor
b) Self excited DC generator
c) Alternator
d) Battery bank
Answer: c
Explanation: When the supply is gone, the diesel engine is started, which runs the alternator and the alternator supplies power to the mains. Non-rotating type UPS are not used anymore.
6. Static UPS requires __________
a) only rectifier
b) only inverter
c) both inverter and rectifier
d) none of the mentioned
Answer: c
Explanation: Rectifier to converter the dc from the battery to ac. Inverter to charge the battery from mains.
7. No discontinuity is observed in case of
a) short break static UPS configuration
b) long break static UPS configuration
c) no break static UPS configuration
d) rotating type UPS configuration
Answer: c
Explanation: No dip or discontinuity is observed in case of no break static UPS configuration, as the battery inverter set immediately takes over the mains.
8. Usually __________ batteries are used in the UPS systems.
a) NC
b) Li-On
c) Lead acid
d) All of the mentioned
Answer: c
Explanation: Lead acid batteries are cheaper and have certain advantages over the other types. NC batteries would however be the best, but are three to four times more expensive than Lead Acid.
9. HVDC transmission has ___________ as compared to HVAC transmission.
a) smaller transformer size
b) smaller conductor size
c) higher corona loss
d) smaller power transfer capabilities
Answer: b
Explanation: The conductor size is smaller as there is no sink effect, and the whole conductor is utilized for transmitting power.
10. The negative polarity is used in the monopolar link because it
a) uses less conductor size
b) is safer
c) produces less radio interference
d) has less resistance
Answer: c
Explanation: The monopolar link uses just a single conductor, which is usually negative as it produces less radio interference and corona.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Applications of Power Electronics-2”.
1. HVDC transmission lines are __________ as compared to HVAC lines.
a) difficult to erect
b) more expensive for long distances
c) more expensive for short distances
d) less expensive for short distances
Answer: c
Explanation: HVDC lines require additional cost of the converting and inverting equipments, hence they are more expensive for short distances as compared to HVAC lines.
2. In HVDC transmission lines
a) both the stations operate as an inverter
b) both the stations operate as a converter
c) one acts as a converter and other as an inverter
d) depends upon the type of the load
Answer: c
Explanation: The supply side station is the converting station, power is transmitted in dc then again converted back to ac at the inverting or receiving station further which it is distributed in ac to houses.
3. Two six pulse converters used for bipolar HVDC transmission system, are rated at 1000 MW, +- 200 kV. What is the dc transmission voltage?
a) 200 kV
b) 400 kV
c) 500 kV
d) 100 kV
Answer: b
Explanation: As the link is bipolar, the total voltage transmitted will 200 + 200 kV.
4. Two six pulse converters used for bipolar HVDC transmission system, are rated at 1000 MW, +- 200 kV. Find the dc current in the transmission line.
a) 500 A
b) 25 A
c) 2500 A
d) 5 A
Answer: c
Explanation: Transmission voltage = 200 + 200 = 400 kV.
I = P/V = 1000 mW/ 400 kV = 2500 A.
5. Two six pulse converters used for bipolar HVDC transmission system, are rated at 1000 MW, +- 200 kV. Find the rms current rating required for the SCRs.
a) 2500 A
b) 1350 A
c) 1445 A
d) none of the mentioned
Answer: c
Explanation: Transmission voltage = 200 + 200 = 400 kV.
I = P/V = 1000 mW/ 400 kV = 2500 A.
As in 3-phase full converter each SCR conducts for 120° for a total period of 360°.
Irms = 2500 x √ = 1443.4 A. Nearly about 1445 A as it is a standard current rating.
6. For high power applications _________ are used as static switches whereas for low power applications __________ are used.
a) Transistors, SCRs
b) SCRs, transistors
c) Diodes, transistors
d) SCRs, diodes
Answer: b
Explanation: As SCR are of higher rating they are preferred in high power applications.
7. _________ can be used as a single phase static ac switch.
a) Diode
b) SCR
c) DIAC
d) TRAIC
Answer: d
Explanation: SCR cannot be used, as it is unidirectional. Diode isn’t a switch nor is the DIAC.
8. ___________ can be used as a dc static switch.
a) GTO
b) Transistor
c) Both GTO and transistor
d) TRIAC
Answer: c
Explanation: Both GTO and transistor can be used as a dc static switch.
9. A single-phase ac switch is used in between a 230 V source and load of 2 kW and 0.8 lagging power factor. Determine the rms current rating required by the SCR. Use the factor of safety = 2.
a) 10.87 A
b) 87 A
c) 21.74 A
d) 32 A
Answer: c
Explanation: I = / = 15.37 A.
I = 15.37/1.414 = 10.87
Required rating = 10.87 x factor of safety = 15.37 x 2 = 21.74 A.
10. Solid State Relays have
a) moving parts
b) no moving parts
c) a coil
d) a contactor
Answer: b
Explanation: SSRs have no moving parts, they simply consist of a LED and a transistor or photo diode.
This set of Power Electronics Multiple Choice Questions & Answers focuses on “Miscellaneous”.
1. Solid State Relays have a
a) coil and contact arrangement
b) optocoupler
c) scr
d) none of the mentioned
Answer: b
Explanation: Coil and contact arrangement is used in mechanical relays, SSRs have a optocoupler which connects the control circuit to the power circuit via light sensitive devices.
2. The converter circuit which employs turn on and turn off when the voltage and/or current through the device is zero at the instant of switching is ____________
a) a conventional converter
b) a resonant converter
c) a zero switching circuit
d) none of the mentioned
Answer: b
Explanation: Resonant converters are used to turn on and turn off when the voltage and/or current through the device is zero at the instant of switching.
3. Induction heating is a ___________ type of heating
a) zero frequency
b) high frequency
c) power frequency
d) none of the mentioned
Answer: b
Explanation: As eddy current is proportional to the square of the supply frequency, induction heating is a high frequency heating.
4. The factors governing the induction heating are
a) resistivity
b) relative permeability
c) magnetic field intensity
d) all of the mentioned
Answer: d
Explanation: Induction heating depends on all of the above given factors.
5. The reverse recovery time of a diode is t rr = 3 μs and the rate of fall of the diode current = 30 A/μs. Determine the storage charge.
a) 145 μs
b) 135 μs
c) 0
d) none of the mentioned
Answer: b
Explanation: Storage charge = x x (t rr ) 2 = 135 μs.
6. For a SCR, conduction angle is 120° when average on-state current is 20 A. When the conduction angle is halved the earlier value, the on-state average current will be?
a) 5 A
b) 40 A
c) 10 A
d) 20 A
Answer: b
Explanation: When the conduction angle is halved, the device will conduct twice then it was conducting earlier. Hence, I = 2x 20 = 40 A.
7. A single-phase full bridge diode rectifier delivers power to a constant load current of 10 A. The average and rms values of the source currents will be respectively.
a) 5 A, 10 A
b) 10 A, 10 A
c) 5 A, 5 A
d) 10 A, 5A
Answer: b
Explanation: As the load current is continuous, Iavg = Irms = 10 A.
8. TRIAC is used in
a) chopper
b) speed control of induction machine
c) speed control of universal motor
d) none of the mentioned
Answer: c
Explanation: TRIAC is used in speed control of universal motor.
9. The ratio V rms / V dc is known as
a) Form factor
b) Ripple factor
c) Utilization factor
d) None of the mentioned
Answer: a
Explanation: V rms / V dc = FF.
10. Determine the loss in the snubber circuit, if C = 0.545 μF and supply is 200 V, 10 kHz.
a) 233 W
b) 133 W
c) 333 W
d) 233 W
Answer: b
Explanation: Snubber loss Ps = x C x V 2 x f = 133.1 W.
