Electric Drives Pune University MCQs

Electric Drives Pune UniversityMCQs

Electric Drives Pune University MCQs

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Dynamics – Types of Loads”.

1. Load torques can be classified into how many types?

a) Three

b) Two

c) Four

d) Five

Answer: b

Explanation: Load torques can have two types. They are active and passive load torques. Active load torques are able to run the motor under equilibrium conditions and their sign remains the same even if the motor rotation changes but passive load torques always opposes the motion by changing their sign with the change in rotation of the motor.

2. Rolling mills exhibit what type of load torque characteristics?

a) Constant torque characteristics

b) Linearly rising torque characteristics

c) Non-Linearly rising torque characteristics

d) Non-Linearly decreasing torque characteristics

Answer: d

Explanation: Rolling mills are an example of non-linearly decreasing torque characteristics because torque and speed exhibits inversely proportional relationships and power are constant.

3. What is the relationship between torque and speed in constant type loads?

a) Torque is independent of speed

b) Torque linearly increases with increase in speed

c) Torque non-linearly increases with an increase in speed

d) Torque non-linearly decreases with an increase in speed

Answer: a

Explanation: Speed hoist is a perfect example of constant type loads in which torque variation is independent of speed. The speed-torque characteristics of this type of load are given by T=K where K is a constant.

4. Torque inversely varies with the speed in the windage load torque component.

a) True

b) False

Answer: b

Explanation: Torque varies with a square of speed in the windage load torque component whereas in coulomb torque component torque is constant.

5. What type of force handles for active torques?

a) Strong nuclear forces

b) Weak nuclear forces

c) Gravitational forces

d) Electrostatic forces

Answer: c

Explanation: Gravitational forces are responsible for active torques. Active torques due to gravitational forces can be obtained in the case of hoists, lifts or elevators and railway locomotives operating on gradients.

6. Passive torques always oppose the motion of the driven machine.

a) True

b) False

Answer: a

Explanation: Passive torques are due to friction or shear and deformation in elastic bodies. They always oppose the motion, restricting the motion of the machine.

7. Among the following which one exhibits linearly rising load torque characteristics?

a) Elevators

b) Rolling Mills

c) Fan load

d) Separately excited dc generator connected to the resistive load

Answer: d

Explanation: Separately excited dc generator connected to the resistive load is an example of linearly rising load torque characteristics as the torque increases linearly with an increase in speed.

8. What is the condition for the steady-state operation of the motor?

a) Load torque > Motor torque

b) Load torque <<<< Motor torque

c) Load torque = Motor torque

d) Load torque < Motor torque

Answer: c

Explanation: According to the dynamic equation of motor, load torque must be equal to motor torque so that motor should run at a uniform speed. If load torque is greater than motor torque, the motor will fail to start and if load torque is less than motor torque, the motor will run at a higher speed which can damage the shaft of the motor.

9. Choose the correct one. .

a) J*d/dt = Load torque – Motor torque

b) J*d/dt = Load torque + Motor torque

c) J*d/dt = Motor torque – Load torque

d) J*d/dt = Load torque * Motor torque

Answer: c

Explanation: J*d/dt = Motor torque – Load torque is the dynamic equation of the motor. Motor torque will try to aid the motion of the motor, but load torque will oppose the motion of motor that’s why it subtracts in the equation.

This set of Electric Drives Interview Questions and Answers focuses on “Dynamics – Quadrantal Diagram of Speed – Torque Characteristics “.

1. Regenerative braking mode can be achieved in which quadrant ?

a) Third

b) Second

c) Fourth

d) First

Answer: b

Explanation: Regenerative braking is only available in second quadrant as power from motor is fed back to source. Back emf generated is more than armature terminal so it works as a generator.

2. Fan type of loads exhibits which type of load torque characteristics?

a) Constant torque characteristics

b) Linearly rising torque characteristics

c) Non-Linearly rising torque characteristics

d) Non-Linearly decreasing torque characteristics

Answer: c

Explanation: Torque produced by the fan is directly proportional to square of speed throughout the range of usable fan speeds. This type of loads exhibits non-linearly rising torque characteristics.

3. Type-A chopper is used for obtaining which type of mode?

a) Motoring mode

b) Regenerative braking mode

c) Reverse motoring mode

d) Reverse regenerative braking mode

Answer: a

Explanation: Only motoring mode is available in case of step-down chopper . Value of output voltage is less than the input voltage in case of step-down chopper.

4. Calculate the value of angular acceleration of motor using the given data: J = 20 kg-m 2 , load torque = 20 N-m, motor torque = 60 N-m.

a) 5 rad/s 2

b) 2 rad/s 2

c) 3 rad/s 2

d) 4 rad/s 2

Answer: b

Explanation: Using the dynamic equation of motor J* = Motor torque – Load torque: 20* = 60-20=40, angular acceleration=2 rad/s 2 .

5. 230V, 10A, 1500rpm DC separately excited motor having resistance of .2 ohm excited from external dc voltage source of 50V. Calculate the torque developed by the motor on full load.

a) 13.89 N-m

b) 14.52 N-m

c) 13.37 N-m

d) 14.42 N-m

Answer: b

Explanation: Back emf developed in the motor during full load can be calculated using equation Eb = Vt- I*Ra = 228 V and machine constant Km = Eb / Wm which is equal to 1.452. Torque can be calculated by using the relation T = Km* I = 1.452*10 = 14.52 N-m.

6. Boost converter is used to _________

a) Step down the voltage

b) Step up the voltage

c) Equalize the voltage

d) Step up and step down the voltage

Answer: b

Explanation: Output voltage of boost converter is Vo = Vin / 1 – D. Value of duty cycle is less than 1 which makes the Vo > Vin as denominator value decreases and becomes less than 1. Boost converter is used to step up voltage.

7. Reverse motoring mode is available in fourth quadrant.

a) True

b) False

Answer: a

Explanation: In reverse motoring mode motor rotates in opposite to original direction as direction of motor torque changes which makes the motor to run in opposite direction and load torque tries to oppose the motion of motor.

8. Calculate the power developed by motor using the given data: Eb = 20V and I = 10 A.

a) 400 W

b) 200 W

c) 300 W

d) 500 W

Answer: b

Explanation: Power developed by motor can be calculated using the formula P = Eb*I = 20*10 = 200 W. If rotational losses are neglected, power developed becomes equal to the shaft power of motor.

9. Which one is an example of variable loss?

a) Windage loss

b) Hysteresis loss

c) Armature copper loss

d) Friction loss

Answer: c

Explanation: Armature copper losses are variable losses as they depend on armature current which further depends on load. As load changes armature current changes hence armature copper losses (I 2 * r) also changes.

This set of Electric Drives Assessment Questions and Answers focuses on “Dynamics – Load Torques that Depend on the Path or Position Taken by the Load During Motion”.

1. What is the empirical formula for the tractive force required to overcome curve resistance?

a) 710×W÷R

b) 700×W÷R

c) 720×W÷R

d) 750×W÷R

Answer: b

Explanation: F c = 700×W÷R is the tractive force required to overcome curve resistance where W is the weight of the body in Kg and R be the radius of curvature in meters.

2. Force resisting the upward motion of a body on an inclined plane is given by .

a) F = W×sin

b) F = W×cosec

c) F = W×sec

d) F = W×cos

Answer: a

Explanation: When a body is moving upward on an inclined plane its weight can be resolved in two perpendicular components that are W×sin and W×cos. W×cos is the component that is opposite to normal of the inclined plane and W×sin is the component that opposes the upward motion of the body.

3. The unit of the torque is ______

a) N-m

b) N-m 2

c) N-m/sec

d) N-Hz

Answer: a

Explanation: Torque is defined as the vector product of displacement and force. The unit of the force is Newton and of the displacement is a meter so the unit of torque in N-m.

4. Calculate the value of the torque when 10 N force is applied perpendicular to a 10 m length of rod fixed at the center.

a) 200 N-m

b) 300 N-m

c) 100 N-m

d) 400 N-m

Answer: b

Explanation: Torque can be calculated using the relation T = × = r×F×sin90. F is given as 10 N and r is 10 m then torque is 10×10 = 100 N-m.

5. What is the dimensional formula for torque?

a) [ML 2 T -2 ]

b) [MLT -2 ]

c) [M 1 L 2 T -3 ]

d) [LT -2 ]

Answer: a

Explanation: Torque is a vector product of force and displacement. Dimensional formula for force is [MLT -2 ] and displacement is [L] so dimensional formula for torque is [MLT -2 ] [L] = [ML 2 T -2 ].

6. Buck converter is used to _________

a) Step down the voltage

b) Step up the voltage

c) Equalize the voltage

d) Step up and step down the voltage

Answer: a

Explanation: The output voltage of the buck converter is V o = D×V in . The value of the duty cycle is less than one which makes the V o < V in . The buck converter is used to step down voltage. V in is a fixed voltage and V o is a variable voltage.

7. If the starting torque of the motor is less than the load torque, the motor will fail to start.

a) True

b) False

Answer: a

Explanation: J×d÷d = Motor torque – Load torque is the dynamic equation of motor. If starting torque is less than the load torque then d÷d <0, acceleration <0 so the motor will decelerate and fails to start.

8. Torque is a scalar quantity.

a) True

b) False

Answer: b

Explanation: Scalar quantity has only magnitude whereas vector quantity has both directions and magnitude. Torque is a force applied on a body perpendicularly. As the force is a vector quantity, the torque must be treated as a vector quantity.

9. 250V, 15A, 1100 rpm separately excited dc motor with armature resistance (R a ) equal to 2 ohms. Calculate back emf developed in the motor when it operates on half of the full load.

a) 210V

b) 240V

c) 230V

d) 235V

Answer: d

Explanation: Back emf developed in the motor can be calculated using the relation E b = V t -I×R a . In question, it is asking for half load, but the data is given for full load so current becomes half of the full load current = 15÷2 = 7.5 A. 250V is terminal voltage it is fixed so E b = 250-7.5×2 = 235V.

This set of Tough Electric Drives Questions and Answers focuses on “Dynamics – Load Torques that Vary with Angle of Displacement of the Shaft”.

1. Duty cycle is _______

a) Ton÷Toff

b) Ton÷

c) Ton÷2×

d) Ton÷2×Toff

Answer: b

Explanation: Duty cycle is defined as the ratio of time for which system is active to the total time period. It is also known as the power cycle. It has no unit.

2. A 220 V, 1000 rpm, 60 A separately-excited dc motor has an armature resistance of .5 ω. It is fed from single-phase full converter with an ac source voltage of 230 V, 50Hz. Assuming continuous conduction, the firing angle for rated motor torque at rpm is _________

a) 122.4°

b) 117.6°

c) 130.1°

d) 102.8°

Answer: d

Explanation: During rated operating conditions of the motor, E b = V t -Ia×Ra = 220-60×.5=190 V. As E b =K m w m so K m =190×60÷ = 1.8152 V-s/rad. Back emf at is K m w m = 1.8152×)÷60 = -76 V. Now V t = -76+60×.5 = -46 V. Average voltage of single-phase full converter is 2×V m ×cos÷3.14. The output of the converter is connected to the input terminal of the motor so α = cos -1 = 102.8 o .

3. The unit of angular acceleration is rad/s 2 .

a) True

b) False

Answer: a

Explanation: Angular acceleration is defined as a derivate of angular velocity with respect to time. It is generally written as α. The unit of angular velocity is rad/sec and of time is second so the unit of angular acceleration is rad/s 2 .

4. Calculate the value of the angular acceleration of the motor using the given data: J= 50 kg-m 2 , load torque = 40 N-m, motor torque = 10 N-m.

a) -.7 rad/s 2

b) -.6 rad/s 2

c) -.3 rad/s 2

d) -.4 rad/s 2

Answer: b

Explanation: Using the dynamic equation of motor J* = Motor torque – Load torque: 50* = 10-40 = -30, angular acceleration=-.6 rad/s 2 . The motor will decelerate and will fail to start.

5. The principle of step-up chopper can be employed for the ________

a) Motoring mode

b) Regenerative mode

c) Plugging

d) Reverse motoring mode

Answer: b

Explanation: The step-down chopper is used in motoring mode but a step-up chopper can operate only braking mode because the characteristics are in the second quadrant only.

6. A Buck-Boost converter is used to _________

a) Step down the voltage

b) Step up the voltage

c) Equalize the voltage

d) Step up and step down the voltage

Answer: d

Explanation: The output voltage of the buck-boost converter is V o = D×V in ÷ . It can step up and step down the voltage depending upon the value of the duty cycle. If the value of the duty cycle is less than .5 it will work as a buck converter and for duty cycle greater than .5 it will work as a boost converter.

7. Which of the following converter circuit operations will be unstable for a large duty cycle ratio?

a) Buck converter

b) Boost converter

c) Buck-Boost converter

d) Boost converter and Buck-Boost converter

Answer: d

Explanation: The output voltage of the buck converter and buck-boost converter are V o =V in ÷ and V o = D×V in ÷ respectively. When the value of the duty cycle tends to 1 output voltage tends to infinity.

8. Calculate the shaft power developed by a motor using the given data: E b = 50V and I = 60 A. Assume frictional losses are 400 W and windage losses are 600 W.

a) 4000 W

b) 2000 W

c) 1000 W

d) 1500 W

Answer: b

Explanation: Shaft power developed by the motor can be calculated using the formula P = E b *I- = 50*60 = 3000 – = 2000 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

9. Which one of the following devices have low power losses?

a) MOSFET

b) IGBT

c) SCR

d) BJT

Answer: c

Explanation: SCR is a minority carrier device due to which it experiences conductivity modulation and it’s ON state resistance reduction due to which conduction losses are very low.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Dynamics – Load Torques that Vary with Time”.

1. Servo motors are an example of which type of load?

a) Pulsating loads

b) Short time loads

c) Impact loads

d) Short time intermittent loads

Answer: b

Explanation: Servo motors are motors with control feedback. The motor can be AC or DC. This is an example of short time loads. They have a high torque to inertia ratio and high-speed.

2. Load torque of the crane is independent of _________

a) Speed

b) Seebeck effect

c) Hall effect

d) Thomson effect

Answer: a

Explanation: The Load torque of the crane is independent of speed. They are short time intermittent types of loads. They require constant power for a short period of time.

3. The unit of angular velocity is rad/s 3 .

a) True

b) False

Answer: b

Explanation: Angular velocity is defined as the rate of change of angular displacement with respect to time. Angular displacement is generally expressed in terms of a radian. The unit of angular velocity is rad/s.

4. R.M.S value of the sinusoidal waveform V=V m sin.

a) V m ÷2 ½

b) V m ÷2 ¼

c) V m ÷2 ¾

d) V m ÷3 ½

Answer: a

Explanation: R.M.S value of the sinusoidal waveform is V m ÷2 ½ and r.m.s value of the trapezoidal waveform is V m ÷3 ½ . The peak value of the sinusoidal waveform is V m .

5. Calculate the time period of the waveform x=24sin.

a) .064 sec

b) .047 sec

c) .083 sec

d) .015 sec

Answer: c

Explanation: The fundamental time period of the sine wave is 2π. The time period of x is 2π÷24π=.083 sec. The time period is independent of phase shifting and time shifting.

6. The turn-off times of the devices in the increasing order is ___________

I. MOSFET

II. BJT

III. IGBT

IV. Thyristor

a) I, III, II, IV

b) I, II, III, IV

c) III, I, II, IV

d) III, II, IV, I

Answer: a

Explanation: Increasing turn-off time implies decreasing speed and majority carrier devices do not have any minority charge carrier storage so they have less turn-off time and hence MOSFET has the least turn off time. So, the increasing order of turn-off time is, MOSFET < IGBT < BJT < Thyristor.

7. Which of the following devices should be used as a switch for high power and high voltage application?

a) GTO

b) MOSFET

c) TRIAC

d) Thyristor

Answer: d

Explanation: Thyristor is used for high power applications but it has a limited frequency range and cannot be used at high frequencies. A thyristor is a unidirectional, bipolar and semi-controlled device.

8. Calculate the useful power developed by a motor using the given data: P in = 3000 W, I a = 60 A, R a = .4 Ω. Assume frictional losses are 200 W and windage losses are 400 W.

a) 970 W

b) 960 W

c) 980 W

d) 990 W

Answer: b

Explanation: Useful power is basically the shaft power developed by the motor that can be calculated using the formula P sh = P dev -. P dev = P in -I a 2 R a = 3000-60 2 =1560 W. The useful power developed by the motor is P sh = P dev -=1560 –=960 W.

9. Calculate the phase angle of the sinusoidal waveform y=55sin.

a) π÷8

b) π÷5

c) π÷7

d) π÷4

Answer: a

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Dynamics of Motor-Load Combination”.

1. The axis along which no emf is produced in the armature conductors is called as ____________

a) Geometrical Neutral Axis

b) Magnetic Neutral Axis

c) Axis of rotation

d) Axis of revolution

Answer: b

Explanation: The coil undergoing commutation must lie along the magnetic neutral axis so that no emf is induced and there is no sparking at the time of commutation.

2. The generated e.m.f from 25-pole armature having 200 conductors driven at 10 rev/sec having flux per pole as 20 mWb, with two parallel paths is ___________

a) 400 V

b) 500 V

c) 200 V

d) 300 V

Answer: b

Explanation: The generated can be calculated using the formula E b = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. E b = .02×25×200×600÷60×2= 500 V.

3. The unit of the flux is Weber.

a) True

b) False

Answer: a

Explanation: Flux is the total amount of magnetic field lines passing through a given area. Φ is a dot product of magnetic flux density and area. The unit of the flux is Weber .

4. Which of the following motor can be referred as a universal motor?

a) DC shunt motor

b) DC compound motor

c) Permanent magnet motor

d) DC series motor

Answer: d

Explanation: DC series motor can operate on DC and AC. It is a universal motor. Universal motors are those motors that can operate on both DC and AC. DC shunt motor can only operate on DC because of pulsating torque in AC.

5. The phase difference between voltage and current in the inductor.

a) 45°

b) 90°

c) 80°

d) 55°

Answer: b

Explanation: In the case of an inductor, the voltage leads the current by 90° or the current lags the voltage by 90o. The phase difference between voltage and current is 90°.

6. The phase difference between voltage and current in the resistor.

a) 85°

b) 90°

c) 0°

d) 5°

Answer: c

Explanation: In the case of a resistor, the voltage and current are in the same phase. The phase difference between voltage and current is 0°. The voltage drop in the resistor is given as V=IR.

7. The phase difference between voltage and current in the capacitor.

a) 90°

b) 80°

c) 95°

d) 91°

Answer: a

Explanation: In the case of a capacitor, the voltage lags the current by 90° or the current leads the voltage by 90 o . The phase difference between voltage and current is 90°.

8. The slope of the I-V curve is 30°. Calculate the value of resistance. Assume the relationship between I and V is a straight line.

a) 1.732 Ω

b) 2.235 Ω

c) 1.625 Ω

d) 1.524 Ω

Answer: a

Explanation: The slope of the I-V curve is reciprocal of resistance. The slope given is 30° so R=1÷tan=1.732 Ω. The slope of the V-I curve is resistance.

9. What is a mark to space ratio?

a) Ton÷Toff

b) Ton÷

c) Ton÷2×

d) Ton÷2×Toff

Answer: a

Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. It has no unit.

This set of Electric Drives Questions and Answers for Freshers focuses on “Dynamics – Moment of Inertia Determination”.

1. What is the formula for the moment of inertia?

a) ∑m i r i 2

b) ∑m i r i

c) ∑m i r i 4

d) ∑m i r i 3

Answer: a

Explanation: The moment of inertia is the property by the virtue of which the body withstand the effect of angular acceleration. It depends on the shape and mass distribution of the body.

2. The generated e.m.f from 50-pole armature having 400 conductors driven at 20 rev/sec having flux per pole as 30 mWb, with lap winding is ___________

a) 230 V

b) 140 V

c) 240 V

d) 250 V

Answer: c

3. The unit of the moment of inertia is Kgm 2 .

a) True

b) False

Answer: a

Explanation: The moment of inertia is taken as the sum of the product of the mass of each particle with the square of their distance from the axis of the rotation. The unit of the moment of inertia is kg×m 2 =kgm 2 .

4. Calculate the moment of inertia of the egg having a mass of 7 kg and radius of 44 cm.

a) .968 kgm 2

b) 1.454 kgm 2

c) 1.545 kgm 2

d) 1.552 kgm 2

Answer: d

Explanation: The moment of inertia of the egg can be calculated using the formula I=∑m i r i 2 . The mass of egg and radius is given. I=× 2 =1.552 kgm 2 . It depends upon the orientation of the rotational axis.

5. Which of the theorems helps in the calculation of the moment of inertia?

a) The theorem of Parallel and Perpendicular axes

b) The theorem of Horizontal and Perpendicular axes

c) The theorem of Vertical and Perpendicular axes

d) The theorem of Parallel and Tilted axes

Answer: a

Explanation: The theorem of Parallel and Perpendicular axes helps in the calculation of the moment of inertia. The moment of inertia of the complex bodies can be easily calculated with the help of these theorems.

6. What is the unit of resistance?

a) ohm

b) ohm -1

c) ohm 2

d) ohm 5

Answer: a

Explanation: The resistance is the opposition offered by the body to the flow of current. It is the ratio of voltage and current. It is given in ohms.

7. Calculate the value of the frequency if the time period of the signal is 20 sec.

a) 0.05 Hz

b) 0.04 Hz

c) 0.02 Hz

d) 0.03 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. F = 1÷T=1÷20=.05 Hz.

8. The slope of the V-I curve is 60°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 1.732 Ω

b) 1.608 Ω

c) 1.543 Ω

d) 1.648 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 60° so R=tan=1.732 Ω. The slope of the V-I curve is resistance.

9. Calculate mark to space ratio if the system is on for 5 sec and off for 10 sec.

a) .5

b) .4

c) .2

d) .6

Answer: c

Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=5÷10=2.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Dynamics – Steady State Stability of Electric Drive”.

1. A 4-pole lap wound generator with 720 armature conductors and a flux per pole of .003 Wb has an armature current of 50 A. The developed torque is _________

a) 17.25 N-m

b) 17.19 N-m

c) 16.54 N-m

d) 16.89 N-m

Answer: b

Explanation: The developed torque in the motor is K m ×I. The value of machine constant(K m ) is E b ÷ω m = Φ×Z×P÷2×3.14×A = .003×720×4÷2×3.14×4 = .3438 Vs/rad. The developed torque is .3438×50 = 17.19 N-m.

2. The generated e.m.f from 70-pole armature having 100 conductors driven at 10 rev/sec having flux per pole as 20 mWb, with wave winding is ___________

a) 730 V

b) 740 V

c) 700 V

d) 690 V

Answer: c

Explanation: The generated e.m.f can be calculated using the formula E b = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. In wave winding number of parallel paths are 2. E b = .02×70×100×600÷60×2 = 700 V.

3. The unit of current is Tesla.

a) True

b) False

Answer: b

Explanation: The current is the amount of charge that can flow through an area in a given amount of time. It is mathematically represented as d/d. It is expressed in terms of Ampere.

4. Calculate the moment of inertia of the sphere having a mass of 12 kg and radius of 78 cm.

a) 7.888 kgm 2

b) 7.300 kgm 2

c) 7.545 kgm 2

d) 7.552 kgm 2

Answer: b

Explanation: The moment of inertia of the egg can be calculated using the formula I=Σm i r i 2 . The mass of egg and radius is given. I=× 2 =7.300 kgm 2 . It depends upon the orientation of the rotational axis.

5. The most suitable servo-motor application is __________

a) AC series motor

b) DC series motor

c) AC two-phase induction motor

d) DC shunt motor

Answer: d

Explanation: DC shunt motor has definite no-load speed, so they don’t ‘run away’ when the load is suddenly thrown off provided the field circuit remains closed. The speed for any load within the operating range of the motor can be readily obtained.

6. In a DC series motor, the electromagnetic torque developed is proportional to ______

a) I a

b) I a 2

c) I a 3

d) I a .5

Answer: b

Explanation: In a DC series motor, the electromagnetic torque developed is equal to K m ΦI a . In a DC series, the motor field winding is connected in series with the armature so the flux in the field winding is proportional to current. T = K m ΦI a α I a 2 .

7. Calculate the value of the time period if the frequency of the signal is 70 sec.

a) 0.014 sec

b) 0.013 sec

c) 0.017 sec

d) 0.079 sec

Answer: a

Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷70=.014 sec.

8. The slope of the V-I curve is 90°. Calculate the value of resistance.

a) 1.732 Ω

b) 1.608 Ω

c) 1.543 Ω

d) 1.648 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 90° so R=tan=infinite Ω. It behaves as an open-circuit.

9. In a DC shunt motor, the electromagnetic torque developed is proportional to ______

a) I a

b) I a 2

c) I a 3

d) I a .5

Answer: a

Explanation: In a DC shunt motor, the electromagnetic torque developed is equal to K m ΦI a . In a DC shunt, the motor field windings are connected separately and excited by a constant DC voltage. T = K m ΦI a α I a .

This set of Tricky Electric Drives Questions and Answers focuses on “Dynamics – Transient Stability of an Electric Drive”.

1. What is the formula for the active power in the cylindrical rotor synchronous machine? (E b represents armature emf, V t represents terminal voltage, δ represents rotor angle, X represents reactance)

a) E b ×V t ×sinδ÷X

b) E b ×V t 2 ×sinδ÷X

c) E b 2 ×V t ×sinδ÷X

d) E b ×V t ×sinδ÷X2

Answer: a

Explanation: The real power in the cylindrical rotor machine is E b ×V t ×sinδ÷X. It is inversely proportional to the reactance. The stability of the machine is decided by the maximum power transfer capability.

2. Salient pole machines are more stable than cylindrical rotor machines.

a) True

b) False

Answer: a

Explanation: Salient pole machines are more stable than cylindrical rotor machines because of the high short circuit ratio and more real power transfer capability. The air gap length in salient pole machines is more as compare to cylindrical rotor machines.

3. The unit of reactive power is VAR.

a) True

b) False

Answer: a

Explanation: The reactive power is the useless power in case of electric circuits. It is the energy trapped that keeps on oscillating between inductive and capacitive element. It plays a vital role in generating flux in electrical machines. It is expressed in Volt Ampere reactive.

4. Calculate the power factor during the resonance condition.

a) .58

b) .42

c) .65

d) 1

Answer: d

Explanation: During the resonance condition, the reactive power generated by the capacitor is completely absorbed by the inductor. Only active power flows in the circuit. Net reactive power is equal to zero and cosΦ=1.

5. Calculate the reactive power in a 5 Ω resistor.

a) 7 VAR

b) 0 VAR

c) 2 VAR

d) 1 VAR

Answer: b

Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. Q = VIsin0 = 0 VAR.

6. What is the unit of the apparent or complex power?

a) VAR

b) VA

c) ohm

d) Volt

Answer: b

Explanation: The apparent power in AC circuits is VI*. It is expressed in volt-amperes . It consists of both active and reactive power. It is the vector sum of the real power and reactive power.

7. Calculate the value of the frequency of the DC supply.

a) 0 Hz

b) 50 Hz

c) 20 Hz

d) 10 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. DC supply magnitude is constant. It does not change with time so the frequency of DC supply is 0 Hz.

8. The slope of the V-I curve is 0°. Calculate the value of resistance. The graph is parallel to the x-axis.

a) 1 Ω

b) 1.8 Ω

c) 0 Ω

d) 2.2 Ω

Answer: c

Explanation: The slope of the V-I curve is resistance. The slope given is 0° so R=tan=0 Ω. The slope of the V-I curve is resistance. It behaves as a short circuit.

9. Calculate the value of the duty cycle if the system is on for 5 sec and off for 10 sec.

a) .333

b) .444

c) .201

d) .642

Answer: a

Explanation: Duty cycle is Ton÷T total . It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D= Ton÷T total =5÷15=.333.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “DC Motors Characteristics – Basic Relations”.

1. Swinburne’s test can be conducted on ___________

a) Series motor

b) Shunt motor

c) Compound motor

d) Shunt and compound motor

Answer: d

Explanation: The test is practically applicable for machines which have flux constant like the shunt and compound machines as this is a no-load test and DC series motor should not be run at no-load because of high speed.

2. The generated e.m.f from 20-pole armature having 800 conductors driven at 30 rev/sec having flux per pole as 60 mWb, with 16 parallel paths is ___________

a) 1900 V

b) 1840 V

c) 1700 V

d) 1800 V

Answer: d

3. The unit of active power is Watt.

a) True

b) False

Answer: a

Explanation: The active power in the electrical circuits is a useful power. It determines the power factor of the system. It is expressed in terms of Watt. P=VIcosΦ.

4. Calculate the mass of the ball having a moment of inertia 4.5 kgm 2 and radius of 14 cm.

a) 229.59 kg

b) 228.56 kg

c) 228.54 kg

d) 227.52 kg

Answer: a

Explanation: The moment of inertia of the ball can be calculated using the formula I=∑m i r i 2 . The moment of inertia of ball and radius is given. M=÷ 2 = 229.59 kg. It depends upon the orientation of the rotational axis.

5. The field control method is suitable for constant torque drives.

a) True

b) False

Answer: b

Explanation: Field control method is generally used for obtaining the speeds greater than the base speed. It is also known as flux weakening method. It is suitable for constant power drives.

6. What is the unit of the intensity?

a) Watt/m 2

b) Watt/m

c) Watt/m 4

d) Watt/m 3

Answer: a

Explanation: Intensity is defined as the amount of power incident on a particular area. It is mathematically expressed as I = Power incident ÷Area(m 2 ).

7. Calculate the value of the frequency if the signal completes half of the cycle in 70 sec. Assume signal is periodic.

a) 0.00714 Hz

b) 0.00456 Hz

c) 0.00845 Hz

d) 0.00145 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. The given signal completes half of the cycle in 70 seconds then it will complete a full cycle in 140 seconds. F = 1÷T=1÷140=.00714 Hz.

8. The slope of the V-I curve is 26°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .487 Ω

b) .482 Ω

c) .483 Ω

d) .448 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 26° so R=tan = .487 Ω. The slope of the V-I curve is resistance.

9. For large DC machines, the yoke is usually made of which material?

a) Cast steel

b) Cast iron

c) Iron

d) Cast steel or cast iron

Answer: a

Explanation: Yoke in DC machines is made up of cast steel. Yoke provides structural support and mechanical strength to the machine. It helps in carrying the flux from the North pole to South pole.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “DC Motors – Basic Characteristics”.

1. Calculate the terminal voltage of Permanent Magnet DC motor having a resistance of 2 Ω and a full load current of 5 A with 20 V back e.m.f.

a) 30 V

b) 25 V

c) 20 V

d) 31 V

Answer: a

Explanation: Permanent Magnet DC motor is a special type of motor in which flux remains constant. The terminal voltage can be calculated using the relation V t = E b +I a R a = 20+5×2 = 30 V.

2. Armature reaction is demagnetizing in nature due to purely lagging load.

a) True

b) False

Answer: a

Explanation: Due to purely lagging load, armature current is in opposite phase with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in the opposite phase with the field flux. It will try to reduce the net magnetic field.

3. The unit of Magento-motive force is Ampere-turns.

a) True

b) False

Answer: a

Explanation: The magneto-motive force is defined as the product of current and turns. It is mathematically expressed as F=NI.

4. Calculate the velocity of the wheel if the angular speed is 25 rad/s and radius is 10 m.

a) 250 m/s

b) 260 m/s

c) 270 m/s

d) 240 m/s

Answer: a

Explanation: The velocity of the wheel can be calculated using the relation V=ω×r. The velocity is the vector product of angular speed and radius. V=Ω×r = 25×10 = 250 m/s.

5. Displacement is a ____________ quantity.

a) Scalar

b) Vector

c) Scalar and Vector

d) Tensor

Answer: b

Explanation: Displacement is a vector quantity. It depends on the initial and final position of the body. It has both direction and magnitude. Distance is a scalar quantity.

6. When a UJT is used for triggering an SCR, then the wave shape of voltage obtained from the UJT circuit will be ____________

a) Square

b) Pulse

c) Trapezoidal

d) Saw-tooth

Answer: b

Explanation: UJT relaxation oscillator using RC circuit is used for SCR triggering and the wave shape obtained is an exponentially decaying pulse. So, we can say pulse waveform is obtained.

7. Calculate the value of the frequency if the time period of the signal is 0 sec.

a) infinity

b) 0.4 Hz

c) 0.2 Hz

d) 0.78 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. F = 1÷0 = infinity. It signifies signal is aperiodic.

8. The slope of the I-V curve is 87°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 0.0524 Ω

b) 0.0254 Ω

c) 0.0543 Ω

d) 0.0648 Ω

Answer: a

Explanation: The slope of the I-V curve is reciprocal of resistance. The slope given is 87° so R=1÷tan=.0524 Ω.

9. The most suitable device for high-frequency inversion in switching mode power supply is ______

a) GTO

b) BJT

c) MOSFET

d) IGBT

Answer: c

Explanation: MOSFET is fastest among all the power semiconductor devices and has the highest frequency range. MOSFET stands for metal oxide silicon field effect transistor.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Modified Speed Torque Characteristics of DC Shunt Motors”.

1. In DC chopper, the waveform for input and output voltages is respectively __________

a) Discontinuous and Continuous

b) Continuous and Discontinuous

c) Both continuous

d) Both discontinuous

Answer: b

Explanation: Chopper has a perfect DC at the input which is chopped into pulses which means the output voltage is discontinuous and by varying the duty cycle we can vary the average output voltage.

2. A chopper behaves as a __________

a) DC equivalent of AC switching device

b) DC equivalent of AC transformer

c) DC equivalent of AC relay

d) AC equivalent of circuit breaker

Answer: b

Explanation: A chopper is used to step up or step down the DC voltage whereas the transformer is used to step up or step down AC voltage so chopper is DC equivalent of AC transformer.

3. A DC chopper feeds an RLE load. If the value of E is increased by 20%, the current ripple ________

a) increases by 20%

b) decreases by 20%

c) increases only 20%

d) remains the same

Answer: d

Explanation: For a buck converter current ripple doesn’t depend on the value of E. ΔI L = V dc ×D××T÷L. The expression is independent of the value of E.

4. The conduction loss versus device current characteristic of a power MOSFET is best approximated by a ________

a) Straight line

b) Rectangular hyperbola

c) Parabola

d) Exponential decaying functions

Answer: c

Explanation: A MOSFET in ON state behaves as resistance so the conduction power loss is given by, P=I 2 R. Hence, the power vs current curve will be a parabola.

5. Which of the following device is NOT suitable for parallel operation?

a) MOSFET

b) BJT

c) IGBT

d) TRIAC

Answer: b

Explanation: BJT has a negative temperature coefficient of resistance. If it is operated in parallel operation thermal run-away will take place and the device will damage.

6. SCR is uni-directional in nature.

a) True

b) False

Answer: a

Explanation: SCR is uni-directional in nature. It only allows current to flow from anode to cathode. If the current is greater than the latching current then only it will work in forwarding conduction mode.

7. Which of the following device should be used as a switch in a low power switched mode power supply ?

a) GTO

b) BJT

c) MOSFET

d) TRIAC

Answer: c

Explanation: MOSFET has a low power rating and high-frequency rating and so it can be used in low power switch mode power supply. MOSFET stands for metal oxide silicon field effect transistor.

8. The slope of the V-I curve is 78°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 4.732 Ω

b) 4.608 Ω

c) 4.543 Ω

d) 4.648 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 78° so R=tan=4.732 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate mark to space ratio if the system is on for 17 sec and time period is 30 sec.

a) 1.307

b) 1.457

c) 1.478

d) 1.146

Answer: a

Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=17÷ = 1.307.

This set of Electric Drives Interview Questions and Answers for freshers focuses on “Modified Speed Torque Characteristics of DC Series Motors”.

1. Turn-on and turn-off times of transistor depend on _________

a) Static Characteristic

b) Junction Capacitance

c) Current Gain

d) Voltage Gain

Answer: b

Explanation: The depletion layer capacitance and diffusion capacitance affects the turn-on and turn-off behavior of transistors. Due to these internal capacitances, transistors do not turn on instantly.

2. The generated e.m.f from 45-pole armature having 400 turns driven at 70 rev/sec having flux per pole as 90 mWb, with 17 parallel paths is ___________

a) 13341.17 V

b) 12370.14 V

c) 14700.89 V

d) 15690.54 V

Answer: a

Explanation: The generated e.m.f can be calculated using the formula E b = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. One turn is equal to two conductors. E b = .09×45×400×2×4200÷60×17 = 13341.17 V.

3. The unit of Magnetic flux density is Tesla.

a) True

b) False

Answer: a

Explanation: Magnetic Flux density is defined as the number of magnetic lines passing through a certain point or a surface. It is generally expressed in terms of Tesla. Its C.G.S unit is Gauss.

4. Calculate the moment of inertia of the hollow cylinder having a mass of 78 kg and radius of 49 cm.

a) 9.363 kgm 2

b) 9.265 kgm 2

c) 9.787 kgm 2

d) 9.568 kgm 2

Answer: a

Explanation: The moment of inertia of the hollow cylinder can be calculated using the formula I=m i r i 2 ÷2. The mass of the hollow cylinder and radius is given. I=×.5× 2 =9.363 kgm 2 . It depends upon the orientation of the rotational axis.

5. Calculate the value of the angular acceleration of the motor using the given data: J = 81 kg-m 2 , load torque = 74 N-m, motor torque = 89 N-m.

a) .195 rad/s 2

b) .182 rad/s 2

c) .183 rad/s 2

d) .185 rad/s 2

Answer: d

Explanation: Using the dynamic equation of motor J× = Motor torque – Load torque: 81× = 89-74=15, angular acceleration=.185 rad/s 2 .

6. 340 V, 45 A, 1400 rpm DC separately excited motor having a resistance of .7 ohm excited by an external dc voltage source of 90 V. Calculate the torque developed by the motor on full load.

a) 94.73 N-m

b) 94.52 N-m

c) 93.37 N-m

d) 94.42 N-m

Answer: a

Explanation: Back emf developed in the motor during the full load can be calculated using equation E b = V t -I×R a = 308.5 V and machine constant K m = E b ÷W m which is equal to 2.1053. Torque can be calculated by using the relation T = K m × I = 2.1053×45 = 94.73 N-m.

7. Calculate the value of the frequency if the time period of the signal is 99 sec.

a) 0.08 Hz

b) 0.02 Hz

c) 0.01 Hz

d) 0.04 Hz

Answer: c

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. F = 1÷T=1÷99=.01 Hz.

8. The slope of the V-I curve is 31°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 0.600 Ω

b) 0.607 Ω

c) 0.543 Ω

d) 0.648 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 31° so R=tan=0.600 Ω. The resistance is the ratio of voltage and current.

9. Calculate the radius of the circular ring having a moment of inertia 59 kgm 2 and mass of 69 kg.

a) .924 m

b) .928 m

c) .934 m

d) .944 m

Answer: a

Explanation: The moment of inertia of the circular ring can be calculated using the formula I=∑m i r i 2 . The moment of inertia of a circular ring and mass is given. R=÷) .5 = .924 m. It depends upon the orientation of the rotational axis.

10. Calculate the power developed by a motor using the given data: E b = 48 V and I= 86 A

a) 4128 W

b) 4150 W

c) 4140 W

d) 4170 W

Answer: a

Explanation: Power developed by the motor can be calculated using the formula P = E b ×I = 48×86 = 4128 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

11. 780 V, 97 A, 1360 rpm separately excited dc motor with armature resistance (R a ) equal to 9 ohms. Calculate back emf developed in the motor when it operates on one-fourth of the full load.

a) 564.75 V

b) 561.75 V

c) 562.45 V

d) 565.12 V

Answer: b

Explanation: Back emf developed in the motor can be calculated using the relation E b = V t -I×R a . In question, it is asking for one-fourth load, but the data is given for full load so current becomes one-fourth of the full load current = 97÷4 = 24.25 A. 250 V is terminal voltage it is fixed so E b = 780-24.25×9 = 561.75 V.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “DC Motors – Application of Modified Characteristics”.

1. Calculate the time period of the waveform y=74cos.

a) .024 sec

b) .027 sec

c) .023 sec

d) .025 sec

Answer: a

Explanation: The fundamental time period of the cosine wave is 2π. The time period of y is 2π÷81π=.024 sec. The time period is independent of phase shifting and time shifting.

2. The generated e.m.f from 42-pole armature having 74 turns driven at 64 rev/sec having flux per pole as 21 mWb, with wave winding is ___________

a) 4177.171 V

b) 4177.152 V

c) 4100.189 V

d) 4190.454 V

Answer: b

Explanation: The generated e.m.f can be calculated using the formula E b = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. One turn is equal to two conductors. In wave winding the number of parallel paths is equal to two. E b =.021×42×74×2×3840÷60×2=4177.152 V.

3. Calculate the phase angle of the sinusoidal waveform x=20sin.

a) π÷9

b) π÷5

c) π÷7

d) π÷4

Answer: c

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the solid sphere having a mass of 28 kg and diameter of 15 cm.

a) 0.01575 kgm 2

b) 0.01875 kgm 2

c) 0.01787 kgm 2

d) 0.01568 kgm 2

Answer: a

Explanation: The moment of inertia of the solid sphere can be calculated using the formula I=2×m i r i 2 ÷5. The mass of the solid sphere and diameter is given. I =×.4× 2 =.01575 kgm 2 . It depends upon the orientation of the rotational axis.

5. R.M.S value of the trapezoidal waveform V=V m sin.

a) V m ÷2 ½

b) V m ÷2 ¼

c) V m ÷2 ¾

d) V m ÷3 ½

Answer: d

Explanation: R.M.S value of the sinusoidal waveform is V m ÷2 ½ and r.m.s value of the trapezoidal waveform is V m ÷3 ½ . The peak value of the sinusoidal waveform is V m .

6. What is the unit of the admittance?

a) ohm

b) ohm -1

c) ohm 2

d) ohm .5

Answer: b

Explanation: The admittance measures how easily current can flow in the circuit. It is the ratio of current and voltage. It is given in ohm -1 . It is reciprocal of impedance.

7. Calculate the value of the frequency if the inductive reactance is 45 Ω and the value of the inductor is 15 H.

a) 0.477 Hz

b) 0.544 Hz

c) 0.465 Hz

d) 0.412 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation X L = 2×3.14×f×L. F = X L ÷2×3.14×L = 45÷2×3.14×15 = .477 Hz.

8. The slope of the V-I curve is 19°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .3254 Ω

b) .3608 Ω

c) .3543 Ω

d) .3443 Ω

Answer: d

Explanation: The slope of the V-I curve is resistance. The slope given is 19° so R=tan=.3443 Ω. The slope of the V-I curve is resistance.

9. Calculate the active power in a 41 H inductor.

a) 2 W

b) 1 W

c) 0 W

d) .5 W

Answer: c

Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W. Voltage leads the current in case of the inductor.

10. Calculate the active power in a 19 F capacitor.

a) 7.8 W

b) 0 W

c) 5.4 W

d) 1.5 W

Answer: b

Explanation: The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the capacitor so the angle between V & I is 90°. P = VIcos90 = 0 W. Current leads the voltage in case of the capacitor.

11. Calculate the active power in a 241 H inductor.

a) 21 W

b) 11 W

c) 0 W

d) .51 W

Answer: c

Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90o in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W.

12. Calculate the active power in a 5 Ω resistor with 5 A current flowing through it.

a) 125 W

b) 110 W

c) 115 W

d) 126 W

Answer: a

This set of Electric Drives Puzzles focuses on “DC Motors – Direct Control of Armature-Terminal Voltage”

1. Calculate the frequency of the waveform x=45sin.

a) 24 Hz

b) 27 Hz

c) 23 Hz

d) 20 Hz

Answer: d

Explanation: The fundamental time period of the sine wave is 2π. The frequency of x is 40π÷2π=20 Hz. The frequency is independent of phase shifting and time shifting.

2. The generated e.m.f from 16-pole armature having 57 turns driven at 78 rev/sec having flux per pole as 5 mWb, with lap winding is ___________

a) 44.16 V

b) 44.15 V

c) 44.46 V

d) 44.49 V

Answer: c

Explanation: The generated e.m.f can be calculated using the formula E b = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. One turn is equal to two conductors. In lap winding, the number of parallel paths is equal to a number of poles. E b = .005×16×57×2×4680÷60×16=44.46 V.

3. Calculate the phase angle of the sinusoidal waveform x=42sin.

a) 2π÷9

b) 2π÷5

c) 2π÷7

d) 2π÷3

Answer: d

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the rod about its end having a mass of 39 kg and length of 88 cm.

a) 9.91 kgm 2

b) 9.92 kgm 2

c) 9.96 kgm 2

d) 9.97 kgm 2

Answer: c

Explanation: The moment of inertia of the rod about its end can be calculated using the formula I=ML 2 ÷3. The mass of the rod about its end and length is given. I=×.33× 2 =9.96 kgm 2 . It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the rod about its center having a mass of 11 kg and length of 29 cm.

a) .091 kgm 2

b) .072 kgm 2

c) .076 kgm 2

d) .077 kgm 2

Answer: d

Explanation: The moment of inertia of the rod about its center can be calculated using the formula I=ML 2 ÷12. The mass of the rod about its center and length is given. I=×.0833× 2 =.077 kgm 2 . It depends upon the orientation of the rotational axis.

6. Calculate the shaft power developed by a motor using the given data: E b = 404V and I = 25 A. Assume frictional losses are 444 W and windage losses are 777 W.

a) 8879 W

b) 2177 W

c) 8911 W

d) 8897 W

Answer: a

Explanation: Shaft power developed by the motor can be calculated using the formula P = E b *I- = 404*25- = 8879 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

7. Calculate the value of the frequency if the capacitive reactance is 13 Ω and the value of the capacitor is 71 F.

a) .0001725 Hz

b) .0001825 Hz

c) .0001975 Hz

d) .0001679 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation X c =1÷2×3.14×f×C. F=1÷X c ×2×3.14×C = 1÷13×2×3.14×71 = .0001725 Hz.

8. The slope of the V-I curve is 27°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .384 Ω

b) .509 Ω

c) .354 Ω

d) .343 Ω

Answer: b

Explanation: The slope of the V-I curve is resistance. The slope given is 27° so R=tan=.509 ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the active power in a 7481 H inductor.

a) 1562 W

b) 4651 W

c) 0 W

d) 4654 W

Answer: c

Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P=VIcos90 = 0 W. Voltage leads the current in case of the inductor.

10. Calculate the active power in a 457 F capacitor.

a) 715 W

b) 565 W

c) 545 W

d) 0 W

Answer: d

11. Calculate the active power in a 181 H inductor.

a) 2448 W

b) 1789 W

c) 4879 W

d) 0 W

Answer: d

12. Calculate the active power in a 17 ω resistor with 18 A current flowing through it.

a) 5508 W

b) 5104 W

c) 5554 W

d) 5558 W

Answer: a

This set of Electric Drives Multiple Choice Questions & Answers focuses on “DC Motors – Three Phase Induction Motors”.

1. A three-phase slip ring induction motor is fed from the rotor side with the stator winding short-circuited. The frequency of the current flowing in the short-circuited stator is ____________

a) Slip frequency

b) Supply frequency

c) The frequency corresponding to rotor speed

d) Zero

Answer: a

Explanation: The relative speed between rotor magnetic field and stator conductors is sip speed and hence the frequency of induced e.m.f is equal to slip frequency.

2. An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 720 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 2

b) 4

c) 3

d) 1

Answer: a

Explanation: Given a number of poles = 8. Supply frequency is 50 Hz. Rotor speed is 720 rpm. N s = 120×f÷P=120×50÷8 = 750 rpm. S=N s -N r ÷N s = 750 – 720÷750 = .04. F 2 =sf=.04×50=2 Hz.

3. Calculate the phase angle of the sinusoidal waveform z=78sin.

a) π÷39

b) 2π÷5

c) π÷74

d) 2π÷4

Answer: a

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the disc having a mass of 54 kg and diameter of 91 cm.

a) 5.512 kgm 2

b) 5.589 kgm 2

c) 5.487 kgm 2

d) 5.018 kgm 2

Answer: b

Explanation: The moment of inertia of the disc can be calculated using the formula I=mr 2 ×.5. The mass of the disc and diameter is given. I=×.5× 2 =5.589 kgm 2 . It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the thin spherical shell having a mass of 73 kg and diameter of 36 cm.

a) 1.56 kgm 2

b) 1.47 kgm 2

c) 1.38 kgm 2

d) 1.48 kgm 2

Answer: a

Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr 2 ×.66. The mass of the thin spherical shell and diameter is given. I=×.66× 2 =1.56 kgm 2 . It depends upon the orientation of the rotational axis.

6. A 50 Hz, 4poles, a single phase induction motor is rotating in the clockwise direction at a speed of 1425 rpm. The slip of motor in the direction of rotation & opposite direction of the motor will be respectively.

a) 0.05, 0.95

b) 0.04, 1.96

c) 0.05, 1.95

d) 0.05, 0.02

Answer: c

Explanation: Synchronous speed, N s =120×50÷4=1500 rpm. Given a number of poles = 4. Supply frequency is 50 Hz. Rotor speed is 1425 rpm. S=N s -N r ÷N s =1500-1425÷1500=.05. S b =2-s=1.95.

7. The frame of an induction motor is made of _________

a) Aluminum

b) Silicon steel

c) Cast iron

d) Stainless steel

Answer: c

Explanation: The frame of an induction motor is made of cast iron. The power factor of an induction motor depends upon the air gap between stator and rotor.

8. The slope of the V-I curve is 5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .3254 Ω

b) .3608 Ω

c) .3543 Ω

d) .3443 Ω

Answer: d

Explanation: The slope of the V-I curve is resistance. The slope given is 5° so R=tan=.3443 ω. The slope of the I-V curve is reciprocal of resistance.

9. In an induction motor, when the number of stator slots is equal to an integral number of rotor slots _________

a) There may be a discontinuity in torque slip characteristics

b) A high starting torque will be available

c) The maximum torque will be high

d) The machine may fail to start

Answer: d

Explanation: When the number of stator slots is an integral multiple of a number of rotor slots the machine fails to start and this phenomenon is called cogging.

10. A 3-phase induction motor runs at almost 1000 rpm at no load and 950 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?

a) 30 revolution per minute

b) 40 revolution per minute

c) 60 revolution per minute

d) 50 revolution per minute

Answer: d

Explanation: Supply frequency=50 Hz. No-load speed of motor = 1000 rpm. The full load speed of motor=950 rpm. Since the no-load speed of the motor is almost 1000 rpm, hence synchronous speed near to 1000 rpm. Speed of rotor field=1000 rpm. Speed of rotor field with respect to rotor = 1000-950 = 50 rpm.

11. Calculate the active power in a 487 H inductor.

a) 2482 W

b) 1545 W

c) 4565 W

d) 0 W

Answer: d

12. Calculate the active power in a 788 ω resistor with 178 A current flowing through it.

a) 24.96 MW

b) 24.44 MW

c) 24.12 MW

d) 26.18 MW

Answer: a

This set of Electric Drives Multiple Choice Questions & Answers focuses on “DC Motors – Three Phase Synchronous Motors”.

1. The direct axis is taken along ________

a) Inter-polar axis

b) Rotor pole axis

c) In between inter polar and rotor axis

d) Parallel to interpolar axis

Answer: b

Explanation: The direct axis is oriented along the rotor pole axis and the quadrature axis is 90° electrical to rotor pole axis. The direct axis is not oriented along the inter-polar axis.

2. A 4-pole, 3-phase, 60 Hz induction motor is operating at a speed of 1500 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 5

b) 4

c) 2

d) 7

Answer: b

Explanation: Given a number of poles = 4. Supply frequency is 60 Hz. Rotor speed is 1500 rpm. N s = 120×f÷P = 120×60÷4 = 1800 rpm. S=N s -N r ÷N s = 1800-1500÷1800 = .166. F 2 =sf=.166×60=4 Hz.

3. Calculate the phase angle of the sinusoidal waveform z=8cos.

a) 2π÷39

b) 2π÷15

c) π÷4

d) 2π÷44

Answer: b

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference.

4. The leakage flux paths are ________ on the angular position of the rotor.

a) Dependent

b) Proportional

c) Independent

d) Dependent and independent

Answer: c

Explanation: Leakage flux and leakage reactance are constant irrespective of rotor angular position. Armature reaction though is dependent on the angular position of the rotor in salient pole machine but in cylindrical rotor machine, both the quantities are independent of rotor position.

5. The hunting phenomenon in a synchronous motor is also referred to as _________

a) Surging

b) Phase swinging

c) Cogging

d) Surging and phase swinging

Answer: d

Explanation: During hunting, power oscillates so power surges are observed and hence it can be called as surging. Also, the rotor phase angle oscillates and it is called as phase swinging.

6. Which of the following are used in preventing the hunting phenomenon in synchronous generators?

a) Damper bars

b) Short pitch chords

c) Distributed winding

d) Damper bars and short pitch chords

Answer: a

Explanation: Damper bars try to maintain synchronism between the rotating magnetic field and the rotor so they help in preventing hunting. It produces surges in the machine.

7. In a synchronous machine, the phase sequence can be reversed by reversing the _________

a) Rotor direction

b) Field polarities

c) Armature terminal

d) Rotor direction and armature terminal

Answer: a

Explanation: In synchronous generator, the phase sequence is governed by the direction of rotation of the rotor and in a synchronous motor, the phase sequence governs the direction of rotation of the rotor.

8. The slope of the V-I curve is 7°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .122 Ω

b) .360 Ω

c) .377 Ω

d) .578 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 7° so R=tan=.122 Ω. The slope of the I-V curve is reciprocal of resistance.

9. In an induction motor, when the number of stator slots is not equal to an integral number of rotor slots _________

a) There may be a discontinuity in torque slip characteristics

b) A high starting torque will be available

c) The machine performs better

d) The machine may fail to start

Answer: c

Explanation: When the number of stator slots is not an integral multiple of a number of rotor slots the machine will not fail to start. It does not cause the cogging phenomenon.

10. A 3-phase induction motor runs at almost 1500 rpm at no load and 900 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?

a) 300 revolution per minute

b) 400 revolution per minute

c) 600 revolution per minute

d) 500 revolution per minute

Answer: c

Explanation: Supply frequency=50 Hz. No-load speed of motor = 1500 rpm. The full load speed of motor=900 rpm. Since the no-load speed of the motor is almost 1500 rpm, hence synchronous speed near to 1500 rpm. Speed of rotor field=1500 rpm. Speed of rotor field with respect to rotor = 1500-900 = 600 rpm.

11. For a practical synchronous motor, the pull-out torque will occur when the torque angle is nearly equal to ________

a) 0°

b) 30°

c) 45°

d) 75°

Answer: d

Explanation: In a practical synchronous motor, the armature resistance cannot be neglected and hence the pull-out occurs at delta=beta which is the impedance angle and is practically 75°.

12. Calculate the active power in an 8 Ω resistor with 8 A current flowing through it.

a) 512 W

b) 514 W

c) 512 W

d) 518 W

Answer: a

This set of Electric Drives Objective Questions & Answers focuses on “Effect of Starting on Power Supply, Motor and Load “.

1. The ferrite cores are used for ____________ transformers.

a) Small transformers

b) Medium transformers

c) Large transformers

d) Medium and small transformers

Answer: a

Explanation: Ferrite cores are used for cores of small transformers used in communication circuits at high frequencies and low energy levels. Because ferrites have high resistivity they will have lower eddy current losses.

2. A 2-pole, 3-phase, 50 Hz induction motor is operating at a speed of 400 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 43.33

b) 42.54

c) 43.11

d) 41.47

Answer: a

Explanation: Given a number of poles = 2. Supply frequency is 50 Hz. Rotor speed is 400 rpm. N s = 120×f÷P=120×50÷2 = 3000 rpm. S=N s -N r ÷N s = 3000-400÷3000 = .8666. F 2 =sf=.8666×50=43.33 Hz.

3. Calculate the phase angle of the sinusoidal waveform w=.45sin.

a) 2π÷39

b) 8π÷787

c) 5π÷74

d) 42π÷4

Answer: b

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the disc having a mass of 1 kg and radius of 1 m.

a) 1 kgm 2

b) .5 kgm 2

c) 2 kgm 2

d) 3 kgm 2

Answer: b

Explanation: The moment of inertia of the disc can be calculated using the formula I=mr 2 ×.5. The mass of the disc and diameter is given. I=×.5× 2 =.5 kgm 2 . It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the thin spherical shell having a mass of 3 kg and diameter of 6 cm.

a) .0178 kgm 2

b) .0147 kgm 2

c) .0398 kgm 2

d) .0144 kgm 2

Answer: a

Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr 2 ×.66. The mass of the thin spherical shell and diameter is given. I=×.66× 2 =.0178 kgm 2 . It depends upon the orientation of the rotational axis.

6. The power factor of a synchronous motor __________

a) Improves with an increase in excitation and may even become leading at high excitations

b) Decreases with increase in excitation

c) Is independent of its excitation

d) Increase with loading for a given excitation

Answer: a

Explanation: From inverted V-curve we can see when the power factor is leading power factor decreases when the excitation increases and it is over-excited conditions and when power factor is lagging if the motor power factor is increasing if excitation increases.

7. The frame of a synchronous motor is made of _________

a) Aluminum

b) Silicon steel

c) Cast iron

d) Stainless steel

Answer: c

Explanation: The frame of a synchronous motor is made of cast iron. The power factor of a synchronous motor depends upon maximum power transfer capability.

8. The slope of the V-I curve is 45°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 2 Ω

b) 3 Ω

c) 4 Ω

d) 1 Ω

Answer: d

Explanation: The slope of the V-I curve is resistance. The slope given is 45° so R=tan=1 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Which one of the following methods would give a higher than the actual value of regulation of the alternator.

a) ZPF method

b) MMF method

c) EMF method

d) ASA method

Answer: c

Explanation: EMF method is a pessimistic method of voltage regulation as it gives higher than the actual value of voltage regulation. EMF method will the values that are greater than the actual value.

10. A 3-phase induction motor runs at almost 1100 rpm at no load and 640 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?

a) 430 revolution per minute

b) 440 revolution per minute

c) 460 revolution per minute

d) 450 revolution per minute

Answer: c

Explanation: Supply frequency=50 Hz. No-load speed of motor = 1100 rpm. The full load speed of motor=640 rpm. Since the no-load speed of the motor is almost 1100 rpm, hence synchronous speed near to 1100 rpm. Speed of rotor field=1100 rpm. Speed of rotor field with respect to rotor = 1100-640 = 460 rpm.

11. Which of the following core has linear characteristics?

a) Air core

b) Steel core

c) CRGO core

d) Iron core

Answer: a

Explanation: Air core coils have linear magnetization characteristics that are they do not saturate. Open circuit characteristics graph is linear in case of synchronous machine.

12. Calculate the active power in a 1 Ω resistor with 2 A current flowing through it.

a) 2 W

b) 4 W

c) 7 W

d) 1 W

Answer: b

13. The slope of the V-I curve is 13°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .2544 Ω

b) .7771 Ω

c) .2308 Ω

d) .5788 Ω

Answer: c

Explanation: The slope of the V-I curve is resistance. The slope given is 13° so R=tan=.2308 Ω. The slope of the I-V curve is reciprocal of resistance.

14. A 2-pole lap wound generator with 44 armature conductors and a flux per pole of .07 Wb has an armature current of 80 A. The developed torque is _________

a) 39.7 N-m

b) 39.2 N-m

c) 38.4 N-m

d) 37.2 N-m

Answer: b

Explanation: The developed torque in the motor is K m ×I. The value of machine constant(K m ) is E b ÷ωm = Φ×Z×P÷2×3.14×A = .07×44×2÷2×3.14×2 = .49 Vs/rad. The developed torque is .49×80 = 39.2 N-m.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Methods of Starting Electric Motors”.

1. Calculate the active power in a .45 H inductor.

a) 0.11 W

b) 0.14 W

c) 0.15 W

d) 0 W

Answer: d

2. A 10-pole, 3-phase, 60 Hz induction motor is operating at a speed of 100 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 52.4

b) 54.8

c) 51.66

d) 51.77

Answer: c

Explanation: Given a number of poles = 10. Supply frequency is 60 Hz. Rotor speed is 100 rpm. N s = 120×f÷P=120×60÷10 = 720 rpm. S=N s -N r ÷N s = 720-100÷720=.86. F 2 =sf=.86×60=51.66 Hz.

3. Calculate the phase angle of the sinusoidal waveform z=.99sin.

a) π÷3

b) 2π

c) π÷7

d) π

Answer: d

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, Ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the disc having a mass of 4 kg and diameter of 1458 cm.

a) 106.288 kgm 2

b) 104.589 kgm 2

c) 105.487 kgm 2

d) 107.018 kgm 2

Answer: a

Explanation: The moment of inertia of the disc can be calculated using the formula I=mr 2 ×.5. The mass of the disc and diameter is given. I=×.5× 2 =106.288 kgm 2 . It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the thin spherical shell having a mass of 703 kg and diameter of 376 cm.

a) 1639.89 kgm 2

b) 1628.47 kgm 2

c) 1678.12 kgm 2

d) 1978.19 kgm 2

Answer: a

Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr 2 ×.66. The mass of the thin spherical shell and diameter is given. I=×.66× 2 =1639.89 kgm 2 . It depends upon the orientation of the rotational axis.

6. Calculate the value of the torque when 89 N force is applied perpendicular to a 78 m length of stick fixed at the center.

a) 6942 N-m

b) 3000 N-m

c) 1000 N-m

d) 4470 N-m

Answer: a

Explanation: Torque can be calculated using the relation T = × = r×F×sin90. F is given as 89 N and r is 78 m then torque is 89×78 = 6942 N-m. .

7. 100 V, 2 A, 90 rpm separately excited dc motor with armature resistance (R a ) equal to 8 ohms. Calculate back emf developed in the motor when it operates on 3 th /4 of the full load.

a) 100 V

b) 87 V

c) 88 V

d) 90 V

Answer: c

Explanation: Back emf developed in the motor can be calculated using the relation E b = V t -I×R a . In question, it is asking for 3 th /4 load, but the data is given for full load so current becomes 3 th /4 of the full load current = 2÷1.33 = 1.5 A. 100 V is terminal voltage it is fixed so E b = 100-1.5×8 = 88 V.

8. The slope of the V-I curve is 16.8°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .324 Ω

b) .301 Ω

c) .343 Ω

d) .398 Ω

Answer: b

Explanation: The slope of the V-I curve is resistance. The slope given is 16.8° so R=tan=.301 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the value of the torque when 1 N force is applied perpendicular to a 1 m length of chain fixed at the center.

a) 1 N-m

b) 3 N-m

c) 2 N-m

d) 4 N-m

Answer: a

Explanation: Torque can be calculated using the relation T = × = r×F×sin90. F is given as 1 N and r is 1 m then torque is 1×1 = 6942 N-m. .

10. A 3-phase induction motor runs at almost 140 rpm at no load and 50 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?

a) 20 revolution per minute

b) 80 revolution per minute

c) 90 revolution per minute

d) 70 revolution per minute

Answer: c

Explanation: Supply frequency=50 Hz. No-load speed of motor= 140 rpm. The full load speed of motor=50 rpm. Since the no-load speed of the motor is almost 140 rpm, hence synchronous speed near to 140 rpm. Speed of rotor field=140 rpm. Speed of rotor field with respect to rotor = 140-50 = 90 rpm.

11. Calculate the active power in a .7889 H inductor.

a) .123 W

b) .155 W

c) 0 W

d) .487 W

Answer: c

12. Calculate the active power in a .8 Ω resistor with 2 A current flowing through it.

a) 2.4 W

b) 3.4 W

c) 2.2 W

d) 3.2 W

Answer: d

13. The unit of angular frequency is Hz.

a) True

b) False

Answer: a

Explanation: Angular frequency is defined as the rate of change of angular displacement with respect to time. Angular displacement is generally expressed in terms of a radian. The unit of angular frequency is Hz. ω=2×3.14×f.

14. Calculate the active power in a .56 F capacitor.

a) 37.8 W

b) 0 W

c) 15.4 W

d) 124.5 W

Answer: b

Explanation: The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the capacitor so the angle between V & I is 90°. P = VIcos90° = 0 W. Current leads the voltage in case of the capacitor.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Starting – Acceleration Time”.

1. Calculate the value of the angular acceleration of the motor using the given data: J = .01 kg-m 2 , load torque= 790 N-m, motor torque= 169 N-m.

a) -62100 rad/s 2

b) -62456 rad/s 2

c) -34056 rad/s 2

d) -44780 rad/s 2

Answer: a

Explanation: Using the dynamic equation of motor J* = Motor torque – Load torque: .01* = 169-790=-621, angular acceleration=-62100 rad/s 2 .The motor will decelerate and will fail to start.

2. A 14-pole, 3-phase, 50 Hz induction motor is operating at a speed of 99 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 39.5

b) 40

c) 38.45

d) 39.9

Answer: c

Explanation: Given a number of poles = 14. Supply frequency is 50 Hz. Rotor speed is 699 rpm. N s =120×f÷P=120×50÷14 = 428.57 rpm. S=N s -N r ÷N s =428.57-99÷428.57=.769. F 2 =sf=.769×50=38.45 Hz.

3. Calculate the phase angle of the sinusoidal waveform z=18cos.

a) 25π÷39

b) 25π÷5

c) 25π÷1

d) π÷4

Answer: c

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, Ω represents angular frequency, α represents a phase difference.

4. Calculate the mass of the solid sphere having a moment of inertia 17 kgm 2 and radius of 4 cm.

a) 10624 kg

b) 10625 kg

c) 10628 kg

d) 10626 kg

Answer: b

Explanation: The moment of inertia of the ball can be calculated using the formula I=Σm i r i 2 . The moment of inertia of ball and radius is given. M=÷ 2 = 10625 kg. It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the thin spherical shell having a mass of 3 kg and diameter of 66 cm.

a) .2156 kgm 2

b) .2147 kgm 2

c) .2138 kgm 2

d) .2148 kgm 2

Answer: a

Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr 2 ×.66. The mass of the thin spherical shell and diameter is given. I=×.66× 2 =.2156 kgm 2 . It depends upon the orientation of the rotational axis.

6. Calculate the value of the time period if the frequency of the signal is 1 sec.

a) 1 sec

b) 2 sec

c) .5 sec

d) 1.5 sec

Answer: a

Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷1=1 sec.

7. The slope of the V-I curve is 270°. Calculate the value of resistance.

a) 112 Ω

b) 178 Ω

c) infinite Ω

d) 187 Ω

Answer: c

Explanation: The slope of the V-I curve is resistance. The slope given is 270° so R=tan=infinite Ω. It behaves as an open-circuit.

8. The slope of the V-I curve is 23.56°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .464 Ω

b) .436 Ω

c) .443 Ω

d) .463 Ω

Answer: b

Explanation: The slope of the V-I curve is resistance. The slope given is 23.56° so R=tan=.436 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the reactive power in a 23 Ω resistor.

a) 45 VAR

b) 10 VAR

c) 245 VAR

d) 0 VAR

Answer: d

10. A 3-phase induction motor runs at almost 50 rpm at no load and 25 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?

a) 25 revolution per minute

b) 20 revolution per minute

c) 10 revolution per minute

d) 30 revolution per minute

Answer: a

Explanation: Supply frequency=50 Hz. No-load speed of motor= 50 rpm. The full load speed of motor=25 rpm. Since the no-load speed of the motor is almost 50 rpm, hence synchronous speed near to 50 rpm. Speed of rotor field=50 rpm. Speed of rotor field with respect to rotor = 50-25 = 25 rpm.

11. Calculate the value of the frequency of the 220 V DC supply.

a) 10 Hz

b) 0 Hz

c) 20 Hz

d) 90 Hz

Answer: b

12. Calculate the value of the duty cycle if the system is on for 48 sec and off for 1 sec.

a) .979

b) .444

c) .145

d) .578

Answer: a

Explanation: Duty cycle is Ton÷T total . It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷T total = 48÷49 = .979.

13. Calculate the value of the frequency if the signal completes half of the cycle in 30 sec. Assume signal is periodic.

a) 0.028 Hz

b) 0.016 Hz

c) 0.054 Hz

d) 0.045 Hz

Answer: b

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. The given signal completes half of the cycle in 30 seconds then it will complete a full cycle in 60 seconds. F = 1÷T = 1÷60 = .016 Hz.

14. Calculate the velocity of the disc if the angular speed is 5 rad/s and radius is 2 m.

a) 25 m/s

b) 20 m/s

c) 25 m/s

d) 10 m/s

Answer: d

Explanation: The velocity of the disc can be calculated using the relation V=ω×r. The velocity is the vector product of angular speed and radius. V = ω×r = 5×2 = 10 m/s.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Energy Relations During Starting”.

1. 440 V, 77 A, 700 rpm DC separately excited motor having a resistance of 0.11 ohms excited by an external dc voltage source of 24 V. Calculate the torque developed by the motor on full load.

a) 453.51 N-m

b) 451.24 N-m

c) 440.45 N-m

d) 452.64 N-m

Answer: a

Explanation: Back emf developed in the motor during the full load can be calculated using equation E b = V t -I×R a = 431.53 V and machine constant K m = E b ÷W m which is equal to 5.88. Torque can be calculated by using the relation T = K m × I = 5.88×77 = 453.51 N-m.

2. Calculate the power developed by a motor using the given data: E b = 55 V and I = 6 A.

a) 440 W

b) 220 W

c) 330 W

d) 550 W

Answer: c

Explanation: Power developed by the motor can be calculated using the formula P = E b ×I = 55×6 = 330 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

3. Calculate the value of the angular acceleration of the motor using the given data: J = 36 kg-m 2 , load torque = 66 N-m, motor torque = 26 N-m.

a) 1.11 rad/s 2

b) 2.22 rad/s 2

c) 3.33 rad/s 2

d) 4.44 rad/s 2

Answer: a

Explanation: Using the dynamic equation of motor J× = Motor torque – Load torque: 36× = 66-26 = 40, angular acceleration = 1.11 rad/s 2 .

4. Calculate the moment of inertia of the apple having a mass of .4 kg and diameter of 12 cm.

a) .0008 kgm 2

b) .0007 kgm 2

c) .0009 kgm 2

d) .0001 kgm 2

Answer: b

Explanation: The moment of inertia of the apple can be calculated using the formula I=mr 2 ×.5. The mass of the apple and diameter is given. I=×.5× 2 = .0007 kgm 2 . It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the thin spherical shell having a mass of 3.3 kg and diameter of .6 cm.

a) .00125 kgm 2

b) .00196 kgm 2

c) .00145 kgm 2

d) .00178 kgm 2

Answer: b

Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr 2 ×.66. The mass of the thin spherical shell and diameter is given. I=×.66× 2 =.00196 kgm 2 . It depends upon the orientation of the rotational axis.

6. Calculate the time period of the waveform y=7cos.

a) .055 sec

b) .037 sec

c) .023 sec

d) .017 sec

Answer: b

Explanation: The fundamental time period of the cosine wave is 2π. The time period of y is 2π÷54π=.037 sec. The time period is independent of phase shifting and time shifting.

7. Calculate the useful power developed by a motor using the given data: P in = 1500 W, I a = 6 A, R a =.2 Ω. Assume frictional losses are 50 W and windage losses are 25 W.

a) 1400 W

b) 1660.5 W

c) 1417.8 W

d) 1416.7 W

Answer: c

Explanation: Useful power is basically the shaft power developed by the motor that can be calculated using the formula P sh = P dev -. P dev = P in -I a 2 R a = 1500-6 2 =1492.8 W. The useful power developed by the motor is P sh = P dev - = 1492.8 – = 1417.8 W.

8. The slope of the V-I curve is 15.5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .277 Ω

b) .488 Ω

c) .443 Ω

d) .457 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 15.5° so R=tan=.277 Ω. The slope of the I-V curve is reciprocal of resistance.

9. The generated e.m.f from 2-pole armature having 2 conductors driven at 3000 rpm having flux per pole as 4000 mWb, with 91 parallel paths is ___________

a) 8.64 V

b) 8.56 V

c) 8.12 V

d) 8.79 V

Answer: d

10. A 3-phase induction motor runs at almost 888 rpm at no load and 500 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?

a) 388 revolution per minute

b) 400 revolution per minute

c) 644 revolution per minute

d) 534 revolution per minute

Answer: a

Explanation: Supply frequency=50 Hz. No-load speed of motor = 888 rpm. The full load speed of motor=500 rpm. Since the no-load speed of the motor is almost 888 rpm, hence synchronous speed near to 888 rpm. Speed of rotor field=888 rpm. Speed of rotor field with respect to rotor = 888-500 = 388 rpm.

11. Calculate the active power in a 157.1545 H inductor.

a) 4577 W

b) 4567 W

c) 4897 W

d) 0 W

Answer: d

12. Calculate the active power in a 1.2 Ω resistor with 1.8 A current flowing through it.

a) 3.88 W

b) 3.44 W

c) 3.12 W

d) 2.18 W

Answer: a

13. Calculate the total heat dissipated in a resistor of 12 Ω when 9.2 A current flows through it.

a) 2.01 KW

b) 3.44 KW

c) 1.01 KW

d) 2.48 KW

Answer: c

14. Calculate mark to space ratio if the system is on for 9 sec and time period is 11 sec.

a) 4.6

b) 4.8

c) 4.5

d) 4.9

Answer: c

Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=9÷ = 4.5.

This set of Electric Drives Questions & Answers for Exams focuses on “Methods to Reduce the Energy Loss During Starting”.

1. Calculate the angular frequency of the waveform y=69sin.

a) 40π Hz

b) 60π Hz

c) 70π Hz

d) 20π Hz

Answer: a

Explanation: The fundamental time period of the sine wave is 2π. The sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, Ω represents angular frequency, α represents a phase difference. Ω can be directly calculated by comparing the equations. Ω = 40π Hz.

2. The generated e.m.f from 22-pole armature having 75 turns driven at 78 rpm having flux per pole as 400 mWb, with lap winding is ___________

a) 76 V

b) 77 V

c) 78 V

d) 79 V

Answer: c

Explanation: The generated e.m.f can be calculated using the formula E b = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. One turn is equal to two conductors. In lap winding, the number of parallel paths is equal to a number of poles. E b = .4×22×75×2×78÷60×22 = 78 V.

3. Calculate the phase angle of the sinusoidal waveform u=154sin.

a) -78π÷9

b) -12π÷5

c) -π÷89

d) -2π÷888

Answer: c

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, Ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the stick about its end having a mass of 22 kg and length of 22 cm.

a) .088 kgm 2

b) .087 kgm 2

c) .089 kgm 2

d) .086 kgm 2

Answer: b

Explanation: The moment of inertia of the stick about its end can be calculated using the formula I=ML 2 ÷3. The mass of the stick about its end and length is given. I = ×.33× 2 =.087 kgm 2 . It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the stick about its center having a mass of 1.1 kg and length of 2.9 m.

a) .66 kgm 2

b) .77 kgm 2

c) .88 kgm 2

d) .47 kgm 2

Answer: b

Explanation: The moment of inertia of the stick about its center can be calculated using the formula I=ML 2 ÷12. The mass of the stick about its center and length is given. I=×.0833× 2 =.77 kgm 2 . It depends upon the orientation of the rotational axis.

6. Calculate the useful power developed by a motor using the given data: E b = 4V and I = 52 A. Assume frictional losses are 3 W and windage losses are 2 W.

a) 203 W

b) 247 W

c) 211 W

d) 202 W

Answer: a

Explanation: Useful power developed by the motor can be calculated using the formula P = E b *I - = 4*52 – = 203 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

7. Calculate the value of the frequency if the capacitive reactance is .1 Ω and the value of the capacitor is .02 F.

a) 71.25 Hz

b) 81.75 Hz

c) 79.61 Hz

d) 79.54 Hz

Answer: c

Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation X c = 1÷2×3.14×f×C. F = 1÷X c ×2×3.14×C = 1÷.1×2×3.14×.02 = 79.61 Hz.

8. The slope of the V-I curve is 6.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .38 Ω

b) .59 Ω

c) .34 Ω

d) .12 Ω

Answer: d

Explanation: The slope of the V-I curve is resistance. The slope given is 6.9° so R=tan=.12 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the active power in an 8764 H inductor.

a) 8645 W

b) 6485 W

c) 0 W

d) 4879 W

Answer: c

Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90° = 0 W. Voltage leads the current in case of the inductor.

10. Calculate the active power in a 543 F capacitor.

a) 581 W

b) 897 W

c) 0 W

d) 892 W

Answer: c

Explanation: The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the capacitor so the angle between V & I is 90°. P=VIcos90°= 0 W. Current leads the voltage in case of the capacitor.

11. Calculate the active power in a 32 H inductor.

a) 28 W

b) 189 W

c) 4 W

d) 0 W

Answer: d

12. Calculate the active power in an 8965 Ω resistor with .23 A current flowing through it.

a) 547.12 W

b) 474.24 W

c) 554.78 W

d) 123.88 W

Answer: b

13. Which one of the following methods would give a lower than the actual value of regulation of the alternator?

a) ZPF method

b) MMF method

c) EMF method

d) ASA method

Answer: b

Explanation: MMF method is an optimistic method of voltage regulation as it gives lower than the actual value of voltage regulation. MMF method will give the values that are lesser than the actual value.

This set of Electric Drives Question Paper focuses on “Speed Control of Shunt or Separately Excited Motors”.

1. The advantage of the double squirrel cage induction motor over a single cage rotor is that its _______

a) Efficiency is higher

b) Power factor is higher

c) Slip is larger

d) Starting current is lower

Answer: d

Explanation: The starting current flows through the outer cage which has higher resistance and hence starting current is lower. This is one the important advantage of the double squirrel cage induction motor over a single cage rotor.

2. A 16-pole, 3-phase, 60 Hz induction motor is operating at a speed of 150 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 20

b) 40

c) 30

d) 10

Answer: b

Explanation: Given a number of poles = 16. Supply frequency is 60 Hz. Rotor speed is 150 rpm. N s =120×f÷P=120×60÷16 = 450 rpm. S=N s -N r ÷N s =450-150÷450=.666. F 2 =sf=.666×60=40 Hz.

3. Calculate the amplitude of the sinusoidal waveform z=715sin.

a) 710

b) 715

c) 716

d) 718

Answer: b

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, Ω represents angular frequency, α represents a phase difference.

4. R.M.S value of the sinusoidal waveform V=48sin.

a) 33.94 V

b) 33.56 V

c) 33.12 V

d) 33.78 V

Answer: a

Explanation: R.M.S value of the sinusoidal waveform is V m ÷2 ½ = 48÷2 ½ = 33.94 V and r.m.s value of the trapezoidal waveform is V m ÷3 ½ . The peak value of the sinusoidal waveform is V m .

5. The short circuit test on a 3-φ induction motor is conducted at a rotor speed of _______

a) Zero

b) < N s

c) > N s

d) N s

Answer: a

Explanation: Short-circuit test in an induction motor is also called a Blocked rotor test so it is conducted at zero speed. Net input power taken is equal to the variable losses.

6. If induction motor air gap power is 10 KW and mechanically developed power is 8 KW, then rotor ohmic loss will be _________ KW.

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=10-8=2 KW.

7. A 3-phase induction machine draws active power P and reactive power Q from the grid. If it is operated as a generator, then P and Q will be _________

a) Positive and Negative

b) Negative and Positive

c) Positive and Positive

d) Negative and Negative

Answer: b

Explanation: Induction generator will be delivering power to the bus to generate flux it will consume reactive power from the bus. Since active power is delivered the active power drawn will be negative but reactive power is absorbed and hence active power absorbed is positive.

8. The slope of the V-I curve is 39.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .81 Ω

b) .36 Ω

c) .75 Ω

d) .84 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 39.1° so R=tan=.81 Ω. The slope of the I-V curve is reciprocal of resistance.

9. If induction motor air gap power is 48 KW and gross developed power is 28 KW, then rotor ohmic loss will be _________ KW.

a) 10

b) 20

c) 30

d) 40

Answer: b

10. The power factor of a squirrel cage induction motor is ___________

a) Low at light load only

b) Low at heavy loads only

c) Low at the light and heavy loads both

d) Low at rate load only

Answer: a

Explanation: At light loads, the current drawn is largely a magnetizing current due to the air gap and hence the power factor is low.

11. Calculate the active power in a .154 H inductor.

a) 22 W

b) 14 W

c) 45 W

d) 0 W

Answer: d

12. Calculate the time period of the waveform z=24sin+4sin.

a) 2 sec

b) 3 sec

c) 4 sec

d) 1 sec

Answer: a

Explanation: The fundamental time period of the sine wave is 2π. The time period of z is L.C.M {2,1}=2 sec. The time period is independent of phase shifting and time shifting.

13. Calculate the total heat dissipated in a resistor of 50 Ω when 1.4 A current flows through it.

a) 98 W

b) 92 W

c) 91 W

d) 93 W

Answer: a

14. Calculate mark to space ratio if the system is on for 26.3 sec and off for 24.2 sec.

a) 1.086

b) 1.042

c) 1.214

d) 1.876

Answer: a

Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=26.3÷24.2=1.086.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Speed Control of Series Motor”.

1. A particular current is made up of two components: a 10 A and a sine wave of peak value 14.14 A. The average value of current is _________

a) Zero

b) 24.14 A

c) 10 A

d) 14.14 A

Answer: c

Explanation: Average value of DC current is 10 A. Average value of AC current is 0 A as it is alternating in nature. The average value of current is 10+0 = 10 A.

2. A 38-pole, 3-phase, 80 Hz induction motor is operating at a speed of 12 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 75.2

b) 76.1

c) 79.2

d) 79.6

Answer: b

Explanation: Given a number of poles = 8. Supply frequency is 50 Hz. Rotor speed is 720 rpm. N s =120×f÷P=120×80÷38 = 252.63 rpm. S=N s -N r ÷N s = 252.63-12÷252.63 = .952. F 2 = sf = .952×80 = 76.1 Hz.

3. Calculate the phase angle of the sinusoidal waveform i=sin.

a) 100π÷88

b) 100π÷8

c) 100π÷8

d) π÷88

Answer: a

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the satellite having a mass of 79 kg and diameter of 83 cm.

a) 13.65 kgm 2

b) 13.60 kgm 2

c) 12.67 kgm 2

d) 13.82 kgm 2

Answer: b

Explanation: The moment of inertia of the satellite can be calculated using the formula I=mr 2 . The mass of the satellite and diameter is given. I=× 2 =13.60 kgm 2 . It depends upon the orientation of the rotational axis.

5. A particular voltage is made up of two components: a 5 A and a cosine wave of peak value 7.8 A. The average value of voltage is _________

a) Zero

b) 35.14 A

c) 78 A

d) 5 A

Answer: d

Explanation: Average value of DC voltage is 5 A. Average value of AC current is 0 A as it is alternating in nature. The average value of current is 5+0 = 5 A.

6. Armature reaction is demagnetizing in nature due to a purely capacitive load in the synchronous generator.

a) True

b) False

Answer: b

Explanation: Due to a purely capacitive load, armature current is in phase with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in phase with the field flux. It will try to increase the net magnetic field.

7. Armature reaction is magnetizing in nature due to a purely resistive load in the synchronous generator.

a) True

b) False

Answer: b

Explanation: Due to a purely resistive load, armature current is in quadrature with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in quadrature with the field flux. It will try to increase the net magnetic field.

8. The slope of the V-I curve is 4.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .03 Ω

b) .08 Ω

c) .04 Ω

d) .07 Ω

Answer: b

Explanation: The slope of the V-I curve is resistance. The slope given is 4.9° so R=tan=.08 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the velocity of the satellite if the angular speed is 87 rad/s and radius is 7.4 m.

a) 643.8 m/s

b) 642.4 m/s

c) 641.9 m/s

d) 643.2 m/s

Answer: a

Explanation: The velocity of the satellite can be calculated using the relation V=Ω×r. The velocity is the vector product of angular speed and radius. V=Ω×r = 87×7.4 = 643.8 m/s.

10. A 3-phase induction motor runs at almost 70 rpm at no load and 50 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?

a) 20 revolution per minute

b) 30 revolution per minute

c) 40 revolution per minute

d) 50 revolution per minute

Answer: a

Explanation: Supply frequency=50 Hz. No-load speed of motor = 70 rpm. The full load speed of motor=50 rpm. Since the no-load speed of the motor is almost 70 rpm, hence synchronous speed near to 70 rpm. Speed of rotor field=70 rpm. Speed of rotor field with respect to rotor=70-50= 20 rpm.

11. Calculate the active power in a 9.854 H inductor.

a) 4.98 W

b) 0 W

c) 8.59 W

d) 1 W

Answer: b

12. Calculate the reactive power in a 45 Ω resistor with 1.78 A current flowing through it.

a) 28.8 VAR

b) 23.4 VAR

c) 25.82 VAR

d) 0 VAR

Answer: d

13. Calculate the value of the frequency if the inductive reactance is 72 Ω and the value of the inductor is 7 H.

a) 1.63 Hz

b) 1.54 Hz

c) 1.78 Hz

d) 1.32 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation X L = 2×3.14×f×L. F = X L ÷2×3.14×L = 72÷2×3.14×7 = 1.63 Hz.

14. Calculate the active power in a 2 Ω resistor with 8 A current flowing through it.

a) 125 W

b) 128 W

c) 123 W

d) 126 W

Answer: b

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Induction Motors – Basics Principles of Speed Control”.

1. In the chopper circuit, commutation times are _______

a) Current commutation is equal to voltage commutation

b) Current commutation is less as compared to that of voltage commutation

c) Current commutation is more as compared to that of voltage commutation

d) Both commutation techniques are not comparable

Answer: b

Explanation: In current commutation, commutation time=CV r ÷I o . In voltage commutation, commutation time= CV s ÷I o . Hence, the commutation time of the current commutation is less as compared to voltage commutation.

2. The generated e.m.f from 4-pole armature having 1 conductors driven at 1 rev/sec having flux per pole as 10 Wb, with wave winding is ___________

a) 30 V

b) 40 V

c) 70 V

d) 20 V

Answer: d

3. The unit of voltage is Pascal.

a) True

b) False

Answer: b

Explanation: The voltage is equal to one volt when 1 A of current flows through 1 Ω resistor. It is mathematically represented as I×R. It is expressed in terms of a volt.

4. Calculate the moment of inertia of the sphere having a mass of 8.4 kg and radius of 61 cm.

a) 3.124 kgm 2

b) 3.125 kgm 2

c) 4.545 kgm 2

d) 5.552 kgm 2

Answer: b

Explanation: The moment of inertia of the egg can be calculated using the formula I=Σm i r i 2 . The mass of egg and radius is given. I=× 2 =3.125 kgm 2 . It depends upon the orientation of the rotational axis.

5. The most suitable control-motor application is __________

a) AC shunt motor

b) DC separately motor

c) AC one-phase induction motor

d) DC shunt motor

Answer: b

Explanation: DC separately motor has definite full-load speed, so they don’t ‘run away’ when the load is suddenly thrown off provided the field circuit remains closed. The speed for any load within the operating range of the motor can be readily obtained.

6. In a DC series motor, the e.m.f developed is proportional to _______

a) N×I a

b) N×I a 2

c) N×I a 3

d) N×I a .5

Answer: a

Explanation: In a DC series motor, the e.m.f developed is equal to K m ΦN. In a DC series, the motor field winding is connected in series with the armature so the flux in the field winding is proportional to current. E b = K m ΦN α I a ×N.

7. Calculate the value of the time period if the frequency of the signal is .07 sec.

a) 14.28 sec

b) 14.31 sec

c) 14.23 sec

d) 14.78 sec

Answer: a

Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷.07=14.28 sec.

8. The slope of the V-I curve is 13.89°. Calculate the value of resistance.

a) .247 Ω

b) .345 Ω

c) .231 Ω

d) .222 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 13.89° so R=tan=.247 Ω. It behaves like a normal resistor.

9. In a DC shunt motor, the e.m.f developed is proportional to ___________

a) I a

b) I a 2

c) I a 3

d) I a o

Answer: d

Explanation: In a DC shunt motor, the e.m.f developed is equal to K m ΦN. In a DC shunt, the motor field windings are connected separately and excited by a constant DC voltage. E = K m ΦN α Ia°.

10. Calculate the power factor angle during the resonance condition.

a) 0°

b) 10°

c) 80°

d) 90°

Answer: d

11. Calculate the value of the duty cycle if the system is on for 5 sec and off for inf sec.

a) 0

b) .4

c) .2

d) .1

Answer: a

Explanation: Duty cycle is Ton÷T total . It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷T total =5÷inf=0.

12. Calculate the value of the frequency of the AC supply in India.

a) 0 Hz

b) 50 Hz

c) 49 Hz

d) 60 Hz

Answer: b

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. AC supply magnitude is variable. It changes with time so the frequency of AC supply is 50 Hz.

13. DC series motor cannot run under no load.

a) True

b) False

Answer: a

Explanation: DC series motor cannot be run under no load condition because at no-load speed of the motor is very which can damage the shaft of the motor. There should be some load that should be connected to it.

14. Calculate the value of the frequency if the time period of the signal is .2 sec.

a) 5 Hz

b) 4 Hz

c) 2 Hz

d) 3 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. F = 1÷T = 1÷.2 = 5 Hz.

This set of Electric Drives Questions and Answers for Experienced people focuses on “Induction Motors – Controlling Speed by Adjusting the Supply Frequency”.

1. 40 V, 7 A, 70 rpm DC separately excited motor having a resistance of 0.3 ohms excited by an external dc voltage source of 4 V. Calculate the torque developed by the motor on half load.

a) 18.10 N-m

b) 4.24 N-m

c) 40.45 N-m

d) 52.64 N-m

Answer: a

Explanation: Back emf developed in the motor during the full load can be calculated using equation E b = V t -I×R a = 37.9 V and machine constant K m = E b ÷W m which is equal to 5.17. Torque can be calculated by using the relation T = K m × I = 5.17×3.5 = 18.10 N-m.

2. Calculate the active power developed by a motor using the given data: E b = 5.5 V and I = .5 A.

a) 2.75 W

b) 2.20 W

c) 5.30 W

d) 5.50 W

Answer: a

Explanation: Power developed by the motor can be calculated using the formula P = E b ×I = 5.5×.5 = 2.75 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

4. Calculate the value of the angular acceleration of the motor using the given data: J = .1 kg-m 2 , load torque = 45 N-m, motor torque = 55 N-m.

a) 100 rad/s 2

b) 222 rad/s 2

c) 300 rad/s 2

d) 400 rad/s 2

Answer: a

Explanation: Using the dynamic equation of motor J× = Motor torque – Load torque: .1× = 55-45=10, angular acceleration = 100 rad/s 2 .

4. Calculate the moment of inertia of the tennis ball having a mass of 7 kg and diameter of 152 cm.

a) 3.55 kgm 2

b) 4.47 kgm 2

c) 2.66 kgm 2

d) 1.41 kgm 2

Answer: c

Explanation: The moment of inertia of the tennis ball can be calculated using the formula I=mr 2 ×.5. The mass of the ball and diameter is given. I=×.5× 2 =2.66 kgm 2 . It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the thin spherical shell having a mass of 7.8 kg and diameter of 145.6 cm.

a) 2.72 kgm 2

b) 5.96 kgm 2

c) 5.45 kgm 2

d) 2.78 kgm 2

Answer: a

Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr 2 ×.66. The mass of the thin spherical shell and diameter is given. I=×.66× 2 =2.72 kgm 2 . It depends upon the orientation of the rotational axis.

6. Calculate the time period of the waveform y=87cos.

a) 2 sec

b) 37 sec

c) 3 sec

d) 1 sec

Answer: a

Explanation: The fundamental time period of the cosine wave is 2π. The time period of y is a 2π÷π=2 sec. The time period is independent of phase shifting and time shifting.

7. Calculate the useful power developed by a motor using the given data: P in = 10 W, I a = .6 A, R a =.2 Ω. Assume frictional losses are 2 W and windage losses are 3 W.

a) 4.928 W

b) 1.955 W

c) 1.485 W

d) 1.488 W

Answer: a

Explanation: Useful power is basically the shaft power developed by the motor that can be calculated using the formula P sh = P dev -. P dev = P in -I a 2 R a = 10-.6 2 =9.92 W. The useful power developed by the motor is P sh = P dev - = 9.92 – = 4.928 W.

8. The slope of the V-I curve is 86.5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 16.34 Ω

b) 15.88 Ω

c) 48.43 Ω

d) 54.57 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 16.34° so R=tan=16.34 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the active power in a 7481 H inductor.

a) 1562 W

b) 4651 W

c) 0 W

d) 4654 W

Answer: c

10. Calculate the active power in a 56 F capacitor.

a) 6.45 W

b) 0 W

c) 15.45 W

d) 14.23 W

Answer: b

Explanation: The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the capacitor so the angle between V & I is 90°. P=VIcos90 = 0 W. Current leads the voltage in case of the capacitor.

11. Calculate the active power in .an 18.064 H inductor.

a) 4.48 W

b) 17.89 W

c) 0 W

d) 25.45 W

Answer: c

12. Calculate the active power in a 1.7 Ω resistor with 1.8 A current flowing through it.

a) 5.5 W

b) 5.1 W

c) 5.4 W

d) 5.7 W

Answer: a

13. Calculate the value of the time period if the frequency of the signal is .48 sec.

a) 2 Hz

b) 3 Hz

c) 7 Hz

d) 9 Hz

Answer: a

Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷.48=2 Hz.

14. Choose the correct in the case of V/F control.

a) N s -N r =constant

b) N s *N r =constant

c) N s %N r =constant

d) N s +N r =constant

Answer: a

Explanation: In variable frequency control N s -N r remains constant. V/f control is part of the synchronous speed changing technique. It is the most used technique in controlling the induction motor.

This set of Electric Drives written test Questions & Answers focuses on “Induction Motors Speed Control – Rotor Voltage Injection”.

1. A 32-pole, 3-phase, 70 Hz induction motor is operating at a speed of 112 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 40.2

b) 46.1

c) 40.1

d) 40.6

Answer: c

Explanation: Given a number of poles = 32. Supply frequency is 70 Hz. Rotor speed is 112 rpm. N s = 120×f÷P=120×70÷32 = 262.5 rpm. S=N s -N r ÷N s = 262.5-112÷262.5=.573. F 2 =sf=.573×70=40.1 Hz.

2. A 20-pole, 3-phase, 90 Hz induction motor is operating at a speed of _______ rpm. The frequency of the rotor current of the motor in Hz is 20.

a) 418.56

b) 420.12

c) 421.23

d) 422.45

Answer: b

Explanation: Given a number of poles = 20. Supply frequency is 90 Hz. N s =120×f÷P=120×90÷20 = 540 rpm. S=20÷90 = .222 Hz. N r =N s = 420.12 rpm. Rotor speed is 420.12 rpm.

3. Calculate the amplitude of the sinusoidal waveform z=.27sin.

a) .287

b) .270

c) .216

d) .287

Answer: b

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference. By comparing the waveform z with the generalized sinusoidal expression we can see V m =.27 and ω=.369π rad/s.

4. R.M.S value of the sinusoidal waveform q=2.11cos.

a) 1.49 V

b) 1.56 V

c) 1.12 V

d) 1.78 V

Answer: a

Explanation: R.M.S value of the sinusoidal waveform is V m ÷2 ½ and r.m.s value of the trapezoidal waveform is V m ÷3 ½ . The peak value of the sinusoidal waveform is V m . The r.m.s value is V rms = 2.11÷2 ½ = 1.49 V.

5. The no-load circuit test on a 3-Φ induction motor is conducted at a rotor speed of _______

a) Zero

b) < Ns

c) > Ns

d) Ns

Answer: d

Explanation: No-load circuit test in an induction motor is also called an open circuit test so it is conducted at synchronous speed. Net input power taken is equal to the no-load rotational losses.

6. If induction motor air gap power is 67 KW and mechanically developed power is 26 KW, then rotor ohmic loss will be _________ KW.

a) 41

b) 42

c) 43

d) 44

Answer: a

7. Calculate the line voltage in star connection when phase voltage=45 V.

a) 77.9 V

b) 77.6 V

c) 77.2 V

d) 77.8 V

Answer: a

Explanation: The line voltage in case of star connection is 1.73 times of phase voltage. It leads the phase voltage by an angle of 30°. V L-L =1.73×45=77.9 V.

8. The slope of the V-I curve is 1.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .019 Ω

b) .036 Ω

c) .075 Ω

d) .084 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 1.° so R=tan=.019 Ω. The slope of the I-V curve is reciprocal of resistance.

9. If induction motor rotor power is 157.5 KW and gross developed power is 79.9 KW, then rotor ohmic loss will be _________ KW.

a) 77.5

b) 77.6

c) 76.9

d) 77.1

Answer: b

10. The power factor of a squirrel cage induction motor generally is ___________

a) .6-.8

b) .1-.2

c) .2-.4

d) .5-.7

Answer: a

Explanation: At light loads, the current drawn is largely a magnetizing current due to the air gap and hence the power factor is low. The power factor of a squirrel cage induction motor generally is .6-.8.

11. Calculate the active power in a .89 H inductor.

a) 1.535 W

b) 0 W

c) 2.484 W

d) 1.598 W

Answer: b

12. Calculate the time period of the waveform i=sin+sin.

a) 2 sec

b) 4 sec

c) 5 sec

d) 3 sec

Answer: a

Explanation: The fundamental time period of the sine wave is 2π. The time period of i is L.C.M {2,2}=2 sec. The time period is independent of phase shifting and time shifting.

13. Calculate the total heat dissipated in a rotor resistor of 14.23 Ω when .65 A current flows through it.

a) 6.45 W

b) 6.01 W

c) 6.78 W

d) 6.98 W

Answer: b

Explanation: The rotor resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. P=I 2 R=.65×.65×14.23=6.01 W.

14. Calculate mark to space ratio if the system is on for 4.3 sec and off for 78.2 sec.

a) .054

b) .047

c) .039

d) .018

Answer: a

Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff = 4.3÷78.2 = .054.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Induction Motors Speed Control – Slip Energy Recovery”.

1. In the rotor voltage injection method, when an external voltage source is in phase with the main voltage then speed will ___________

a) Increase

b) Decrease

c) Remain unchanged

d) First increases then decrease

Answer: a

Explanation: In the rotor injection method, when an external voltage is in phase with the main voltage net voltage increases and the value of slip decreases and the value of rotor speed increases.

2. A 2-pole, 3-phase, ______ Hz induction motor is operating at a speed of 550 rpm. The frequency of the rotor current of the motor in Hz is 2.

a) 9.98

b) 9.71

c) 9.12

d) 9.37

Answer: d

Explanation: Given a number of poles = 2. Rotor speed is 550 rpm. N s =120×f÷P=120×f÷2 = 60f rpm. S=N s -N r ÷N s . F 2 =sf. S=F 2 ÷f. Supply frequency is 9.37 Hz.

3. Calculate the average value of the sinusoidal waveform x=848sin.

a) 0

b) 78 V

c) 15 V

d) 85 V

Answer: a

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, Ω represents angular frequency, α represents a phase difference. The average value of a sine wave is zero because of equal and opposite lobes areas.

4. R.M.S value of the periodic square waveform of amplitude 72 V.

a) 72 V

b) 56 V

c) 12 V

d) 33 V

Answer: a

Explanation: R.M.S value of the periodic square waveform is V m and r.m.s value of the trapezoidal waveform is V m ÷3 ½ . The peak value of the periodic square waveform is V m .

5. In the rotor voltage injection method, when an external voltage source is in opposite phase with the main voltage then speed will ___________

a) Increase

b) Decrease

c) Remain unchanged

d) First increases then decrease

Answer: b

Explanation: In the rotor injection method, when an external voltage is in opposite phase with the main voltage net voltage decreases and the value of slip increases and the value of rotor speed decreases.

6. The rotor injection method is a part of the slip changing technique.

a) True

b) False

Answer: a

Explanation: Rotor injection method comes under slip changing technique. It uses an external voltage source to change the slip value. The load torque remains constant here.

7. The slip recovery scheme is a part of the synchronous speed changing technique.

a) True

b) False

Answer: a

Explanation: Slip recovery scheme comes under the synchronous speed changing technique. It uses an external induction machine to change the frequency value. The load torque remains constant here.

8. The slope of the V-I curve is 9.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .16 Ω

b) .26 Ω

c) .25 Ω

d) .44 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 9.1° so R=tan=.16 Ω. The slope of the I-V curve is reciprocal of resistance.

9. If induction motor air gap power is 1.8 KW and gross developed power is .1 KW, then rotor ohmic loss will be _________ KW.

a) 1.7

b) 2.7

c) 3.7

d) 4.7

Answer: a

10. The power factor of a squirrel cage induction motor is ___________

a) High at light load only

b) High at heavy loads only

c) Low at the light and heavy loads both

d) Low at rate load only

Answer: b

Explanation: At heavy loads, the current drawn is high due to which active power component increases. Increase in active power component increases the power factor of the machine.

11. At low values of slip, the electromagnetic torque is directly proportional to ___________

a) s

b) s 2

c) s 3

d) s 4

Answer: a

Explanation: At low values of slip, the electromagnetic torque is directly proportional to slip value. Due to heavy loading slip value decreases which increases the ratio of R 2 ÷s.

12. Calculate the time period of the waveform v=12sin+144sin+ 445sin.

a) 8 sec

b) 4 sec

c) 7 sec

d) 3 sec

Answer: b

Explanation: The fundamental time period of the sine wave is 2π. The time period of z is L.C.M {4,1,2}=4 sec. The time period is independent of phase shifting and time shifting.

13. Calculate the total heat dissipated in a resistor of 44 Ω when 0 A current flows through it.

a) 0 W

b) 2 W

c) 1.5 W

d) .3 W

Answer: a

14. The value of slip at which maximum torque occurs ________

a) R 2 ÷X 2

b) 4R 2 ÷X 2

c) 2R 2 ÷X 2

d) R 2 ÷3X 2

Answer: a

Explanation: The maximum torque occurs when the slip value is equal to R 2 ÷X 2 . Maximum torque is also known as breakdown torque, stalling torque and pull-out torque.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Induction Motors – Controlling Speed Using Inductance”.

1. Calculate the voltage regulation in the synchronous machine if the no-load voltage is 12 V and the full load voltage is 15V.

a) -20%

b) -40%

c) -60%

d) -80%

Answer: a

Explanation: Voltage regulation is defined as the fluctuation in the load voltage when the load is varied from no-load to full load. V.R = ÷ Full load voltage=12-15÷15 = -20%.

2. Calculate the condition for maximum voltage regulation in the synchronous machine.

a) Φ=ϴ s

b) Φ=2ϴ s

c) Φ=4ϴ s

d) Φ=8ϴ s

Answer: a

Explanation: Voltage regulation is maximum in case of inductive load. The condition for maximum voltage regulation is when the power factor angle of the load becomes equal to the impedance angle. V.R=R p.u cos+X p.u sin. Differentiate V.R with respect to Φ and put it equal to zero. We will get tan = X p.u ÷R p.u =tan(ϴ s ) then Φ=ϴ s .

3. Zero voltage regulation can be only achieved in leading power factor load.

a) True

b) False

Answer: a

Explanation: Zero voltage regulation only occurs during a leading power factor load. Condition for zero voltage regulation occurs when Φ+ϴ s > 90 o . For example – Capacitive load.

4. R.M.S value of the sinusoidal waveform v=211sin.

a) 149.19 V

b) 156.23 V

c) 116.57 V

d) 178.64 V

Answer: a

Explanation: R.M.S value of the sinusoidal waveform is Vm÷2 ½ = 211÷2 ½ = 149.19 V and r.m.s value of the trapezoidal waveform is V m ÷3 ½ . The peak value of the sinusoidal waveform is V m .

5. Calculate the peak value of sinusoidal waveform if the r.m.s value is 21 V.

a) 29.69 V

b) 48.74 V

c) 69.23 V

d) 25.74 V

Answer: a

Explanation: R.M.S value of the sinusoidal waveform is V peak ÷2 ½ and r.m.s value of the trapezoidal waveform is V m ÷3 ½ . The peak value of the sinusoidal waveform is V r.m.s ×2 ½ . V peak = V r.m.s ×2 ½ =21×1.414=29.69 V.

6. If induction motor air gap power is 2 KW and mechanically developed power is 1 KW, then rotor ohmic loss will be _________ KW.

a) 1

b) 2

c) 3

d) 4

Answer: a

7. Calculate the line voltage in the delta connection when phase voltage=45 V.

a) 46 V

b) 47 V

c) 45 V

d) 78 V

Answer: c

Explanation: The line voltage in case of delta connection is phase voltage. Line current leads the phase current by an angle of 30°. V L-L =V ph = 45 V.

8. The slope of the V-I curve is 35.48°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 0.452 Ω

b) 0.462 Ω

c) 0.752 Ω

d) 0.712 Ω

Answer: d

Explanation: The slope of the V-I curve is resistance. The slope given is 35.48° so R=tan=.712 Ω. The slope of the I-V curve is reciprocal of resistance.

9. If induction motor rotor power is 157.5 KW and gross developed power is 79.9 KW, then rotor ohmic loss will be _________ KW.

a) 77.5

b) 77.6

c) 76.9

d) 77.1

Answer: b

10. Calculate the line current in the delta connection when phase current=17 A.

a) 29.44 A

b) 24.64 A

c) 23.48 A

d) 26.56 A

Answer: a

Explanation: The line voltage in case of delta connection is phase voltage. Line current leads the phase current by an angle of 30°. I L-L = 1.73×I ph = 29.44 A.

11. Calculate the active power in a 168.12 H inductor.

a) 65 W

b) 0 W

c) 68 W

d) 64 W

Answer: b

12. Calculate the maximum value of slip when rotor resistance is 2 Ω and rotor reactance is 3 Ω.

a) 0.66

b) 0.33

c) 0.44

d) 0.99

Answer: a

Explanation: The maximum torque occurs when the slip value is equal to R 2 ÷X 2 . Maximum torque is also known as breakdown torque, stalling torque and pull-out torque. The maximum value of slip is R 2 ÷X 2 =2/3=.66.

13. Calculate the total heat dissipated in a rotor resistor of 21 Ω when .81 A current flows through it.

a) 13.77 W

b) 12.56

c) 16.78 W

d) 13.98 W

Answer: a

14. Calculate the active power in an 8.12 F capacitor.

a) 89 W

b) 41 W

c) 0 W

d) 48 W

Answer: c

This set of Electric Drives Problems focuses on “Induction Motors – Controlling Speed by Adjusting the Stator Voltage”.

1. The stator voltage control method is a part of the slip changing technique.

a) True

b) False

Answer: a

Explanation: Stator voltage control method comes under slip changing technique. The load torque remains constant here. SV 2 =constant.

2. Calculate the value of new slip using the given data: V 1 =12 V, S 1 =.1, V 2 =5.

a) 0.576

b) 0.247

c) 0.487

d) 0.987

Answer: a

Explanation: This question is based on the concept of stator voltage control method. The load torque remains constant. S 1 V 1 2 =S 2 V 2 2 =constant. S 2 =.576.

3. Calculate the average value of the sinusoidal waveform y=4.56cos.

a) 41 V

b) 0 V

c) 48 V

d) 78 V

Answer: b

Explanation: Sinusoidal waveform is generally expressed in the form of V=V m sin where V m represents peak value, ω represents angular frequency, α represents a phase difference. The average value of a sine wave is zero because of equal and opposite lobes areas. Since sine wave is an odd function then the net area of the waveform over a period is Net area = A+ = 0. The average value is Net area÷Time=0.

4. R.M.S value of the periodic square waveform of amplitude 20 V is _______

a) 20 V

b) 18 V

c) 17 V

d) 13 V

Answer: a

Explanation: R.M.S value of the periodic square waveform is V m and r.m.s value of the trapezoidal waveform is V m ÷3 ½ . The peak value of the periodic square waveform is V m . V m =20 V.

5. Calculate the time period of the waveform y=sin+cos.

a) 20 sec

b) 30 sec

c) 40 sec

d) 10 sec

Answer: a

Explanation: The fundamental time period of the sine and cosine wave is 2π. The time period of y is L.C.M {20,10}=20 sec. The time period is independent of phase shifting and time shifting.

6. The V/F control is a part of the synchronous speed changing technique.

a) True

b) False

Answer: b

Explanation: V/F control comes under the synchronous speed changing technique. Speed above or below synchronous speed can be achieved using V/F control.

7. The slope of the V-I curve is 56.489°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 1.5 Ω

b) 1.6 Ω

c) 2.5 Ω

d) 1.4 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 56.489° so R=tan=1.5 Ω. The slope of the I-V curve is reciprocal of resistance.

8. If induction motor air gap power is 28.63 KW and gross developed power is 18.8 KW, then rotor ohmic loss will be _________ KW.

a) 9.83

b) 10.55

c) 15.54

d) 4.74

Answer: a

9. Calculate the value of inductive reactance if F=50 Hz and L=12 H.

a) 3768 Ω

b) 2578 Ω

c) 2477 Ω

d) 2456 Ω

Answer: a

Explanation: Inductive reactance can be calculated using the relation X L =2×3.14×f×L. The value of inductive reactance is X L =2×3.14×50×12=3768 Ω.

10. Calculate the value of capacitive reactance if F=60 Hz and C=14 H.

a) 189.5 mΩ

b) 252.4 mΩ

c) 244.5 mΩ

d) 244.8 mΩ

Answer: a

Explanation: Capacitive reactance can be calculated using the relation X c =1÷2×3.14×f×L. The value of capacitive reactance is X c =1÷2×3.14×50×12=189.5 mΩ.

11. Calculate the time period of the waveform v=.6sin+17cos+ 4tan.

a) 88 sec

b) 20 sec

c) 70 sec

d) 43 sec

Answer: b

Explanation: The fundamental time period of the sine wave is 2π. The time period of v is L.C.M {20,1,20}=20 sec. The time period is independent of phase shifting and time shifting.

12. Calculate the value of resistance if total heat dissipated is 78 W when 5 A current flows through it.

a) 3.12 Ω

b) 2.24 Ω

c) 1.45 Ω

d) 5.13 Ω

Answer: a

13. The value of slip at the starting of an induction motor is ________

a) 0

b) 1

c) 2

d) 3

Answer: b

Explanation: The maximum torque occurs when the slip value is equal to R 2 ÷X 2 . At the starting, the rotor speed is equal to zero so slip value is 1.

14. The value of the slip of an induction motor during full load condition is ________

a) 0.99

b) .1

c) .8

d) 0

Answer: d

Explanation: During full load condition the speed of the rotor is nearly synchronous speed. The value of slip is nearly zero during full load condition.

This set of Electric Drives Questions and Answers for Experienced people focuses on “Induction Motors – Controlling Speed by Adjusting the Supply Frequency”.

1. 40 V, 7 A, 70 rpm DC separately excited motor having a resistance of 0.3 ohms excited by an external dc voltage source of 4 V. Calculate the torque developed by the motor on half load.

a) 18.10 N-m

b) 4.24 N-m

c) 40.45 N-m

d) 52.64 N-m

Answer: a

2. Calculate the active power developed by a motor using the given data: E b = 5.5 V and I = .5 A.

a) 2.75 W

b) 2.20 W

c) 5.30 W

d) 5.50 W

Answer: a

4. Calculate the value of the angular acceleration of the motor using the given data: J = .1 kg-m 2 , load torque = 45 N-m, motor torque = 55 N-m.

a) 100 rad/s 2

b) 222 rad/s 2

c) 300 rad/s 2

d) 400 rad/s 2

Answer: a

Explanation: Using the dynamic equation of motor J× = Motor torque – Load torque: .1× = 55-45=10, angular acceleration = 100 rad/s 2 .

4. Calculate the moment of inertia of the tennis ball having a mass of 7 kg and diameter of 152 cm.

a) 3.55 kgm 2

b) 4.47 kgm 2

c) 2.66 kgm 2

d) 1.41 kgm 2

Answer: c

5. Calculate the moment of inertia of the thin spherical shell having a mass of 7.8 kg and diameter of 145.6 cm.

a) 2.72 kgm 2

b) 5.96 kgm 2

c) 5.45 kgm 2

d) 2.78 kgm 2

Answer: a

Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr 2 ×.66. The mass of the thin spherical shell and diameter is given. I=×.66× 2 =2.72 kgm 2 . It depends upon the orientation of the rotational axis.

6. Calculate the time period of the waveform y=87cos.

a) 2 sec

b) 37 sec

c) 3 sec

d) 1 sec

Answer: a

Explanation: The fundamental time period of the cosine wave is 2π. The time period of y is a 2π÷π=2 sec. The time period is independent of phase shifting and time shifting.

7. Calculate the useful power developed by a motor using the given data: P in = 10 W, I a = .6 A, R a =.2 Ω. Assume frictional losses are 2 W and windage losses are 3 W.

a) 4.928 W

b) 1.955 W

c) 1.485 W

d) 1.488 W

Answer: a

8. The slope of the V-I curve is 86.5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 16.34 Ω

b) 15.88 Ω

c) 48.43 Ω

d) 54.57 Ω

Answer: a

Explanation: The slope of the V-I curve is resistance. The slope given is 16.34° so R=tan=16.34 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the active power in a 7481 H inductor.

a) 1562 W

b) 4651 W

c) 0 W

d) 4654 W

Answer: c

10. Calculate the active power in a 56 F capacitor.

a) 6.45 W

b) 0 W

c) 15.45 W

d) 14.23 W

Answer: b

Explanation: The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the capacitor so the angle between V & I is 90°. P=VIcos90 = 0 W. Current leads the voltage in case of the capacitor.

11. Calculate the active power in .an 18.064 H inductor.

a) 4.48 W

b) 17.89 W

c) 0 W

d) 25.45 W

Answer: c

12. Calculate the active power in a 1.7 Ω resistor with 1.8 A current flowing through it.

a) 5.5 W

b) 5.1 W

c) 5.4 W

d) 5.7 W

Answer: a

13. Calculate the value of the time period if the frequency of the signal is .48 sec.

a) 2 Hz

b) 3 Hz

c) 7 Hz

d) 9 Hz

Answer: a

Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷.48=2 Hz.

14. Choose the correct in the case of V/F control.

a) N s -N r =constant

b) N s *N r =constant

c) N s %N r =constant

d) N s +N r =constant

Answer: a

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Induction Motors – Voltage/Frequency Control”.

1. Calculate the minimum value of the active power in the cylindrical rotor synchronous machine? (E b represents armature emf, V t represents terminal voltage, δ represents rotor angle, X represents reactance)

a) E b ×V t ×sinδ÷X

b) 0

c) E b 2×V t ×sinδ÷X

d) E b ×V t ÷X

Answer: b

2. Cylindrical pole machines are less stable than salient rotor machines.

a) True

b) False

Answer: a

Explanation: Cylindrical pole machines are less stable than salient rotor machines because of the less short circuit ratio and less real power transfer capability. The air gap length in cylindrical pole machines is less as compare to salient rotor machines.

3. The unit of area is m 2 .

a) True

b) False

Answer: a

Explanation: Area is defined as the product of length and breadth. The length and breadth are expressed in the meter. The unit of area is m 2 .

4. Calculate the power factor if the power angle is 45°.

a) .707

b) .407

c) .608

d) 1

Answer: a

Explanation: Power factor is the ratio of the real power to the apparent power. It measures the useful power contained in the total power cosΦ=cos=.707.

5. Calculate the reactive power in a 451.26 Ω resistor.

a) 1 VAR

b) .6 VAR

c) 0 VAR

d) .9 VAR

Answer: c

6. What is the unit of displacement?

a) m/s

b) m

c) atm/m

d) Volt

Answer: a

Explanation: Displacement is the difference between the final and initial point. It is a vector quantity. It is expressed in the meter. It is not a tensor quantity. It has direction.

7. Calculate the reactive power in a 725.45 Ω resistor.

a) 122.1 VAR

b) 261.1 VAR

c) 0 VAR

d) 199.7 VAR

Answer: c

8. The slope of the I-V curve is 180°. Calculate the value of resistance.

a) inf Ω

b) .81 Ω

c) 45 Ω

d) 41.2 Ω

Answer: a

Explanation: The slope of the I-V curve is the reciprocal of the resistance. The slope given is 180° so R=1÷tan=inf Ω. The slope of the V-I curve is resistance. It behaves as an open circuit.

9. Calculate the value of the duty cycle if the system is on for 1 sec and off for inf sec.

a) 0

b) .89

c) .148

d) .46

Answer: a

Explanation: Duty cycle is Ton÷T total . It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷T total =1÷inf = 0.

10. The phase difference between voltage and current in the purely inductive coil.

a) 20°

b) 90°

c) 40°

d) 88°

Answer: b

Explanation: In the case of a purely inductive coil, the voltage leads the current by 90° or the current lags the voltage by 90°. The phase difference between voltage and current is 90°.

11. Armature reaction is purely demagnetizing in nature due to a capacitive load in the synchronous motor.

a) True

b) False

Answer: a

Explanation: Due to a capacitive load, armature current is in opposite and 90° phase with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in 180° phase and 90° phase with the field flux. The nature of the armature reaction is partially demagnetizing and cross magnetizing.

12. Armature reaction is cross-magnetizing in nature due to a resistive load in the synchronous generator.

a) True

b) False

Answer: a

13. Calculate the value of the short circuit ratio if V oc =78 V, I sc =15 A with field current = 5 A.

a) 5.2

b) 4.8

c) 3.2

d) 1.8

Answer: a

Explanation: The value of the short circuit ratio is the ratio of open circuit voltage to the short circuit current with same field current. SCR=V oc ÷I sc =78÷15=5.2.

14. SCR determines the stability of the synchronous machine.

a) True

b) False

Answer: a

Explanation: The value of the short circuit ratio determines the stability of the synchronous machine. It is inversely proportional to the per unit value of reactance. Stability of the synchronous machine depends upon the maximum power transfer capability.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Induction Motors – Current Source Speed Control”.

1. Calculate the value of the short circuit ratio if the per unit value of synchronous reactance is 1.5 Ω.

a) 0.66

b) 0.33

c) 0.17

d) 0.12

Answer: a

Explanation: The value of the short circuit ratio determines the stability of the synchronous machine. It is inversely proportional to the per unit value of reactance. SCR=1÷X s =1÷1.5=.66.

2. A higher value of armature reaction means the lower value of the short circuit ratio.

a) True

b) False

Answer: a

Explanation: SCR is inversely proportional to the per unit value of synchronous reactance. Armature reaction depends upon the value of the synchronous reactance. A higher value of armature reaction means the lower value of the short circuit ratio.

3. The unit of SCR is Ω.

a) True

b) False

Answer: b

Explanation: SCR is the ratio of the field currents required to produce open circuit voltage and short circuit current. It is a ratio. It has no unit.

4. Calculate the power factor if the power angle is 12°.

a) .87

b) .47

c) .97

d) 0

Answer: c

Explanation: Power factor is the ratio of the real power to the apparent power. It measures the useful power contained in the total power cosΦ=cos=.97.

5. Calculate the reactive power in an 84 MΩ resistor.

a) 1.47 VAR

b) 45.6 VAR

c) 0.59 VAR

d) 0 VAR

Answer: d

6. Calculate the value of the resistance of a wire of length = 12 m, area = 2 m 2 and Ρ = 6 Ω-m.

a) 36 Ω

b) 6 Ω

c) 26 Ω

d) 10 Ω

Answer: a

Explanation: Resistance is the opposition offered by the body to the flow of the current. It is expressed in terms of Ω. R = ΡL/A = 36 Ω.

7. Calculate the reactive power in a 29.6 Ω resistor.

a) 45.1 VAR

b) 41.1 VAR

c) 46 VAR

d) 0 VAR

Answer: d

8. The slope of the I-V curve is 160°. Calculate the value of resistance.

a) .36 Ω

b) .41 Ω

c) .89 Ω

d) 2.74 Ω

Answer: d

Explanation: The slope of the I-V curve is the reciprocal of the resistance. The slope given is 160° so R=1÷tan=2.74 Ω. The slope of the V-I curve is resistance.

9. Calculate the value of the power if voltage=4.2 V and current=10 A.

a) 42 W

b) 12 W

c) 16 W

d) 18 W

Answer: a

Explanation: Power is the product of the voltage and current. It is generally expressed in term of W. P=vi=4.2×10=42 W. 1 horsepower=746 W.

10. 1 calorie is equal to ________

a) 4.18 J

b) 5.23 J

c) 6.23 J

d) 8 J

Answer: a

Explanation: The calorie is the unit of energy. It is equal to 4.18 Joule. It is the amount of heat energy required to raise the temperature of one 1 g of water by 1° Celsius.

11. 1 Horse-power is equal to _______

a) 745.7 W

b) 741 W

c) 747.7 W

d) 740 W

Answer: a

Explanation: 1 Horse-power is equal to 745.7 W. It is equal to the rate at which work is done. Horse-power is the amount of work done by horse in carrying the weight of 90 kg for 50 meters in 60 seconds.

12. Calculate the value of the time period if the frequency of the signal is 58 sec.

a) .017 sec

b) .014 sec

c) .045 sec

d) .077 sec

Answer: a

Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T=1÷F=1÷58=.017 sec.

13. Calculate the value of the short circuit ratio if the per unit value of synchronous reactance is 4.2 Ω.

a) 0.26

b) 0.23

c) 0.67

d) 0.72

Answer: b

Explanation: The value of the short circuit ratio determines the stability of the synchronous machine. It is inversely proportional to the per unit value of reactance. SCR=1÷X s =1÷4.2=.23.

14. Calculate the value of the short circuit ratio if the per unit value of synchronous reactance is 1 Ω. Assume armature resistance is 0 ohm.

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: The value of the short circuit ratio determines the stability of the synchronous machine. It is inversely proportional to the per unit value of reactance. SCR=1÷X s =1÷1=1.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Electric Motors – Regenerative Braking”.

1. Regenerative braking is not possible in a series motor.

a) True

b) False

Answer: a

Explanation: In regenerative braking, the motor acts as a generator. The back emf is more than the terminal voltage in case of regenerative braking.

2. Regenerative is the best electrical braking among all braking techniques.

a) True

b) False

Answer: a

Explanation: In regenerative braking, the power is fed back to the source from the battery. The kinetic energy of the motor shaft is converted into electrical energy.

3. Full form of SCIM.

a) Squirrel cage induction motor

b) Solid cage induction motor

c) Square cage induction motor

d) Squirrel cage inverter motor

Answer: a

Explanation: SCIM stands for squirrel cage induction motor. SCIM rotor is made up of aluminum, copper bars. There are no armature conductors involved in it.

4. Wound rotor induction motor has better ________ characteristics than Squirrel cage induction motor.

a) Starting

b) Running

c) Modified

d) Quasi-state

Answer: a

Explanation: Wound rotor induction motor has better starting characteristics because external resistance can be connected to it using slip rings to start the motor. In the case of SCIM, no external resistance can be connected to the rotor side.

5. All circuits are always _________

a) Networks

b) Resistors

c) Capacitors

d) Inductors

Answer: a

Explanation: The network is defined as the interconnection of electrical elements that may or may not has a closed path. The circuit is defined as the interconnection of an element that must have at least one closed path. All circuits are always networks but vice-versa is not true.

6. The dead network does not have any _________

a) Dependent source

b) Independent source

c) Resistor

d) Capacitor

Answer: b

Explanation: In the case of the dead network there are no independent sources available in the electrical circuit. The Thevenin voltage of the circuit is zero because no force is available to drive the current in the circuit.

7. During short circuit condition, the voltage is equal to ________

a) 2 V

b) 0 V

c) 1 V

d) 3 V

Answer: b

Explanation: During the short circuit condition, the voltage is equal to zero volts. According to Ohm’s law, V=IR=0 V. The value of the current can be positive, negative or zero.

8. When 5 A current flows into the positive terminal of voltage source 6 V. Calculate the power delivered by the source.

a) -30 W

b) 36 W

c) -43 W

d) 50 W

Answer: a

Explanation: When the current enters the positive terminal of an element it will always absorb the power and when the current leaves the positive terminal it will deliver the power. Power delivered is -5×6=-30 W.

9. When 10 A current leaves the positive terminal of resistance 5Ω. Calculate the power absorbed by the resistance.

a) -308 W

b) 500 W

c) -63 W

d) 60 W

Answer: b

10. Calculate the equivalent inductance when two inductors are connected in series of values 4 H and 9 H.

a) 13 H

b) 10 H

c) 11 H

d) 12 H

Answer: a

Explanation: When two inductors are connected in series their equivalent inductance is equal to the sum of the individual inductances. L eq =L 1 +L 2 =4+9=13 H.

11. Calculate the equivalent resistance when three resistances are connected in series of values 8 Ω, 3 Ω, 1 Ω.

a) 12 Ω

b) 18 Ω

c) 15 Ω

d) 16 Ω

Answer: a

Explanation: When three resistances are connected in series their equivalent resistance is equal to the sum of the individual resistances. R eq =R 1 + R 3 +R 2 =12 Ω.

12. Calculate the active power in an inf Ω resistor with 0 current flowing through it.

a) inf MW

b) 4.44 MW

c) 12 MW

d) 6.18 MW

Answer: a

13. When 15 A current flows into the positive terminal of voltage source 60 V. Calculate the power delivered by the source.

a) -900 W

b) 936 W

c) -943 W

d) 950 W

Answer: a

14. Source transformation is not valid for ___________

a) Ideal voltage source

b) Ideal current and voltage source

c) Practical voltage source

d) Practical current source

Answer: b

Explanation: Source transformation is not valid for an ideal current and voltage source. The internal resistance for ideal voltage and current source is zero due to which source transformation is not valid.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Electric Motors – Dynamic Braking”.

1. Which braking method is the best method for obtaining high braking torque?

a) Regenerative braking

b) Plugging

c) Dynamic braking

d) Rheostatic braking

Answer: b

Explanation: Plugging is the best braking method among all braking techniques. In plugging the value of the armature current reverses and the mechanical energy is extracted. A very high braking torque is produced in case of plugging.

2. The polarity of back e.m.f changes in which of the method?

a) Plugging

b) Regenerative braking

c) Dynamic braking

d) Rheostatic braking

Answer: a

Explanation: In case of Plugging braking technique polarity of the back e.m.f voltage changes due to which current direction changes and mechanical, electrical energy are wasted in the form of heat in resistors.

3. Full form of WRIM.

a) Wound round induction motor

b) World cage induction motor

c) Wolf cage induction motor

d) Squirrel cage inverter motor

Answer: a

Explanation: WRIM stands for wound round induction motor. WRIM rotor is made up of armature conductors. There are no copper bars, rods involved in it.

4. Calculate the energy stored in the capacitor if the voltage across the capacitor is 20 V and capacitance value is 2 F.

a) 400 J

b) 200 J

c) 100 J

d) 50 J

Answer: a

Explanation: The energy stored in the capacitor is .5×C×V 2 . It is the total amount of energy stored in the capacitor in the steady state condition. E=.5×C×V 2 =.5×2×20×20=400 J.

5. Calculate the energy stored in the inductor if the current value is 4 A and inductance value is 1 H.

a) 4 J

b) 2 J

c) 8 J

d) 5 J

Answer: c

Explanation: The energy stored in the inductor is .5×L×I 2 . It is the total amount of energy stored in the inductor in the steady state condition. E=.5×L×I 2 =.5×1×4×4=8 J.

6. All networks are always circuits.

a) True

b) False

Answer: b

Explanation: The network is defined as the interconnection of electrical elements that may or may not has a closed path. The circuit is defined as the interconnection of an element that must have at least one closed path. All networks are not always circuits but vice-versa is true.

7. The Thevenin voltage of a dead circuit is zero volts.

a) True

b) False

Answer: a

8. During the open circuit condition, the current is equal to ______

a) 1 A

b) 0 A

c) 4 A

d) 7 A

Answer: b

Explanation: During the open circuit condition, the current is equal to zero amperes. According to Ohm’s law, I=V/R=0 A. The value of the voltage can be positive, negative or zero.

9. The minimum value of the impedance in case of the series RLC network is ___________

a) R

b) 0

c) X l

d) X c

Answer: a

Explanation: The minimum value of the series impedance in case of the series RLC network is R. During resonance condition X l =X c so Z=(R 2 +(X l -X C ) 2 ) .5 =R.

10. Calculate the quality factor for the purely inductive coil.

a) inf

b) 10

c) 3

d) 6

Answer: a

Explanation: The quality factor is defined as the ratio of the reactive power to the active power consumed. The purely inductive coil always absorb reactive power. The value of active power consumed is 0. Quality factor=Q/0=inf.

11. Calculate the quality factor for the pure capacitor.

a) inf

b) 0

c) 1

d) 2

Answer: a

Explanation: The quality factor is defined as the ratio of the reactive power to the active power consumed. The pure capacitor always absorbs reactive power. The value of active power consumed is 0. Quality factor=Q/0=inf.

12. Calculate the equivalent inductance when n inductors are connected in series of values L.

a) n 2 L H

b) nL H

c) .5L H

d) 2 H

Answer: b

Explanation: When two inductors are connected in series their equivalent inductance is equal to the sum of the individual inductances. L eq =L+L+…..+n times L=nL H.

13. Calculate the equivalent resistance when two resistances are connected in series of values 12 Ω, 10 Ω.

a) 32 Ω

b) 22 Ω

c) 47 Ω

d) 17 Ω

Answer: b

Explanation: When two resistances are connected in series their equivalent resistance is equal to the sum of the individual resistances. R eq =R 1 +R 2 =22 Ω.

14. Reactive power is positive in the case of the inductor.

a) True

b) False

Answer: a

Explanation: The reactive power is defined as the product of the voltage, current, and sine of the difference between the phase angle of voltage and current. In the case of an inductor, the voltage leads the current. The value of sine is positive and reactive power is positive.

15. Reactive power is positive in the case of the capacitor.

a) True

b) False

Answer: b

Explanation: The reactive power is defined as the product of the voltage, current, and sine of the difference between the phase angle of voltage and current. In the case of a capacitor, the voltage lags the current. The value of sine is negative and reactive power is negative.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Electric Motors – Countercurrent Braking”.

1. The value of unit step function at t=5 sec is ________

a) 0

b) 1

c) -1

d) 2

Answer: b

Explanation: Unit step function is discontinuous at t=0. The value of the unit step function is 1 for t>0 and zero for t<0. It is right handed and bounded signal.

2. The value of impulse at t=0 is __________

a) 0

b) 2

c) 1

d) infinite

Answer: d

Explanation: Impulse function is neither an energy signal nor a power signal. It is a pulse of infinite amplitude and zero width. The value of impulse is 0 for time not equal to zero.

3. Full form of IFOC.

a) Indirect field oriented control

b) Inverter field oriented control

c) Isolated field oriented control

d) Insight field oriented control

Answer: a

Explanation: IFOC stands for indirect field oriented control. DTC and IFOC are the techniques used for controlling the speed and torque of a 3-phase induction motor.

4. The characteristics shown by an element in the I-V curve is a straight line passing through the origin with a positive slope. The nature of the element is _______

a) Non-linear, Bilateral

b) Linear, Unilateral

c) Linear, Bilateral

d) Non-linear, Unilateral

Answer: c

Explanation: The nature of the element is linear and bilateral. It follows the principle of homogeneity and law of additivity. For bilateral nature, it should be symmetrical in the first and third quadrant.

5. Calculate the total energy stored in the inductor if the current value is 4 A and inductance values are 1 H and 2 H.

a) 72 J

b) 22 J

c) 82 J

d) 24 J

Answer: d

Explanation: The energy stored in the inductor is .5×L eq ×I 2 . The value of L eq =1+2=3 H. The total amount of energy stored in the inductor in the steady state condition. E=.5× eq ×I 2 =.5×3×4×4=24 J.

6. Mesh analysis is only applied to the planar circuits.

a) True

b) False

Answer: a

Explanation: The mesh analysis is one of the network tools to find the responses in the circuits. It is only applicable to planar circuits. Planar circuits are 2-D circuits as mesh analysis involves mesh circulating currents.

7. All loops are meshes.

a) True

b) False

Answer: b

Explanation: Loop is defined as the closed path which can contain small closed paths. Mesh does not contain any closed path. All meshes are loops but vice-versa is not true.

8. The nodal method is better than mesh analysis for solving circuit problems.

a) True

b) False

Answer: a

Explanation: The nodal method is one of the best methods to solve any circuit problem. It is based on the KCL equation. It is applicable for planar and non-planar circuits. Mesh analysis cannot be applicable for non-planar circuits.

9. The minimum value of the admittance in case of the parallel RLC network is ___________

a) R

b) 1÷R

c) X l -X c

d) X c

Answer: b

Explanation: The minimum value of the admittance in case of the parallel RLC network is R. During resonance condition X l =X c so Y=1÷Z=(R 2 +(X l -X c ) 2 ) .5 =1÷R.

10. Calculate the value of the inductive reactance during resonance if the value of capacitive reactance is 2 ohm.

a) 8 Ω

b) 2 Ω

c) 3 Ω

d) 1 Ω

Answer: b

Explanation: Resonance is defined as the phenomenon in which energy of any element changes from one form to another. During resonance condition X l =X c =2 Ω.

11. Calculate the quality factor for the resistor.

a) inf

b) 0

c) 1

d) .7

Answer: b

Explanation: The quality factor is defined as the ratio of the reactive power to the active power consumed. The resistor always absorbs active power. The value of reactive power consumed is 0. Quality factor=0/P=0.

12. Selectivity value increases for lower values of bandwidths.

a) True

b) False

Answer: a

Explanation: Bandwidth is defined as the range of frequencies for which the signal exists. Selectivity is inversely proportional to the bandwidth. Lower the bandwidth lower will be the cost.

13. Calculate the equivalent resistance when two resistances are connected in parallel of values 2 Ω, 2 Ω.

a) 1 Ω

b) 2 Ω

c) 4 Ω

d) 7 Ω

Answer: a

Explanation: When two resistances are connected in parallel their equivalent resistance is equal to the harmonic mean of the individual resistances. R eq =R 1 .R 2 ÷(R 1 +R 2 )=22 Ω.

14. Calculate the equivalent capacitance when two capacitors are connected in parallel of values 8 F, 7 F.

a) 15 F

b) 20 F

c) 40 F

d) 7 F

Answer: a

Explanation: When two capacitances are connected in parallel their equivalent capacitance is equal to the sum of the individual capacitances. C eq =(C 1 +C 2 )=15 F.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Regenerative Braking of DC Shunt Motors”.

1. A circuit consists of 3 F capacitor and 5 H inductor. Determine the order of the circuit.

a) 2

b) 1

c) 3

d) 0

Answer: a

Explanation: The order of the circuit is the number of memory/storing elements which are non-separable present in the circuit. In mathematics, the order is defined as the highest order derivate in the differential equation. The order of the circuit is 2.

2. The forced response is due to a source present in the circuit.

a) True

b) False

Answer: a

Explanation: The steady-state response is a part of the forced response. The forced response is due to the electrical source present in the circuit. Its mathematical equation involves the source present in the circuit.

3. Full form of DTC.

a) Direct torque control

b) Digital torque control

c) Discrete torque control

d) Distribution torque control

Answer: a

Explanation: DTC stands for Direct torque control. DTC and IFOC are the techniques used for controlling the speed and torque of a 3-phase induction motor.

4. The characteristics shown by an element in the I-V curve is V=I 2 . The nature of the element is _______

a) Non-linear, Bilateral, Passive

b) Linear, Unilateral, Active

c) Linear, Bilateral, Passive

d) Non-linear, Unilateral, Active

Answer: d

Explanation: The nature of the element is non-linear, unilateral and active. The shape of the characteristic is parabolic. For bilateral nature, it should be symmetrical in the first and third quadrant. Its slope is negative in the second quadrant which determines its active nature.

5. Transient response is a temporary response.

a) True

b) False

Answer: a

Explanation: Transient response is a primary response in the circuit. It is a temporary response which dies out at t=infinity. It consists of exponential decaying functions.

6. Calculate the steady state value for x=4(1-e -3t ).

a) 5

b) 4

c) 3

d) 2

Answer: b

Explanation: The steady state value is obtained at t=∞. The value of x at t=∞ is 4(1-e -∞ )=4=4. The term e -3t is an exponentially decaying function.

7. The natural response is due to _________ conditions present in the circuit.

a) Initial

b) Final

c) Zero

d) Negative

Answer: a

Explanation: The natural response is due to the initial conditions present in the circuit. The natural response is a complete part of the transient response. It is mathematically represented as y=y o ×e -at .

8. Calculate the value of the coefficient of coupling for the isolated coils.

a) 0

b) 1

c) 5

d) 7

Answer: a

Explanation: The coefficient of coupling expresses how the two coils are magnetically coupled. It is mathematically represented as K=M÷√L 1 .L 2 . For the isolated coils, the value of the mutual inductance is 0. The value of the coefficient of coupling is 0.

9. Calculate the value of equivalent inductance for series aiding of two coils whose self inductances are 5 H, 2 H, and mutual inductance value is 4 H.

a) 15 H

b) 12 H

c) 13 H

d) 11 H

Answer: a

Explanation: The equivalent inductance for series aiding of two coils is L eq =L 1 +L 2 +2M=5+2+8=15 H. The value of equivalent inductance increases due to mutually induced e.m.f in case of the magnetically coupled circuit.

10. Calculate the value of equivalent inductance for series subtracting polarity of two coils whose self inductances are 14 H, 5 H, and mutual inductance value is 1 H.

a) 15 H

b) 17 H

c) 12 H

d) 10 H

Answer: b

Explanation: The equivalent inductance for series subtracting of two coils is L eq =L 1 +L 2 -2M=14+5-2=17 H. The value of equivalent inductance decreases due to negatively mutually induced e.m.f in case of the magnetically coupled circuit.

11. Calculate the value of the capacitive reactance during resonance if the value of the inductor is 5 H and supply frequency is 20 rad/sec.

a) 100 Ω

b) 200 Ω

c) 300 Ω

d) 700 Ω

Answer: a

Explanation: Resonance is defined as the phenomenon in which energy of any element changes from one form to another. During resonance condition X c =X l =ΩL=20×5=100 Ω.

12. Calculate the quality factor for the R-L circuit if R=28 Ω and L=2 H.

a) 14

b) 16

c) 10

d) 17

Answer: a

Explanation: The quality factor is defined as the ratio of the reactive power to the active power consumed. The resistor always absorbs active power and inductor absorbs the reactive power. Quality factor=R÷L=28÷2=14.

13. The quality factor is calculated for ___________

a) Power factor loads

b) Inductive coils

c) Lagging loads

d) Leading loads

Answer: b

Explanation: Quality factor is calculated for inductive coils, not for power factor loads. Its value determines the quality of the inductor or capacitor. It shows how good an inductor or capacitor can absorb reactive power.

14. Calculate the equivalent resistance when two resistances are connected in parallel of values 8 Ω, 8 Ω.

a) 3 Ω

b) 2 Ω

c) 4 Ω

d) 7 Ω

Answer: c

Explanation: When two resistances are connected in parallel their equivalent resistance is equal to the harmonic mean of the individual resistances. R eq =R 1 .R 2 ÷(R 1 +R 2 )=8×8÷=4 Ω.

15. Calculate the quality factor for the R-C circuit if R=2 Ω and C=1 F.

a) 0.2

b) 0.4

c) 0.6

d) 0.5

Answer: d

Explanation: The quality factor is defined as the ratio of the reactive power to the active power consumed. The resistor always absorbs active power and capacitor absorbs the reactive power. Quality factor=1÷RC=1÷2=.5.

This set of Basic Electric Drives Questions and Answers focuses on “Regenerative Braking of DC Series Motors”.

1. What is the condition for maximum power transfer theorem in DC circuits?

a) R L =2R th

b) R L ≠R th

c) R L ≫R th

d) R L =R th

Answer: d

Explanation: The condition for maximum power transfer theorem is load should be variable and R L =R th . During maximum power transfer condition voltage across load becomes half of Thevenin voltage and efficiency becomes 50 %.

2. Instantaneous power in the 3-φ system is constant.

a) True

b) False

Answer: a

Explanation: The 3-φ system is more economical than 2-∅ system because of no vibrations in power waveform. The instantaneous power in the 3-∅ system is constant. P=3V p I p cos=√3V L I L cos.

3. The length of phasor is ___________

a) R.M.S

b) Average

c) Peak to Peak

d) Minimum

Answer: a

Explanation: Phasors are the rotating values which rotate with some angular frequency Ω. The length of the phasor value is r.m.s value.

4. Transient analysis is only applicable to bounded systems.

a) True

b) False

Answer: a

Explanation: Bounded systems are those systems which have some finite maximum value and are decaying functions. Transient analysis is only applicable to bounded signals as they achieve steady state after some time.

5. Calculate the velocity of the ball if the angular speed is 7 rad/s and radius is .2 m.

a) 2.5 m/s

b) 1.4 m/s

c) 4.5 m/s

d) 1.0 m/s

Answer: b

Explanation: The velocity of the ball can be calculated using the relation V=Ω×r. The velocity is the vector product of angular speed and radius. V = Ω×r = 7×.2 = 1.4 m/s.

6. Calculate the value of the time period if the frequency of the signal is .001 Hz.

a) 1000 sec

b) 2000 sec

c) 5000 sec

d) 1500 sec

Answer: a

Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷.001=1000 sec.

7. Calculate the value of the frequency of the 440 V DC supply.

a) 100 Hz

b) 0 Hz

c) 200 Hz

d) 500 Hz

Answer: b

8. Calculate the value of power factor if the values of R and Z are 2 Ω and 10 Ω.

a) 0.8

b) 0.5

c) 0.2

d) 0.4

Answer: c

Explanation: The power factor is defined as the ratio of active power to the apparent power. Cos=R÷Z=2÷10=.2. It has no unit.

9. Calculate the value of capacitor voltage during resonance condition if the value of supply voltage is 20 V and the quality factor is 3.

a) 60 V

b) 50 V

c) 10 V

d) 30 V

Answer: a

Explanation: During the resonance condition X L =X c . The value of the capacitor voltage is Q×V s . The power factor of the circuit is one. V c =20×3=60 V.

10. When 30 A current flows into the positive terminal of current source 8 V. Calculate the power delivered by the source.

a) -240 W

b) 360 W

c) -430 W

d) 500 W

Answer: a

11. The slope of the V-I curve is 17.587 o . Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .322 Ω

b) .360 Ω

c) .316 Ω

d) .778 Ω

Answer: c

Explanation: The slope of the V-I curve is resistance. The slope given is 17.587 o so R=tan(17.587 o )=.316 Ω. The slope of the I-V curve is reciprocal of resistance.

12. Calculate the equivalent inductance when two inductors are connected in parallel of values 12 H and 12 H.

a) 6 H

b) 10 H

c) 12 H

d) 8 H

Answer: a

Explanation: When two inductors are connected in parallel their equivalent inductance is equal to the harmonic mean of the inductances. L eq =L 1 ×L 2 ÷(L 1 +L 2 )=6 H.

13. Calculate the active power in an 0 Ω resistor with 0 current flowing through it.

a) inf MW

b) 0 MW

c) 2 MW

d) 18 MW

Answer: b

14. When 1 A current flows out of the positive terminal of voltage source 6 V. Calculate the power delivered by the source.

a) 6 W

b) 7 W

c) -9 W

d) -5 W

Answer: a

15. Calculate the value of inductor voltage during resonance condition if the value of supply voltage is 7 V and the quality factor is 9.

a) 59 V

b) 63 V

c) 73 V

d) 33 V

Answer: b

Explanation: During the resonance condition X L =X c . The value of the inductor voltage is Q×V s . The power factor of the circuit is one. V c =7×9=63 V.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Dynamic Braking of DC Shunt Motors”.

1. The sine function is an odd function.

a) True

b) False

Answer: a

Explanation: The odd functions are those functions which are symmetric about the origin. The sine function is an odd function whose time period is 2π.

2. The cosine function is an even function.

a) True

b) False

Answer: a

Explanation: The even functions are those functions which are a mirror image of the y-axis. The cosine function is an even function whose time period is 2π.

3. Full form of NENO.

a) Neither even nor odd

b) Neither energy nor odd

c) Neither even nor original

d) Neither even nor orthogonal

Answer: a

Explanation: NENO stands for Neither even nor odd. The functions which are not the mirror image of the y-axis and nor symmetric about the origin are NENO functions.

4. The characteristics shown by an element in the V-I curve is V=I s (1-e -V/K ). The nature of the element is _______

a) Non-linear, Bilateral, Passive

b) Linear, Unilateral, Active

c) Linear, Bilateral, Passive

d) Non-linear, Unilateral, Passive

Answer: d

Explanation: The nature of the element is non-linear, unilateral and passive. The shape of the characteristic is exponential rising. For bilateral nature, it should be symmetrical in the first and third quadrant. Its slope is positive in the first quadrant which determines its passive nature.

5. Calculate the resonant frequency if the values of the capacitor and inductor are 2 F and 2 H.

a) .5 rad/sec

b) .6 rad/sec

c) .8 rad/sec

d) .9 rad/sec

Answer: a

Explanation: During resonance condition X L =X c . The value of the resonant frequency is 1÷√LC=1÷√4=.5 rad/sec. The V o ltage across the capacitor and inductor becomes equal.

6. Calculate the steady state value for x=7e -9t .

a) 0

b) 8

c) 3

d) 1

Answer: a

Explanation: The steady state value is obtained at t=∞. The value of x at t=∞ is 7e -∞ =4=0. The term e -9t is an exponentially decaying function.

7. The maximum V o ltage across the capacitor V c =V o (1- e -t ) is __________

a) V o

b) 2V o

c) 3V o

d) -V o

Answer: a

Explanation: The V c =V o (1- e -t ) is V o is the transient equation of the capacitor Voltage. At the steady state V c =V o (1- e -∞ ) is V o . The maximum Voltage across the capacitor is V o .

8. Calculate the value of the coefficient of coupling for the tightly coupled coils.

a) 0

b) 1

c) 3

d) 2

Answer: b

Explanation: The coefficient of coupling expresses how the two coils are magnetically coupled. It is mathematically represented as K=M÷√L 1 .L 2 . For tightly coupled coils, the value of the mutual inductance is √L 1 .L 2 . The value of the coefficient of coupling is 1.

9. The maximum current in the inductor I L =I o (1 – e -t/α ) is __________

a) I o e -t/α

b) I o

c) 2I o

d) -I o

Answer: b

Explanation: The I L =I o (1 – e -t/α ) is I o is the transient equation of the inductor current. At the steady state I L =I o is I o . The maximum current in the inductor is I o .

10. 20 V, 10 A, 10 rpm separately excited dc motor with armature resistance (R a ) equal to .8 ohms. Calculate back emf developed in the motor when it operates on the full load.

a) 12 V

b) 14 V

c) 13 V

d) 11 V

Answer: a

Explanation: Back emf developed in the motor can be calculated using the relation E b = V t – I×R a . In question, it is asking for a full load. 20 V is terminal Voltage it is fixed so E b = 20-10×.8 = 12 V.

11. Speed of DC shunt motor is directly proportional to___________

a) E b

b) Φ

c) V t

d) I a .R a

Answer: a

Explanation: The back e.m.f in case of DC shunt motor is E b =V t -I a .R a . The speed in DC shunt motor is

E b ÷K v . The speed is directly proportional to E b .

12. Calculate the value of the angular acceleration of the DC shunt motor using the given data: J = 1 kg-m 2 , load torque = 1 N-m, motor torque = 2 N-m.

a) 1 rad/s 2

b) 2 rad/s 2

c) 3 rad/s 2

d) 5 rad/s 2

Answer: a

Explanation: Using the dynamic equation of motor J× = Motor torque – Load torque: 1× = 2-1=1, angular acceleration=1 rad/s 2 .

13. Calculate the quality factor for the R-L circuit if R=16 Ω and X L =8 Ω with supply frequency is 1 rad/sec.

a) 2

b) 6

c) 0

d) 7

Answer: a

14. Calculate the equivalent resistance when two armature resistances are connected in parallel of values 6 Ω, 3 Ω.

a) 3 Ω

b) 2 Ω

c) 4 Ω

d) 7 Ω

Answer: b

Explanation: When two resistances are connected in parallel their equivalent resistance is equal to the harmonic mean of the individual resistances. R eq =R 1 .R 2 ÷(R 1 +R 2 )=6×3÷=2 Ω.

15. Calculate the quality factor for the R-C circuit if R=1 Ω and C=1 F.

a) 2

b) 4

c) 1

d) 5

Answer: c

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Dynamic Braking of DC Series Motors”.

1. Dynamic braking requires a strong _________

a) Magnetic field

b) Electrostatic field

c) Nuclear field

d) Gravitational field

Answer: a

Explanation: Dynamic braking requires a strong magnetic field to convert energy from the magnetic domain into the electrical domain. The kinetic energy is wasted at a higher rate.

2. In DC series machine ∅ is directly proportional to I a .

a) True

b) False

Answer: a

Explanation: In series DC machine, the field winding is in series with the armature winding. The armature current flows through the field winding. F=NI a =R∅. I a ∝∅.

3. For faster braking, the series field is separated from the armature circuit.

a) True

b) False

Answer: a

Explanation: For faster braking, the series field is separated from the armature circuit. The series field is excited by a different voltage source.

4. At the starting, the field is very strong but it decreases with time because of the ________

a) Reduction in armature current

b) Increase in armature current

c) Reduction in speed

d) Increase in speed

Answer: a

Explanation: At the starting, the field is very strong but it decreases with time because of the reduction in armature current. ∅∝I a .

5. Calculate the time period of the waveform x=1cot.

a) 0.5 sec

b) 1 sec

c) 3 sec

d) 5 sec

Answer: b

Explanation: The fundamental time period of the cot wave is π. The time period of x is a π÷π=1 sec. The time period is independent of phase shifting and time shifting.

6. In dynamic braking, DC series motor behaves as a ____________

a) Separately excited motor

b) Transformer

c) Induction motor

d) Thyristor

Answer: a

Explanation: In dynamic braking, DC series motor behaves as a separately excited motor as the motor armature is disconnected from the source and connected across resistance R B .

7. Calculate the flux produced by the DC series field winding using the following data: I a =5 A, N=23, R=23.

a) 5 Wb

b) 6 Wb

c) 9 Wb

d) 8 Wb

Answer: a

Explanation: In DC series motor, field windings are connected in series with the armature circuit. F=NI a =R∅. I a ∝∅. ∅=F÷R=5 wb.

8. The shape of the speed-armature current characteristics in DC series motor are __________

a) Hyperbolic

b) Parabola

c) Circle

d) Ellipse

Answer: a

Explanation: The shape of the speed-armature current characteristics in DC series motor is hyperbolic. The motor equation is E b =V t -I a R a =K m ∅ω m . N=(V t /K m I a )-R a /K m . N∝1÷I a .

9. The shape of the speed-torque current characteristics in DC series motor are __________

a) Rectangular Hyperbolic

b) Parabola

c) Circle

d) Ellipse

Answer: a

Explanation: The shape of the speed-torque characteristics in the DC series motor is a rectangular hyperbola. The motor equation is E b =V t -I a R a =K m ∅ω m . N=(V t /K m I a )-R a /K m . I a ∝√T .N∝1÷√T.

10. DC series motors are used where the high ________ torque is required.

a) Maximum

b) Minimum

c) Starting

d) Stalling

Answer: c

Explanation: DC series motors are used where the high starting torque is required. At starting the torque produced by DC series motor is very high. To start a motor torque should be greater than load torque.

11. The shape of the current-torque current characteristics in DC series motor is a __________

a) Rectangular Hyperbolic

b) Parabola

c) Circle

d) Ellipse

Answer: b

Explanation: The shape of the current-torque characteristics in the DC series motor is a parabola. The motor equation is E b =V t -I a R a =K m ∅ω m . T=K m ∅I a . ∅∝I a . T∝I a 2 .

12. Calculate the current in DC series field winding using the following data: I a =3 A.

a) 4 A

b) 2 A

c) 3 A

d) 6 A

Answer: c

Explanation: In DC series motor, field windings are connected in series with the armature circuit. The armature current will also flow through the series winding. I f =I a =3 A.

13. For a given torque, increasing diverter resistance of a DC series motor causes the ________

a) decrease in speed and less armature current

b) decrease in speed but armature current remains the same

c) increase in speed and more armature current

d) increase in speed but armature current remains the same

Answer: a

Explanation: In DC series motor, increasing diverter resistance increases the field current and field flux and also decreases the armature current and therefore decreases the speed.

14. The field diverter resistance for a DC series motor is kept ________

a) High

b) Low

c) zero

d) one

Answer: b

Explanation: The field diverter resistance for a DC series motor is kept low in order to increase the speed of the motor. This will reduce the field flux and increase the armature current.

This set of Advanced Electric Drives Questions and Answers focuses on “Countercurrent Braking of DC Shunt Motors”.

1. Plugging is suitable for __________ type loads.

a) Gravitational

b) Electrostatic

c) Weightless

d) Leading

Answer: a

Explanation: Plugging is suitable for gravitational type loads. It halts the motor operation. It can reverse the direction of rotation of the motor. It can hold the motor at zero speed.

2. In DC shunt machine ∅ is directly proportional to I a .

a) True

b) False

Answer: b

Explanation: In shunt DC machine, the field winding is in parallel with the armature winding. The armature current flows through the armature winding. I F ∝∅∝V t .

3. Braking resistance should be selected in order to limit the braking current.

a) True

b) False

Answer: a

Explanation: During braking, a high current flows in the armature circuit. To limit the braking current a proper value of braking resistance should be selected.

4. Torque developed in the motor is ___________

a) K m ΦI a

b) K m ΦI a 3

c) 2K m ΦI a

d) K m ΦI a 2

Answer: a

Explanation: The torque developed in the motor is directly proportional to the flux, machine constant, armature current. It is mathematically represented as T= K m ΦI a .

5. Full form of TVR is _________

a) Terminal voltage reversal

b) Total voltage reversal

c) Terminal voltage redirect

d) Total voltage reversal

Answer: a

Explanation: TVR stands for terminal voltage reversal. It is one of the methods of countercurrent braking of the DC shunt motor. It halts the motor rapidly and reverses its direction of rotation.

6. Calculate the current in DC shunt field winding using the following data: I a =15 A, I L =21, R=22, N=484.

a) 4 A

b) 2 A

c) 7 A

d) 6 A

Answer: d

Explanation: In DC shunt motor, field windings are connected in parallel with the armature circuit. I F =I L -I a =11-5=6 A. Motor current is the sum of field and armature current.

7. Calculate the flux produced by the DC shunt field winding using the following data: I a =5 A, I L =11, R=2, N=4.

a) 14 Wb

b) 12 Wb

c) 17 Wb

d) 18 Wb

Answer: b

Explanation: In DC shunt motor, field windings are connected in parallel with the armature circuit. I F =I L -I a =11-5=6 A. F=NI F =R∅. I F ∝∅. ∅=F÷R=12 wb.

8. The shape of the speed-armature current characteristics in DC shunt motor is __________

a) Hyperbolic

b) Parabola

c) Negative slope linear line

d) Ellipse

Answer: c

Explanation: The shape of the speed-armature current characteristics in DC shunt motor is negative slope linear line. The motor equation is E b =V t -I a R a =K m ∅ω m . N=(V t -IR a )/K m . N vs I a is a negative slope line with slope = -R a /K m .

9. The shape of the speed-torque characteristics in DC shunt motor is __________

a) Rectangular Hyperbolic

b) Whole x-y plane

c) Circle

d) Ellipse

Answer: a

Explanation: The shape of the speed-torque characteristics in the DC shunt motor is a rectangular hyperbola. The motor equation is E b =V t -I a R a =K m ∅ω m . N=(V t /K m I a )-R a /K m . I a ∝√T .N∝1÷√T.

10. DC shunt motors are used where the high _________ is required.

a) Starting torque

b) Maximum torque

c) Minimum torque

d) Breakdown torque

Answer: a

Explanation: DC series motors are used where the high starting torque is required. At starting the torque produced by DC series motor is very high.

11. The shape of the current-torque characteristics in DC shunt motor is __________

a) Rectangular Hyperbola

b) Parabola

c) Straight line

d) Ellipse

Answer: c

Explanation: The shape of the current-torque characteristics in the DC shunt motor is a straight line. The motor equation is E b =V t -I a R a =K m ∅ω m . T=K m ∅I a . ∅∝I f . Torque ∝ I a .

12. The relationship between the torque and power developed in the DC shunt motor is ___________

a) T ∝ √P

b) T ∝ P 2

c) T ∝ ∛P

d) T ∝ P

Answer: d

Explanation: The torque developed in the DC shunt motor is the ratio of power developed and angular

speed of the motor. T=E b I a ÷ω m =P÷ω m . T ∝ P.

13. The value of current at time of starting is ___________

a) High

b) Low

c) Very low

d) 0

Answer: a

Explanation: At the time of starting the motor is at rest. The e.m.f developed in the motor is zero. I a ×R a = V t -E b . I a =V t ÷R a . The starting current is very high about 10-15 times of full load current.

15. Calculate the bandwidth in series RLC circuit if the frequency is 20 Hz and the quality factor is 5.

a) 2 Hz

b) 4 Hz

c) 6 Hz

d) 8 Hz

Answer: b

Explanation: Bandwidth is defined as the range of frequencies for which the signal exists. Selectivity is inversely proportional to the bandwidth. B.W=f÷Q=20÷5=4 Hz.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Induction Motors – Regenerative Braking ”.

1. Power input to an induction motor is given by _________

a) 3V p I p cosΦ

b) 2V p I p cosΦ

c) V p I p cosΦ

d) 5V p I p cosΦ

Answer: a

Explanation: Power input to a 3-Φ induction motor is 3V p I p cosΦ. The power remains the same for Delta and star connection. The power is constant and has no vibrations.

2. For motoring mode, the phase angle between the stator phase voltage and stator phase current should be _______

a) < 90°

b) > 90°

c) < 40°

d) 180°

Answer: a

Explanation: For motoring mode, the phase angle between the stator phase voltage and stator phase current should be < 90°. The rotor magnetic field tries to catch up the stator magnetic field.

3. The relative speed between the stator magnetic field and rotor in the induction motor is ______

a) N s

b) N s -N r

c) N s +N r

d) 0

Answer: b

Explanation: In induction motor rotor tries to catch up the stator magnetic field in order to satisfy Lenz law. Effect opposes the cause. Stator magnetic field rotates at the speed of synchronous speed.

4. The relative speed between the stator magnetic field and rotor in the synchronous generator is ______

a) N s

b) N s -N r

c) N s +N r

d) 0

Answer: d

Explanation: In synchronous generator rotor and stator magnetic field rotates at speed of synchronous speed in order to produce the steady state torque.

5. The relative speed between the stator magnetic field and stator in the induction motor is ______

a) N s

b) N s -N r

c) N s +N r

d) 0

Answer: a

Explanation: In Induction motor stator magnetic field rotates at the speed of synchronous speed. The stator is made up of Silicon steel and remains at rest. The relative speed is N s -0 = N s .

6. The relative speed between the rotor and stator in the induction motor is ______

a) N r

b) N s -N r

c) N s +N r

d) 0

Answer: a

Explanation: In Induction motor rotor tries to catch up the stator magnetic field in order to satisfy Lenz law. Effect opposes the cause. Stator remains at rest. The relative speed is N r -0=N r .

7. What will happen if we increase the air gap in the induction motor?

a) Power factor will reduce

b) Power factor will increase

c) Reduction in harmonics

d) Speed will increase

Answer: a

Explanation: If we increase the air gap in the induction motor, the machine will draw more magnetizing current to maintain flux in the air gap. The power factor of the machine will reduce.

8. The ratio of the rotor copper loss and air gap power is _______

a) s

b) 1-s

c) 1+s

d) 0

Answer: a

Explanation: The ratio of the rotor copper loss and air gap power is s. The air gap power in the induction motor is P g and the rotor copper loss is s×P g . The ratio is s.

9. Magnetizing current in the induction motor is ______

a) more than the transformer

b) equal as a transformer

c) less than the transformer

d) 0

Answer: a

Explanation: The magnetizing current in the induction motor is % of the full load current. The magnetizing current in the transformer is 3-5% of the full load.

10. Single phase induction motor is self-starting.

a) True

b) False

Answer: b

Explanation: The single-phase induction motor is not self-starting because of the equal and opposite forward & backward magnetic field. The net torque developed is zero.

11. What is the speed of the 6 th order time harmonics in the induction motor?

a) N s

b) N s /6

c) 6×N s

d) 0

Answer: c

Explanation: The time harmonics are present in the supply voltage. The speed of the 6 th order time harmonics in the induction motor is N s /6.

12. What is the speed of the 11 th order harmonics in the induction motor?

a) N s

b) N s /11

c) 2×N s

d) N r

Answer: b

Explanation: The speed of the 6k±1 is N s ÷6k±1. The speed of the 11 th order time harmonics in the induction motor is N s /11.

13. A 3 – phase, 4 pole IM is supplied from a 40 Hz source. Calculate rotor frequency when the rotor runs 100 rpm.

a) 35 Hz

b) 36.6 Hz

c) 9 Hz

d) 8 Hz

Answer: b

Explanation: The slip of the induction motor can be calculated using the relation s = N s -N r ÷ N s . N s =120F÷P=1200 rpm. The value of slip is .91. The rotor frequency is 36.6 Hz.

14. Calculate the three-phase power using the following data: V p =23 V, I p =23 A, ∅=60.

a) 793.5 W

b) 790 W

c) 792 W

d) 791 W

Answer: a

Explanation: Power input to a 3-Φ induction motor is 3V p I p cosΦ. The power remains the same for Delta and star connection. P=3×23×23×co60°=793.5 W.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Induction Motors – Dynamic Braking”.

1. Order of the circuit is the number of memory elements present in the circuit.

a) True

b) False

Answer: a

Explanation: The order of the circuit is the number of memory/storing elements which are non-separable present in the circuit. The examples of memory elements are capacitor, inductor.

2. In Induction motor rotor laminations are thicker than stator laminations.

a) True

b) False

Answer: a

Explanation: The lamination thickness basically depends upon the frequency. The rotor frequency is s×f where f is stator supply frequency. The rotor laminations are thicker than stator laminations because rotor frequency is less.

3. Choose the Induction motor with peak speed.

a) 10 Pole

b) 12 Pole

c) 14 Pole

d) 16 Pole

Answer: a

Explanation: Synchronous speed of Induction motor is N s =120×f÷P. N s ∝. The synchronous speed is inversely proportional to the pole. The maximum speed will be achieved with a minimum number of poles.

4. Which starting method is the best method in Induction motor?

a) Direct online starting

b) Autotransformer starting

c) Reactance starting

d) Star-Delta starting

Answer: d

Explanation: The Star-Delta starting method is one of the best methods for the starting of the Induction motor. This method is cheaper and the line current in the star is 1÷3 times of in delta connection.

5. Induction motor is analogous to __________

a) Transformer

b) DC machine

c) Synchronous motor

d) Synchronous generator

Answer: a

Explanation: The Induction motor is analogous to the transformer. It is also known as a rotating transformer. Both works on the principle of electromagnetic induction.

6. Which type of Induction motor is best for pole changing method?

a) SCIM

b) WRIM

c) Single-phase IM

d) Linear IM

Answer: a

Explanation: Pole changing method is used to control the speed of Induction motor. It is only applicable for SCIM because SCIM rotor is made up of aluminum, copper bars.

7. _________ motor operates at high power factor.

a) Capacitor start

b) Capacitor run

c) Shaded pole

d) Reluctance

Answer: b

Explanation: In the capacitor run, the capacitor is permanently connected with auxiliary windings. The capacitor is used for reactive power compensation. It improves the power factor of the machine.

8. The use of the capacitor banks in the 3-phase Induction motor is ________

a) To increase the power factor

b) To decrease the speed

c) To increase the speed

d) To decrease the power factor

Answer: a

Explanation: Capacitor banks are used with 3-phase Induction motor in order to improve the power factor of the machine. The inductor motor generally operates with .6-.8 power factor range.

9. The concept of the slip is used in ___________

a) Synchronous machine

b) Induction machine

c) DC machine

d) Transformer

Answer: b

Explanation: The slip in the Induction motor is s = N s -N r ÷ N s . The slip existence is due to the relative speed between the stator magnetic field and rotor.

10. Which type of Induction motor is best for e.m.f injection method?

a) SCIM

b) WRIM

c) Single-phase IM

d) Linear IM

Answer: b

Explanation: E.M.F injection method is used to control the speed of Induction motor. It is only applicable for WRIM because WRIM rotor consists of copper windings connected in star connection. The external e.m.f can be connected with it using the slip rings.

11. The slots used in the Induction motor are ___________

a) Semi-closed

b) Closed

c) Vacuum

d) Open

Answer: a

Explanation: The slots used in the Induction motor are semi-closed. Open slots are used in the synchronous machine. Semi-closed slots have moderate air gap and lesser flux leakage.

12. The slots used in the Synchronous machines are ___________

a) Closed

b) Semi-open

c) Vacuum

d) Open

Answer: d

Explanation: The slots used in the Synchronous machine are open. Open slots have higher air gap and minimum flux leakage. The maximum power transfer capability increase due to a decrease in the leakage reactance.

13. The unbalanced set of 3-phase voltages causes ________

a) Positive sequence current

b) Negative sequence current

c) Zero sequence current

d) Half sequence current

Answer: b

Explanation: The unbalanced set of 3-phase voltages causes a negative sequence current. The magnetic field generated due to negative sequence current is in the opposite direction to the fundamental magnetic field. This induces double the frequency current in the rotor side which causes overheating of the motor.

14. The ratio of the rotor gross power and air gap power is _______

a) s

b) 1-s

c) 1+s

d) 0

Answer: b

Explanation: The ratio of the rotor gross power and air gap power is 1-s. The air gap power in the induction motor is P g and the rotor gross power is ×P g . The ratio is 1-s.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Motors – Heating Effects”.

1. Iron losses occur in the machine are variable losses.

a) True

b) False

Answer: b

Explanation: Iron losses consist of eddy current and hysteresis losses. These are constant losses. Iron losses are independent of the load current. The copper losses are variable losses.

2. Which of the following are the classes of insulating material in electrical machines?

a) Γ, A, E, F, H, C

b) Γ, E, B, F, H, C

c) A, E, B, F, H, C

d) Γ, A, E, F, C

Answer: a

Explanation: Depending upon the temperature limits, insulating materials are classified into seven classes Γ, A, E, B, F, H, C. The insulation has the lowest temperature limit.

3. Cooling air travels 10 m distance in 5 sec from point A to point B. Calculate the velocity of the cooling air.

a) 2 m/s

b) 4 m/s

c) 6 m/s

d) 10 m/s

Answer: a

Explanation: The velocity of the cooling air can be calculated using the ratio of displacement and time. The value of the displacement is 10 m. The value of the velocity is 10÷5=2 sec.

4. Calculate the heating time constant of the machine if the thermal capacity of the machine is 24 watts/℃ and heat dissipation constant value is 3 watts/℃.

a) 8 sec

b) 7 sec

c) 6 sec

d) 9 sec

Answer: a

Explanation: The heating time constant of the machine is the ratio of the thermal capacity of the machine and heat dissipation constant value. Τ = c÷D=24÷3 = 8 sec.

5. Inadequate rating of the equipment can ___________

a) Increase the speed

b) Decrease the speed

c) Damage the winding

d) Increase the efficiency

Answer: c

Explanation: The inadequate rating of the equipment can damage the winding and stops the machine operation. The windings can burn out.

6. What is heating time constant of the machine?

a) The ratio of the thermal capacity and heat dissipation constant value

b) The ratio of the thermal capacity and heat dissipation constant value

c) The ratio of the thermal capacity and heat dissipation constant value

d) The ratio of the thermal capacity and heat dissipation constant value

Answer: a

Explanation: The heating time constant of the machine is the ratio of the thermal capacity of the machine and heat dissipation constant value. Τ = c÷D. It has a unit of second.

7. Insufficient rating of the machine can damage the windings.

a) True

b) False

Answer: a

Explanation: The inadequate rating of the equipment can damage the winding and halt the machine operation. The windings can burn out and lead to the breakdown of the machine.

8. A circuit consists of two 7 F capacitors connected in series and 12 H inductor. Determine the order of the circuit.

a) 1

b) 2

c) 0

d) 3

Answer: b

9. Calculate the steady state temperature value for T=1.5(1-e -17t ).

a) 1.5

b) 4.8

c) 3.9

d) 2.4

Answer: a

Explanation: The steady state temperature value is obtained at t=∞. The value of T at t=∞ is 1.5=1.5=1.5. The term e -17t is an exponentially decaying function.

10. Calculate the quality factor for the R-C circuit connected in the cooling system if R=1 Ω and C=10 F.

a) 0.1

b) 0.2

c) 0.3

d) 0.4

Answer: a

11. Calculate the heat absorbed by a 3 Ω resistor when 6 V is applied across it.

a) 12 W

b) 10 W

c) 8 W

d) 11 W

Answer: a

Explanation: The resistor always absorbs active power and dissipates in the form of the heat. Power absorbed by a 3 Ω resistor is 6×6÷3=12 W.

12. What is the unit of power?

a) Joule/sec

b) Joule/sec 2

c) Joule/sec 3

d) Joule/sec 4

Answer: a

Explanation: The power is defined as the rate at which work is done. The ratio of energy and time is power. It is expressed in Watt or Joule/sec.

13. Convert 20 ℃ into Fahrenheit.

a) 68 ℉

b) 58 ℉

c) 78 ℉

d) 48 ℉

Answer: a

Explanation: Temperature can be expressed in Celsius and Fahrenheit. The Standard unit of the temperature is the Kelvin. ℃ + 32 = 68℉.

14. Convert 50℉ into Celsius.

a) 20 ℃

b) 10 ℃

c) 30 ℃

d) 40 ℃

Answer: a

Explanation: Temperature can be expressed in Celsius and Fahrenheit. The basic unit of temperature in SI is the Kelvin. The formula used is ℉ × 5/9 = 10℃.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Motors Rating and Heating – Loading Conditions and Classes of Duty”.

1. Which duty cycle has on load and off-loads period?

a) Intermittent duty

b) Short time duty

c) Continuous duty with constant load

d) Continuous duty with variable load

Answer: a

Explanation: Intermittent duty cycle has frequent on load and off-loads period. The on-load period and off-loads periods are too short to achieve steady-state thermal condition.

2. Which motor is preferred for Jaw crushers?

a) WRIM

b) SCIM

c) Belt slip ring IM

d) DC shunt motor

Answer: c

Explanation: Belt slip ring IM is preferred for the operation of Jaw crushers. A high starting is provided by the Belt ring IM. Slip rings are used to add external resistance.

3. A high starting torque is offered by the Belt Conveyors.

a) True

b) False

Answer: a

Explanation: The Belt Conveyors can provide continuous-flow transportation. The amount of material transported on a conveyor system depends upon belt speed, belt width.

4. Which starting method is the worst method in Induction motor?

a) Direct online starting

b) Autotransformer starting

c) Reactance starting

d) Star-Delta starting

Answer: c

Explanation: The Reactance starting method is the worst method in an Induction motor. It uses the concept of the potential divider. The losses are very high in case of Reactance starting method.

5. Induction motor is also known as __________

a) Frequency changer

b) Frequency remover

c) Time period remover

d) DC machine

Answer: a

Explanation: The Induction motor is also known as frequency changer. The stator frequency can be changed using the slip concept. The rotor frequency of IM is slip×.

6. The capacity of the Crane is expressed in terms of tonnes.

a) True

b) False

Answer: a

Explanation: Cranes are used to carry or lift the heavy loads. The capacity of a Crane is expressed in terms of tonnes. 1 tonne is equal to 1000 Kg.

7. 1800 second rated motors are used for __________

a) Heavy duty cranes

b) Light duty cranes

c) Medium duty cranes

d) Intermittent duty cranes

Answer: c

Explanation: Medium duty crane motor is used to develop high starting torque. They provide the output for a specified interval of time without exceeding a specified temperature.

8. Calculate the frequency of the rotor side of the IM if the value of slip is 0.34 and the supply frequency is 70 Hz.

a) 23.8 Hz

b) 22.7 Hz

c) 24.5 Hz

d) 23.1 Hz

Answer: a

Explanation: The frequency of the rotor side is low as compared to supply frequency. The rotor frequency is slip×=.34×70=23.8 Hz.

9. Calculate the frequency of the stator side of the IM if the value of slip is 0.04 and rotor frequency is 2 Hz.

a) 50 Hz

b) 49 Hz

c) 45 Hz

d) 51 Hz

Answer: a

Explanation: The frequency of the stator side is high as compared to the rotor side. The stator frequency is ÷ slip = 2÷.04 = 50 Hz.

10. Calculate the resonant frequency if the values of the capacitor and inductor are 4 F and 4 H.

a) .25 rad/sec

b) .26 rad/sec

c) .28 rad/sec

d) .29 rad/sec

Answer: a

Explanation: During resonance condition X L =X c . The value of the resonant frequency is 1÷√LC=1÷√16=.25 rad/sec. The voltage across the capacitor and inductor becomes equal.

11. Calculate the value of the angular acceleration of the Heavy duty crane using the given data: J = 5 kg-m 2 , load torque = 30 N-m, motor torque = 80 N-m.

a) 10 rad/s 2

b) 20 rad/s 2

c) 30 rad/s 2

d) 50 rad/s 2

Answer: a

Explanation: Using the dynamic equation of motor J× = Motor torque – Load torque: 5× = 80-30=50, angular acceleration=10 rad/s 2 .

12. Calculate the value of the angular acceleration of the machine using the given data: J = .541 kg-m 2 , load torque = 10 N-m, motor torque = 2 N-m.

a) 45 rad/s 2

b) The machine will fail to start

c) 30 rad/s 2

d) 51 rad/s 2

Answer: b

Explanation: Using the dynamic equation of motor J× = Motor torque – Load torque: .541× = 2-10=-8, angular acceleration=-14.78 rad/s 2 . The load torque is greater than the motor torque. The motor will fail to start.

13. The direction of the single-phase IM can be reversed by using which one of the method?

a) By interchanging the supply terminals

b) By removing the capacitor

c) By changing the direction of auxiliary winding current

d) By removing the main winding

Answer: a

Explanation: The direction of the single-phase IM can be reversed by changing the direction of auxiliary winding or main winding current.The net torque developed in the IM is I m I a sin.

14. Which motor is preferred for overhead traveling cranes?

a) Intermittent periodic motor

b) Continuous duty motor

c) Slow speed duty motor

d) Short time rated motor

Answer: a

Explanation: Intermittent periodic motor is preferred for overhead traveling cranes. Cranes require high starting torque for a short interval of time. Intermittent periodic motors have low synchronous speed.

This set of Electric Drives Interview Questions and Answers for Experienced people focuses on “Power Rating Determination of Electric Motors for Different Applications”.

1. Which one of the following motor is a 1-Φ AC motor?

a) Synchronous motor

b) Series motor

c) Shunt motor

d) Capacitor run

Answer: d

Explanation: Capacitor run motor is a 1-Φ AC motor. The capacitor is used to provide a 90° phase difference between the currents of main and auxiliary winding.

2. Current rating of the IM depends upon the cross-section area of the conductors.

a) True

b) False

Answer: a

Explanation: The current rating of the Induction motor depends upon the cross-section area of the conductors. Resistance ∝ 1÷Area, I ∝Area.

3. Voltage rating of the IM depends upon the insulation level.

a) True

b) False

Answer: a

Explanation: The voltage rating of the IM depends upon the insulation level. The insulation level is also decided on the basis of the over-voltages.

4. 450-seconds rated motor is preferable for __________

a) Light duty cranes

b) Medium duty cranes

c) Intermittent duty cranes

d) Heavy duty cranes

Answer: a

Explanation: Light duty cranes are used in power plants, compressor plants. 450-seconds rated motor is preferable for light-duty cranes. They can safely operate for 450-seconds without exceeding the specified temperature.

5. Calculate the 3-∅ power developed in the IM using the following data: V L =20 V, I L =5 A, cosΦ=.8.

a) 138.56 W

b) 128.41 W

c) 163.78 W

d) 177.59 W

Answer: a

Explanation: The power developed in the IM is √3V L I L cos∅. The power is independent of time and frequency. P=√3V L I L cos∅=√3×20×5×.8=138.56 W.

6. The unit of the moment of inertia is Joule.

a) True

b) False

Answer: b

7. Calculate mark to space ratio if the system is on for 20 sec and off for 3 sec.

a) 6.66

b) 4.48

c) 8.2

d) 5.6

Answer: a

Explanation: Mark to space ratio is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=20÷3=6.66.

8. Calculate the average power dissipated if the value of the duty cycle is .5 and P max =12 W.

a) 6 W

b) 5 W

c) 4 W

d) 3 W

Answer: a

Explanation: The average power dissipated can be calculated using the relation P avg =D×P max . The value of the average power is .5×12=6 W.

9. Full form of PWM.

a) Pulse width modulation

b) Pulse run modulation

c) Power width modulation

d) Pulse wisdom mode

Answer: a

Explanation: PWM stands for pulse width modulation. This technique is used to alter the voltage level of the output of the inverter. We can control the voltage level using the PWM technique.

10. What is the correct relationship between the phase voltage and line voltage in Delta connection with the positive sequence?

a) V L =V p

b) V L =√3V p

c) V L =√2V p

d) I L =√2I p

Answer: a

Explanation: In the delta connection line voltage is the same as the phase voltage. The line current is √3 times of the phase current and leads by 30°.

11. What is the relationship between line and phase current in Star connection with the positive sequence?

a) V L =V p

b) V L =√9V p

c) V L =√2V p

d) I L =I p

Answer: d

Explanation: In the star connection line current is the same as the phase current. The line voltage is √3 times of the phase voltage and leads by 30°.

12. Choose the correct relationship in delta connection with the positive sequence.

a) I L =I p

b) V L =√3V p

c) V L =V p

d) I L =√3I p

Answer: d

Explanation: In the delta connection line voltage is the same as the phase voltage. The line current is √3 times of the phase current and leads by 30°.

13. Choose the correct relationship in Star connection with the positive sequence.

a) V L =V p

b) V L =√3V p

c) V L =√2V p

d) I L =√3I p

Answer: b

Explanation: In the star connection line current is the same as the phase current. The line voltage is √3 times of the phase voltage and leads by 30°.

14. Calculate the 3-∅ power developed in the IM connected in star using the following data: V p =7 V, I L =7 A, cosΦ=.6.

a) 88.5 W

b) 88.4 W

c) 88.8 W

d) 88.2 W

Answer: d

Explanation: The power developed in the IM is √3V L I L cos∅. The power is independent of time and frequency. In star connection V L =√3V p =12.12 V. P=√3V L I L cos∅=3×20×5×.8=88.2 W.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Motors Rating and Heating – Load Inertia Effect”.

1. Hooke’s law is applicable within the elastic limit.

a) True

b) False

Answer: a

Explanation: Hooke’s law is only applicable within the elastic limit. It states that within the elastic limit the strain is directly proportional to the stress.

2. Calculate the load stress if the value of force is 20 N and the area of cross-section is 2 m 2 .

a) 10 N/m 2

b) 20 N/m 2

c) 30 N/m 2

d) 40 N/m 2

Answer: a

Explanation: The stress is defined as the ratio of the internal restoring force and area of the cross-section. Stress=F res ÷A=20÷2=10 N/m 2 .

3. What is the effect of impurities on elasticity?

a) It will increase

b) Remains the same

c) It will decrease

d) It will slightly increase

Answer: d

Explanation: The Young modulus of elasticity is slightly increased by the addition of impurities. The intermolecular forces inside the wire effectively increase by impurity due to this external force can be easily opposed.

4. Which of the following are the types of Stress?

a) Longitudinal stress, Volume stress, Tangential stress

b) Longitudinal stress, Volume stress

c) Longitudinal stress, Lateral stress, Tangential stress

d) Longitudinal stress, Shear stress

Answer: a

Explanation: There are three types of stress that are Longitudinal stress, Volume stress, Tangential or Shear stress. Stress is the ratio of restoring force and cross-section area.

5. The relationship between Young’s modulus , Bulk modulus and Poisson’s ratio is _____________

a) Y=K

b) Y=3K

c) Y=5K

d) Y=4K

Answer: b

Explanation: The relationship between Young’s modulus , Bulk modulus and Poisson’s ratio is Y=3K. The Young modulus of elasticity is defined as the ratio of Longitudinal stress and Longitudinal strain.

6. Calculate the work done in stretching the wire of volume=22 m 3 and the values of stress and strain are 2 N/m 2 and 4.

a) 66 J

b) 44 J

c) 22 J

d) 88 J

Answer: d

Explanation: The work done in stretching a wire is .5×Stress×Strain×Volume. The value of the work done is .5×22×4×2=88 J.

7. Plasticity is the ability of the material to deform without breaking.

a) True

b) False

Answer: a

Explanation: Plasticity is the ability of the material to deform without breaking. Plastic deformation occurs in metal forming processes.

8. _________ motor operates at low power factor.

a) Capacitor run

b) Capacitor start

c) Shaded pole

d) Reluctance

Answer: c

Explanation: Shaded pole motors operate at a low power factor. There is a weak magnetic field develops in the case of Shaded pole motor. It is a single-phase AC Induction motor.

9. Calculate the compensator rating required for cos=.7.

a) 0.71 P.U

b) 0.21 P.U

c) 0.47 P.U

d) 0.12 P.U

Answer: a

Explanation: The compensator rating can be calculated using the relation Q P.U =√1-cos 2 =√1-.49=.71. This per unit value VAR compensator is required to improve the power factor of the system.

10. Full form of SPWM.

a) Sinusoidal pulse width modulation

b) State per width modulation

c) State pulse width modulation

d) Sinusoidal per width modulation

Answer: a

Explanation: SPWM stands for sinusoidal pulse width modulation. SPWM is a high-frequency PWM technique. This technique is used to achieve variable voltage level.

11. Which harmonic is not present in 3-Φ fully controlled rectifier?

a) 3 rd harmonic

b) 5 th harmonic

c) 7 th harmonic

d) 11 th harmonic

Answer: a

Explanation: Triplen harmonics are absent in the case of 3-phase fully controlled rectifier. Because of the absence of triplen harmonics THD of the rectifier 31%.

12. Lowest order harmonic present in 3-Φ fully controlled rectifier is __________

a) 5 th

b) 7 th

c) 3 rd

d) 2 nd

Answer: a

Explanation: Only 6k±1 order harmonics are present in 3-Φ fully controlled rectifier. The lowest order harmonic is 5 th harmonic for k=1. Third order harmonics are absent in 3-Φ fully controlled rectifier.

13. Calculate the moment of inertia of the circular hollow disk having a mass of 2 kg and an internal radius of 5 m and an external radius of 10 m.

a) 125 kgm 2

b) 127 kgm 2

c) 122 kgm 2

d) 123 kgm 2

Answer: a

Explanation: The moment of inertia of the circular hollow disk can be calculated using the formula I=M(R 1 2 +R 2 2 )÷2. The mass of circular hollow disk and radius is given. I=×{ 2 + 2 }=125 kgm 2 . It depends upon the orientation of the rotational axis.

14. Calculate the mass of the solid cylinder having a mass of 4 kgm 2 and radius of 4 m & length of 12 m.

a) 64 kgm 2

b) 68 kgm 2

c) 67 kgm 2

d) 69 kgm 2

Answer: a

Explanation: The moment of inertia of the solid cylinder can be calculated using the formula I= M(l 2 ÷12+R 2 2 ÷4). The mass of solid cylinder and radius, the length is given. I={}= 64 kgm 2 . It depends upon the orientation of the rotational axis.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Motors Rating and Heating – Load Equalisation”.

1. Calculate the moment of inertia of the flywheel of mass 45 kg and radius of 2 m.

a) 185 kg-m 2

b) 180 kg-m 2

c) 175 kg-m 2

d) 170 kg-m 2

Answer: b

Explanation: The moment of inertia of the flywheel can be calculated using the formula I=M(R 2 ). The mass of flywheel and radius is given. I=× 2 =180 kg-m 2 . It depends upon the orientation of the rotational axis.

2. Electric motor and control system are the basic elements of the electric drive.

a) True

b) False

Answer: a

Explanation: Electric motor and control system are the basic elements of the electric drive. The electric drive uses the control system methods to control the speed and torque of the motor.

3. Capacitor start capacitor run single-phase Induction motor is used in ceiling fan.

a) True

b) False

Answer: a

Explanation: Capacitor start capacitor run single-phase Induction motor is used in ceiling fan. The motor operation is silent and non-vibrational. It provides a high starting torque.

4. Which of the following are the types of Longitudinal stress?

a) Tensile stress, Compressive stress

b) Tensile stress, Volume stress

c) Tensile stress, Shear stress

d) Tensile stress, Lateral stress

Answer: a

Explanation: Longitudinal stress is one the type of stress. Longitudinal stress is of two types. It consists of Tensile stress and Compressive stress. It has a unit of N/m 2 .

5. Which type of motor is used for milling and grinding operations?

a) SCIM

b) WRIM

c) Split phase

d) Reluctance motor

Answer: a

Explanation: Squirrel cage Induction motor is used for milling and grinding operations. SCIM has a rigid structure and high tolerating capability. It provides good running characteristics.

6. Calculate the work done in stretching the wire of length=2 m and the values of stress and strain are 16 N/m 2 and 24.

a) 1566 J

b) 1644 J

c) 1522 J

d) 1536 J

Answer: d

Explanation: The work done in stretching a wire is .5×Stress×Strain×Volume. The value of the work done is .5×8×16×24=1536 J. The volume of wire is 2×2×2=8 m 3 .

7. Which type of force phenomenon is used in Sugar mills?

a) Centripetal force

b) Centrifugal force

c) Electrostatic force

d) Nuclear forces

Answer: b

Explanation: In sugar mills, a centrifugal is used to separate out sugar crystals from the syrup by the action of centrifugal forces. Centrifugal force acts away from the center.

8. The Speed control by variation of field flux results in _____________

a) Constant torque drive

b) zero speed

c) Variable torque drive

d) Variable power drive

Answer: c

Explanation: Field flux control method is used to achieve the speed above the base speed. It provides constant power drive and variable torque drive.

9. Calculate the compensator rating required for sin=.8.

a) 0.9 P.U

b) 0.8 P.U

c) 0.7 P.U

d) 0.2 P.U

Answer: b

Explanation: The compensator rating can be calculated using the relation Q P.U =√1-cos 2 =sin=.8. This per unit value VAR compensator is required to improve the power factor of the system.

10. Calculate the heating time constant of the machine if the thermal capacity of the machine is 16 watts/℃ and heat dissipation constant value is 4 watts/℃.

a) 6 sec

b) 7 sec

c) 4 sec

d) 9 sec

Answer: c

Explanation: The heating time constant of the machine is the ratio of the thermal capacity of the machine and heat dissipation constant value. τ = c÷D = 16÷4 = 4 sec.

11. Calculate the heat absorbed by a 20 Ω resistor when 40 V is applied across it.

a) 120 W

b) 100 W

c) 80 W

d) 110 W

Answer: c

Explanation: The resistor always absorbs active power and dissipates in the form of the heat. Power absorbed by a 20 Ω resistor is 40×40÷20=80 W.

12. Full form of VFD is _____________

a) Variable frequency drive

b) Variable force drive

c) Value frequency drive

d) Variable force diverter

Answer: a

Explanation: VFD stands for variable frequency drive. It is a controller for AC or DC motors to achieve the desired speed. It can perform the four quadrant operation.

13. What is Load Equalisation?

a) Process of smoothing the fluctuating load

b) Process of removing the load

c) Process of changing the voltage level

d) Process of changing the speed

Answer: a

Explanation: The Load Equalisation is the process of smoothing the fluctuating load. Load fluctuation causes a large voltage dip in the system and can damage the device.

14. Which device is used to store the energy during load equalization?

a) Flywheel

b) Capacitor

c) Inductor

d) Resistor

Answer: a

Explanation: Flywheel is used to store the energy during load equalization. It acts as a reservoir. It stores the energy during light load and utilizes the stored energy during peak load conditions.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Motors Rating and Heating – Environmental Factors “.

1. Which test is used to determine the conditions where electric motors are being used?

a) Thermography

b) Geography

c) Seismography

d) Anthropology

Answer: a

Explanation: Thermography is used to detect heat patterns. It uses an Infrared rays camera. This test is used to determine the conditions where electric motors are being used.

2. What is the range of size of the motor for efficiency nearly equals to 90%?

a) 15-150

b) 0-2

c) 3-15

d) 150-250

Answer: a

Explanation: As the size of the motor increases its speed decreases. The range of size of the motor for efficiency nearly equals to 90 % is 15-150. The efficiency increases as the size increases.

3. Calculate the heat loss in the electric motor for time interval 0-5 sec using the following data: I a =2 A, R a =.8Ω.

a) 16 J-sec

b) 14 J-sec

c) 12 J-sec

d) 10 J-sec

Answer: a

Explanation: The heat loss can be calculated using the Joule law of heating effect. The value of the heat loss is I a 2 ×R a ×t=4×4=16 J-sec.

4. 1 Btu/h is equal to ___________

a) .641

b) .334

c) .293

d) .417

Answer: c

Explanation: One British thermal unit is equal to .293 Watt. 1 Btu/h is the amount of energy requires to increase the temperature of one pound of water by 1℉.

5. Overloading can be prevented using _____________

a) Over-current protection

b) Speed protection

c) Over frequency protection

d) Oversize protection

Answer: a

Explanation: Overloading can be prevented using over-current protection. It will detect the over-current and interrupt the supply. IDMT over-current relay can be used for the protection purpose.

6. Full form of IDMT is _________

a) Inverse definite minimum time

b) Inverter definite minimum time

c) Inverter definite maximum time

d) Insert definite minimum time

Answer: a

Explanation: IDMT stands for Inverse definite minimum time over-current relay. It has both the characteristics of inverse over-current and definite time over-current relay.

7. The most common cause of motor failure is ____________

a) Overloading

b) Low resistance

c) Contamination

d) Over-Heating

Answer: b

Explanation: The most common cause of motor failure is low resistance. Overheating, corrosion can damage the insulation of the windings which leads to low resistance.

8. Full form of THD is ____________

a) Total harmonic diverter

b) Total harmonic digital

c) Total harmonic distortion

d) Total harmonic discrete

Answer: c

Explanation: THD stands for total harmonic distortion. It measures the harmonic distortion in the signal. According to IEEE-1992, only 5% of THD is valid.

9. Calculate the compensator rating required for a sec=1.

a) 0.64 P.U

b) 0.21 P.U

c) 0.57 P.U

d) 0 P.U

Answer: d

Explanation: The compensator rating can be calculated using the relation Q P.U =√1-cos 2 =0. This per unit value VAR compensator is required to improve the power factor of the system.

10. Calculate the value of Voltage ripple factor if the Form factor is 1.1.

a) 45.8 %

b) 46.2 %

c) 47.5 %

d) 49.9 %

Answer: a

Explanation: Voltage ripple factor measures the harmonics on the DC side of the converter. The value of voltage ripple factor is √F.F 2 -1=45.8 %. The form factor is V r.m.s ÷V avg .

11. Calculate the value of THD if the distortion factor is .8.

a) 76 %

b) 75 %

c) 74 %

d) 73 %

Answer: b

Explanation: THD measures the harmonics distortion in the signal. The value of THD is √ 2 -1=75 %. The value of the distortion factor is I or.m.s ÷I R.m.s .

12. Full form of VRF.

a) Voltage ripple factor

b) Voltage revert factor

c) Volume ripple factor

d) Volume revert factor

Answer: a

Explanation: Voltage ripple factor measures the harmonics on the DC side of the converter. The value of voltage ripple factor is √F.F 2 -1. The form factor is V r.m.s ÷V avg .

13. Second lowest order harmonic present in 3-Φ fully controlled rectifier is __________

a) 5 th

b) 7 th

c) 3 rd

d) 2 nd

Answer: b

Explanation: Only 6k±1 order harmonics are present in 3-Φ fully controlled rectifier. The second lowest order harmonic is 7 th harmonic for k=1. Third order harmonics are absent in 3-Φ fully controlled rectifier.

14. Electric motors are placed in __________ to protect them from dust and contamination.

a) Enclosures

b) Open environment

c) Moisturizing environment

d) Hot environment

Answer: a

Explanation: Electric motors are placed in enclosures to protect them from dust and contamination. There are seven most common types of enclosures. For example- Open Drip Proof.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid State Controlled Drives – DC Motor Systems”.

1. RLE load is a voltage stiff load.

a) True

b) False

Answer: b

Explanation: RLE load is a current stiff load because of the presence of the inductor. The load current does not suddenly with a change in voltage. Inductor opposes the change of current.

2. In Half-wave uncontrolled rectifier calculate the average value of the voltage if the supply is 23sin.

a) 7.32 V

b) 8.32 V

c) 9.32 V

d) 7.60 V

Answer: a

Explanation: In Half-wave uncontrolled rectifier, the average value of the voltage is V m ÷π=23÷π=7.32 V. The diode will conduct only for the positive half cycle. The conduction period of the diode is π.

3. In Half-wave controlled rectifier calculate the average value of the voltage if the supply is 10sin and firing angle is 30°.

a) 2.32 V

b) 2.97 V

c) 4.26 V

d) 5.64 V

Answer: b

Explanation: In Half-wave controlled rectifier, the average value of the voltage is V m )÷2π=10÷6.28=2.97 V. The thyristor will conduct from ∝ to π.

4. Calculate the extinction angle in purely inductive load if the firing angle is π÷4.

a) 315°

b) 145°

c) 345°

d) 285°

Answer: a

Explanation: The extinction angle in the purely inductive load is 2π-∝=360°-45°=315°. The extinction angle is the angle at which current in the circuit becomes zero. The average value of the voltage in a purely inductive load is zero.

5. Calculate the conduction angle in purely inductive load if the firing angle is π÷2.

a) 205°

b) 175°

c) 180°

d) 195°

Answer: c

Explanation: The conduction angle in the purely inductive load is β-α=2=2. The conduction angle is the angle for which the current exists in the circuit. The average value of the voltage in a purely inductive load is zero.

6. RLE load is also known as DC motor load.

a) True

b) False

Answer: a

Explanation: RLE load is also known as DC motor load because the armature circuit consists of back e.m.f, inductive coils, and armature resistance.

7. In single phase RLE load, calculate the voltage across the thyristor when current decays to zero using the data: (V s ) r.m.s =220 V, f=40 Hz, R=1 Ω, E=90 V, β=230°.

a) -328.33 V

b) -325.48 V

c) -254.85 V

d) -284.48 V

Answer: a

Explanation: In single phase RLE load the voltage across the thyristor when current decays to zero V T =V m sin-E=220×√2sin-90=-328.33 V.

8. Calculate the displacement factor if the fundamental voltage is 24sin and fundamental current is 47sin.

a) -0.5

b) -0.7

c) 0.9

d) 0.4

Answer: a

Explanation: The displacement factor is the cosine of the angle difference between the fundamental voltage and fundamental current. D.F=cos=-0.5.

9. Calculate the PIV for the Mid-point configuration of Full-wave rectifier if the peak value of the supply voltage is 311.

a) 622 V

b) 620 V

c) 624 V

d) 626 V

Answer: a

Explanation: The peak inverse voltage for the Mid-point configuration of Full wave rectifier is 2V m =2×311=622 V. The peak inverse is the maximum negative voltage across the thyristor.

10. Calculate the average value of the current through the thyristor in case of 1-Φ Full wave bridge rectifier if the value of the load current is 42 A.

a) 21 A

b) 12 A

c) 14 A

d) 16 A

Answer: a

Explanation: The average value of the current through the thyristor in case of 1-∅ full wave bridge rectifier is I o ÷2=21 A. Each thyristor conducts for 180°.

11. Calculate the r.m.s value of the current through the thyristor in case of 1-Φ Full wave bridge rectifier if the value of the load current is 2 A.

a) 1.414 A

b) 1.214 A

c) 1.347 A

d) 1.657 A

Answer: a

Explanation: The r.m.s value of the current through the thyristor in case of 1-∅ full wave bridge rectifier is I o ÷√2=√2 A=1.414 A. Each thyristor conducts for 180°.

12. Calculate the value of THD value for 1-Φ Full wave bridge rectifier.

a) 48.43 %

b) 47.25 %

c) 49.26 %

d) 50.48 %

Answer: a

Explanation: The value of the distortion factor is .9. The value of THD value for 1-Φ Full wave bridge rectifier is √ 2 -1=48.43 %. THD measures the amount of harmonic distortion.

13. Calculate the value of the Input power factor for 1-Φ Full wave bridge rectifier if the firing angle value is 45°.

a) .65

b) .64

c) .61

d) .63

Answer: d

Explanation: The value of the input power factor for 1-Φ Full wave bridge rectifier is .9cos=.63. The input power factor is a product of distortion factor and displacement factor.

14. Calculate the value of the fundamental displacement factor for 1-Φ Full wave bridge rectifier if the firing angle value is 60°.

a) .5

b) .4

c) .2

d) .8

Answer: a

Explanation: Fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. D.F=cos=cos=0.5.

15. Calculate the value of the fundamental displacement factor for 1-Φ Full wave semi-converter if the firing angle value is 20°.

a) .82

b) .98

c) .74

d) .26

Answer: b

Explanation: Fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. The fundamental displacement factor for 1-Φ Full wave semi-converter is cos=cos=.98.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid State Controlled Drives – AC Motor Systems”.

1. Calculate the output frequency for the two-pulse converter if the supply frequency is 20 Hz.

a) 40 Hz

b) 20 Hz

c) 60 Hz

d) 90 Hz

Answer: a

Explanation: The output of a two-pulse converter consists of two pulses in one cycle. The output frequency of the two pulse converter is 2×supply frequency=2×20=40 Hz.

2. Calculate the pulse number if the supply frequency is 2π and the output frequency is π÷3.

a) 4

b) 2

c) 6

d) 8

Answer: c

Explanation: The pulse number can be calculated using the ratio of input frequency to the output frequency. The value of pulse number is 2π÷π÷3=6. It is a six-pulse converter.

3. In 3-Φ Fully controlled rectifier calculate the average value of the voltage if the supply is 400 V and firing angle is 15°.

a) 521.2 V

b) 522 V

c) 523 V

d) 524 V

Answer: b

Explanation: In 3-Φ Fully controlled rectifier, the average value of the voltage is 3V ml )÷π=3×400×√2)÷3.14=522 V.

4. Calculate the value of the Input power factor for 3-Φ Fully controlled rectifier if the firing angle value is 70°.

a) .32

b) .38

c) .31

d) .33

Answer: a

Explanation: The value of the input power factor for 3-Φ Fully controlled rectifier is .95cos=.32. The input power factor is a product of distortion factor and displacement factor.

5. Calculate the value of THD value for 3-Φ Fully controlled rectifier.

a) 48.43 %

b) 47.25 %

c) 39.26 %

d) 31 %

Answer: d

Explanation: The value of the distortion factor is .95. The value of THD value for 3-Φ Fully controlled rectifier is √ 2 -1=31 %. THD measures the amount of harmonic distortion.

6. In 3-Φ Semi-controlled rectifier calculate the average value of the voltage if the supply is 440 V and firing angle is 22°.

a) 571.5 V

b) 572.8 V

c) 548.3 V

d) 524.1 V

Answer: b

Explanation: In 3-Φ Semi-controlled rectifier, the average value of the voltage is 3V ml )÷2π=3×440×√2)÷6.28=572.8 V.

7. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 20° and supply frequency is 60 Hz.

a) 8.8 msec

b) 7.4 msec

c) 10.1 msec

d) 6.5 msec

Answer: c

Explanation: The circuit turn-off time for 3-Φ Fully controlled rectifier is ÷ω. The value of circuit turn-off time for ∝ < 60° is ÷6.28×60=10.1 msec.

8. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 110° and supply frequency is 50 Hz.

a) 3.8 msec

b) 5.2 msec

c) 9.3 msec

d) 8.7 msec

Answer: a

Explanation: The circuit turn-off time for 3-Φ Fully controlled rectifier is ÷ω;. The value of circuit turn-off time for ∝ ≥ 60° is ÷6.28×50=3.8 msec.

9. Calculate peak-peak voltage if V max =80 V and V min =20 V.

a) 60 V

b) 50 V

c) 70 V

d) 10 V

Answer: a

Explanation: Peak-Peak voltage is equal to the difference between the maximum and minimum voltage. It is mathematically represented as V p-p =V max -V min =80-20=60 V.

10. Calculate the value of Crest factor if V peak =12 V and V r.m.s =24 V.

a) .2

b) .3

c) .4

d) .5

Answer: d

Explanation: The value of the crest factor is V peak ÷V r.m.s =12÷24=.5. It signifies the peak value is .5 times than the r.m.s value.

11. Calculate the output voltage of the Buck converter if the supply voltage is 11 V and duty cycle value is .4.

a) 4.4 V

b) 2.2 V

c) 4.8 V

d) 6.4 V

Answer: a

Explanation: The output voltage of the buck converter is V o = V in ×=11×.4=4.4 V. The value of the duty cycle is less than one which makes the V o < V in . The buck converter is used to step down the voltage.

12. Calculate the output voltage of the Boost converter if the supply voltage is 8 V and duty cycle value is .6.

a) 40 V

b) 20 V

c) 48 V

d) 51 V

Answer: b

Explanation: The output voltage of the boost converter is V o = V in ÷ = 8÷.4 = 20 V. The value of the duty cycle is less than one which makes the V o > V in as denominator value decreases and becomes less than one. The boost converter is used to step up the voltage.

13. Calculate the output voltage of the Buck-Boost converter if the supply voltage is 78 V and duty cycle value is .1.

a) 7.2 V

b) 4.5 V

c) 8.6 V

d) 5.1 V

Answer: c

Explanation: The output voltage of the buck-boost converter is V o = D×V in ÷ =.1÷.9=8.6 V. It can step up and step down the voltage depending upon the value of the duty cycle. If the value of the duty cycle is less than .5 it will work as a buck converter and for duty cycle greater than .5 it will work as a boost converter.

14. The principle of Boost converter can be applied for the regenerative braking.

a) True

b) False

Answer: a

Explanation: The Buck converter is used in motoring mode but a Boost converter can operate only braking mode because the characteristics are in the second quadrant only.

15. The unit of angular acceleration is Joule.

a) True

b) False

Answer: b

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid State Controlled Drives – Switch-Reluctance Motor Drives”.

1. Reluctance motors are singly-excited.

a) True

b) False

Answer: a

Explanation: Reluctance motors are singly-excited. Stator side of the motor is excited by an AC source. Due to excitation, a revolving magnetic field is a created and rotor rotates with the synchronous speed.

2. Reluctance motor can operate on DC supply.

a) True

b) False

Answer: a

Explanation: Reluctance motor can operate on AC supply. It cannot operate on DC supply. Reluctance motors are singly-excited.

3. An infinite bus-bar has _____________

a) Constant voltage

b) Constant frequency

c) Constant voltage and frequency

d) constant speed

Answer: c

Explanation: An infinite bus-bar has constant voltage and constant speed. It has many types like single bus-bar, double bus-bar. It is connected to the grid.

4. Calculate the value of the Input power factor for 3-Φ Fully controlled rectifier if the firing angle value is 90°.

a) 0

b) .2

c) .3

d) .4

Answer: a

Explanation: The value of the input power factor for 3-Φ Fully controlled rectifier is .95cos=0. The input power factor is a product of distortion factor and displacement factor.

5. Which of the following motor is used in recording instruments?

a) Reluctance motor

b) Induction motor

c) Synchronous motor

d) DC motor

Answer: a

Explanation: Reluctance motor is used in recording instruments. It is used for many constant speed applications. The starting torque of the motor depends upon the rotor position.

6. Increasing the stator poles ___________ the speed of Reluctance motor.

a) Decreases

b) Increases

c) Remains the same

d) Negative

Answer: a

Explanation: Increasing the stator poles decreases the speed of Reluctance motor. The rotating magnetic field is developed which rotates at the speed of N s . N s ∝ .

7. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 27° and supply frequency is 50 Hz.

a) 11.8 msec

b) 14.4 msec

c) 16.1 msec

d) 16.5 msec

Answer: a

Explanation: The circuit turn-off time for 3-Φ Fully controlled rectifier is ÷ω. The value of circuit turn-off time for ∝ < 60° is ÷6.28×50=11.8 msec.

8. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 120° and supply frequency is 60 Hz.

a) 1.8 msec

b) 3.2 msec

c) 6.3 msec

d) 2.7 msec

Answer: d

Explanation: The circuit turn-off time for 3-Φ Fully controlled rectifier is ÷ω. The value of circuit turn-off time for ∝ ≥ 60° is ÷6.28×60=2.7 msec.

9. Calculate peak-peak voltage if V max =180 V and V min =60 V.

a) 120 V

b) 150 V

c) 170 V

d) 110 V

Answer: a

Explanation: Peak-Peak voltage is equal to the difference between the maximum and minimum voltage. It is mathematically represented as V p-p =V max -V min =180-60=120 V.

10. Calculate the value of Crest factor if V peak =798 V and V r.m.s =489 V.

a) 1.63

b) 1.54

c) 1.59

d) 1.26

Answer: a

Explanation: The value of the crest factor is V peak ÷V r.m.s =798÷489=1.63 V. It signifies the peak value is 1.63 times than the r.m.s value.

11. Reluctance motor operates at power factor of __________

a) .8

b) .2

c) .3

d) .9

Answer: a

Explanation: Reluctance motor operates at a power factor of .8. The Reluctance motor works on the principle of an Induction motor.

12. Calculate the reactive power in a 24.5 Ω resistor.

a) 0 VAR

b) 5 VAR

c) 8 VAR

d) 9 VAR

Answer: a

13. Calculate the value of the duty cycle if the system is on for 1 sec and off for 17 sec.

a) .055

b) .781

c) .484

d) .947

Answer: a

Explanation: Duty cycle is Ton÷T total . It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷T total =1÷18=.055.

14. Calculate mark to space ratio if the system is on for 4 sec and off time is 9 sec.

a) .54

b) .44

c) .45

d) .46

Answer: b

Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff = 4÷ = .44.

15. Calculate the value of the frequency if the inductive reactance is 60 Ω and the value of the inductor is 4 H.

a) 2.38 Hz

b) 5.54 Hz

c) 4.65 Hz

d) 9.42 Hz

Answer: a

Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation X L = 2×3.14×f×L. F = X L ÷2×3.14×L = 60÷2×3.14×4 = 2.38 Hz.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid State Controlled Drives – Brushless DC Motors”.

1. The Hall effect sensor is used as the rotor position sensor for the BLDC motor.

a) True

b) False

Answer: a

Explanation: The Hall effect sensor is used as the rotor position sensor for the BLDC motor. It is used to measure the strength of the magnetic field.

2. In BLDC motor armature windings are placed on the stator side.

a) True

b) False

Answer: a

Explanation: In BLDC motor armature windings are placed on the stator side. It is more economical and less maintenance is required. Permanent magnets are placed on the rotor side.

3. BLDC motor is analogous to ______________

a) Permanent magnet synchronous motor

b) DC motor

c) Rotating Transformer

d) Single-phase Induction motor

Answer: a

Explanation: BLDC motor is analogous to Permanent magnet synchronous motor. In BLDC motor armature windings are placed on the stator side. It is more economical and less maintenance is required. Permanent magnets are placed on the rotor side.

4. The speed of a BLDC motor can be controlled by __________

a) Changing input DC voltage

b) Changing temperature

c) Changing wind direction

d) Cannot be controlled

Answer: a

Explanation: The speed of a BLDC motor can be controlled by changing the input DC voltage or the current using PWM. It can be achievable by using a transistor and chopper.

5. Which are the advantages of BLDC motor?

I. Low cost

II. Simplicity

III. Reliability

IV. Good performance

a) I, II, III, IV

b) I, II

c) I, II, IV

d) II, III, IV

Answer: a

Explanation: The BLDC motor is used in low power application. They have many advantages like low cost, simplicity, reliability, good performance, long life.

6. Due to low inertia, BLDC motors have __________

a) Faster acceleration

b) Slower acceleration

c) High-cost

d) Low cost

Answer: a

Explanation: Due to low inertia, BLDC motors have faster acceleration. BLDC motors have less weight. They can run at high speed than a conventional DC motor.

7. Calculate the electrical angle for 6 pole machine.

a) 4×θ m

b) 5×θ m

c) 2×θ m

d) 3×θ m

Answer: d

Explanation: The electrical angle for 6 pole machine is 3×θ m . One electrical angle is equal to 180° mechanical angle. θ e =×θ m .

8. Which of the following are the types of BLDC motor?

a) Unipolar, Bipolar

b) Unipolar, PWM

c) Bipolar, PWM

d) Synchronous, Induction

Answer: a

Explanation: Unipolar and Bipolar are the types of BLDC motor. They use the Hall effect rotor position sensor and optical rotor position sensor.

9. Calculate peak-peak voltage if V max =45 V and V min =45 V.

a) 60 V

b) 30 V

c) 50 V

d) 0 V

Answer: d

Explanation: Peak-Peak voltage is equal to the difference between the maximum and minimum voltage. It is mathematically represented as V p-p =V max -V min =45-45=0 V. It signifies the signal is DC signal.

10. Calculate the value of Crest factor if V peak =141 V and V r.m.s =100 V for sinusoidal voltage.

a) 1.41

b) 2.38

c) 4.42

d) 5.58

Answer: a

Explanation: The value of the crest factor is V peak ÷V r.m.s =141÷100=1.41. It signifies the peak value is 1.41 times than the r.m.s value.

11. In the biomedical instruments like artificial heart pumps, the motor used is ____________

a) DC shunt motor

b) DC series motor

c) Induction motor

d) BLDC motor

Answer: d

Explanation: BLDC motors are widely used in various applications of the medical industry. Sensorless BLDC motor and with sensor BLDC motors are used because of easy operation and high reliability compare to conventional motors.

12. Calculate the output power of the Buck converter if the supply voltage is 4 V and duty cycle value is .1 for 4 Ω load.

a) 40 mW

b) 20 mW

c) 50 mW

d) 60 mW

Answer: a

Explanation: The output voltage of the buck converter is V o = V in ×=.4. The value of the duty cycle is less than one which makes the V o < V in . The buck converter is used to step down the voltage. The output power is V o 2 ÷R=40 mW.

13. Calculate the compensator rating required for cos=.1.

a) 0.91 P.U

b) 0.99 P.U

c) 0.97 P.U

d) 0.92 P.U

Answer: b

Explanation: The compensator rating can be calculated using the relation Q P.U =√1-cos 2 =√1-.01=.99. This per unit value VAR compensator is required to improve the power factor of the system.

14. Calculate the average inductor current of the Boost converter if the load current is 7 A and duty cycle value is .8.

a) 36 A

b) 35 A

c) 34 A

d) 31 A

Answer: b

Explanation: The average inductor current of the Boost converter is I o ÷=7÷.2=35 A. The average value of the inductor current can be calculated using ampere-sec balance method.

15. Calculate the output voltage of the Buck-Boost converter if the supply voltage is 14 V and duty cycle value is .85.

a) 79.3 V

b) 45.5 V

c) 86.5 V

d) 54.7 V

Answer: a

Explanation: The output voltage of the buck-boost converter is V o = D×V in ÷ =.85÷.15=79.3 V. It can step up and step down the voltage depending upon the value of the duty cycle. If the value of the duty cycle is less than .5 it will work as a buck converter and for duty cycle greater than .5 it will work as a boost converter.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid-State Devices – Transistors”.

1. The base of a transistor is lightly doped.

a) True

b) False

Answer: a

Explanation: The base of a transistor is lightly doped because it carries the electrons from the emitter and passed it to the collector region. There should be less recombination in the base region.

2. Who is larger in size?

a) Base

b) Collector

c) Emitter

d) P-N Junction

Answer: a

Explanation: Collector is larger in size because it receives the electrons from the base region. If it is not bigger in sizer there will be crowding of electrons which causes high power dissipation.

3. Calculate the collector current using the data: I E =14 mA, I B =7 mA.

a) 7 mA

b) 8 mA

c) 9 mA

d) 10 mA

Answer: a

Explanation: The collector current in the transistor is I E -I B =14-7=7 mA. The KCL equation I E =I B +I c is always in every configuration of the transistor.

4. Calculate the current gain using the data: I c =25 mA, I B =10 mA.

a) 3.5

b) 2.5

c) 1.4

d) 7.8

Answer: b

Explanation: The current gain is represented as β in electronic circuits. The current gain is the ratio of the collector current and base current. β=I c ÷I B =25÷10=2.5.

5. Calculate the value of ∝ using the data: I c =18 mA, I E =17 mA.

a) 1.05

b) 1.47

c) 1.89

d) 2.55

Answer: a

Explanation: The value of ∝ is I c ÷I E . The current gain is the ratio of the collector current and base current. ∝=I c ÷I E =18÷17=1.05.

6. In a transistor, I c =47 mA and I E =53 mA. The value of β is _____________

a) 8.48

b) 7.83

c) 9.26

d) 3.15

Answer: b

Explanation: The value of β is I c ÷I B . The value of I B =I E -I c =53-47=6 mA. The value of β is I c ÷I B =47÷6=7.83.

7. Choose the correct relationship between β and ∝.

a) β=∝/

b) β=1/

c) β=1/

d) β=1/

Answer: a

Explanation: The relationship between β and ∝ is β=∝/. The current gain is the ratio of the collector current and base current. The value of β is I c ÷I B .

8. A Transistor is a current operated device.

a) True

b) False

Answer: a

Explanation: Transistor is a current operated device. The value of collector current is α times of the emitter current. It is a current controlled device.

9. Calculate the value of β for α=.8.

a) 2

b) 8

c) 3

d) 4

Answer: d

Explanation: The relationship between β and ∝ is β=∝/=.8÷.2=4. The current gain is the ratio of the collector current and base current. The value of β is I c ÷I B .

10. Which one of the following is correct?

a) I E =I B +I c

b) I E =I B -I c

c) I E =I c

d) I E =I B

Answer: a

Explanation: The KCL equation that holds correct is I E =I B +I c . At any point emitter current is the sum of the base current and collector current.

11. In a transistor, I B =2 mA and I c =3 mA. The value of β is _____________

a) 1.5

b) 5.8

c) 9.6

d) 3.5

Answer: a

Explanation: The value of β is I c ÷I B . The value of I B and I c are 2 mA, 3 mA. The value of β is I c ÷I B =3÷2=1.5.

12. In a transistor, β=75 and I c =53 mA. The value of I B is _____________

a) .70 mA

b) .25 mA

c) .26 mA

d) .15 mA

Answer: a

Explanation: The value of β is I c ÷I B . The value of I B =I E -I c . The value of I B is I c ÷β=53÷75=.70 mA. The relationship between β and ∝ is β=∝/.

13. Calculate the value of V peak if crest factor=5 and V r.m.s =4 V.

a) 20 V

b) 30 V

c) 40 V

d) 50 V

Answer: a

Explanation: The value of the crest factor is V peak ÷V r.m.s . The value of V peak is × V r.m.s =5×4=20 V. It signifies the ratio of V peak and V r.m.s .

14. Calculate the value of α for β=.2.

a) .85

b) .15

c) .23

d) .16

Answer: d

Explanation: The relationship between β and ∝ is ∝=β/=.2÷1.2=.16. The current gain is the ratio of the collector current and base current. The value of α is I c ÷I E .

15. Calculate the work done in stretching the wire of volume=12 m 3 and the values of stress and strain are 28 N/m 2 and 7.

a) 1176 J

b) 1544 J

c) 2542 J

d) 8788 J

Answer: a

Explanation: The work done in stretching a wire is .5×Stress×Strain×Volume. The value of the work done is .5×12×28×7=1176 J.

16. Calculate the compensator rating required for cos=.35.

a) .936 P.U

b) .154 P.U

c) .481 P.U

d) .289 P.U

Answer: a

Explanation: The compensator rating can be calculated using the relation Q P.U =√1-cos 2 =√1-.1225=.936. This per unit value VAR compensator is required to improve the power factor of the system.

This set of Electric Drives Quiz focuses on “Solid-State Devices – Thyristors”.

1. The value of latching current is more than holding current.

a) True

b) False

Answer: a

Explanation: The latching current is the minimum anode current at which SCR starts conducting. Holding current is the anode below which SCR stops conduction. The latching current is 2-3 times more than the holding current.

2. SCR is a current controlled device.

a) True

b) False

Answer: a

Explanation: SCR is a current controlled device. It can be explained using a two transistor model. When the gate pulse is applied, then it will start conduction.

3. Calculate the value of safety factor using the data: t c =12 msec, t q =6 msec.

a) 4

b) 2

c) 3

d) 6

Answer: b

Explanation: The value of the safety factor is the ratio of the circuit turn-off time and device turn-off time. S.F=t c ÷t q =12÷6=2.

4. For successful commutation circuit turn-off time ______ device turn-off time.

a) <

b) >

c) =

d) ~

Answer: b

Explanation: For successful commutation circuit turn-off time must be greater than the device turn-off time. The excess stored charged carriers should be removed from the gate junction in order to avoid false triggering.

5. SCR is a __________ device.

a) Uncontrolled

b) Semi-controlled

c) Fully controlled

d) Voltage

Answer: b

Explanation: The SCR is a semi-controlled device. Turn-on time can be controlled by using a gate pulse but commutation circuit is required to turn off the device.

6. Holding current of the SCR is ___________

a) > than latching current

b) < than latching current

c) = to latching current

d) ≥ than latching current

Answer: b

Explanation: Holding current of the SCR is less than the latching current. The value of latching current is 2-3 times than holding current.

7. If holding current of an SCR is 6 mA then its latching current should be ___________

a) 4 mA

b) 5 mA

c) 12 mA

d) 6 mA

Answer: c

Explanation: If holding current of an SCR is 6 mA then its latching current should be 12 mA. The latching current is 2-3 times more than the holding current.

8. If latching current of an SCR is 10 mA then its holding current should be ____________

a) 5 mA

b) 7 mA

c) 8 mA

d) 4 mA

Answer: a

Explanation: The latching current is 2-3 times more than the holding current. If a latching current of an SCR is 10 mA then its holding current should be 5 mA. Holding current=÷2=10÷2=5 mA.

9. Which of the following is used to protect SCR from overcurrent?

a) Heat sink

b) Fuse

c) Snubber circuit

d) Zener diode

Answer: b

Explanation: Fuse and circuit breakers are used to protect SCR from overcurrent. When current exceeds the certain limit they cut down the supply of current in the circuit.

10. Which of the following is used to protect SCR from thermal conditions?

a) Heat sink

b) Voltage clamping device

c) Zener diode

d) Snubber circuit

Answer: a

Explanation: SCR is mounted on a heat sink. When temperature exceeded the rating of the SCR excess amount of heat is dissipated through the help of the heat sink.

11. Which of the following is used to protect SCR from d/d condition?

a) Inductor

b) Capacitor

c) Heat sink

d) Fuse

Answer: a

Explanation: Inductor is used to protect SCR from d/d condition. The inductor does not allow the sudden change in current.

12. Which of the following is used to protect SCR from d/d condition?

a) Capacitor

b) Fuse

c) Snubber circuit

d) Inductor

Answer: c

Explanation: Snubber circuit is used to protect SCR from d/d condition. Snubber circuit consists of a Capacitor and resistor. The capacitor does not allow the sudden change in voltage and resistor resists the discharging of the capacitor.

13. A snubber circuit is used with thyristor in ____________

a) Parallel

b) Series

c) Anti-parallel

d) Series or Parallel

Answer: a

Explanation: A snubber circuit is used with thyristor in parallel. Snubber circuit is used to protect SCR from d/d condition. Snubber circuit consists of a Capacitor and resistor.

14. When maximum power loss occurs?

a) Rise time

b) Delay time

c) Spread time

d) Rise time or Delay time

Answer: a

Explanation: Rise time is very low . During rising, time anode voltage and anode current are very high. Power loss is very high.

15. Calculate the value of V peak if crest factor=47 and V r.m.s =6 V.

a) 282 V

b) 300 V

c) 290 V

d) 286 V

Answer: a

Explanation: The value of the crest factor is V peak ÷V r.m.s . The value of V peak is × V r.m.s = 47×6 = 282 V. It signifies the ratio of V peak and V r.m.s .

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid-State Devices – Power Electronic Devices Ratings”.

1. Which one of the following load is suitable for load commutation in Single phase Half-bridge inverter?

a) R-L load

b) R-L-C overdamped

c) R-L-C underdamped

d) L-load

Answer: c

Explanation: R-L-C underdamped are leading power factor loads. They do not require any forced commutation technique. Anti-Parallel diodes help in the commutation process.

2. R-L-C overdamped loads are generally lagging power factor loads.

a) True

b) False

Answer: a

Explanation: R-L-C overdamped loads are generally lagging power factor loads. They require forced commutation. Anti-Parallel diodes do not help in the commutation process.

3. A step-down chopper has input voltage 40 V and output voltage 20 V. Calculate the value of the duty cycle.

a) 0.5

b) 0.2

c) 0.3

d) 0.4

Answer: a

Explanation: The output voltage of the step-down chopper is V o = V in ×. The value of the duty cycle is less than one which makes the V o < V in . The step-down chopper is used to step down the voltage. The value of the duty cycle is 20÷40 = .5.

4. Calculate the De-rating factor if the string efficiency is 20 %.

a) .8

b) .4

c) .5

d) .6

Answer: a

Explanation: De-rating factor is used to measure the reliability of a string. The value of the De-rating factor is 1- = 1-.2 = .8.

5. Calculate the string efficiency if the de-rating factor is .48.

a) 48 %

b) 52 %

c) 50 %

d) 60 %

Answer: b

Explanation: The string efficiency is calculated for series and parallel connection of SCRs. The value of string efficiency is 1 – = 1-.48 = 52 %.

6. 400 V rated 5 SCRs are connected in series. The operation voltage of the string is 500. Calculate the string efficiency.

a) 25 %

b) 40 %

c) 38 %

d) 35 %

Answer: a

Explanation: The string efficiency can be calculated using the formula operation voltage÷ = 500÷ = 25 %.

7. 100 V rated 3 SCRs are connected in series. The operation voltage of the string is 200. Calculate the De-rating factor.

a) .35

b) .33

c) .38

d) .18

Answer: b

Explanation: The string efficiency can be calculated using the formula operation voltage÷ = 200÷ = .66. The De-rating factor value is 1-.66 = .33.

8. Gate triggering is more reliable.

a) True

b) False

Answer: a

Explanation: Gate triggering is more reliable than any other triggering method. The risk of false triggering of SCR reduces in the case of gate triggering.

9. The range of turn-off time is _______

a) .51-3 μsec

b) 3-101 μsec

c) 9-400 μsec

d) 7-787 μsec

Answer: b

Explanation: The range of turn-off time is 3-101 μsec. The time is very low as SCR takes low time to turn-off. Commutation time should be greater than circuit turn-off time to avoid false triggering.

10. The range of On-State voltage drop is _______

a) 1-1.6 V

b) 0-1 V

c) 1.6-2 V

d) 2.4-3 V

Answer: a

Explanation: When SCR is in forwarding blocking mode it does not conduct. When the gate pulse is applied, it starts conducting. The On-state voltage drop is very low around 1-1.6 V.

11. Thyristors are used in series to complete ________

a) High voltage demand

b) High current demand

c) Low voltage demand

d) Low current demand

Answer: c

Explanation: Thyristors are used in series to complete high voltage demand. In series voltage across the components can be the same or different but the value of current remains the same.

12. Thyristors are used in parallel to complete ________

a) High voltage demand

b) High current demand

c) Low current demand

d) Low voltage demand

Answer: c

Explanation: Thyristors are used in parallel to complete high current demand. In parallel current in the components can be the same or different but the value of voltage remains the same.

13. Calculate peak-peak voltage if V max =200 V and V min =190 V.

a) 20 V

b) 10 V

c) 60 V

d) 80 V

Answer: b

Explanation: Peak-Peak voltage is equal to the difference between the maximum and minimum voltage. It is mathematically represented as V p-p =V max -V min =200-190=10 V.

14. Which harmonic is not present in 3-Φ fully controlled rectifier?

a) 27 rd harmonic

b) 15 th harmonic

c) 17 th harmonic

d) 12 th harmonic

Answer: a

Explanation: Triplen harmonics are absent in the case of 3-phase fully controlled rectifier. Because of the absence of triplen harmonics THD of the rectifier 31%.

15. Third lowest order harmonic present in 3-Φ fully controlled rectifier is __________

a) 11 th

b) 17 th

c) 13 th

d) 22 nd

Answer: a

Explanation: Only 6k±1 order harmonics are present in 3-Φ fully controlled rectifier. The lowest order harmonic is 11 th harmonic for k=2. Third order harmonics are absent in 3-Φ fully controlled rectifier.

This set of Electric Drives Questions and Answers for Campus interviews focuses on “Solid-State Devices – di/dt and dv/dt Protection”.

1. Calculate the value of output r.m.s voltage for Single phase Half-bridge inverter if the DC supply is 80 V.

a) 40 V

b) 20 V

c) 10 V

d) 60 V

Answer: a

Explanation: The output voltage of a single phase half bridge inverter is a square wave with amplitude is equal to v D.C ÷2. The r.m.s value of output voltage is v D.C ÷2=80÷2=40 V.

2. High d÷d causes _______________

a) Local hotspots

b) Breakdown

c) High cost

d) Low cost

Answer: a

Explanation: High d÷d causes local hotspots. Local hotspots are the small areas with high temperature due to overcrowdedness of charge generation. This cause high heat generation.

3. Calculate the value of output r.m.s voltage for Single phase Full-bridge inverter if the DC supply is 25 V.

a) 26 V

b) 25 V

c) 27 V

d) 28 V

Answer: b

Explanation: The output voltage of a single phase full bridge inverter is a square wave with amplitude is equal to v D.C . The r.m.s value of output voltage is v D.C =25 V.

4. High d÷d causes _______________

a) Local hotspots

b) False triggering

c) High cost

d) High-speed operation

Answer: b

Explanation: High d÷d causes false triggering. The high value of d÷d causes the breakdown of the junction J 2 . The thyristor will trigger automatically.

5. A step-down chopper has input voltage 12 V and output voltage 2 V. Calculate the value of the duty cycle.

a) 0.26

b) 0.16

c) 0.33

d) 0.41

Answer: b

6. Calculate the De-rating factor if the string efficiency is 90 %.

a) .4

b) .1

c) .7

d) .8

Answer: a

Explanation: De-rating factor is used to measure the reliability of a string. The value of the De-rating factor is 1-=1-.9=.1.

7. Calculate the string efficiency if the de-rating factor is .22.

a) 78 %

b) 42 %

c) 20 %

d) 90 %

Answer: a

Explanation: The string efficiency is calculated for series and parallel connection of SCRs. The value of string efficiency is 1-=1-.22=78 %.

8. 25 V rated n SCRs are connected in series. The operation voltage of the string is 300 V and the string efficiency is 40 %. Calculate the value of n.

a) 27

b) 30

c) 34

d) 36

Answer: b

Explanation: The string efficiency can be calculated using the formula operation voltage÷. The value of n is 300÷=30.

9. The most reliable method for thyristor triggering is thermal triggering.

a) True

b) False

Answer: b

Explanation: Gate triggering is more reliable than any other triggering method. The risk of false triggering of SCR reduces in the case of gate triggering.

10. SCR is unipolar.

a) True

b) False

Answer: b

Explanation: The SCR is uni-directional, bipolar power electronics device. It is bipolar in nature because of its reverse and forward mode blocking capability.

11. What is the formula for thermal voltage?

a) K×e÷T

b) K×T÷e

c) K 2 ×e÷T

d) K×e 2 ÷T

Answer: a

Explanation: The formula for thermal voltage is K×e÷T where K is the Boltzmann constant, e is the charge on the electron, T is the absolute temperature.

12. Device with highest d÷d and d÷d capability is _____________

a) GTO

b) BJT

c) SCR

d) SITH

Answer: d

Explanation: SITH stands for static induction thyristor. It has a buried gate structure in which the gate electrode is placed in the n-base region. They have high d÷d and d÷d capability.

13. ____________ is the switching frequency of SITH.

a) 55 KHz

b) 100 KHz

c) 91 KHz

d) 105 KHz

Answer: b

Explanation: 100 KHz is the switching frequency of static induction thyristor. Static Induction Thyristor has a low frequency because of lower turn-off current gain.

14. Calculate the value of Crest factor if V peak =1.5 V and V r.m.s =2.5 V.

a) .4

b) .6

c) .5

d) .7

Answer: b

Explanation: The value of the crest factor is V peak ÷V r.m.s =1.5÷2.5=.6 V. It signifies the peak value is .6 times than the r.m.s value.

15. Calculate the output voltage of the Boost converter if the supply voltage is 1 V and duty cycle value is .9.

a) 10 V

b) 20 V

c) 40 V

d) 60 V

Answer: a

Explanation: The output voltage of the buck converter is V o = V in ×. The value of the duty cycle is less than one which makes the V o < V in . V o = V in ×=10 V.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “State Switching Circuits – Single Phase, Half-Wave, AC/DC Conversion for Resistive Loads”.

1. In Half-wave controlled rectifier calculate the average value of the voltage if the supply is 13sin and firing angle is 13°.

a) 4.08 V

b) 4.15 V

c) 3.46 V

d) 5.48 V

Answer: a

Explanation: In Half-wave controlled rectifier, the average value of the voltage is V m )÷2π=13)÷6.28=4.08 V. The thyristor will conduct from ∝ to π.

2. Calculate the extinction angle in purely inductive load if the firing angle is 13°.

a) 328°

b) 347°

c) 349°

d) 315°

Answer: b

Explanation: The extinction angle in the purely inductive load is 2π-∝=360°-13°=347°. The extinction angle is the angle at which current in the circuit becomes zero. The average value of the voltage in a purely inductive load is zero.

3. Calculate the conduction angle in purely inductive load if the firing angle is 165°.

a) 78°

b) 55°

c) 30°

d) 19°

Answer: c

Explanation: The conduction angle in the purely inductive load is β-α=2=2=30°. The conduction angle is the angle for which the current exists in the circuit. The average value of the voltage in a purely inductive load is zero.

4. R-L-C underdamped loads are generally lagging power factor loads.

a) True

b) False

Answer: b

Explanation: R-L-C underdamped loads are generally leading power factor loads. They do not require forced commutation. Anti-Parallel diodes help in the commutation process.

5. In Half-wave uncontrolled rectifier calculate the average value of the voltage if the supply is 3sin.

a) .95 V

b) .92 V

c) .93 V

d) .94 V

Answer: a

Explanation: In Half-wave uncontrolled rectifier, the average value of the voltage is V m ÷π=3÷π=.95 V. The diode will conduct only for the positive half cycle. The conduction period of the diode is π.

6. In Half-wave uncontrolled rectifier calculate the r.m.s value of the voltage if the supply is 89sin.

a) 91.5 V

b) 44.5 V

c) 25.1 V

d) 15.1 V

Answer: b

Explanation: In Half-wave uncontrolled rectifier, the r.m.s value of the voltage is V m ÷2=89÷2=44.5 V. The diode will conduct only for the positive half cycle. The conduction period of the diode is π.

7. In Half-wave uncontrolled rectifier calculate the power dissipation across the 8 Ω resistor if the supply is 29sin.

a) 26.2 W

b) 24.2 W

c) 26.1 W

d) 29.1 W

Answer: a

Explanation: In Half-wave uncontrolled rectifier, the r.m.s value of the voltage is V m ÷2=29÷2=14.5 V. The diode will conduct only for the positive half cycle. Power dissipation across the resistor is V 2 r.m.s ÷R=14.5 2 ÷8=26.2 W.

8. The conduction period of diode in Half-wave uncontrolled rectifier for resistive load is ______________

a) π

b) 2π

c) 3π

d) 4π

Answer: a

Explanation: The conduction period of the diode in Half-wave uncontrolled rectifier for the resistive load is π. For the negative A.C supply diode will be reverse biased.

9. In Half-wave uncontrolled rectifier calculate the average value of the current for 3 Ω resistive load if the supply is 34sin.

a) 3.6 A

b) 2.6 A

c) 2.5 A

d) 3.1 A

Answer: a

Explanation: In Half-wave uncontrolled rectifier, the average value of the current is V m ÷πR=34÷3π=3.6 A. The diode will conduct only for the positive half cycle. The conduction period of the diode is π.

10. In Half-wave controlled rectifier calculate the average value of the current for 2.5 Ω resistive load if the supply is sin and firing angle is 26°.

a) 0.8 V

b) 0.15 V

c) 0.12 V

d) 0.21 V

Answer: c

Explanation: In Half-wave controlled rectifier, the average value of the current is V m )÷2πR=)÷6.28×2.5=.12 V. The thyristor will conduct from ∝ to π.

11. Calculate the circuit turn-off time for Half-wave controlled rectifier for a ω=5 rad/sec for resistive load.

a) .62 sec

b) .42 sec

c) .58 sec

d) .64 sec

Answer: a

Explanation: The value of the circuit turn-off for Half-wave controlled rectifier for a ω=5 rad/sec for the resistive load is a π÷Ω=π÷5=.62 sec.

12. Calculate the string efficiency if the de-rating factor is .429.

a) 48.1 %

b) 57.1 %

c) 47.8 %

d) 46.5 %

Answer: b

Explanation: The string efficiency is calculated for series and parallel connection of SCRs. The value of string efficiency is 1-=1-.429=57.1 %.

13. Calculate the output frequency for the six-pulse converter if the supply frequency is 10 Hz.

a) 40 Hz

b) 30 Hz

c) 60 Hz

d) 80 Hz

Answer: c

Explanation: The output of a six-pulse converter consists of six pulses in one cycle. The output frequency of the six pulse converter is 6×supply frequency=6×10=60 Hz.

14. Calculate the pulse number if the supply frequency is 2π and the output frequency is π÷6.

a) 4

b) 12

c) 16

d) 8

Answer: b

Explanation: The pulse number can be calculated using the ratio of input frequency to the output frequency. The value of pulse number is 2π÷π÷6=12. It is a twelve-pulse converter.

15. Volt-sec balance method is based on the principle of the energy of conservation.

a) True

b) False

Answer: a

Explanation: Volt-sec balance method states that the net area over the cycle will be zero. It is based on the principle of the energy of conservation. The amount of charging is equal to discharging.

This set of Electric Drives Questions and Answers for Aptitude test focuses on “State Switching Circuits – Single Phase, Full-Wave, AC/DC Conversion for Resistive Loads “.

1. Calculate the value of the Input power factor for 1-Φ Full wave bridge rectifier if the firing angle value is 39°.

a) .69

b) .59

c) .78

d) .15

Answer: a

Explanation: The value of the input power factor for 1-Φ Full wave bridge rectifier is .9cos=.69. The input power factor is a product of distortion factor and displacement factor.

2. Calculate the value of the fundamental displacement factor for 1-Φ Full wave bridge rectifier if the firing angle value is 38°.

a) .22

b) .78

c) .33

d) .44

Answer: b

Explanation: Fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. D.F=cos=cos=0.78.

3. Calculate the value of the fundamental displacement factor for 1-Φ Full wave semi-converter if the firing angle value is 69°.

a) .48

b) .24

c) .82

d) .88

Answer: c

4. Calculate the fundamental component of source current in 1-Φ Full wave bridge rectifier for load current=3.14 A.

a) 2.82 A

b) 1.45 A

c) 3.69 A

d) 4.78 A

Answer: a

Explanation: The fundamental component of source current in 1-Φ Full wave bridge rectifier is 2√2I o ÷π. It is the r.m.s value of the fundamental component. I s1 = 2√2I o ÷π=2√2=2.82 A.

5. Calculate the circuit turn-off time for 1-Φ Full wave bridge rectifier for α=145°. Assume the value of ω=5 rad/sec.

a) 84.9 msec

b) 94.5 msec

c) 101.2 msec

d) 87.2 msec

Answer: d

Explanation: The circuit turn-off time for 1-Φ Full wave bridge rectifier is ÷ω. The value of circuit turn-off time is ÷5=87.2 msec.

6. Calculate the fundamental component of source current in 1-Φ Full wave bridge rectifier for the load current=78 A.

a) 78 A

b) 45 A

c) 69 A

d) 13 A

Answer: a

Explanation: The fundamental component of source current in 1-Φ Full wave bridge rectifier is I o . It is the r.m.s value of the source current. I s =I o =78 A.

7. Calculate the r.m.s value of source current in 1-Φ Full wave semi-converter for the load current=51.2 A and α=15°.

a) 10.53 A

b) 14.52 A

c) 44.92 A

d) 49.02 A

Answer: d

Explanation: The r.m.s value of source current in 1-Φ Full wave semi-converter is I o √π-α÷π. It is the r.m.s value of the source current. I = I o √π-α÷π = 51.2 = 49.02 A.

8. Calculate the r.m.s value of thyristor current in 1-Φ Full wave semi-converter for the load current=2.2 A and α=155°.

a) .58 A

b) .57 A

c) .51 A

d) .52 A

Answer: b

Explanation: The r.m.s value of source current in 1-Φ Full wave semi-converter is I o √π-α÷2π. It is the r.m.s value of the thyristor current. I = I o √π-α÷2π=2.2=.57 A.

9. Calculate the r.m.s value of diode current in 1-Φ Full wave semi-converter for the load current=5.1 A and α=115°.

a) 4.21 A

b) 4.61 A

c) 4.71 A

d) 4.52 A

Answer: b

Explanation: The r.m.s value of diode current in 1-Φ Full wave semi-converter is I o √π+α÷2π. It is the r.m.s value of the diode current. I = I o √π+α÷π=5.1=4.61 A.

10. Calculate the average value of thyristor current in 1-Φ Full wave semi-converter for the load current=25.65 A and α=18°.

a) 11.54 A

b) 12.15 A

c) 15.48 A

d) 14.52 A

Answer: a

Explanation: The average value of thyristor current in 1-Φ Full wave semi-converter is I o . It is the average value of the thyristor current. I avg = I o =25.65=11.54 A.

11. Calculate the average value of diode current in 1-Φ Full wave semi-converter for the load current=75.2 A and α=41°.

a) 46.16 A

b) 42.15 A

c) 41.78 A

d) 41.18 A

Answer: a

Explanation: The average value of diode current in 1-Φ Full wave semi-converter is I o . It is the average value of the diode current. I avg = I o =75.2=46.16 A.

12. Calculate the average value of diode current in 1-Φ Full wave semi-converter for the load current=5.2 A and α=11°.

a) .32 A

b) .31 A

c) .25 A

d) .27 A

Answer: b

Explanation: The average value of diode current in 1-Φ Full wave semi-converter is I o . It is the average value of the diode current. I avg = I o =5.2=.31 A.

13. Calculate the r.m.s value of diode current in 1-Φ Full wave semi-converter for the load current=.2 A and α=74°.

a) .154 A

b) .248 A

c) .128 A

d) .587 A

Answer: c

Explanation: The r.m.s value of diode current in 1-Φ Full wave semi-converter is I o √. It is the r.m.s value of the diode current. I r.m.s = I o √=.2√=.128 A.

14. Diodes in 1-Φ Full wave semi-converter protects the thyristor from short-circuiting.

a) True

b) False

Answer: a

Explanation: Diodes in 1-Φ Full wave semi-converter protects the thyristor from short-circuiting. They provide the gap from to avoid the conduction of one leg thyristors.

15. The problem of short-circuiting in 1-Φ Full wave semi-converter is very common.

a) True

b) False

Answer: a

Explanation: The problem of short-circuiting in 1-Φ Full wave semi-converter is very common. Diodes protect the thyristor from short-circuiting. They provide the gap from to avoid the conduction of one leg thyristors.

This set of Electric Drives online test focuses on “State Switching Circuits – Single Phase, Half-Wave, AC/DC Conversion for Inductive Loads Without Freewheeling Diode”.

1. Calculate the value of the conduction angle for R-L load if the value of β and α are 19° and 29°.

a) 10°

b) 70°

c) 30°

d) 80°

Answer: a

Explanation: The conduction angle for R-L load is β-α=29°-19°=10°. R-L load is a current stiff type of load. The current in the circuit only flows from α to β.

2. Calculate the V 0 avg for Single-phase Half Wave rectifier for R-L load using the data: V m =24 V, ∝=30°, β=60°.

a) 1.39 V

b) 8.45 V

c) 4.55 V

d) 1.48 V

Answer: a

Explanation: In Half-wave controlled rectifier, the average value of the voltage is V m -cos)÷2π=24- cos)÷6.28=1.39 V. The thyristor will conduct from ∝ to β.

3. Calculate the I 0 avg for Single-phase Half Wave rectifier for R-L load using the data: V m =56 V, ∝=15°, β=30°, R=2 Ω.

a) .56 A

b) .44 A

c) .26 A

d) .89 A

Answer: a

Explanation: In Half-wave controlled rectifier, the average value of the current is V m -cos)÷2πR=56- cos)÷12.56=.44 A. The thyristor will conduct from ∝ to β.

4. Calculate the I 0 r.m.s for Single-phase Half Wave rectifier for R-L load using the data: V m =110 V, ∝=16°, β=31°, R=1 Ω.

a) 4.56 A

b) 1.82 A

c) 4.81 A

d) 9.15 A

Answer: b

Explanation: In Half-wave controlled rectifier, the r.m.s value of the current is V m – cos)÷2πR=110 – cos)÷6.28=1.82 A. The thyristor will conduct from ∝ to β.

5. Calculate the value of the fundamental displacement factor for 1-φ Full wave semi-converter if the firing angle value is 0 o .

a) 1

b) .8

c) .4

d) .2

Answer: a

6. For highly inductive load current remains continuous.

a) True

b) False

Answer: a

Explanation: For highly inductive load current remains continuous. Inductor opposes the change in the current. When current is about to extinct the next conducting cycle starts due to which currently remains constant.

7. Calculate the value of the Input power factor using the data: Fundamental displacement factor=.96, Distortion factor=.97.

a) .93

b) .84

c) .48

d) .89

Answer: a

Explanation: The value of the input power factor is a product of the ×=g×F.D.F=.97×.96=.93.

8. Displacement factor depends upon the shape of the waveform.

a) True

b) False

Answer: a

Explanation: Displacement factor depends upon the shape of the waveform. Displacement factor is the cosine of the angle between the fundamental voltage and fundamental current. It depends upon the type of load, converter type.

9. In single phase RLE load, calculate the voltage across the thyristor when current decays to zero using the data: (V s ) r.m.s =340 V, f=50 Hz, R=2 Ω, E=150 V, β=160°.

a) 45.89 V

b) 74.45 V

c) 54.85 V

d) 84.48 V

Answer: b

Explanation: In single phase RLE load the voltage across the thyristor when current decays to zero V T =V m sin-E=340×√2sin-90=74.45 V.

10. In single phase RLE load, calculate the angle at which conduction starts using the data: (V s )=14sin, f=50 Hz, E=10 V.

a) 45.58°

b) 46.26°

c) 47.26°

d) 49.56°

Answer: a

Explanation: The angle at which conduction starts when V m sin=E. The conduction will remain from ϴ to π-θ. Θ=sin-(E÷V m )=sin-=45.58°.

11. Full form of PIV.

a) Peak inverse voltage

b) Peak insert voltage

c) Paas Inverse volatile

d) Peak insert volatile

Answer: a

Explanation: PIV stands for Peak inverse voltage. It is the maximum negative voltage after which thyristor will breakdown.

12. Calculate the PIV for the Full-wave bridge rectifier if the peak value of the supply voltage is 230.

a) 324.4 V

b) 325.2 V

c) 524.4 V

d) 626.8 V

Answer: a

Explanation: The peak inverse voltage for the Full-wave bridge rectifier is V m =√2×230=325.2 V. The peak inverse is the maximum negative voltage across the thyristor.

13. In single phase RLE load, calculate the Peak inverse voltage using the data: (V s ) r.m.s =89 V, f=50 Hz, E=440 V.

a) 521.2 V

b) 527.8 V

c) 597 V

d) 529 V

Answer: c

Explanation: In a single phase, RLE load the peak inverse voltage across the thyristor V T =V m +E=529 V.

14. Full form of DAC is ___________

a) Digital to Analog converter

b) Discrete to Analog converter

c) Digital to AC converter

d) Discrete to Avalanche converter

Answer: a

Explanation: DAC stands for Digital to Analog converter. It converts the digital signal to an analog signal. It is used in microcontrollers and microprocessors.

15. Full form of ADC is ___________

a) Analog to Digital converter

b) Discrete to Analog converter

c) Analog to Digital converter

d) Discrete to Avalanche converter

Answer: c

Explanation: ADC stands for Analog to Discrete converter. It converts the analog signal to a discrete signal. It is used in microcontrollers and microprocessors.

This set of Electric Drives MCQs focuses on “State Switching Circuits – Single Phase, Half-Wave, AC/DC Conversion for Inductive Loads With Freewheeling Diode”.

1. Calculate the voltage across the freewheeling diode when the output voltage is 24 V.

a) -15 V

b) -24 V

c) 28 V

d) 39 V

Answer: b

Explanation: The freewheeling diode is used to provide a freewheeling path. It is connected in the anti-parallel direction of the load. The voltage across the diode is -V o =-24 V.

2. Calculate the value of the conduction angle for SCR for R-L load with a freewheeling diode if the value of β and α are 65° and 12°.

a) 168°

b) 170°

c) 130°

d) 180°

Answer: a

Explanation: The conduction angle for SCR for R-L load with a freewheeling diode is π-α=180°-12°=168°. R-L load is a current stiff type of load. The current in the SCR only flows from α to π.

3. Calculate the value of the conduction angle for diode for R-L load with a freewheeling diode if the value of α is 45°.

a) 220°

b) 225°

c) 230°

d) 280°

Answer: b

Explanation: The conduction angle for diode for R-L load with a freewheeling diode is π+α=180°+45°=225°. R-L load is a current stiff type of load.

4. Calculate the value of the conduction angle for diode for R-L load with a freewheeling diode if the value of β is 226°.

a) 40°

b) 45°

c) 50°

d) 46°

Answer: b

Explanation: The conduction angle for diode for R-L load with a freewheeling diode is β-π=226°-180°=45°. R-L load is a current stiff type of load.

5. Calculate the value of the conduction angle for R-L load if the value of β and α are 56° and 18°.

a) 48°

b) 38°

c) 57°

d) 15°

Answer: b

Explanation: The conduction angle for R-L load is β-α=56°-18°=38°. R-L load is a current stiff type of load. The current in the circuit only flows from α to β.

6. Calculate the value of the fundamental displacement factor for 1-Φ Full wave bridge rectifier if the firing α=17°.

a) .98

b) .19

c) .56

d) .95

Answer: d

Explanation: Fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. D.F=cos=cos=0.95.

7. Calculate the value of the fundamental displacement factor for 1-Φ Full wave semi-converter if the firing angle value is 95 o .

a) .60

b) .68

c) .62

d) .67

Answer: d

8. Which one of the following load is suitable for lagging power factor load in Single phase Half-bridge inverter?

a) C load

b) R-L-C overdamped

c) R-L-C underdamped

d) L-C load

Answer: b

Explanation: R-L-C overdamped loads are generally lagging power factor loads. They require forced commutation. Anti-Parallel diodes do not help in the commutation process.

9. Which one of the following load is suitable for leading power factor load in Single phase Half-bridge inverter?

a) C load

b) R-L-C overdamped

c) R-L-C underdamped

d) L-C load

Answer: c

Explanation: R-L-C underdamped are leading power factor loads. They do not require any forced commutation technique. Anti-Parallel diodes help in the commutation process.

10. A step-down chopper has input voltage .1 V and output voltage .01 V. Calculate the value of the duty cycle.

a) 0.1

b) 0.2

c) 0.3

d) 0.5

Answer: a

11. Calculate the De-rating factor if the string efficiency is 98%.

a) .04

b) .02

c) .05

d) .03

Answer: b

Explanation: De-rating factor is used to measure the reliability of a string. The value of the De-rating factor is 1-=1-.98=.02.

12. Calculate the string efficiency if the de-rating factor is .50.

a) 28 %

b) 42 %

c) 80 %

d) 50 %

Answer: d

Explanation: The string efficiency is calculated for series and parallel connection of SCRs. The value of string efficiency is 1-=1-.50=50 %.

13. 70 V rated 6 SCRs are connected in series. The operation voltage of the string is 130. Calculate the De-rating factor.

a) .70

b) .73

c) .78

d) .74

Answer: a

Explanation: The string efficiency can be calculated using the formula operation voltage÷=130÷=.30. The De-rating factor value is 1-.30=.70.

14. d÷d is more reliable for SCR triggering.

a) True

b) False

Answer: b

Explanation: Gate triggering is more reliable than any other triggering method. The risk of false triggering of SCR increases in the case of the d÷d triggering.

15. Calculate the output voltage of the Buck converter if the supply voltage is 13 V and duty cycle value is .16.

a) 2.08 V

b) 2.24 V

c) 2.58 V

d) 2.54 V

Answer: a

Explanation: The output voltage of the buck converter is V o = V in ×=13×.16=2.08 V. The value of the duty cycle is less than one which makes the V o < V in . The buck converter is used to step down the voltage.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid-State Switching Circuits – Three Phase, Half-Wave, AC/DC Conversion for Resistive Loads”.

1. Calculate the V o for the 3-Φ phase Half-wave uncontrolled rectifier if the supply value is 440 V.

a) 297.25 V

b) 298.15 V

c) 298.11 V

d) 300.15 V

Answer: a

Explanation: The V o for the 3-Φ phase Half-wave uncontrolled rectifier is 3V ml ÷2π. The value of output voltage is 3V ml ÷2π=3×√2×440÷6.28=297.25 V.

2. Calculate the pulse number if the supply frequency is 2π and the output frequency is 2π÷3.

a) 4

b) 5

c) 6

d) 3

Answer: d

Explanation: The pulse number can be calculated using the ratio of input frequency to the output frequency. The value of pulse number is 2π÷=3. It is a three-pulse converter or 3-Φ phase Half-wave uncontrolled rectifier.

3. Calculate the average value of current for the 3-Φ phase Half-wave uncontrolled rectifier if the supply value is 400 V and resistive load value is 5 Ω.

a) 54.04 A

b) 57.26 A

c) 51.64 A

d) 58.15 A

Answer: a

Explanation: The I o for the 3-Φ phase Half-wave uncontrolled rectifier is 3V ml ÷2πR. The value of output voltage is 3V ml ÷2πR=3×√2×440÷6.28×5=54.04 A.

4. In 3-Φ Fully controlled rectifier calculate the average value of the voltage if the supply is 400 V and firing angle is 45°.

a) 381.15 V

b) 382.16 V

c) 383.19 V

d) 384.25 V

Answer: b

Explanation: In 3-Φ Fully controlled rectifier, the average value of the voltage is 3V ml )÷π=3×400×√2)÷3.14=382.16 V.

5. Calculate the V o for the 3-Φ phase Half-wave uncontrolled rectifier if the supply value is 441 V.

a) 297.25 V

b) 298.15 V

c) 298.11 V

d) 300.15 V

Answer: a

Explanation: The V o for the 3-Φ phase Half-wave uncontrolled rectifier is 3V ml ÷2π. The value of output voltage is 3V ml ÷2π=3×√2×441÷6.28=297.25 V.

6. Calculate the V o for the 3-Φ phase Half-wave controlled rectifier if the supply value is 364 V and the value of firing angle is 2°.

a) 247.62 V

b) 245.76 V

c) 214.26 V

d) 233.26 V

Answer: b

Explanation: The V o for the 3-Φ phase Half-wave controlled rectifier is 3V ml cos÷2π. The value of output voltage is 3V ml cos÷2π=3×√2×364×÷6.28=245.76 V.

7. Calculate the value of the Input power factor for 3-Φ Fully controlled rectifier if the firing angle value is 16°.

a) .96

b) .91

c) .93

d) .94

Answer: b

Explanation: The value of the input power factor for 3-Φ Fully controlled rectifier is .95cos=.91. The input power factor is a product of distortion factor and displacement factor.

8. In 3-Φ Semi-controlled rectifier calculate the average value of the voltage if the supply is 299 V and firing angle is 56.15°.

a) 311.26 V

b) 304.26 V

c) 314.51 V

d) 312.45 V

Answer: c

Explanation: In 3-Φ Semi-controlled rectifier, the average value of the voltage is 3V ml )÷2π=3×299×√2)÷6.28=314.51 V.

9. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 13° and supply frequency is 49.5 Hz.

a) 730.2 msec

b) 740.8 msec

c) 754.5 msec

d) 755.1 msec

Answer: a

Explanation: The circuit turn-off time for 3-Φ Fully controlled rectifier is ÷ω. The value of circuit turn-off time for ∝ < 60° is ÷6.28×49.5=730.2 msec.

10. Calculate the circuit turn-off time for 3-Φ Fully controlled rectifier if the firing angle is 170° and supply frequency is 50.5 Hz.

a) 32.4 msec

b) 35.2 msec

c) 39.3 msec

d) 31.5 msec

Answer: d

Explanation: The circuit turn-off time for 3-Φ Fully controlled rectifier is ÷Ω. The value of circuit turn-off time for ∝ ≥ 60° is ÷6.28×50.5=31.5 msec.

11. Calculate the form factor if V =45 V, V =15 V.

a) 2

b) 5

c) 3

d) 8

Answer: c

Explanation: The form factor is defined as the ratio of the V r.m.s ÷V avg =45÷15=3. The r.m.s value is three times of average value V avg .

12. Calculate the V o for the 3-Φ phase Half-wave controlled rectifier if the supply value is 480 V and the value of firing angle is 92°.

a) 264.02 V

b) 487.26 V

c) 858.26 V

d) 248.25 V

Answer: a

Explanation: The V o for the 3-Φ phase Half-wave controlled rectifier is 3V m )÷2π. The value of output voltage is 3V m )÷2π=3×480×√3×√2)÷2π=264.02 V.

13. Full form of SCR is ____________

a) Silicon controlled rectifier

b) State-controlled rectifier

c) State cover rectifier

d) State-controlled reset

Answer: a

Explanation: SCR stands for silicon controlled rectifier. It is a semi-controlled, bipolar, uni-directional switch. It is a high power rating than other power electronic devices.

14. Full form of TRIAC is __________

a) Triode for Alternating current

b) Tri for Alternating current

c) Triode for Alternating counter

d) Tri for Alternating counters

Answer: a

Explanation: TRIAC stands for Triode for Alternating current. It is a bipolar, bidirectional switch. It is used in fan regulators to control the speed of fans.

15. TRIAC is a unipolar switch.

a) True

b) False

Answer: b

Explanation: TRIAC stands for Triode for Alternating current. It is a bipolar, bidirectional switch. It is used in fan regulators to control the speed of fans.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid-State Switching Circuits – Three Phase, Full-Wave, AC/DC Conversion”.

1. For α > 90°, 3-Φ Full wave bridge rectifier acts as a natural commutated inverter.

a) True

b) False

Answer: a

Explanation: The output voltage of 3-Φ Full wave bridge rectifier is 3V ml )÷π. For α > 90° the output voltage becomes negative. The power flows from DC to the AC side.

2. What is the formula for output voltage for 3-Φ Full wave bridge rectifier for R-L load?

a) 3V ml )÷2π

b) 3V ml )÷π

c) 2V ml )÷π

d) 6V ml )÷π

Answer: b

Explanation: The output voltage of 3-Φ Full wave bridge rectifier for R-L load is 3V ml )÷π. The net area of the output voltage for R-L load remains zero.

3. The output voltage of 3-Φ Full wave bridge rectifier is six times of 3-Φ Half-wave rectifier.

a) True

b) False

Answer: b

Explanation: The output voltage of 3-Φ Full wave bridge rectifier is 3V ml )÷π. The output voltage of 3-Φ Half wave rectifier is 3V ml )÷2π.

4. What is the formula for output voltage for 3-Φ Full wave bridge rectifier for R load for α < 60°?

a) 2V ml )÷π

b) 3V ml )÷2π

c) 3V ml )÷π

d) 6V ml )÷π

Answer: c

Explanation: The output voltage of 3-Φ Full wave bridge rectifier for R load is 3V ml )÷π for α < 60°. Conduction will only remain from 60°+α to 120°+α.

5. ____________ is the boundary for C.C.M and D.C.M mode in 3-Φ Full wave bridge rectifier for R load.

a) 60°

b) 10°

c) 80°

d) 50°

Answer: a

Explanation: 60° is the boundary for C.C.M and D.C.M mode in 3-Φ Full wave bridge rectifier for R load. Conduction will only remain from 60°+α to 120°+α .

6. What is the formula for output voltage for 3-Φ Full wave bridge rectifier for R load for α > 60°?

a) 2V ml )÷π

b) 3V ml )÷2π

c) 3V ml )÷π

d) 6V ml )÷π

Answer: c

Explanation: The output voltage of 3-Φ Full wave bridge rectifier for R load is 3V ml )÷π for α > 60°. Conduction will only remain from 60°+α to 180° .

7. Calculate the r.m.s value of thyristor current in 3-Φ Full wave converter for the load current=4 A and α=12°.

a) 2.3 A

b) 2.5 A

c) 2.7 A

d) 2.9 A

Answer: a

Explanation: The r.m.s value of thyristor current in 3-Φ Full wave converter is I o √1÷3. It is the r.m.s value of the thyristor current. I = I o √1÷3 = 2.3 A.

8. Calculate the average value of thyristor current in 3-Φ Full wave converter for the load current=9 A and α=26°.

a) 4 A

b) 5 A

c) 7 A

d) 3 A

Answer: d

Explanation: The average value of thyristor current in 3-Φ Full wave converter is I o ÷3. It is the average value of the thyristor current. I = I o ÷3 = 3 A.

9. Calculate the De-rating factor if the string efficiency is 16 %.

a) .84

b) .44

c) .5

d) .6

Answer: a

Explanation: De-rating factor is used to measure the reliability of a string. The value of the De-rating factor is 1-=1-.16=.84.

10. Full form of MOSFET is ___________

a) Metal oxide silicon field effect transistor

b) Metal oxide semiconductor field effect transistor

c) Metal oxide settle field effect transistor

d) Metal oriented silicon field effect transistor

Answer: b

Explanation: MOSFET stands for Metal oxide semiconductor field effect transistor. It is a three terminal device consists of source, drain, gate. It is a high switching device.

11. Full form of FET is ___________

a) Field effect transistor

b) Field engage transistor

c) Field effect terminal

d) Fire engage transistor

Answer: a

Explanation: FET stands for Field effect transistor. It is a transistor that depends upon the electric field to control the conductivity of the channel.

12. Calculate the compensator rating required for sin=.13.

a) 0.45 P.U

b) 0.12 P.U

c) 0.13 P.U

d) 0.82 P.U

Answer: c

Explanation: The compensator rating can be calculated using the relation Q P.U =√1-cos 2 =sin=.13. This per unit value VAR compensator is required to improve the power factor of the system.

13. Full form of DIAC is ___________

a) Digital Alternating current

b) Discrete Alternating current

c) Diode for Alternating current

d) Digital Alternating counter

Answer: c

Explanation: DIAC stands for Diode Alternating current. It is a bipolar switch. It will conduct when the voltage across it becomes greater than the breakover voltage.

14. Which harmonic is not present in 3-Φ fully controlled rectifier?

a) 81 st harmonic

b) 15 th harmonic

c) 17 th harmonic

d) 11 th harmonic

Answer: a

Explanation: Triplen harmonics are absent in the case of 3-phase fully controlled rectifier. Because of the absence of triplen harmonics THD of the rectifier 31%.

15. Fifth lowest order harmonic present in 3-Φ fully controlled rectifier is __________

a) 15 th

b) 17 th

c) 13 th

d) 12 th

Answer: b

Explanation: Only 6k±1 order harmonics are present in 3-Φ fully controlled rectifier. The fifth lowest order harmonic is 17 th harmonic for k=3. Third order harmonics are absent in 3-Φ fully controlled rectifier.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid-State Switching Circuits – DC/DC Conversion”.

1. Choppers are used to control the DC voltage level.

a) True

b) False

Answer: a

Explanation: Choppers are used to control the DC voltage level. They can increase, decrease or both the input DC voltage.

2. Which one of the following device is uncontrolled?

a) SCR

b) MOSFET

c) Diode

d) TRIAC

Answer: c

Explanation: Diode is uncontrolled, unidirectional power electronic device. When the voltage across the diode becomes positive it starts conduction otherwise remains off.

3. Cuk-converter is better than Buck converter in terms of the output voltage.

a) True

b) False

Answer: a

Explanation: Cuk-converter is better than Buck converter in terms of the output voltage. The filter is present at the output and input side of the Cuk-converter whereas in Buck there is no filter circuit present at the input side.

4. What is the formula for output voltage for Buck converter?

a) 8D×V in

b) 5D×V in

c) 2D×V in

d) D×V in

Answer: d

5. What is the formula for output voltage for Buck-Boost converter?

a) D×V in

b) V in ÷

c) D×V in ÷

d) D×V in ÷

Answer: c

6. What is the formula for output voltage for Boost converter?

a) 8D×V in

b) 5D×V in

c) 2D×V in

d) D×V in

Answer: d

Explanation: The output voltage of the boost converter is V o = V in ÷ . The value of the duty cycle is less than one which makes the V o > V in . The boost converter is used to step up the voltage. V in is a fixed voltage and V o is a variable voltage.

7. ____________ is the boundary for C.C.M and D.C.M mode in 3-Φ Half wave bridge rectifier for R load.

a) 60°

b) 10°

c) 80°

d) 30°

Answer: d

Explanation: 30° is the boundary for C.C.M and D.C.M mode in 3-Φ Half wave bridge rectifier for R load. Conduction will only remain from 30°+α to 150°+α.

8. Buck-Boost acts as Buck converter for duty cycle is equal to _________

a) .9

b) .7

c) .6

d) .4

Answer: d

9. Buck-Boost acts as Boost converter for duty cycle is equal to _________

a) .8

b) .1

c) .2

d) .4

Answer: a

Explanation: The output voltage of the buck-boost converter is V o = D×V in ÷ . It can step up and step down the voltage depending upon the value of the duty cycle. If the value of the duty cycle is more than .5 it will work as a boost converter. The answer is .8.

10. Inductor and Capacitor in Buck converter are used to ___________

a) Increase the cost

b) Decrease the cost

c) Filter out the harmonics

d) Increase the harmonics

Answer: c

Explanation: Inductor and capacitor in Buck converter are used to filter out the harmonics. They remove the ripple from the output voltage.

11. Calculate peak-peak voltage if V max =1 V and V min =-1 V.

a) 6 V

b) 2 V

c) 3 V

d) 1 V

Answer: b

Explanation: Peak-Peak voltage is equal to the difference between the maximum and minimum voltage. It is mathematically represented as V p-p =V max -V min =1+1=2 V.

12. Calculate the value of Crest factor if V peak =0 V and V r.m.s =24 V.

a) 0

b) 3

c) 5

d) 8

Answer: a

Explanation: The value of the crest factor is V peak ÷V r.m.s =0÷24=0 V. It signifies the peak value is 0 times than the r.m.s value.

13. Calculate the output voltage of the Buck converter if the supply voltage is 789 V and duty cycle value is .9.

a) 711.1 V

b) 710.1 V

c) 722.2 V

d) 713.2 V

Answer: b

Explanation: The output voltage of the buck converter is V o = V in ×=789×.9=710.1 V. The value of the duty cycle is less than one which makes the V o < V in . The buck converter is used to step down the voltage.

14. Calculate the average value of thyristor current in 3-Φ Full wave converter for the load current=27 A and α=6°.

a) 9 A

b) 4 A

c) 5 A

d) 9 A

Answer: d

Explanation: The average value of thyristor current in 3-Φ Full wave converter is I o ÷3. It is the average value of the thyristor current. I = I o ÷3=9 A.

15. Calculate the output voltage of the Boost converter if the supply voltage is 156 V and duty cycle value is .4.

a) 260 V

b) 264 V

c) 261 V

d) 268 V

Answer: a

Explanation: The output voltage of the boost converter is V o = V in ×=156×1.66=260 V. The value of the duty cycle is less than one which makes the V o > V in . The boost converter is used to step up the voltage.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid-State Switching Circuits – DC/AC Conversion”.

1. Inverters are used to convert DC power into variable AC power.

a) True

b) False

Answer: a

Explanation: Inverters are used to convert DC power into variable AC power. The variable AC power means variable frequency and variable voltage level.

2. Short circuit problem is severe in case of Voltage source inverter.

a) True

b) False

Answer: a

Explanation: Short-circuit problem is severe in case of voltage source inverter. The output voltage of VSI is fixed but the output current depends upon the load. When Z=0, I=V÷Z=∞.

3. Switches used in VSI are ___________

a) Unipolar & Unidirectional

b) Bipolar & Bidirectional

c) Unipolar & Bidirectional

d) Unidirectional

Answer: c

Explanation: Switches used in VSI are unipolar and bidirectional in nature. MOSFET, BJT are the examples of switches used in VSI. The output voltage of VSI is fixed but the output current depends upon the load.

4. Switches used in CSI are ____________

a) Bipolar & Unidirectional

b) Bipolar & Bidirectional

c) Unipolar & Unidirectional

d) Bidirectional

Answer: a

Explanation: Switches used in CSI are bipolar and unidirectional in nature. GTO, thyristors are the examples of switches used in CSI. The output voltage of CSI is variable but the output current is fixed.

5. Full form of VSI is ___________

a) Voltage source inverter

b) Volume source inverter

c) Voltage severe inverter

d) Voltage source inserter

Answer: a

Explanation: VSI stands for Voltage source inverter. The output voltage of VSI is fixed but the output current depends upon the load. Switches used in VSI are unipolar and bidirectional in nature.

6. Full form of CSI is ___________

a) Counter source inverter

b) Counter severe inverter

c) Current source inverter

d) Current severe inverter

Answer: c

Explanation: CSI stands for Current source inverter. The output voltage of CSI is variable but the output current is fixed.

7. Number of switches turned on in 3-phase 180° VSI at a single time are __________

a) 3

b) 2

c) 1

d) 4

Answer: b

Explanation: In 3-phase 180° VSI each phase conducts for 180° and only two switches from different phases conduct at a single time.

8. Number of switches turned on in 3-phase 120° VSI at a single time are __________

a) 2

b) 4

c) 5

d) 3

Answer: d

Explanation: In 3-phase 120° VSI each phase conducts for 120° and only three switches from different phases conduct at a single time.

9. Calculate the phase voltage for 3-∅ 180° VSI if the value of DC supply is 20 V.

a) 9.40 V

b) 9.42 V

c) 9.18 V

d) 9.78 V

Answer: b

Explanation: The phase voltage for 3-∅ 180° VSI is ×V DC =.471×20=9.42 V. In 3-phase 180° VSI each phase conducts for 180° and only two switches from different phases conduct at a single time.

10. Calculate the line voltage for 3-∅ 180° VSI if the value of DC supply is 22 V.

a) 18.52 V

b) 18.12 V

c) 17.96 V

d) 15.48 V

Answer: c

Explanation: The line voltage for 3-∅ 180° VSI is ×V DC =.81×22=17.96 V. In 3-phase 180° VSI each phase conducts for 180° and only two switches from different phases conduct at a single time.

11. Calculate the line voltage for 3-∅ 120° VSI if the value of DC supply is 24 V.

a) 16.52 V

b) 16.97 V

c) 12.96 V

d) 12.48 V

Answer: b

Explanation: The line voltage for 3-∅ 120° VSI is (V DC ÷√2)=.707×24=16.97 V. In 3-phase 120° VSI each phase conducts for 120° and only three switches from different phases conduct at a single time.

12. Calculate the phase voltage for 3-∅ 120° VSI if the value of DC supply is 26 V.

a) 10.52 V

b) 10.97 V

c) 10.96 V

d) 10.61 V

Answer: d

Explanation: The line voltage for 3-∅ 120° VSI is (V DC ÷√6)=10.61 V. In 3-phase 120° VSI each phase conducts for 120° and only three switches from different phases conduct at a single time.

13. Calculate power absorbed by 5 Ω load in 3-∅ 120° VSI if the value of DC supply is 16 V.

a) 17.06 W

b) 15.15 W

c) 18.25 W

d) 19.26 W

Answer: a

Explanation: The power absorbed by 5 Ω load in 3-∅ 120° VSI if the value of DC supply is 16 V is V ph 2 ÷R= 2 ÷5=17.06 W.

14. Calculate power absorbed by 10 Ω load in 3-∅ 180° VSI if the value of DC supply is 36 V.

a) 20.6 W

b) 24.5 W

c) 26.4 W

d) 28.8 W

Answer: d

Explanation: The power absorbed by 10 Ω load in 3-∅ 180° VSI if the value of DC supply is 36 V is V ph 2 ÷R= 2 ÷10=28.8 W.

15. T.H.D value for 3-∅ 180° VSI is ________________

a) 36 %

b) 31 %

c) 39 %

d) 46 %

Answer: b

Explanation: T.H.D value for 3-∅ 180° VSI is 31 %. Triplen harmonics are absent in 3-∅ 180° VSI and 3-∅ 120° VSI. Only 6k±1 harmonics are present.

This set of Electric Drives Questions and Answers for Entrance exams focuses on “Solid-State Switching Circuits – Three-Phase Energy Recovery Systems”.

1. Rectifiers are used to convert DC power into variable AC power.

a) True

b) False

Answer: b

Explanation: Inverters are used to convert DC power into variable AC power. The variable AC power means variable frequency and variable voltage level.

2. Short circuit problem is severe in case of current source inverter.

a) True

b) False

Answer: b

Explanation: Short-circuit problem is severe in case of voltage source inverter. The output voltage of VSI is fixed but the output current depends upon the load. When Z=0, I=V÷Z=∞.

3. Full form of KERS is ___________

a) The kinetic energy recovery system

b) The kinetic ever recovery system

c) Kin ever recovery system

d) The kinetic energy recovery store

Answer: a

Explanation: Full form of KERS is the Kinetic energy recovery system. The systems recover the kinetic energy from the heat generated by the car’s braking.

4. Full form of IGBT is ___________

a) Insulated gate bipolar transistor

b) Insulated gate bipolar transducer

c) Insulator gate bidirectional transducer

d) Insulated gate bidirectional transistor

Answer: a

Explanation: Full form of IGBT is Insulated gate bipolar transistor. It is a 3-terminal device, bipolar in nature. It can be used as a switch.

5. Flywheel is analogous to ___________

a) Chemical battery

b) Mechanical battery

c) Nuclear battery

d) Electrostatic battery

Answer: b

Explanation: A flywheel is a heavy wheel that stores a high amount of rotational kinetic energy. It requires a lot of force to spin the wheel.

6. Energy is stored in Flywheel in the form of __________

a) Kinetic energy

b) Electrical energy

c) Solar energy

d) Wind energy

Answer: a

Explanation: Energy is stored in Flywheel in the form of kinetic energy. It stores the rotational kinetic energy.

7. Calculate the coefficient of fluctuation of speed using the data: V max =25 m/sec, V mean =12 m/sec.

a) 2.08

b) 2.16

c) 2.34

d) 2.74

Answer: a

Explanation: The coefficient of fluctuation of speed is the ratio of the maximum fluctuation speed and mean speed. C.F = V max ÷ V mean = 25÷12 = 2.08.

8. Calculate the maximum fluctuation of speed using the data: V max =23 m/sec, V min =13 m/sec.

a) 12

b) 10

c) 14

d) 15

Answer: b

Explanation: The maximum fluctuation of speed is the difference between the maximum speed and minimum speed. Maximum fluctuation of speed= V max – V min = 23 – 13 = 10.

9. Calculate Kinetic energy of the object of mass = 7 Kg moving in a straight line with a velocity of 12 m/s.

a) 504 J

b) 502 J

c) 503 J

d) 509 J

Answer: a

Explanation: The Kinetic energy of the object of mass = 7 Kg moving in a straight line with a velocity of 12 m/s is .5×m×v 2 = .5×7×12×12 = 504 J.

10. Calculate the rotational kinetic energy of the object using the data: I=5 Kg-m 2 , ω=25 rad/sec.

a) 1.46 KJ

b) 1.56 KJ

c) 2.58 KJ

d) 3.15 KJ

Answer: b

Explanation: The Kinetic energy of the object having a moment of inertia=5 kg-m 2 rotating with an angular velocity equals to 25 rad/sec is .5×I×ω 2 = .5×5×25×25 = 1.56 KJ.

11. Calculate the line voltage for 3-∅ 120° VSI if the value of DC supply is 96 V.

a) 68.25 V

b) 67.87 V

c) 66.25 V

d) 62.14 V

Answer: b

Explanation: The line voltage for 3-∅ 120° VSI is (V DC ÷√2)=.707×96=67.87 V. In 3-phase 120° VSI each phase conducts for 120° and only three switches from different phases conduct at a single time.

12. Calculate the phase voltage for 3-∅ 120° VSI if the value of DC supply is 78 V.

a) 30.45 V

b) 15.18 V

c) 31.84 V

d) 48.15 V

Answer: c

Explanation: The line voltage for 3-∅ 120° VSI is (V DC ÷√6)=31.84 V. In 3-phase 120° VSI each phase conducts for 120° and only three switches from different phases conduct at a single time.

13. Calculate the r.m.s value of thyristor current in 3-Φ Full wave converter for the load current=2 A and α=59°.

a) 1.15 A

b) 2.55 A

c) 2.07 A

d) 2.89 A

Answer: a

Explanation: The r.m.s value of thyristor current in 3-Φ Full wave converter is I o √1÷3. It is the r.m.s value of the thyristor current. I = I o √1÷3=1.15 A.

14. Calculate the average value of thyristor current in 3-Φ Full wave converter for the load current=729 A and α=3°.

a) 254 A

b) 285 A

c) 247 A

d) 243 A

Answer: d

Explanation: The average value of thyristor current in 3-Φ Full wave converter is I o ÷3. It is the average value of the thyristor current. I = I o ÷3=243 A.

15. 1600 V rated 8 SCRs are connected in series. The operation voltage of the string is 4400. Calculate the string efficiency.

a) 34.37 %

b) 30.15 %

c) 38.48 %

d) 35.84 %

Answer: a

Explanation: The string efficiency can be calculated using the formula

operation voltage ÷ = 4400÷ = 34.37 %.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solid-State Switching Circuits – Current Source Inverter”.

1. The use of inductor in CSI is to make source current constant.

a) True

b) False

Answer: a

Explanation: The inductor in CSI is used to make the source current constant. The input current of CSI should be constant.

2. The output voltage of CSI depends upon the load.

a) True

b) False

Answer: a

Explanation: The output voltage of CSI depends upon the load. The output current of CSI is fixed as the input source current is fixed.

3. Which one of the element is used to make R load commutating?

a) Inductor

b) Capacitor

c) Resistor

d) Coil

Answer: b

Explanation: Capacitor is used to make the R load commutating. It is used in parallel with the resistive load.

4. Calculate the minimum time period for ASCI using the data: R=12 kΩ, C=45 μF.

a) 2.58 sec

b) 2.16 sec

c) 2.23 sec

d) 2.48 sec

Answer: b

Explanation: The minimum time period for ASCI is a 4×R×C = 4×12×45×.001 = 2.16 sec. The total time is the sum of charging and discharging time.

5. Full form of ASCI is ___________

a) Auto sequential commutated inverter

b) Auto starter commutated inverter

c) Auto sequential current inverter

d) Auto sequential current inserter

Answer: a

Explanation: ASCI stands for Auto sequential commutated inverter. No external forced commutation is required for ASCI. The capacitor itself works as a load commutated inverter.

6. Calculate the maximum frequency for ASCI using the data: R=15 kΩ, C=5 μF.

a) 3.51 Hz

b) 3.69 Hz

c) 3.33 Hz

d) 3.98 Hz

Answer: c

Explanation: The maximum frequency for ASCI is a 1÷ = 1÷ = 3.33 Hz. The total frequency is the sum of charging and discharging frequency.

7. The output current waveform for R load in CSI is ___________

a) Square

b) Triangular

c) Constant

d) Vertical

Answer: a

Explanation: The output current of CSI for R load is square in nature. The amplitude of the output current varies from I to -I.

8. T.H.D value for CSI is ________________

a) 48.43 %

b) 41.15 %

c) 39.15 %

d) 46.48 %

Answer: a

Explanation: T.H.D value for VSI is 48.43 %. Triplen harmonics are present in CSI due to which harmonic content increases.

9. What is the distortion factor value in CSI?

a) .85

b) .90

c) .95

d) .88

Answer: b

Explanation: The distortion value in CSI is .90. The r.m.s value of the fundamental current is 2√2I÷π and r.m.s value of the output current is I. The distortion factor is ÷I = .90.

10. Number of switches turned on in CSI at a single time are __________

a) 2

b) 4

c) 5

d) 3

Answer: a

Explanation: In CSI two switches conduct at a single time. The common switch used in CSI is thyristor. Thyristors are unidirectional, bipolar in nature.

11. Calculate the r.m.s value of the output current having an amplitude of 2 A for CSI.

a) 3 A

b) 4 A

c) 2 A

d) 1 A

Answer: b

Explanation: The r.m.s value of the output current having an amplitude of 2 A for CSI is 2A. The waveform for output current is a square wave. The r.m.s value of the square wave is its amplitude.

12. Calculate the r.m.s value of the fundamental component of the output current having an amplitude of 3 A for CSI.

a) 2.80 A

b) 2.70 A

c) 2.54 A

d) 3.48 A

Answer: b

Explanation: The r.m.s value of the fundamental component of the output current is 2√2I÷π=8.48÷3.14=2.70 A. The r.m.s value of the output current is I.

13. Calculate the string efficiency if the de-rating factor is .76.

a) 24 %

b) 25 %

c) 26 %

d) 27 %

Answer: a

Explanation: The string efficiency is calculated for series and parallel connection of SCRs. The value of string efficiency is 1-=1-.76=24 %.

14. 20 V rated 8 SCRs are connected in series. The operation voltage of the string is 50. Calculate the De-rating factor.

a) .67

b) .68

c) .69

d) .70

Answer: b

Explanation: The string efficiency can be calculated using the formula operation voltage÷=50÷=.31. The De-rating factor value is 1-.31=.68.

15. Forced commutated CSI require _________

a) Diode

b) Resistor

c) Inductor

d) Capacitor

Answer: d

Explanation: Forced commutated CSI requires a capacitor. The capacitor acts as a commutating element. It is connected in parallel with the load.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solar Panels”.

1. A solar cell converts light energy into __________

a) Electrical energy

b) Thermal energy

c) Sound energy

d) Heat energy

Answer: a

Explanation: A solar cell converts light energy into electrical energy. The light energy excites the electron of the solar cell which further flows in the circuit and constitutes the electric current.

2. There are three types of the solar cells.

a) True

b) False

Answer: a

Explanation: There are three types of solar cells. Single crystal, polycrystal, and amorphous silicon cells are the major types of solar cells.

3. Series and parallel combination of the solar cell is known as _________

a) Solar array

b) Solar light

c) Solar sight

d) Solar eye

Answer: a

Explanation: Series and parallel combination of the solar cell is known as Solar array. Shunt diodes are used to avoid the circulating current.

4. Full form of FF in the solar field is ____________

a) Form factor

b) Fill factor

c) Face factor

d) Fire factor

Answer: b

Explanation: FF stands for Fill factor. It is the ratio of the maximum obtainable power to the product of the open-circuit voltage and short circuit current.

5. Calculate Fill factor using the data: P max =15 W, V oc =18 V, I sc =4 A.

a) .65

b) .59

c) .20

d) .98

Answer: c

Explanation: Fill factor is the ratio of the maximum obtainable power to the product of the open-circuit voltage and short circuit current. F.F=P max ÷(V oc ×I sc )=15/72=.20.

6. Permanent magnet DC motor is more expensive than an Induction motor.

a) True

b) False

Answer: a

Explanation: Permanent magnet DC motor is more expensive than an Induction motor because of the magnetic used in it. Magnets are made up of ferromagnetic material which results in high cost.

7. Calculate the line voltage in star connection when phase voltage=311 V.

a) 548.29 V

b) 538.66 V

c) 587.28 V

d) 185.58 V

Answer: b

Explanation: The line voltage in case of star connection is 1.73 times of phase voltage. It leads the phase voltage by an angle of 30°. V L-L = 1.73×311 = 538.66 V.

8. The slope of the V-I curve of a solar cell is 66.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 3.5 Ω

b) 2.2 Ω

c) 2.5 Ω

d) 2.9 Ω

Answer: b

Explanation: The slope of the V-I curve of a solar cell is resistance. The slope given is 66.1° so R=tan=2.2 Ω. The slope of the I-V curve is reciprocal of resistance.

9. SCIM has a _________

a) Shorter life

b) Medium life

c) Longer life

d) Infinite life

Answer: c

Explanation: SCIM stands for squirrel cage induction motor. It has high efficiency, maintenance-free operation, and long life because of rigidness nature.

10. The power factor of a SCIM generally is ___________

a) .1-.2

b) .6-.8

c) .3-.4

d) .5-.7

Answer: b

Explanation: At light loads, the current drawn is largely a magnetizing current due to the air gap and hence the power factor is low. The power factor of a SCIM generally is .6-.8.

11. Calculate the active power in a 11710 H inductor.

a) 8245 W

b) 1781 W

c) 0 W

d) 1964 W

Answer: c

12. Calculate the frequency of the waveform i = sin+sin.

a) .5 Hz

b) .8 Hz

c) .1 Hz

d) .9 Hz

Answer: a

Explanation: The fundamental time period of the sine wave is 2π. The time period of i is L.C.M {2,2}=2 sec. The frequency of the waveform is 1/2 = .5 Hz. The time period is independent of phase shifting and time-shifting.

13. Full form of CEL is ____________

a) Central Electronics Limited

b) Central Electrical Limited

c) Circle Electronics Limited

d) Circle Electrical Limited

Answer: a

Explanation: The full form of CEL is Central Electronics Limited. It is a Government of India enterprise founded in 1974. It is a publicly listed company.

14. Full form of BEL is ____________

a) Busy Electronics Limited

b) Burden Electrical Limited

c) Bharat Electronics Limited

d) Brahma Electrical Limited

Answer: c

Explanation: The full form of BEL is Bharat Electronics Limited. It is a Government of India enterprise founded in 1954. It is a publicly listed company.

15. Material used for making solar cell is _________

a) Silicon

b) Carbon

c) Sodium

d) Magnesium

Answer: a

Explanation: Material used for making solar cells is Silicon. It is a naturally obtained semi-conductor. It has a lower cut-off voltage and minimum energy bandgap.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solar Powered Pump Drives”.

1. The output power of the centrifugal pump is proportional to the _________ of the speed.

a) Square

b) Cube

c) Square-root

d) Cube-root

Answer: b

Explanation: The output power of the centrifugal pump is proportional to the cube of the speed. They require a small torque to start.

2. There are four types of Water pumps.

a) True

b) False

Answer: b

Explanation: There are two types of Water pumps. Centrifugal and reciprocating are the major types of water pumps.

3. The output of the Solar panel is DC.

a) True

b) False

Answer: a

Explanation: The output of the solar panel is DC. Inverters are used at the output terminals of the solar panel in order to convert DC into AC.

4. Which of the relationship is correct in Permanent magnet DC motor?

a) T ∝ (I a ) 2

b) T ∝ (I a ) 0

c) T ∝ I a

d) T 2 ∝ I a

Answer: c

Explanation: The value of flux remains constant in Permanent magnet DC motor. The torque in the PMMC motor is directly proportional to the armature current.

5. Calculate Fill factor using the data: P max =5 W, V oc =5 V, I sc =1 A.

a) 5

b) 3

c) 2

d) 1

Answer: d

Explanation: Fill factor is the ratio of the maximum obtainable power to the product of the open-circuit voltage and short circuit current. F.F=P max ÷(V oc ×I sc )=5/5=1.

6. Full form of ICE is ____________

a) Internal Combustion Engine

b) International Combustion Engine

c) Iterative Combustion Engine

d) Iterative Commercial Engine

Answer: a

Explanation: Full form of ICE is International Combustion Engine. It consists of pistons inside the closed chamber in order to drive the machine.

7. Which of the relationship is correct in Permanent magnet DC motor?

a) E ∝ 2

b) E ∝ 0

c) E ∝ N

d) E2 ∝ N

Answer: c

Explanation: The value of flux remains constant in Permanent magnet DC motor. The back e.m.f in the PMMC motor is directly proportional to the speed.

8. Full form of SSS is ____________

a) Saur Sujala Yojana

b) Saur Solar Yojana

c) Solar Sun Yojana

d) Sun Sujala Yojana

Answer: a

Explanation: Full form of SSS is Saur Sujala Yojana. It is an initiative by the Government of India to encourage people to use solar energy as an energy source.

9. Who made the first solar cell?

a) Gerald Pearson

b) Tim Bell

c) Tim Cook

d) Nicola Tesla

Answer: a

Explanation: The first solar cell was made by Gerald Pearson in 1954 in Bell laboratories. It was a lithium-silicon photovoltaic cell.

10. When was the first solar cell invented?

a) 1950

b) 1954

c) 1965

d) 1952

Answer: b

Explanation: The first solar cell was made by Gerald Pearson in 1954 in Bell laboratories. It was a lithium-silicon photovoltaic cell.

11. Who invented the first Solar car?

a) Ian Bell

b) Trent Boult

c) Jon Woakes

d) William G. Cobb

Answer: d

Explanation: The first solar car was invented by William G. Cobb in 1955. It was a 15-inch vehicle. It was made by General Motors.

12. The power factor of a WRIM generally is ___________

a) .1-.2

b) .6-.8

c) .3-.4

d) .5-.7

Answer: d

Explanation: At light loads, the current drawn is largely a magnetizing current due to the air gap and hence the power factor is low. The power factor of a WRIM generally is .5-.7.

13. Full form of BHEL is ____________

a) Bharat Heavy Electronics Limited

b) Bharat Heavy Electrical Limited

c) Bharat Hero Electronics Limited

d) Busy Hero Electrical Limited

Answer: b

Explanation: Full form of BHEL is Bharat Heavy Electrical Limited. It is a Government of India enterprise founded in 1964. It is a publicly listed company.

14. Calculate peak-peak voltage if V max =0 V and V min =9 V.

a) -5 V

b) -6 V

c) -9 V

d) -10 V

Answer: c

Explanation: Peak-Peak voltage is equal to the difference between the maximum and minimum voltage. It is mathematically represented as V p-p =V max -V min =0-9=-9 V.

15. Calculate the De-rating factor if the string efficiency is 5 %.

a) .95

b) .55

c) .65

d) .85

Answer: a

Explanation: De-rating factor is used to measure the reliability of a string. The value of the De-rating factor is 1-=1-.05=.95.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Solar Powered Vehicles”.

1. When was the first electric car invented?

a) 1830

b) 1985

c) 1832

d) 1945

Answer: c

Explanation: The first electric car was developed by Robert Anderson in 1832. It was the first crude electric vehicle. Electric cars are the future of the automobile industry.

2. The moving coil instrument measures the ___________ of the signal.

a) Average value

b) R.M.S value

c) Zero value

d) Half value

Answer: b

Explanation: The moving coil instrument measures the R.M.S value of the signal. The moving coil instrument can be used to measure the DC value.

3. The PMMC instrument measures the average value of the signal.

a) True

b) False

Answer: a

Explanation: The Permanent magnetic moving coil instrument measures the average value of the signal. It is not suitable for AC measurement as pointer fluctuates in case of AC measurement.

4. Which of the relationship is correct in Ferromagnetic motor?

a) T ∝ (I a ) 2

b) T ∝ (I a ) 0

c) T ∝ I a

d) T 2 ∝ I a

Answer: c

Explanation: The value of flux remains constant in Ferromagnetic motor. The torque in the Ferromagnetic motor is directly proportional to the armature current.

5. Calculate Fill factor using the data: P max =5 W, V oc =0 V, I sc =1 A.

a) ∞

b) 3

c) 2

d) 1

Answer: a

Explanation: Fill factor is the ratio of the maximum obtainable power to the product of the open-circuit voltage and short circuit current. F.F = P max ÷ (V oc ×I sc ) = 5/0 = ∞.

6. Electric Vehicles are generally powered by __________

a) Aluminum batteries

b) Lead-acid batteries

c) Sodium batteries

d) Magnesium batteries

Answer: b

Explanation: Electric vehicles are generally powered by Lead-acid batteries. They consist of lead electrodes with H 2 SO 4 as an electrolyte.

7. Calculate the line voltage in star connection when phase voltage=315 V.

a) 545.5 V

b) 548.6 V

c) 547.8 V

d) 585.8 V

Answer: a

Explanation: The line voltage in the case of star connection is 1.73 times of phase voltage. It leads the phase voltage by an angle of 30°. V L-L =1.73×315=545.5 V.

8. Who coined the term battery?

a) George Franklin

b) Benjamin Fernandes

c) Benjamin Franklin

d) George Bush

Answer: c

Explanation: Benjamin Franklin was the first who coined the term battery. Batteries were the main source of electrical energy before the invention of generators.

9. Who invented the battery?

a) Alessandro Volta

b) Alexander Bell

c) Alessandro Bell

d) Tim Southee

Answer: a

Explanation: A battery converts the chemical energy into electrical energy. Alessandro Volta invented the battery in 1800. He was an Italian scientist.

10. Combination of cells is known as the battery.

a) True

b) False

Answer: a

Explanation: A battery is a combination of cells connected in parallel and series. It converts chemical energy into electrical energy.

11. Calculate the active power in a 315 F capacitor.

a) 45 W

b) 81 W

c) 0 W

d) 64 W

Answer: c

12. Calculate the frequency of the waveform v = sin + sin + sin.

a) .1 Hz

b) .8 Hz

c) .5 Hz

d) .2 Hz

Answer: c

Explanation: The fundamental time period of the sine wave is 2π. The time period of i is L.C.M {2,2,2}=2 sec. The frequency of the waveform is 1/2=.5 Hz. The time period is independent of phase shifting and time-shifting.

13. The diode conducts only when current flows from anode to cathode.

a) True

b) False

Answer: a

Explanation: Diode is a unidirectional device. It starts conduction when current flows from cathode to anode of the diode. It is a bipolar device.

14. Full form of EV is ____________

a) Energy voltage

b) Electric vehicles

c) Electric voltage

d) Energy vehicles

Answer: b

Explanation: Full form of EV is Electric Vehicles. It uses an electric motor for the machine propulsion. It does not cause pollution as there is no tail-to-pipe emission.

15. Full form of ICV is ____________

a) Internal combustion vehicles

b) Internet combustion vehicles

c) Internally combined vehicles

d) Internet combined vehicles

Answer: b

Explanation: Full form of ICV is Internal combustion vehicle. It requires fuel for propulsion. It consists of a piston attached in a cylinder chamber.

This set of Electric Drives test focuses on “Electric Traction Services”.

1. Electric trains are of two types.

a) True

b) False

Answer: a

Explanation: Mainline trains and suburban trains are the two types of trains. They run on fixed rails. They are powered by electricity.

2. The shape of pantograph collector is _________________

a) Square

b) Pentagon

c) Circle

d) Hexagon

Answer: b

Explanation: The shape of the pantograph collector is a pentagon. It has a conducting strip which is pressed against contact wire with the help of springs.

3. There are typically three types of pantograph.

a) True

b) False

Answer: a

Explanation: Open frame, faiveley, crossed arm are typically three types of the pantograph. The function of the pantograph is to maintain pressure and prevent vertical vibrations.

4. Full form of EMU is ____________

a) Electronics Multiple Unit

b) Electrical Multiple Unit

c) Electrical Multiple Usage

d) Electrical Multiple User

Answer: b

Explanation: Full form of EMU is Electrical Multiple Unit. The trains having motor coaches and trailer coaches are known as EMU trains. They provide flexibility.

5. Who is known as the Father of Indian Railways?

a) Lord Dalhousie

b) Lord Ripon

c) Lord Boult

d) Lord Hero

Answer: a

Explanation: Lord Dalhousie is known as the father of the Indian Railways. The first railway line connecting Bombay with Thane was laid in 1853.

6. Who built the first train?

a) Richardson

b) Rampa

c) Richard Trevithick

d) John Carry

Answer: c

Explanation: Richard Trevithick built the first train in 1804. It was a steam locomotive train that used to carry 10 tonnes of iron from one place to another.

7. Who built the first bullet train?

a) Hideo Rampa

b) Hideo Shima

c) Hideo Carry

d) Hideo Stokes

Answer: b

Explanation: Hideo Shima built the first bullet train. He was a Japenese engineer. He was born in 1901 and died in 1998. He was the man behind the concept of a bullet train.

8. The supply conductor in electric trains is also known as ____________

a) Contact Wire

b) Contact Distance

c) Contact Direction

d) Supply Wire

Answer: a

Explanation: The supply conductor in electric trains is also known as Contact wire. Driving motor and power modulators are housed in the locomotive.

9. Calculate the value measured by PMMC when a sinusoidal voltage signal is applied to it.

a) 15 V

b) 0 V

c) 16 V

d) 18 V

Answer: b

Explanation: PMMC instrument measures the average of the input signal. The sinusoidal voltage signal is symmetric about the origin. The average value measured by PMMC is zero.

10. Calculate the value measured by Moving Iron when a sinusoidal voltage signal V=20sin is applied to it.

a) 14.14 V

b) 20.15 V

c) 16.18 V

d) 22.18 V

Answer: a

Explanation: Moving iron instrument measures the r.m.s value of the input signal. The r.m.s value of the sinusoidal signal is 20÷√2=14.14 V.

11. Full form of PMMC is ____________

a) Permanent Magnet Moving Coil

b) Premature Magnet Moving Coil

c) Permanent Magnet Motion Coil

d) Peterson Magnet Motion Coil

Answer: a

Explanation: Full form of PMMC is Permanent Magnet Moving Coil. It measures the average value of the applied input signal.

12. Fuel used in the train is ___________

a) Diesel

b) Kerosene

c) Petrol

d) Ice

Answer: a

Explanation: Fuel used in the train is Diesel. Diesel fuel in trains provides more efficiency and more economy. It is more economical than air travel.

13. How much fuel does a train use per Km?

a) 7.85 L/Km

b) 7.97 L/Km

c) 6.59 L/Km

d) 7.15 L/Km

Answer: b

Explanation: The average fuel consumed by a train is 7.97 L/Km. Consumption is the same for passenger and cargo trains.

14. Which motor is used in the Electric train?

a) DC shunt motor

b) DC series motor

c) Synchronous motor

d) Stepper motor

Answer: b

Explanation: DC series motor is used in Electric train. They provide a high starting torque. DC series motor is a universal motor that can run on DC as well as on AC.

15. Calculate the minimum time period for ASCI using the data: R=1 kΩ, C=1 μF.

a) 4 msec

b) 8 msec

c) 5 msec

d) 3 msec

Answer: a

Explanation: The minimum time period for ASCI is a 4×R×C = 4×1×1×.001 = 4 msec. The total time is the sum of charging and discharging time.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Nature of Traction Load”.

1. There are four types of friction.

a) True

b) False

Answer: b

Explanation: There are two types of friction. Internal and external friction are types of friction. Coefficient of friction is independent of the force of friction.

2. Calculate the value of the coefficient of friction using the data: F=24 N, Normal force=12 N.

a) 2

b) 4

c) 8

d) 9

Answer: a

Explanation: Coefficient of friction is the ratio of the friction force and normal force. It is dependent on the smoothness of the surface area. μ=24÷12=2.

3. If the friction force value is doubled, calculate the new value of the coefficient of friction.

a) 4 times of the original value

b) 2 times of the original value

c) 7 times of the original value

d) Remains the same

Answer: d

Explanation: The value of the coefficient of friction will not change if the friction force is doubled. It only depends upon the nature of the surface.

4. The frictional force always acts in the _________ direction of motion of the body.

a) Opposite

b) Same

c) Perpendicular

d) Vertical

Answer: a

Explanation: The friction of force always acts in the opposite direction of the motion of the body. It is a non-conservative force. It helps in the motion of the body.

5. Calculate the coefficient of friction if the angle of inclination is 60°.

a) 2.73

b) 1.73

c) 2.48

d) 1.41

Answer: b

Explanation: The coefficient of friction is the tangent of the angle of the inclined plane. The value of μ is the tan=1.73. The coefficient of friction depends on the nature of the surface.

6. Which one of the following is the correct relationship?

a) F=μR

b) F=4μR

c) F=3μR

d) F=2μR

Answer: a

Explanation: The value of the friction force is the product of the coefficient of friction and normal reaction. The correct value is F=μR.

7. Calculate the velocity of the bottom point of the wheel for perfect rolling using the data: r=20 cm, ω=100 rad/sec.

a) 20 m/sec

b) 40 m/sec

c) 60 m/sec

d) 80 m/sec

Answer: a

Explanation: The velocity of the bottom point of the wheel for perfect rolling is the product of the radius and angular velocity. It is expressed in terms of m/sec. V = r×ω = .2×100=20 m/sec.

8. The coefficient of adhesion is maximum when rails are dusty.

a) True

b) False

Answer: b

Explanation: The coefficient of adhesion is maximum when rails are dry. The coefficient of adhesion depends on the nature of the material.

9. Calculate the value of the coefficient of adhesion for dusty rails if the value of the coefficient of adhesion for dry rails is .55.

a) 0.65

b) 0.89

c) 0.97

d) 0.48

Answer: d

Explanation: The value of the coefficient of adhesion for dusty rails is .48. The value of the coefficient of adhesion for dry rails is maximum.

10. Calculate the value measured by Moving Iron when a sinusoidal voltage signal V = √2sin is applied to it.

a) 2 V

b) 1 V

c) 3 V

d) 4 V

Answer: b

Explanation: Moving iron instrument measures the r.m.s value of the input signal. The r.m.s value of the sinusoidal signal is √2÷√2 = 1 V.

11. The value of area under a velocity-time graph is _________

a) Acceleration

b) Displacement

c) Force

d) Momentum

Answer: b

Explanation: The value of the area under a velocity-time graph is displacement. The velocity is the ratio of displacement and time.

12. The value of area under acceleration-time graph is _________

a) Velocity

b) Displacement

c) Force

d) Impulse

Answer: a

Explanation: The value of the area under an acceleration-time graph is velocity. The acceleration is the ratio of velocity and time.

13. The value of area under force-time graph is _________

a) Reaction

b) Change in momentum

c) Force

d) Torque

Answer: b

Explanation: The value of the area under a force-time graph is a change in momentum. The force is the ratio of change in momentum with respect to time.

14. Full form of HVDC is _________

a) High Voltage Direct Current

b) High Voltage Distribution Combination

c) High-Value Direct Current

d) High-Value Distribution Combination

Answer: a

Explanation: Full form of HVDC is High voltage direct current. HVDC is more preferred than HVAC because of minimum losses.

15. Full form of HVAC is _________

a) High Voltage Alternative Current

b) High Voltage Alternating Current

c) High-Value Alternating Combination

d) High-Value Alternative Counter

Answer: b

Explanation: Full form of HVAC is High voltage alternating current. It is mostly preferred for smaller distances. HVAC has higher losses than HVDC.

This set of Electric Drives online quiz focuses on “Traction Drives – Power Factor and Harmonics”.

1. Full form of SVC is ___________

a) Static Var Compensator

b) Static Variable Counter

c) State Value Counter

d) Static Value Compensator

Answer: a

Explanation: The full form of SVC is Static Var Compensator. It is a FACTS controller used for reactive power compensation. It maintains the voltage profile.

2. Full form of STATCOM is ___________

a) Static Synchronous Counter

b) Static Synchronous Compensator

c) State Synchronized Compensator

d) State System Counter

Answer: b

Explanation: The full form of STATCOM is a Static synchronous compensator. It is a voltage source converter. It maintains the voltage profile of the system. It is a FACTS controller.

3. The circuit which operates signals is called as ___________

a) Track circuit

b) Fellow circuit

c) Ground circuit

d) Unit circuit

Answer: a

Explanation: The circuit which operates signals is called track circuit. Its supply consists of low AC or DC voltage supply. Running rail is used as the return line for the track circuit.

4. SVC is used to maintain the power factor above __________

a) .2

b) .5

c) .8

d) .6

Answer: c

Explanation: SVC stands for static var compensator. It is the FACTS controller used for reducing the harmonics from the system. It is used for reactive power compensation.

5. Which is the most severe type of fault?

a) LL

b) LLG

c) LLL

d) LLLG

Answer: d

Explanation: LLLG fault is the most severe type of fault. The order of severity is maximum for LLLG fault. LG fault is more severe at the generator terminals.

6. Skin effect increases at higher frequencies.

a) True

b) False

Answer: a

Explanation: Skin effect increases with the increase in frequency. The value of skin depth decreases with the increase in frequency.

7. If induction motor air gap power is 11 KW and mechanically developed power is 2 KW, then rotor ohmic loss will be _________ KW.

a) 8

b) 10

c) 9

d) 11

Answer: c

Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses = Air gap power – Mechanical developed power = 11-2 = 9 KW.

8. If induction motor rotor power is 26 KW and gross developed power is 21 KW, then rotor ohmic loss will be _________ KW.

a) 5

b) 6

c) 7

d) 8

Answer: a

9. Calculate the active power in 14 μH inductors if the THD value is 58.1%.

a) 5 W

b) 9 W

c) 4 W

d) 0 W

Answer: d

10. Calculate the power factor using the data: P=15 W, S=25 VA.

a) 0.5

b) 0.9

c) 0.4

d) 0.6

Answer: d

Explanation: The power factor is the ratio of active power and apparent power. The value of the power factor is P÷S=15÷25=.6.

11. Calculate the active power in 56 μF capacitor if the VRF value is 5.1%.

a) 6 W

b) 0 W

c) 9 W

d) 2 W

Answer: b

12. Calculate the apparent power using the data: P=3 W, Q=4 VAR.

a) 5 VA

b) 6 VA

c) 9 VA

d) 2 VA

Answer: a

Explanation: The apparent power is the product of the voltage and current conjugate phasor. The value of apparent power is .5 = 5 VA.

13. Calculate the value of impedance using the data: R=2 Ω, X L =4 Ω.

a) 5.5 Ω

b) 5.4 Ω

c) 2.5 Ω

d) 4.4 Ω

Answer: d

Explanation: The value of impedance is { 2 +(X L ) 2 ) .5 = .5 = 4.4 Ω. It is total impedance offered by an AC circuit. It is expressed in Ω.

14. Which type of grounding is used to suppress the capacitive effect?

a) Solid grounding

b) Reactance grounding

c) Resistance grounding

d) Peterson coil

Answer: d

Explanation: Peterson coil or resonance grounding is used to suppress the capacitive effect. The frequency in resonance grounding is ω 2 = 1÷.

This set of Electric Drives Multiple Choice Questions & Answers focuses on “Electrical Drive Systems”.

1. Calculate the minimum anode current required to turn off SCR if the latching current value is 2 A.

a) 2 A

b) 3 A

c) 4 A

d) 5 A

Answer: a

Explanation: The minimum anode current required to turn off the SCR is the latching current. The value of the latching current is 2 A.

2. Calculate the value of the coefficient of static friction using the data: F=4 N, Normal force=16 N.

a) .24

b) .25

c) .15

d) .14

Answer: b

Explanation: Coefficient of friction is the ratio of the friction force and normal force. It is dependent on the smoothness of the surface area. μ = 4÷16 = .25.

3. Full form of TVS is __________

a) Transient voltage suppression

b) Transient voltage surge

c) Total voltage surge

d) True voltage surge

Answer: a

Explanation: The full form of TVS is Transient voltage suppression. It is one of the types of light surge suppressors. It is used to suppress transient voltages.

4. The frictional force always acts in the perpendicular direction of motion of the body.

a) True

b) False

Answer: b

Explanation: The friction of force always acts in the opposite direction of the motion of the body. It is a non-conservative force. It helps in the motion of the body.

5. Calculate the coefficient of friction if the angle of inclination is 0°.

a) 2

b) 1

c) 4

d) 0

Answer: d

Explanation: The coefficient of friction is the tangent of the angle of the inclined plane. The value of μ is the tan = 0. The coefficient of friction depends on the nature of the surface.

6. When was the first train built?

a) 1804

b) 1810

c) 1814

d) 1820

Answer: a

Explanation: Richard Trevithick built the first train in 1804. It was a steam locomotive train that used to carry 10 tonnes of iron from one place to another.

7. Who invented the first circuit breaker?

a) Albert Einstein

b) Thomas Edison

c) Nicola Tesla

d) Trent Boult

Answer: b

Explanation: Thomas Edison was the inventor of the circuit breaker. It was designed in 1879. It was an early form of the circuit breaker.

8. The power conductor in electric trains is called a contact wire.

a) True

b) False

Answer: a

Explanation: The supply conductor in electric trains is also known as Contact wire. Driving motor and power modulators are housed in the locomotive.

9. Full form of LSS is ____________

a) Line Surge Suppressors

b) Local Surge System

c) Local Surge Suppressors

d) Line Surge System

Answer: a

Explanation: Full form of LSS is Line surge suppressors. They are used to protect the semiconductor converter against voltage spikes produced in line.

10. Calculate the value measured by PMMC when a sinusoidal voltage signal V=20sin is applied to it.

a) 1 V

b) 2 V

c) 6 V

d) 0 V

Answer: d

Explanation: Permanent magnet moving coil instrument measures the average value of the input signal. The average value of the sinusoidal signal is 0 V.

11. Full form of MCC is ____________

a) Motor control center

b) Motor counter center

c) Machine control center

d) Motor coil counter

Answer: a

Explanation: The full form of MCC is Motor control center. It controls the power to system drives. There can many MCCs exist in an area.

12. Full form of PCC is ____________

a) Producer control center

b) Poor counter center

c) Power control center

d) Power coil counter

Answer: c

Explanation: The full form of PCC is Power control center. They receive power from the main power distribution center.

13. Current Chopping is most severe in which one of the following?

a) Air Blast Circuit Breaker

b) Oil Circuit Breaker

c) Air Break Circuit Breaker

d) Vacuum Circuit Breaker

Answer: a

Explanation: Current chopping is most severe in the case of Air blast circuit breaker. Resistance switching is used to avoid the problem of current chopping.

14. Oil circuit breaker is mainly used in _____________

a) Synchronous machine

b) Induction machine

c) DC machine

d) Transformer

Answer: d

Explanation: Oil circuit breaker is mainly used in transformers. The oil acts a coolant for the transformer. When a fault occurs in the transformer oil decomposes in the gases.

15. Full form of PLC is ____________

a) Programmable Logic Controller

b) Programmable Language Controller

c) Peterson Language Counter

d) Programmable Logic Counter

Answer: a

Explanation: The full form of PLC is Programmable logic controller. It is a computer control system that monitors input states and makes decisions based upon a program to control output states.

This set of Electric Drives Question Bank focuses on “Electrical Drive Systems and Components – Sequence Operations and Protection”.

1. Relay is an electromechanical switch.

a) True

b) False

Answer: a

Explanation: Relay is an electromechanical switch. It uses electrical signals as input and mechanical operation as an output.

2. Mho relay is used in the medium transmission line.

a) True

b) False

Answer: b

Explanation: Mho relay is an inherent directional relay. It is used in the long transmission line. It is less affected by power surges.

3. Full form of LED is _____________

a) Light Emitting Diode

b) Light Emission Digital

c) Light Energy Diode

d) Light Energy Digital

Answer: a

Explanation: The full form of LED is a light-emitting diode. It converts light energy into electrical energy. It works in the reverse bias region.

4. Inter turn fault can be detected by using __________

a) Oil circuit breaker

b) Split phase relay

c) SF 6 circuit breaker

d) Vertical breaker

Answer: b

Explanation: Inter turn fault mainly occurs in the winding of the rotor side of the alternator. They can be detected by using the split-phase relaying technique.

5. Calculate the value of the fault impedance if the fault voltage is 2 V and fault current is 4 A.

a) .5 ohm

b) .2 ohm

c) .4 ohm

d) .6 ohm

Answer: a

Explanation: The value of the fault impedance is the ratio of fault voltage and fault current. It is expressed in ohms. Z=V/I=2/4=.5 ohm.

6. Reactance relay is used in _______

a) Medium line

b) Short transmission line

c) Lossless line

d) Long transmission line

Answer: b

Explanation: Reactance relay is used in a short transmission line. Reactance relay is independent of the resistance value. It protects the short transmission line from any fault.

7. Load interrupter operated at ________

a) No-load

b) Full load

c) Half of the full load

d) One-fourth of the full load

Answer: b

Explanation: Load interrupter is used to avoid overloading conditions. It separates the under-stressed area from the line. It operates at full load.

8. Calculate the maximum frequency for ASCI using the data: R=1 kΩ, C=1 μF.

a) 250 Hz

b) 280 Hz

c) 300 Hz

d) 320 Hz

Answer: a

Explanation: The minimum time period for ASCI is a 4×R×C=4×1×1×.001=4 msec. The total time is the sum of charging and discharging time. The maximum frequency is 1÷f=250 Hz.

9. Calculate the PIV for the Full-wave bridge rectifier if the peak value of the supply voltage is 50.

a) 72.5 V

b) 72.8 V

c) 70.7 V

d) 76.1 V

Answer: c

Explanation: The peak inverse voltage for the Full-wave bridge rectifier is V m = √2×230=70.7 V. The peak inverse is the maximum negative voltage across the thyristor.

10. In single-phase, RLE load, calculate the Peak inverse voltage using the data: (V s ) r.m.s = 9 V, f = 50 Hz, E = 40 V.

a) 52.2 V

b) 52.7 V

c) 59.7 V

d) 52.9 V

Answer: b

Explanation: In a single phase, RLE load the peak inverse voltage across the thyristor V T = V m +E = 52.7 V.