#### 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 torquecharacteristics because torque and speed exhibits inversely proportionalrelationships 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 torquevariation is independent of speed. The speed-torque characteristics of this type ofload 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 torquecomponent 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 torquesdue to gravitational forces can be obtained in the case of hoists, lifts or elevators andrailway 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 elasticbodies. They always oppose the motion, restricting the motion of the machine.

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

**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 anexample of linearly rising load torque characteristics as the torque increases linearlywith 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 equalto motor torque so that motor should run at a uniform speed. If load torque is greatthan motor torque, the motor will fail to start and if load torque is less than motortorque, the motor will run at a higher speed which can damage the shaft of themotor.

**9. Choose the correct one. (* stands for multiplication, J represents the moment of**

**inertia, w represents angular speed).**

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

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

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

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

**Answer: C**

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

**10. Regenerative braking mode can be achieved in which quadrant (V-I curve)?
**

a) Third

**b) Second**

c) Fourth

d) First

**Answer: b**

Explanation: Regenerative braking is only available in second quadrant as powerfrom motor is fed back to source. Back emf generated (Eb) is more than armatureterminal (Vt) so it works as a generator.

11. 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.

#### 12. 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 (Type-A chopper). Value of output voltage (Vo) is less than the input voltage (Vin) in case of step-down chopper.

#### 13. Calculate the value of angular acceleration of motor using the given data: J = 20 kg-
m2, load torque = 20 N-m, motor torque = 60 N-m.

a) 5 rad/s2

**b) 2 rad/s2**

c) 3 rad/s2

d) 4 rad/s2

**Answer: b**

Explanation:Using the dynamic equation of motor J*(angular acceleration) = Motor torque – Load torque: 20*(angular acceleration) = 60-20=40, angular acceleration=2 rad/s2

#### 14. 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.

#### 15. 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.

16. 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 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.

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

A. (Assume rotational losses are neglected)

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 becomesequal to the shaft power of motor.

18. 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 (I2 * r) also changes.

#### 19. What is the empirical formula for the tractive force required to overcome curve
resistance? (W-the weight of the body, R – radius of curvature)

a) 710×W÷R

**b) 700×W÷R**

c) 720×W÷R

d) 750×W÷R

**Answer: b**

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

#### 20. Force resisting the upward motion of a body on an inclined plane is given by (alpha
– the angle of inclination, W- the weight of the body).

**a) F = W×sin(alpha)**

b) F = W×cosec(alpha)

c) F = W×sec(alpha)

d) F = W×cos(alpha)

**Answer: a**

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

21. The unit of the torque is ______

**a) N-m**

b) N-m2

c) N-m/sec

d) N-Hz

**Answer: a**

Explanation:Torque is defined as the vector product of displacement and force. Theunit of the force is Newton(N) and of the displacement is a meter (m) so the unit oftorque in N-m.

#### 22. 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 = (length of rod) × (Force applied) = r×F×sin90. F is given as 10 N and r is 10 m then torque is 10×10 = 100 N- m. (the angle between F and r is 90 degrees)

#### 23. What is the dimensional formula for torque?

**a) [ML2T -2 ]**

b) [MLT-2
]

c) [M1L
2T -3 ]

d) [LT-2
]

**Answer: a**

Explanation: Torque is a vector product of force and displacement. Dimensionalformula for force is [MLT-2] and displacement is [L] so dimensional formula for torqueis [MLT-2] [L] = [ML2T -2 ].

#### 24. 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 Vo = D×Vin. The value of theduty cycle is less than one which makes the Vo < Vin. The buck converter is used tostep down voltage. Vin is a fixed voltage and Vo is a variable voltage.

#### 25. 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(w)÷d(t) = Motor torque – Load torque is the dynamic equation of motor. If starting torque (motor torque) is less than the load torque then d(w)÷d(t) <0, acceleration <0 so the motor will decelerate and fails to start.

#### 26. 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.

#### 27. 250V, 15A, 1100 rpm separately excited dc motor with armature resistance (Ra)

equal to 2 ohms. Calculate back emf developed in the motor when it operates on half

of the full load. (Assume rotational losses are neglected)

a) 210V

b) 240V

c) 230V

**d) 235V**

**Answer: d**

Explanation:Back emf developed in the motor can be calculated using the relation Eb = Vt-I×Ra. 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 Eb = 250-7.5×2 = 235V.

28. Duty cycle (D) is _______

a) Ton÷Toff

b) Ton÷(Ton+ Toff)

c) Ton÷2×(Ton+ Toff)

**d) Ton÷2×Toff**

**Answer: b**

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

#### 29. A 220 V, 1000 rpm, 60 A separately-excited dc motor has an armature resistance of .5 Ï‰. 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 (-

400) rpm is _________

a) 122.4°

b) 117.6°

c) 130.1°

**d) 102.8°**

**Answer: d**

Explanation:During rated operating conditions of the motor, Eb = Vt-Ia×Ra = 220- 60×.5=190 V. As Eb=Kmwm so Km=190×60÷(2×3.14×1000) = 1.8152 V-s/rad. Back emf at (-400 rpm) is Kmwm = 1.8152×(2×3.14×(-400))÷60 = -76 V. Now Vt = -76+60×.5 = -46 V. Average voltage of single-phase full converter is2×Vm×cos(Î±)‚3.14. The output of the converter is connected to the input terminal of the motor so Î± = cos-1 (- 46×3.14÷2×230×1.414) = 102.8o

#### 30. The unit of angular acceleration is rad/s2

**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 Î±. The unit of angular velocity is rad/sec and of time is second so the unit of angular acceleration is rad/s2.

#### 31. Calculate the value of the angular acceleration of the motor using the given data: J= 50 kg-m2, load torque = 40 N-m, motor torque = 10 N-m.

a) -.7 rad/s2

**b) -.6 rad/s2**

c) -.3 rad/s2

d) -.4 rad/s2

**Answer: b**

Explanation:Using the dynamic equation of motor J*(angular acceleration) = Motortorque – Load torque: 50*(angular acceleration) = 10-40 = -30, angular acceleration=-.6 rad/s2. The motor will decelerate and will fail to start.

#### 32. 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.

#### 33. 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 Vo = D×Vin ÷ (1-D). Itcan step up and step down the voltage depending upon the value of the duty cycle. Ifthe value of the duty cycle is less than .5 it will work as a buck converter and for dutycycle greater than .5 it will work as a boost converter.

34. 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 areVo=Vin ÷ (1-D) and Vo = D×Vin ÷ (1-D) respectively. When the value of the duty cycletends to 1 output voltage tends to infinity.

#### 35. Calculate the shaft power developed by a motor using the given data: Eb = 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 = Eb*I-(rotational losses) = 50*60 = 3000 – (600+400) = 2000 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

36. 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 experiencesconductivity modulation and it’s ON state resistance reduction due to whichconduction losses are very low.

37. 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 ACor DC. This is an example of short time loads. They have a high torque to inertiaratio and high-speed.

#### 38. Load torque of the crane is independent of _________

**a) Speed**

b) Seeback effect

c) Hall effect

d) Thomson effect

**Answer: a**

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

#### 39. The unit of angular velocity is rad/s3

a) True

**b) False**

**Answer: b**

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

#### 40. R.M.S value of the sinusoidal waveform V=Vmsin(Ï‰t+Î±).

**a) Vm÷2½**

b) Vm÷2¼

c) Vm÷2¾

d) Vm÷3½

**Answer: a**

Explanation:R.M.S value of the sinusoidal waveform is Vm÷2½ and r.m.s value of the trapezoidal waveform is Vm÷3½ . The peak value of the sinusoidal waveform is Vm.

#### 41. Calculate the time period of the waveform x(t)=24sin(24Ï€t+Ï€‚4).

a) .064 sec

b) .047 sec

**c) .083 sec**

d) .015 sec

**Answer: c**

Explanation:The fundamental time period of the sine wave is 2Ï€. The time period ofx(t) is 2Ï€‚24Ï€=.083 sec. The time period is independent of phase shifting and timeshifting.

42. 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 carrierdevices do not have any minority charge carrier storage so they have less turn-offtime and hence MOSFET has the least turn off time. So, the increasing order of turn-off time is, MOSFET < IGBT < BJT < Thyristor.

#### 43. 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 limitedfrequency range and cannot be used at high frequencies. A thyristor is aunidirectional, bipolar and semi-controlled device.

44. Calculate the useful power developed by a motor using the given data: Pin = 3000 W,
Ia = 60 A, Ra = .4 Î©. 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 thatcan be calculated using the formula Psh = Pdev-(rotational losses). Pdev = Pin-Ia 2Ra = 3000-602 (.4)=1560 W. The useful power developed by the motor is Psh = Pdev- (rotational losses)=1560-(200+400)=960 W.

45. Calculate the phase angle of the sinusoidal waveform y(t)=55sin(4Ï€t+Ï€‚8).

**a) Ï€‚8**

b) Ï€‚5

c) Ï€‚7

d) Ï€‚4

**Answer: a**

Explanation:Sinusoidal waveform is generally expressed in the form of V=Vmsin(Ï‰t+Î±) where Vm represents peak value, Ï‰ represents angular frequency, Î± represents a phase difference.

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

a) Geometrical Neutral Axis (GNA)

**b) Magnetic Neutral Axis (MNA)**

c) Axis of rotation

d) Axis of revolution

**Answer: b**

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

#### 47. 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 Eb = Î¦×Z×N×P‚60×A, Î¦ 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. Eb = .02×25×200×600÷60×2= 500 V.

#### 48. 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. Î¦ is a dot product of magnetic flux density and area. The unit of the flux is Weber (Wb).

#### 49. 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 shuntmotor can only operate on DC because of pulsating torque in AC.

#### 50. 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 thecurrent lags the voltage by 90o. The phase difference between voltage and current is90°.

#### 51. 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.

#### 52. 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 thecurrent leads the voltage by 90°. The phase difference between voltage and current is 90°.

53. 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 Î©**

b) 2.235 Î©

c) 1.625 Î©

d) 1.524 Î©

**Answer: a**

Explanation:The slope of the I-V curve is reciprocal of resistance. The slope given is 30° so R=1‚tan(30°)=1.732 Î©. The slope of the V-I curve is resistance.

#### 54. What is a mark to space ratio?

**a) Ton÷Toff**

b) Ton÷(Ton- Toff)

c) Ton÷2×(Ton*Toff)

d) Ton÷2×Toff

**Answer: a**

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

#### 55. What is the formula for the moment of inertia? (m – a mass of the body, r – distance
from the axis of the rotation)

**a) ∑miri 2**

b) ∑miri

c) ∑miri
4

d) ∑miri
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.