This set of Electric Drives Multiple Choice Questions & Answers (MCQs) 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
Explanation: Back emf developed in the motor during the full load can be calculated using equation Eb = Vt-I×Ra = 431.53 V and machine constant Km = Eb÷Wm which is equal to 5.88. Torque can be calculated by using the relation T = Km × I = 5.88×77 = 453.51 N-m.
2. Calculate the power developed by a motor using the given data: Eb = 55 V and I = 6 A.
a) 440 W
b) 220 W
c) 330 W
d) 550 W
Explanation: Power developed by the motor can be calculated using the formula P = Eb×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-m2, load torque = 66 N-m, motor torque = 26 N-m.
a) 1.11 rad/s2
b) 2.22 rad/s2
c) 3.33 rad/s2
d) 4.44 rad/s2
Explanation: Using the dynamic equation of motor J×(angular acceleration) = Motor torque – Load torque: 36×(angular acceleration) = 66-26 = 40, angular acceleration = 1.11 rad/s2.
4. Calculate the moment of inertia of the apple having a mass of .4 kg and diameter of 12 cm.
a) .0008 kgm2
b) .0007 kgm2
c) .0009 kgm2
d) .0001 kgm2
Explanation: The moment of inertia of the apple can be calculated using the formula I=mr2×.5. The mass of the apple and diameter is given. I=(.4)×.5×(.06)2 = .0007 kgm2. 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 kgm2
b) .00196 kgm2
c) .00145 kgm2
d) .00178 kgm2
Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr2×.66. The mass of the thin spherical shell and diameter is given. I=(3.3)×.66×(.03)2=.00196 kgm2. It depends upon the orientation of the rotational axis.
6. Calculate the time period of the waveform y(t)=7cos(54πt+2π÷4).
a) .055 sec
b) .037 sec
c) .023 sec
d) .017 sec
Explanation: The fundamental time period of the cosine wave is 2π. The time period of y(t) 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: Pin= 1500 W, Ia= 6 A, Ra=.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
Explanation: Useful power is basically the shaft power developed by the motor that can be calculated using the formula Psh = Pdev-(rotational losses). Pdev = Pin-Ia2Ra = 1500-62(.2)=1492.8 W. The useful power developed by the motor is Psh = Pdev-(rotational losses) = 1492.8 –(50+25) = 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
Explanation: The slope of the V-I curve is resistance. The slope given is 15.5° so R=tan(15.5°)=.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
Explanation: The generated e.m.f can be calculated using the formula Eb = 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. Eb = 4×2×3000×2÷60×91 = 8.79 V.
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
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
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.
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
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 90o. P=I2R=1.8×1.8×1.2=3.88 W.
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
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°. P=I2R=9.2×9.2×12=1.01 kW.
14. Calculate mark to space ratio if the system is on for 9 sec and time period is 11 sec.
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÷(11-9) = 4.5.
Sanfoundry Global Education & Learning Series – Electric Drives.
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