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
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
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. One turn is equal to two conductors. Eb = .09×45×400×2×4200÷60×17 = 13341.17 V.
3. The unit of Magnetic flux density is Tesla.
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 kgm2
b) 9.265 kgm2
c) 9.787 kgm2
d) 9.568 kgm2
Explanation: The moment of inertia of the hollow cylinder can be calculated using the formula I=miri2÷2. The mass of the hollow cylinder and radius is given. I=(78)×.5×(.49)2=9.363 kgm2. 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-m2, load torque = 74 N-m, motor torque = 89 N-m.
a) .195 rad/s2
b) .182 rad/s2
c) .183 rad/s2
d) .185 rad/s2
Explanation: Using the dynamic equation of motor J×(angular acceleration) = Motor torque – Load torque: 81×(angular acceleration) = 89-74=15, angular acceleration=.185 rad/s2.
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
Explanation: Back emf developed in the motor during the full load can be calculated using equation Eb = Vt-I×Ra = 308.5 V and machine constant Km = Eb÷Wm which is equal to 2.1053. Torque can be calculated by using the relation T = Km × 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
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
Explanation: The slope of the V-I curve is resistance. The slope given is 31° so R=tan(31°)=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 kgm2 and mass of 69 kg.
a) .924 m
b) .928 m
c) .934 m
d) .944 m
Explanation: The moment of inertia of the circular ring can be calculated using the formula I=∑miri2. The moment of inertia of a circular ring and mass is given. R=((59)÷(69)).5 = .924 m. It depends upon the orientation of the rotational axis.
10. Calculate the power developed by a motor using the given data: Eb= 48 V and I= 86 A (Assume rotational losses are neglected.)
a) 4128 W
b) 4150 W
c) 4140 W
d) 4170 W
Explanation: Power developed by the motor can be calculated using the formula P = Eb×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 (Ra) equal to 9 ohms. Calculate back emf developed in the motor when it operates on one-fourth of the full load. (Assume rotational losses are neglected)
a) 564.75 V
b) 561.75 V
c) 562.45 V
d) 565.12 V
Explanation: Back emf developed in the motor can be calculated using the relation Eb = Vt-I×Ra. 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 Eb = 780-24.25×9 = 561.75 V.
Sanfoundry Global Education & Learning Series – Electric Drives.
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