This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “Starting – Acceleration Time”.
1. Calculate the value of the angular acceleration of the motor using the given data: J = .01 kg-m2, load torque= 790 N-m, motor torque= 169 N-m.
a) -62100 rad/s2
b) -62456 rad/s2
c) -34056 rad/s2
d) -44780 rad/s2
Explanation: Using the dynamic equation of motor J*(angular acceleration) = Motor torque – Load torque: .01*(angular acceleration) = 169-790=-621, angular acceleration=-62100 rad/s2.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 __________
Explanation: Given a number of poles = 14. Supply frequency is 50 Hz. Rotor speed is 699 rpm. Ns=120×f÷P=120×50÷14 = 428.57 rpm. S=Ns-Nr÷Ns=428.57-99÷428.57=.769. F2=sf=.769×50=38.45 Hz.
3. Calculate the phase angle of the sinusoidal waveform z(t)=18cos(1546πt+1900π÷76).
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm 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 kgm2 and radius of 4 cm.
a) 10624 kg
b) 10625 kg
c) 10628 kg
d) 10626 kg
Explanation: The moment of inertia of the ball can be calculated using the formula I=Σmiri2. The moment of inertia of ball and radius is given. M=(17)÷(.04)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 kgm2
b) .2147 kgm2
c) .2138 kgm2
d) .2148 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)×.66×(.33)2=.2156 kgm2. 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
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
Explanation: The slope of the V-I curve is resistance. The slope given is 270° so R=tan(270°)=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
Explanation: The slope of the V-I curve is resistance. The slope given is 23.56° so R=tan(23.56°)=.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
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.
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
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
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.
12. Calculate the value of the duty cycle if the system is on for 48 sec and off for 1 sec.
Explanation: Duty cycle is Ton÷Ttotal. 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÷Ttotal = 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
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
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.
Sanfoundry Global Education & Learning Series – Electric Drives.
To practice all areas of Electric Drives, here is complete set of 1000+ Multiple Choice Questions and Answers.
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