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
The duty cycle is Ton÷Ttotal. It is the ratio of the time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷Ttotal=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
The frequency is defined as the number of oscillations per second. It is reciprocal to the time period. AC supply magnitude is variable. It changes with time so the frequency of the AC supply is 50 Hz.
13. DC series motor cannot run under no load.
A. True
B. False
Answer: A
DC series motor cannot be run under no-load condition because at a 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
The frequency is defined as the number of oscillations per second. It is reciprocal to the time period. It is expressed in Hz.
F = 1÷T = 1÷.2 = 5 Hz.
15. 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
Back emf developed in the motor during the full load can be calculated using the equation
Eb = Vt-I×Ra = 37.9 V
Machine constant Km = Eb÷Wm which is equal to 5.17.
Torque can be calculated by using the relation
T = Km× I = 5.17×3.5 = 18.10 N-m.
16. Calculate the active power developed by a motor using the given data: Eb = 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
Power developed by the motor can be calculated using the formula
P = Eb×I = 5.5×.5 = 2.75 W.
If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.
17. Calculate the value of the angular acceleration of the motor using the given data: J = .1 kg-m2, load torque = 45 N-m, motor torque = 55 N-m.
A. 100 rad/s2
B. 222 rad/s2
C. 300 rad/s2
D. 400 rad/s2
Answer: A
Using the dynamic equation of motor J×(angular acceleration)
= Motor torque – Load torque: .1×(angular acceleration)
= 55-45=10, angular acceleration = 100 rad/s2.
18. Calculate the moment of inertia of the tennis ball having a mass of 7 kg and a diameter of 152 cm. [/bg_collapse]
A. 3.55 kgm2
B. 4.47 kgm2
C. 2.66 kgm2
D. 1.41 kgm2
Answer: C
The moment of inertia of the tennis ball can be calculated using the formula
I=mr2×.5.
The mass of the ball and diameter is given
I=(7)×.5×(.76)2=2.66 kgm2.
It depends upon the orientation of the rotational axis.
19. 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 kgm2
B. 5.96 kgm2
C. 5.45 kgm2
D. 2.78 kgm2
Answer: A
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=(7.8)×.66×(.728)2=2.72 kgm2.
It depends upon the orientation of the rotational axis.
20. Calculate the time period of the waveform y(t)=87cos(πt+289π÷4).
A. 2 sec
B. 37 sec
C. 3 sec
D. 1 sec
Answer: A
The fundamental time period of the cosine wave is 2π.
The time period of y(t) is 2π÷π=2 sec.
The time period is independent of phase shifting and time-shifting.
