1. 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.

2. 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.

4. 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

Using the dynamic equation of motor J×(angular acceleration)

= Motor torque – Load torque: .1×(angular acceleration)

= 55-45=10, angular acceleration = 100 rad/s^{2}.

4. Calculate the moment of inertia of the tennis ball having a mass of 7 kg and a diameter of 152 cm.

A. 3.55 kgm^{2}
B. 4.47 kgm^{2}
C. 2.66 kgm^{2}
D. 1.41 kgm^{2}

Answer: C

The moment of inertia of the tennis ball can be calculated using the formula

I=mr^{2}×.5.

The mass of the ball and diameter is given. I=(7)×.5×(.76)2=2.66 kgm^{2}.

It depends upon the orientation of the rotational axis.

5. 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=mr^{2}×.66.

The mass of the thin spherical shell and diameter is given.

I=(7.8)×.66×(.728)^{2}=2.72 kgm^{2}.

It depends upon the orientation of the rotational axis.

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6. 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.

7. Calculate the useful power developed by a motor using the given data: Pin = 10 W, Ia = .6 A, Ra=.2 Ω. Assume frictional losses are 2 W and windage losses are 3 W.

A. 4.928 W
B. 1.955 W
C. 1.485 W
D. 1.488 W

Answer: A

Useful power is basically the shaft power developed by the motor that can be calculated using the formula

Psh = Pdev-(rotational losses).

Pdev = Pin-Ia^{2}Ra = 10-.6^{2}(.2)=9.92 W.

The useful power developed by the motor is

Psh = Pdev-(rotational losses) = 9.92 –(5) = 4.928 W.

8. The slope of the V-I curve is 86.5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. 16.34 Ω
B. 15.88 Ω
C. 48.43 Ω
D. 54.57 Ω

Answer: A

The slope of the V-I curve is resistance.

The slope given is 16.34° so R=tan(16.34°)=16.34 Ω.

The slope of the I-V curve is reciprocal to resistance.

9. Calculate the active power in a 7481 H inductor.

A. 1562 W
B. 4651 W
C. 0 W
D. 4654 W

Answer: C

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 the case of the inductor so the angle between V & I is 90°.

P = VIcos90 = 0 W.

Voltage leads the current in the case of the inductor.

10. Calculate the active power in a 56 F capacitor.

A. 6.45 W
B. 0 W
C. 15.45 W
D. 14.23 W

Answer: B

The capacitor 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 the case of the capacitor so the angle between V & I is 90°.

P=VIcos90 = 0 W.

Current leads the voltage in the case of the capacitor.