1. Calculate the time period of the waveform v(t)=.6sin(.1πt+8π)+17cos(2πt+π)+ 4tan(.1πt).
a) 88 sec
b) 20 sec
c) 70 sec
d) 43 sec
Explanation: The fundamental time period of the sine wave is 2π. The time period of v(t) is L.C.M {20,1,20}=20 sec. The time period is independent of phase shifting and time shifting.
2. The value of slip at the starting of an induction motor is ________
a) 0
b) 1
c) 2
d) 3
Explanation: The maximum torque occurs when the slip value is equal to R2÷X2. At the starting, the rotor speed is equal to zero so slip value is 1.
3. 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
Explanation: 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
Explanation: Using the dynamic equation of motor J×(angular acceleration) = Motor torque – Load torque: .1×(angular acceleration) = 55-45=10, angular acceleration = 100 rad/s2.
5. 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
Explanation: The fundamental time period of the cosine wave is 2π. The time period of y(t) is a 2π÷π=2 sec. The time period is independent of phase shifting and time shifting.
6. 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
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 = 10-.62(.2)=9.92 W. The useful power developed by the motor is Psh = Pdev-(rotational losses) = 9.92 –(5) = 4.928 W.
7. 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
Explanation: 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 of resistance
8. Calculate the active power in a 7481 H inductor.
a) 1562 W
b) 4651 W
c) 0 W
d) 4654 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. Voltage leads the current in case of the inductor.
9. Calculate the active power in a 56 F capacitor.
a) 6.45 W
b) 0 W
c) 15.45 W
d) 14.23 W
Explanation: 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 case of the capacitor so the angle between V & I is 90°. P=VIcos90 = 0 W. Current leads the voltage in case of the capacitor.
10. Calculate the active power in a 1.7 Ω resistor with 1.8 A current flowing through it.
a) 5.5 W
b) 5.1 W
c) 5.4 W
d) 5.7 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 0°. P=I2R=1.8×1.8×1.7=5.5 W.