1. Calculate the useful power developed by a motor using the given data: Pin= 1500 W, Ia= 6 A, Ra=.2 Ω. Assume frictional losses are 50 W and windage losses are 25 W.
a) 1400 W
b) 1660.5 W
c) 1417.8 W
d) 1416.7 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 = 1500-62(.2)=1492.8 W. The useful power developed by the motor is Psh = Pdev-(rotational losses) = 1492.8 –(50+25) = 1417.8 W.
2. The slope of the V-I curve is 15.5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .277 Ω
b) .488 Ω
c) .443 Ω
d) .457 Ω
Explanation: The slope of the V-I curve is resistance. The slope given is 15.5° so R=tan(15.5°)=.277 Ω. The slope of the I-V curve is reciprocal of resistance.
3. The generated e.m.f from 2-pole armature having 2 conductors driven at 3000 rpm having flux per pole as 4000 mWb, with 91 parallel paths is ___________
a) 8.64 V
b) 8.56 V
c) 8.12 V
d) 8.79 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. Eb = 4×2×3000×2÷60×91 = 8.79 V
4. A 3-phase induction motor runs at almost 888 rpm at no load and 500 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) 388 revolution per minute
b) 400 revolution per minute
c) 644 revolution per minute
d) 534 revolution per minute
Explanation: Supply frequency=50 Hz. No-load speed of motor = 888 rpm. The full load speed of motor=500 rpm. Since the no-load speed of the motor is almost 888 rpm, hence synchronous speed near to 888 rpm. Speed of rotor field=888 rpm. Speed of rotor field with respect to rotor = 888-500 = 388 rpm.
5. Calculate the active power in a 157.1545 H inductor.
a) 4577 W
b) 4567 W
c) 4897 W
d) 0 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.
6. Calculate the active power in a 1.2 Ω resistor with 1.8 A current flowing through it.
a) 3.88 W
b) 3.44 W
c) 3.12 W
d) 2.18 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 90o. P=I2R=1.8×1.8×1.2=3.88 W
7. Calculate the total heat dissipated in a resistor of 12 Ω when 9.2 A current flows through it.
a) 2.01 KW
b) 3.44 KW
c) 1.01 KW
d) 2.48 KW
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°. P=I2R=9.2×9.2×12=1.01 kW.
8. Calculate mark to space ratio if the system is on for 9 sec and time period is 11 sec.
a) 4.6
b) 4.8
c) 4.5
d) 4.9
Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=9÷(11-9) = 4.5.
9. Calculate the angular frequency of the waveform y(t)=69sin(40πt+4π).
a) 40π Hz
b) 60π Hz
c) 70π Hz
d) 20π Hz
Explanation: The fundamental time period of the sine wave is 2π. The 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. Ω can be directly calculated by comparing the equations. Ω = 40π Hz.
10. The generated e.m.f from 22-pole armature having 75 turns driven at 78 rpm having flux per pole as 400 mWb, with lap winding is ___________
a) 76 V
b) 77 V
c) 78 V
d) 79 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. In lap winding, the number of parallel paths is equal to a number of poles. Eb = .4×22×75×2×78÷60×22 = 78 V.