1. Calculate the angular frequency of the waveform y(t)=69sin(40πt+4π).
A. 40π Hz
B. 60π Hz
C. 70π Hz
D. 20π Hz
Answer: A
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 the angular frequency
α represents a phase difference
Ω can be directly calculated by comparing the equations.
Ω = 40π Hz.
2. 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
Answer: C
The generated e.m.f can be calculated using the formula
Eb = Φ×Z×N×P÷60×A
where
Φ represents 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 equals the number of poles.
Eb = .4×22×75×2×78÷60×22 = 78 V.
3. Calculate the phase angle of the sinusoidal waveform u(t)=154sin(9.85πt-π÷89).
A. -78π÷9
B. -12π÷5
C. -π÷89
D. -2π÷888
Answer: C
The sinusoidal waveform is generally expressed in the form of
V=Vmsin(Ωt+α)
where
Vm represents the peak value
Ω represents the angular frequency
α represents a phase difference.
4. Calculate the moment of inertia of the stick about its end having a mass of 22 kg and a length of 22 cm.
A. .088 kgm2
B. .087 kgm2
C. .089 kgm2
D. .086 kgm2
Answer: B
The moment of inertia of the stick about its end can be calculated using the formula
I=ML2÷3.
The mass of the stick about its end and length is given.
I = (22)×.33×(.11)2 =.087 kgm2.
It depends upon the orientation of the rotational axis.
5. Calculate the moment of inertia of the stick about its center having a mass of 1.1 kg and a length of 2.9 m.
A. .66 kgm2
B. .77 kgm2
C. .88 kgm2
D. .47 kgm2
Answer: B
The moment of inertia of the stick about its center can be calculated using the formula
I=ML2÷12.
The mass of the stick about its center and length are given.
I=(1.1)×.0833×(2.9)2=.77 kgm2.
It depends upon the orientation of the rotational axis.
6. Calculate the useful power developed by a motor using the given data: Eb = 4V and I = 52 A. Assume frictional losses are 3 W and windage losses are 2 W.
A. 203 W
B. 247 W
C. 211 W
D. 202 W
Answer: A
Useful power developed by the motor can be calculated using the formula
P = Eb*I -(rotational losses) = 4*52 – (5) = 203 W.
If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.
7. Calculate the value of the frequency if the capacitive reactance is .1 Ω and the value of the capacitor is .02 F.
A. 71.25 Hz
B. 81.75 Hz
C. 79.61 Hz
D. 79.54 Hz
Answer: C
The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation
Xc = 1÷2×3.14×f×C. F
= 1÷Xc×2×3.14×C
= 1÷.1×2×3.14×.02 = 79.61 Hz.
8. The slope of the V-I curve is 6.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
A. .38 Ω
B. .59 Ω
C. .34 Ω
D. .12 Ω
Answer: D
The slope of the V-I curve is resistance.
The slope given is 6.9° so R=tan(6.9°)=.12 Ω.
The slope of the I-V curve is reciprocal to resistance.
9. Calculate the active power in an 8764 H inductor.
A. 8645 W
B. 6485 W
C. 0 W
D. 4879 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 543 F capacitor.
A. 581 W
B. 897 W
C. 0 W
D. 892 W
Answer: C
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 to the voltage in the case of the capacitor.
