# Electric Drives Questions and Answers – DC Motors – Application of Modified Characteristics

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This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “DC Motors – Application of Modified Characteristics”.

1. Calculate the time period of the waveform y(t)=74cos(81πt+π).
a) .024 sec
b) .027 sec
c) .023 sec
d) .025 sec

Explanation: The fundamental time period of the cosine wave is 2π. The time period of y(t) is 2π÷81π=.024 sec. The time period is independent of phase shifting and time shifting.

2. The generated e.m.f from 42-pole armature having 74 turns driven at 64 rev/sec having flux per pole as 21 mWb, with wave winding is ___________
a) 4177.171 V
b) 4177.152 V
c) 4100.189 V
d) 4190.454 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 wave winding the number of parallel paths is equal to two. Eb=.021×42×74×2×3840÷60×2=4177.152 V.

3. Calculate the phase angle of the sinusoidal waveform x(t)=20sin(9πt+π÷7).
a) π÷9
b) π÷5
c) π÷7
d) π÷4

Explanation: 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.

4. Calculate the moment of inertia of the solid sphere having a mass of 28 kg and diameter of 15 cm.
a) 0.01575 kgm2
b) 0.01875 kgm2
c) 0.01787 kgm2
d) 0.01568 kgm2

Explanation: The moment of inertia of the solid sphere can be calculated using the formula I=2×miri2÷5. The mass of the solid sphere and diameter is given. I =(28)×.4×(.0375)2=.01575 kgm2. It depends upon the orientation of the rotational axis.

5. R.M.S value of the trapezoidal waveform V=Vmsin(Ωt+α).
a) Vm÷2½
b) Vm÷2¼
c) Vm÷2¾
d) Vm÷3½

Explanation: R.M.S value of the sinusoidal waveform is Vm÷2½ and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm.

6. What is the unit of the admittance?
a) ohm
b) ohm-1
c) ohm2
d) ohm.5

Explanation: The admittance measures how easily current can flow in the circuit. It is the ratio of current and voltage. It is given in ohm-1. It is reciprocal of impedance.

7. Calculate the value of the frequency if the inductive reactance is 45 Ω and the value of the inductor is 15 H.
a) 0.477 Hz
b) 0.544 Hz
c) 0.465 Hz
d) 0.412 Hz

Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation XL = 2×3.14×f×L. F = XL÷2×3.14×L = 45÷2×3.14×15 = .477 Hz.

8. The slope of the V-I curve is 19°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .3254 Ω
b) .3608 Ω
c) .3543 Ω
d) .3443 Ω

Explanation: The slope of the V-I curve is resistance. The slope given is 19° so R=tan(19°)=.3443 Ω. The slope of the V-I curve is resistance.

9. Calculate the active power in a 41 H inductor.
a) 2 W
b) 1 W
c) 0 W
d) .5 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.

10. Calculate the active power in a 19 F capacitor.
a) 7.8 W
b) 0 W
c) 5.4 W
d) 1.5 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.

11. Calculate the active power in a 241 H inductor.
a) 21 W
b) 11 W
c) 0 W
d) .51 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 90o in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W.

12. Calculate the active power in a 5 Ω resistor with 5 A current flowing through it.
a) 125 W
b) 110 W
c) 115 W
d) 126 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 90°. P=I2R=5×5×5=125 W.

Sanfoundry Global Education & Learning Series – Electric Drives.

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