Electric Drives Questions and Answers – Induction Motors – Controlling Speed Using Inductance

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This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “Induction Motors – Controlling Speed Using Inductance”.

1. Calculate the voltage regulation in the synchronous machine if the no-load voltage is 12 V and the full load voltage is 15V.
a) -20%
b) -40%
c) -60%
d) -80%
View Answer

Answer: a
Explanation: Voltage regulation is defined as the fluctuation in the load voltage when the load is varied from no-load to full load. V.R(%) = (No-load voltage-Full load voltage) ÷ Full load voltage=12-15÷15 = -20%.
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2. Calculate the condition for maximum voltage regulation in the synchronous machine.
a) Φ=ϴs
b) Φ=2ϴs
c) Φ=4ϴs
d) Φ=8ϴs
View Answer

Answer: a
Explanation: Voltage regulation is maximum in case of inductive load. The condition for maximum voltage regulation is when the power factor angle of the load becomes equal to the impedance angle. V.R=Rp.ucos(Φ)+Xp.usin(Φ). Differentiate V.R with respect to Φ and put it equal to zero. We will get tan(Φ) = Xp.u÷Rp.u=tan(ϴs) then Φ=ϴs.

3. Zero voltage regulation can be only achieved in leading power factor load.
a) True
b) False
View Answer

Answer: a
Explanation: Zero voltage regulation only occurs during a leading power factor load. Condition for zero voltage regulation occurs when Φ+ϴs > 90o. For example – Capacitive load.

4. R.M.S value of the sinusoidal waveform v(t)=211sin(7.25t+7π÷78.3).
a) 149.19 V
b) 156.23 V
c) 116.57 V
d) 178.64 V
View Answer

Answer: a
Explanation: R.M.S value of the sinusoidal waveform is Vm÷2½ = 211÷2½ = 149.19 V and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm.

5. Calculate the peak value of sinusoidal waveform if the r.m.s value is 21 V.
a) 29.69 V
b) 48.74 V
c) 69.23 V
d) 25.74 V
View Answer

Answer: a
Explanation: R.M.S value of the sinusoidal waveform is Vpeak÷2½ and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vr.m.s×2½. Vpeak = Vr.m.s×2½=21×1.414=29.69 V.
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6. If induction motor air gap power is 2 KW and mechanically developed power is 1 KW, then rotor ohmic loss will be _________ KW.
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=2-1=1 KW.

7. Calculate the line voltage in the delta connection when phase voltage=45 V.
a) 46 V
b) 47 V
c) 45 V
d) 78 V
View Answer

Answer: c
Explanation: The line voltage in case of delta connection is phase voltage. Line current leads the phase current by an angle of 30°. VL-L=Vph = 45 V.

8. The slope of the V-I curve is 35.48°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) 0.452 Ω
b) 0.462 Ω
c) 0.752 Ω
d) 0.712 Ω
View Answer

Answer: d
Explanation: The slope of the V-I curve is resistance. The slope given is 35.48° so R=tan(35.48°)=.712 Ω. The slope of the I-V curve is reciprocal of resistance.

9. If induction motor rotor power is 157.5 KW and gross developed power is 79.9 KW, then rotor ohmic loss will be _________ KW.
a) 77.5
b) 77.6
c) 76.9
d) 77.1
View Answer

Answer: b
Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=157.5-79.9=77.6 KW.
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10. Calculate the line current in the delta connection when phase current=17 A.
a) 29.44 A
b) 24.64 A
c) 23.48 A
d) 26.56 A
View Answer

Answer: a
Explanation: The line voltage in case of delta connection is phase voltage. Line current leads the phase current by an angle of 30°. IL-L = 1.73×Iph = 29.44 A.

11. Calculate the active power in a 168.12 H inductor.
a) 65 W
b) 0 W
c) 68 W
d) 64 W
View Answer

Answer: b
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.

12. Calculate the maximum value of slip when rotor resistance is 2 Ω and rotor reactance is 3 Ω.
a) 0.66
b) 0.33
c) 0.44
d) 0.99
View Answer

Answer: a
Explanation: The maximum torque occurs when the slip value is equal to R2÷X2. Maximum torque is also known as breakdown torque, stalling torque and pull-out torque. The maximum value of slip is R2÷X2=2/3=.66.

13. Calculate the total heat dissipated in a rotor resistor of 21 Ω when .81 A current flows through it.
a) 13.77 W
b) 12.56
c) 16.78 W
d) 13.98 W
View Answer

Answer: a
Explanation: The rotor 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=.81×.81×21=13.77 W.
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14. Calculate the active power in an 8.12 F capacitor.
a) 89 W
b) 41 W
c) 0 W
d) 48 W
View Answer

Answer: c
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.

Sanfoundry Global Education & Learning Series – Electric Drives.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn