# Electrical Drive System Protection MCQ [Free PDF] – Objective Question Answer for Electrical Drive System Protection Quiz

1. Relay is an electromechanical switch.

A. True
B. False

Relay is an electromechanical switch. It uses electrical signals as input and mechanical operation as output.

2. Mho relay is used in the medium transmission line.

MCQ on Virus - Examsegg Biology

A. True
B. False

Mho relay is an inherent directional relay. It is used in the long transmission line. It is less affected by power surges.

3. Full form of LED is _____________

A. Light Emitting Diode
B. Light Emission Digital
C. Light Energy Diode
D. Light Energy Digital

The full form of LED is a light-emitting diode. It converts light energy into electrical energy. It works in the reverse bias region.

4. Inter turn fault can be detected by using __________

A. Oil circuit breaker
B. Split phase relay
C. SF6 circuit breaker
D. Vertical breaker

Inter-turn fault mainly occurs in the winding of the rotor side of the alternator. They can be detected by using the split-phase relaying technique.

5. Calculate the value of the fault impedance if the fault voltage is 2 V and the fault current is 4 A.

A. .5 ohm
B. .2 ohm
C. .4 ohm
D. .6 ohm

The value of the fault impedance is the ratio of fault voltage and fault current. It is expressed in ohms.

Z=V/I=2/4=.5 ohm.

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6. Reactance relay is used in _______

A. Medium line
B. Short transmission line
C. Lossless line
D. Long transmission line

Reactance relay is used in a short transmission line. The reactance relay is independent of the resistance value. It protects the short transmission line from any fault.

7. Load interrupter operated at ________

C. Half of the full load
D. One-fourth of the full load

8. Calculate the maximum frequency for ASCI using the data: R=1 kΩ, C=1 μF.

A. 250 Hz
B. 280 Hz
C. 300 Hz
D. 320 Hz

The minimum time period for ASCI is a

4×R×C=4×1×1×.001=4 msec.

The total time is the sum of charging and discharging time. The maximum frequency is 1÷f=250 Hz.

9. Calculate the PIV for the Full-wave bridge rectifier if the peak value of the supply voltage is 50.

A. 72.5 V
B. 72.8 V
C. 70.7 V
D. 76.1 V

The peak inverse voltage for the Full-wave bridge rectifier is

Vm = √2×230=70.7 V.

The peak inverse is the maximum negative voltage across the thyristor.

10. In single-phase, RLE load, calculate the Peak inverse voltage using the data: (Vs)r.m.s = 9 V, f = 50 Hz, E = 40 V.

A. 52.2 V
B. 52.7 V
C. 59.7 V
D. 52.9 V