Switchgear and Protection MCQs Part 4
Contents
- 1 A fuse performs
- 2 When oil circuit breaker operates under fault conditions, 1 cm³ of oil forms about
- 3 A 50 Hz, 11 kV, 3-phase alternator with earthed neutral has reactance of 7 ohms per phase and is connected to bus-bar through a circuit breaker. Distributed capacitance upto circuit breaker between phase and neutral is 0.01 µF. The average rate of rise of re-striking voltage upto first peak is
- 4 The fuse in motor circuit provides protection against
- 5 The breaking capacity of fuse as compared to circuit breaker is
A fuse performs
A fuse performs
- Detection function only
- Interruption function only
- Both of these
- None of these
Correct answer: 3. Both of these
When oil circuit breaker operates under fault conditions, 1 cm³ of oil forms about
When oil circuit breaker operates under fault conditions, 1 cm³ of oil forms about:
- 1 cm³ of gas
- 10 cm³ of gas
- 100 cm³ of gas
- 1000 cm³ of gas
Correct answer: 3. 100 cm³ of gas
A 50 Hz, 11 kV, 3-phase alternator with earthed neutral has reactance of 7 ohms per phase and is connected to bus-bar through a circuit breaker. Distributed capacitance upto circuit breaker between phase and neutral is 0.01 µF. The average rate of rise of re-striking voltage upto first peak is
A 50 Hz, 11 kV, 3-phase alternator with earthed neutral has reactance of 7 ohms per phase and is connected to bus-bar through a circuit breaker. Distributed capacitance upto circuit breaker between phase and neutral is 0.01 µF. The average rate of rise of re-striking voltage upto first peak is:
- 89 * 10^3 kV/sec
- 95.3 * 10^3 kV/sec
- 102.9 * 10^3 kV/sec
- 121 * 10^3 kV/sec
Correct answer: 4. 121 * 10^3 kV/sec
Solution:
Maximum value of recovery voltage (phase to neutral – Emax is)
Emax = √2 * 11 / √3 = 8.98 kV
————————–
Peak restriking voltage = 2 Emax = 2 * 8.98 = 17.96 kV
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Inductance per phase L = XL/2πf = 7/2π50 = 0.0222 H
Capacitance per phase, C = 0.01 µF = 0.01 * 10^-6 F
————————
Peak restriking voltage occurs at time t given by
t = 1/2fn = π√LC
= π√0.0222 H * 0.01 * 10^-6 F = 1.48 * 10^-4 Hz
———————–
Average rate of rise of re-striking voltage = Peak restriking voltage/Time upto first peak = 17.96/1.48 * 10^-4 Hz = 121 * 10^3 kV/sec
The fuse in motor circuit provides protection against
The fuse in motor circuit provides protection against:
- Open circuit
- Short circuit
- Over-load
- None of these
Correct answer: 2. Short circuit
The breaking capacity of fuse as compared to circuit breaker is
The breaking capacity of fuse as compared to circuit breaker is:
- Less
- More
- Almost same
- None of these
Correct answer: 1. Less