Switchgear and Protection MCQs Part 9
Contents
- 1 During which part of fault current wave, the fuse melts
- 2 During short circuit test on circuit breaker following readings were obtained as single frequency transient:
- 3 The making current of a 3-phase circuit breaker rated at 1000 MVA, 66 kV is
- 4 Which of following is responsible for initiation of electric arc at the instant of contact separation in the circuit breakers:
- 5 If relay setting is 80% and CT ratio is 400/5, and fault current is 2000 A, the plug setting multiplier will be
During which part of fault current wave, the fuse melts
During which part of fault current wave, the fuse melts:
- When first peak is reached
- Before first peak is reached
- After first peak is reached
- None of these
Correct answer: 2. Before first peak is reached
During short circuit test on circuit breaker following readings were obtained as single frequency transient:
During short circuit test on circuit breaker following readings were obtained as single frequency transient:
- Time to reach peak restriking voltage = 100 µsec
- Peak restriking voltage = 120 kV
The frequency of oscillations is
- 1000 Hz
- 2500 Hz
- 5000 Hz
- 10000 Hz
Correct answer: 3. 5000 Hz
Explanation:
Natural frequency of oscillations is
fn = 1/(2 * Time to reach peak value)
1/(2 * 100 * 10^-6) = 5000 Hz
The making current of a 3-phase circuit breaker rated at 1000 MVA, 66 kV is
The making current of a 3-phase circuit breaker rated at 1000 MVA, 66 kV is:
- 18955 A
- 20050 A
- 22306 A
- 25382 A
Correct answer: 3. 22306 A
Solution:
Symmeterical breaking current = MVA/√3 * V
= 1000 * 10^6/√3 * 66 * 10^3 = 8747.73 A
—————-
Rated making current = 2.55 * Rated breaking current
Rated making current = 2.55 * 8747.73 A = 22306.71 A
Which of following is responsible for initiation of electric arc at the instant of contact separation in the circuit breakers:
Which of following is responsible for initiation of electric arc at the instant of contact separation in the circuit breakers:
- Field emission of electrons
- Thermionic emission of electrons
- Both of these
- None of these
Correct answer: 3. Both of these
If relay setting is 80% and CT ratio is 400/5, and fault current is 2000 A, the plug setting multiplier will be
If relay setting is 80% and CT ratio is 400/5, and fault current is 2000 A, the plug setting multiplier will be:
- 3
- 5
- 6.25
- 8.75
Correct answer: 3. 6.25
Solution:
Pick-up value = Rated secondary current of C.T * Current setting = 5 * 0.8 = 4 A
Fault current in relay coil = 2000 * 5/400 = 25 A
————————
Plug setting multiplier = 25/4 = 6.25