Question

Arrange the following statements :
• [A.] Abnormal impedance appears on transmission line
• [B.] Relay comparator detects deviation from set value
• [C.] Relay output contacts close
• [D.] Trip coil of circuit breaker is energised
• [E.] Breaker contacts interrupt fault current Choose the correct answer from the options given below :

• A. A, B, C, D, E
• B. A, C, B, D, E
• C. B, A, C, D, E
• D. C, B, A, D, E

Hint:

Protection sequence: \[ \text{Fault Detection} \rightarrow \text{Relay Operation} \rightarrow \text{Trip Coil Energization} \rightarrow \text{Breaker Isolation} \]

Answer 1

The Correct Option is A

Concept: Protection systems in power systems operate in a definite sequence. The sequence starts from: \[ \text{Occurrence of abnormal condition} \] and ends with: \[ \text{Isolation of faulty section} \] The relay and circuit breaker together protect the system.

Step 1: Occurrence of abnormal impedance. During fault on transmission line: \[ \text{Impedance changes abnormally} \] This abnormality is sensed by protective relay. Hence first event is: \[ A \]

Step 2: Relay detects deviation. Relay comparator compares measured quantity with preset value. When deviation exceeds limit: \[ \text{Relay operates} \] Thus second step is: \[ B \]

Step 3: Relay contacts close. After detection: \[ \text{Relay output contacts close} \] This completes trip circuit. Therefore next step is: \[ C \]

Step 4: Trip coil gets energized. Closing of relay contacts energizes: \[ \text{Circuit breaker trip coil} \] Hence: \[ D \] comes next.

Step 5: Circuit breaker interrupts fault current. Finally: \[ \text{Breaker contacts separate} \] and interrupt fault current. Thus last step is: \[ E \]

Step 6: Writing the final sequence. Correct sequence: \[ A,\ B,\ C,\ D,\ E \]

Step 7: Selecting the correct option. Hence correct answer is: \[ \boxed{(1)} \]