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Question

A domestic MCCB is marked 25A, 6 kA, 230V AC. Which fault current will this MCCB safely interrupt?

This question was previously asked in
RRB ALP 2025 CBT 2 Wiremen Question Paper (28-Jul-2026) (Shift 2)
The correct answer is

Any fault current up to 6,000 A

To solve this question, we need to understand the specifications on the MCCB (Molded Case Circuit Breaker) and what they signify:

  1. Rated Current (25A): This is the maximum load current the MCCB can carry without tripping. It indicates that the MCCB can handle continuous currents up to 25 amperes without operating (i.e., tripping).

  2. Breaking Capacity (6 kA): The breaking capacity, or interrupting capacity, defines the maximum fault current that the MCCB can safely interrupt without damage. Therefore, an MCCB marked with 6 kA can interrupt fault currents up to 6,000 amperes safely.

  3. Nominal Voltage (230V AC): This indicates the standard voltage for which the MCCB is designed to operate. Here, 230 volts is the supply voltage for which the device is suitable.

Now, analyzing the options:

  • Option A: "Any current at 230 V AC" - This is incorrect because it does not consider the limits of fault current or overload.
  • Option B: "Any fault current up to 6,000 A" - This matches the breaking capacity specification of 6 kA, thus is correct.
  • Option C: "Any overload up to 25 A" - Incorrect, as overload relates to operating current, not fault interruption capacity.
  • Option D: "Any short-circuit above 25A" - This is incorrect. Fault interruption refers to the ability to interrupt short-circuits up to the breaking capacity, not just above 25A.

Therefore, the MCCB can safely interrupt "Any fault current up to 6,000 A". Thus, the correct answer is:

Any fault current up to 6,000 A

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