The self-inductance of a circuit is numerically equal to (a) the work done in establishing the magnetic flux associated with circuit. (b) twice the work done in establishing the magnetic flux associated with unit current in the circuit. (c) thrice the work done in establishing the magnetic flux associated with unit current in the circuit. (d) the work done in establishing the magnetic flux associated with unit current in the circuit.

Answer: (b). Work done in establishing current I: W = 12 L I². For unit current ( I = 1): W = 12 L ⇒ L = 2W. So the answer is (b) twice the work done in establishing the flux associated with unit current.

The self-inductance of a circuit is numerically equal to | MHT CET 2025 (19 Apr, Shift 1)

Q27MHT CET 2025 · 19 APR · SHIFT 1Easy

Self-Inductance & Energy

The self-inductance of a circuit is numerically equal to

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Answer: (b)

Work done in establishing current $I$ : $W = \tfrac12 L I^2$ .
For unit current ( $I = 1$ ): $W = \tfrac12 L \Rightarrow L = 2W$ .
So the answer is (b) twice the work done in establishing the flux associated with unit current.

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Self-Inductance & Energy question from MHT CET 2025 (19 Apr, Shift 1)

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