Two ideal inductors L1 = L2 = L and the three identical resistors R1 = R2 = R3 = R have been connect

Question

Two ideal inductors L1 = L2 = L and the three identical resistors R1 = R2 = R3 = R have been connected to a DC source of emf E as shown in the circuit. When the key K is kept pressed (closed) for a long time, the current through the resistance R1 on the extreme right is measured to be I. Immediately after releasing (switching off) the key, the current through the resistors is :-

Accepted Answer

[Key Concept: In DC steady state an ideal inductor acts as a plain wire, and its current cannot change instantaneously.]

Step 1: Steady state: all three resistors are across E, each carries I; L₁ carries I and L₂ carries 2I.

Step 2: Just after K opens, L₂ still carries 2I and L₁ still carries I; the source branch carries nothing.

Step 3: Hence R₁ carries I downwards, R₂ carries 2I − I = I downwards, and the 2I loop closes upward through R₃. Options A, B, C are correct; D is false.

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