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A condenser of capacity C is charged to a potential difference of V1. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is

1. (C(V1^2 - V2^2) / L)^1/2
2. (C(V1 - V2)^2 / L)^1/2
3. C(V1^2 - V2^2) / L
4. C(V1 - V2)^2 / L

Correct answer: (C(V1^2 - V2^2) / L)^1/2

Solution

The system forms an LC oscillatory circuit. The total energy in the system is conserved and is shared between the capacitor and the inductor. Initially, the energy is stored in the capacitor as (1/2)C(V1^2). When the potential difference across the capacitor reduces to V2, the remaining energy in the capacitor is (1/2)C(V2^2), and the rest is stored as magnetic energy in the inductor, (1/2)LI^2. Equating the energy difference to the inductor's energy gives the current I = sqrt(C(V1^2 - V2^2) / L).

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