devarshi-dt-logo

Question:

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

(C(V1−V2)2L)12

C(V21−V22)L

C(V21+V22)L

(C(V21−V22)L)12

Solution:

In case of oscillatory discharge of a capacitor through an inductor, charge at instant t is given by
q=q0cosωt
where ω=1√LC
∴cosωt=q/q0=CV2/CV1=V2/V1 (∵q = CV) .. (i)
Also, q=q0cosωt
dq/dt=-q0ωsinωt
i=-q0ωsinωt
The current through the inductor is given by i = dq/dt.
From (i), cosωt = V2/V1
sinωt = √(1-cos²ωt) = √(1-(V2/V1)²) = √((V1²-V2²)/V1²)
Therefore, i = -q0ω √((V1²-V2²)/V1²)
= -CV1ω √((V1²-V2²)/V1²)
i = -Cω√(V1²-V2²) = -C(1/√LC)√(V1²-V2²)
i = -√(C/L)√(V1²-V2²) = -√(C(V1²-V2²)/L)
i = -√(C(V1²-V2²)/L) [Minus sign indicates direction]
|i| = √(C(V1²-V2²)/L) = √(C(V1²-V2²)/L)