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(a) $[\frac{C(V^{2}_{1}-V^{2}_{2})}{L}]^{1/2}$(b) $[\frac{C(V_{1}-V_{2})^{2}}{L}]^{1/2}$(c) $\frac{C(V^{2}_{1}-V^{2}_{2})}{L}$(d) $\frac{C(V_{1}-V_{2})}{L}$

Answer 1

Answer:

B) C (v1²-v2²)/L~1/2

Explanation:

In case of a oscillatory discharge of a capacitor through an inductor, the charge at instant time (t) is given by q = qcosωt

where ω = 1/√LC

cosωt = q/q0 = CV2/CV1 = V2/V1 ( as q = CV)

Current through the inductor

I = dq/dt = d(q0cosωt)/dt

= - q0sinωt

I = CV, 1/√LC ( 1-q0cosωt)

v1 = √C/L (1-(v2/v1)²~1/2

= C (v1²-v2²)/L~1/2

Thus, The current through the inductor when the potential difference across the condenser reduces to V₂ is C (v1²-v2²)/L~1/2