Question

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

Answer 1

Initially only the capacitor was charged, hence energy stored in the capacitor

E = CV₁²/2

When the capacitor is connected to the inductor, then the net energy will be distributed between inductor and capacitor,

Energy in inductor = LI²/2

Energy in capacitor = CV₂²/2

Using the law of conservation of energy,

Initial energy = final energy

=> CV₁²/2 = LI²/2 + CV₂²/2

=> CV₁² = LI² + CV₂²

=> LI² = C(V₁² - V₂²)

=> I² = C(V₁² - V₂²)/L

=> I = √C(V₁² - V₂²)/L

which is the required expression for current.

Answer 2

Thank you for asking this question. Here is your answer:

The options for this question are missing, here are the options:

(a){ [ C (V1 - V2)^2] / L}^1/2

(b) [C ( V1^2 - V2^2)] / L

(c) [ C ( V1^2 + V2^2 )] / L

(d) { [C( V1^2 - V2^2)]/ L } ^1/2

Apply conservation of energy (1/2)C(V1)² = (1/2)LI² + (1/2)C(V2)² so LI²

= C[(V1)² - (V2)²]

then I = √{(C/L)[(V1)²-(V2)²]}

If there is any confusion please leave a comment below.