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 to V₂ is

Options

(a) [ C (V₁² – V₂²) / L ]¹/²
(b) [ C (V₁ – V₂)² / L ]¹/²
(c) C (V₁² – V₂²) / L
(d) C (V₁ – V₂) / L

Correct Answer:

[ C (V₁² – V₂²) / L ]¹/²

Explanation:

q = CV₁ cos wt
i = dq/ dt = – ?Cv₁ sin ?t
Also, ?ￂﾲ = 1 / LC and V = V₁ cos ?t
At t = t₁, V = V₂ and i = – ?CV₁ sin ?t₁
cos ?t₁ = V₂ / V₁ (-ve sign gives direction)
Hence, i = V √(C/L) [ 1 – V₂² / V₁²) ]¹/²
= [ C (V₁² – V₂²) / L ]¹/²