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 an inductor when the potential difference across the condenser reduces to ‘V₂’ is

• A. \(\frac {C(V_1^2 - V_2^2)}{L}\)
• B. \(\frac {C(V_1^2 + V_2^2)}{L}\)
• C. \((\frac {C(V_1^2 - V_2^2)}{L})^{\frac {1}{2}}\)
• D. \((\frac {C(V_1 - V_2)}{L})^{\frac {1}{2}}\)

Answer 1

The Correct Option is C

To derive this expression, we can use the conservation of energy in an LC circuit.The energy stored in the capacitor initially is given by: 
E₁ = \(\frac {1}{2}\) x C x V₁² 
and E₂ = \(\frac {1}{2}\) x C x V₂²
The energy stored in the inductor is: 
E = \(\frac {1}{2}\) x L x I²
Now equate E₁ + E = E₂
\(\frac {1}{2}\) x C x V₁² + \(\frac {1}{2}\) x L x I² = \(\frac {1}{2}\) x C x V₂²
Simplifying we get 
I = \(\sqrt{(\frac {C(V_1^2 - V_2^2)}{L})}\)
Therefore, the current through the inductor when the potential difference across the capacitor reduces to V2 is: \((\frac {C(V_1^2 - V_2^2)}{L})^{\frac {1}{2}}\)