Rankers Physics

Alternating Current: Practice Problem & Solution

29. A condenser of capacity C is charged to a potential difference of $V_1$. 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 $V_2$ is: (2010 Mains)
$ \left( \frac{C(V_1 - V_2)^2}{L} \right)^{\frac{1}{2}} $
$ \frac{C(V_1^2 - V_2^2)}{L} $
$ \frac{C(V_1^2 + V_2^2)}{L} $
$ \left[ \frac{C(V_1^2 - V_2^2)}{L} \right]^{\frac{1}{2}} $

Solution Explained:

To solve this problem, we apply the core principles of Alternating Current. Understanding the underlying formula is key to arriving at the correct answer below:

By conservation of energy, the decrease in electrical energy of the capacitor equals the magnetic energy gained by the inductor.
$\frac{1}{2} C V_1^2 - \frac{1}{2} C V_2^2 = \frac{1}{2} L I^2$.
$C(V_1^2 - V_2^2) = L I^2 \implies I^2 = \frac{C(V_1^2 - V_2^2)}{L}$.
Therefore, the current is $I = \left[ \frac{C(V_1^2 - V_2^2)}{L} \right]^{\frac{1}{2}}$.

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