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)
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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