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

Question

In an a.c. circuit, voltage and current are given by: \[ V = 100 \sin(100t) \, \text{V} \quad \text{and} \quad I = 100 \sin\left(100t + \frac{\pi}{3}\right) \, \text{mA}. \] The average power dissipated in one cycle is:

Answer 1

To solve this problem, we need to understand the energy exchange between the capacitor and the inductor in an LC circuit.

Initially, the energy stored in the charged capacitor is given by:

$E_1 = \frac{1}{2} C V_1^2$

When the potential difference across the capacitor reduces to $V_2$, the energy in the capacitor changes to:

$E_2 = \frac{1}{2} C V_2^2$

The decrease in energy of the capacitor accounts for the increase in energy of the inductor, since energy conservation in the LC circuit is assumed:

$\frac{1}{2} C V_1^2 - \frac{1}{2} C V_2^2 = \frac{1}{2} L I^2$

Solving for the current $I$ through the inductor:

$\frac{1}{2} C (V_1^2 - V_2^2) = \frac{1}{2} L I^2$

This simplifies to:

$C (V_1^2 - V_2^2) = L I^2$

Solving for $I$, we get:

$I = \sqrt{\frac{C (V_1^2 - V_2^2)}{L}}$

Thus, the current through the inductor when the potential difference across the condenser reduces to $V_2$ is:

$\bigg(\frac{C(V_1^2 - V_2^2)}{L}\bigg)^\frac{1}{2}$

Conclusion: The correct option is $\bigg(\frac{C(V_1^2 - V_2^2)}{L}\bigg)^\frac{1}{2}$.

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

The Correct Option is D

Solution and Explanation

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