Q. 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
A) $\left( \frac{C\left( V_{1} - V_{2} \right)^{2}}{L} \right)^{\frac{1}{2}}$
B) $\frac{C(V_{1}^{2} - V_{2}^{2})}{L}$
C) $\frac{C(V_{1}^{2} + V_{2}^{2})}{L}$
D) $\left( \frac{C\left( V_{1}^{2} - V_{2}^{2} \right)}{L} \right)^{\frac{1}{2}}$
Explanation
The correct answer is Option D: $\left( \frac{C\left( V_{1}^{2} - V_{2}^{2} \right)}{L} \right)^{\frac{1}{2}}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.