Q. 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 the inductor when the potential difference across the condenser reduces to V₂ is
A) [C(V₁ − V₂)²/L]^(1/2)
B) C(V₁² − V₂²)/L
C) C(V₁² + V₂²)/L
D) [C(V₁² − V₂²)/L]^(1/2)
Explanation
[Solution from source] In the LC circuit energy is conserved: ½CV₁² = ½CV₂² + ½LI². So I = [C(V₁² − V₂²)/L]^(1/2). Using (V₁ − V₂)² instead of V₁² − V₂² treats the voltage drop wrongly, and omitting the square root gives I² rather than I. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.