JEE Main201815 Apr 2018Morning ShiftPhysicsElectromagnetic InductionActual
An ideal capacitor of capacitance 0.2 F is charged to a potential difference of 10 ~V . The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH . The current at a time when the potential difference across the capacitor is 5 ~V , is:
Options
- A0.17 ~A
- B0.15 ~A
- C0.34 ~A
- D0.25 ~A
Correct answer
A. 0.17 ~A
Step-by-step solution
Given: Capacitance, C=0.2 F =0.2 10⁻⁶ F Inductance L =0.5 mH =0.5 10⁻³ H Current I = ? Using energy conservation 1 2 C V^2= 1 2 C V₁^2+ 1 2 L I^2 1 2 0.2 10⁻⁶ 10^2+0= 1 2 0.2 10⁻⁶ 5^2+ 1 2 0.5 10⁻³ I^2 I= 3 10⁻¹ ~A =0.17 ~A