A capacitor of capacitance 200 F is fully charged using a DC source of 100 V . It is then disconnected from the source and connected to an inductor of inductance 20 mH to form an LC circuit. The current in the circuit at the instant the potential difference across the capacitor drops to 60 V is _____ A .

Step-by-step solution

The total energy of the LC circuit is the initial energy stored in the fully charged capacitor. U_ total = 1 2 CV₀^2 U_ total = 1 2 200 10⁻⁶ (100)^2 = 1 J At the instant the voltage across the capacitor is V = 60 V , the energy remaining in the capacitor is: U_C = 1 2 CV^2 U_C = 1 2 200 10⁻⁶ (60)^2 = 0.36 J By conservation of energy, the energy stored in the inductor at this instant is: U_L = U_ total - U_C = 1 - 0.36 = 0.64 J The energy in the inductor is also given by 1 2 LI^2 . 1 2 LI^2 = 0.64 1 2 20 10⁻³ I^2 =