A capacitor of capacitance 500 μ F is charged completely using a de supply of 100 V . It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be _____ A .

Step-by-step solution

Energy stored in capacitor E = 1 2 C V 2 = 1 2 500 × 10 - 6 × 10 4 ⇒ E = 5 2 J Current will be maximum when whole energy of capacitor becomes energy of inductor. ⇒ 1 2 L I 2 = 5 2 ⇒ I = 5 L = 5 50 × 10 - 3 = 10   A