A capacitor of capacitance 500 $$\mu$$F is charged completely using a DC 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.

We have a capacitor of capacitance $$C = 500 \, \mu\text{F} = 500 \times 10^{-6}$$ F charged to $$V = 100$$ V, now connected to an inductor of inductance $$L = 50$$ mH $$= 50 \times 10^{-3}$$ H to form an LC circuit.

In an LC circuit, energy oscillates between the capacitor and the inductor. By conservation of energy, the maximum energy stored in the inductor equals the initial energy stored in the capacitor: $$\frac{1}{2}LI_{\max}^2 = \frac{1}{2}CV^2$$

Solving for $$I_{\max}$$: $$I_{\max} = V\sqrt{\frac{C}{L}} = 100 \times \sqrt{\frac{500 \times 10^{-6}}{50 \times 10^{-3}}} = 100 \times \sqrt{\frac{500}{50000}} = 100 \times \sqrt{0.01} = 100 \times 0.1 = 10$$ A.

Hence, the correct answer is 10.