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Physics Question on Capacitors and Capacitance

A capacitor of capacitance 500 μF 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 the LC circuit will be ______A.

Answer

q0 = CV
= 500 × 100 × 10-6 C
= 5 × 10-2 C
For imax,
12Lim2=12q02C\frac{1}{2}Li_m^2=\frac{1}{2}\frac{q_0^2}{C}
50×10−3×Im250×10^{−3}×I_m^2

=(5×10−2)2500×10−6=\frac{(5×10^{−2})^2}{500×10^{−6}}
⇒im=5×10−25×10−3⇒i_m=\frac{5×10^−2}{5×10^{−3}}
im=10 A
So, the answer is 10 A.