A capacitor of capacitance `1 mu F` is charged to a potential of 20 volt. The battery is then disconnected and a pure inductive coil of inductance 10 mH is connected across the capacitor so that L-C oscillation are set-up in the circuit. What is the maximum current in the circuit?

Options

A. 0.1 A
B. 0.15 A
C. 0.2 A
D. 0.3 A

Answer

C

Explanation

Maximum electric energy in the capacitor
= maximum magnetic energy in the inductor
`:. 1/2 CV^(2) = 1/2 LI_(0)^(2)`
`:. I_(0)^(2) = (CV^2)/(L) :. I_(0) = sqrt(C/L)V`
`:. I_(0) = 20 xx sqrt((10^(-6))/(10xx10^(-3))) = 20 xx sqrt(10^(-4))`
`:. I_(0) = 20 xx 10^(-2) = 0.2 A`.