Question

An L-C circuit contains 20 mH inductor and a `50 muF` capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. what is the total energy stored initially ? At what times is the total energy shared equally between the inductor and the capacitor ?

Answer 1

Total energy stored initially means energy stored in the capacitor
`Q_(0)^(2)/(2C) = (10 xx 10^(-3))^(2)/(2 xx 50 xx 10^(-6))`= 1 joule
total energy is `Q_(0)^(2)/(2C)`, So, when the total energy is shared equally between the capacitor and inductor, the energy stored in the capacitor that time will be half of the total energy = `1/2(Q_(0)^(2)/(2C))^(2)`