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

Given Inductance `L = 20 mH = 20xx10^(-3)H`
Capacitance of capacitor
`C=50 mu f=50xx10^(-6)F`
Initial charge on the capacitor,
`Q_(i)=10mc =10xx10^(3)C`
Equal sharing of energy between inductor and capacitor means the energy stored in capacitor `= (1)/(2)xx` Maximum energy
`(Q^(2))/(2C)=(1)/(2)xx(Q_(0)^(2))/(2C) " " Q=(Q_(0))/(sqrt(2))` .....(ii)
From `Q=Q_(0)cos omega t=Q_(0)cos.(2pi)/(T)t`
`(Q_(0))/(sqrt(2))=Q_(0).cos.(2pi)/(T)t`
`(1)/(sqrt(2))=cos.(2pi t)/(T)`
or `cos(2n+1)pi//4 = cos.(2pi t)/(T)`
`((2n+1)pi)/(4)=(2pi t)/(T)`
`t=T//8(2n+1) (n=0,1,2,3,....)`
Hence the energy will be shared half on capacitor and half on inductor.
`t=(T)/(8), (3T)/(8), (5T)/(8)`, .......