An LC circuit contains a 20 mH inductor and a `50 mu F` 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.
(a) What is the total energy stored initially ? Is it conserved during the oscillalions?
(b) What is the natural frequency of the circuit?
(c ) At what time is the energy stored? (i) Completely electrical ? (ii) Completely magnetic ?
(d) At what time is the total energy shared equally between the inductor and the capacitor ?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat ?

Explanation

Here, `L = 20 mH = 20 xx 10^(-3) H, C = 50 mu F = 50 xx 10^(-6) F, q_(0) = 10 mC = 10 xx 10^(-3) C = 10^(-2) C`
(a) `E = (q_(0)^(2))/(2 C) = ((10^(-2))^(2))/(2 xx 50 xx 10^(-6)) = 1` joule
Yes, this energy is conserved during the oscillations.
(b) `v = (1)/(2 pi sqrt(LC)) = (1 xx 7)/(2 xx 22 sqrt(20 xx 10^(-3) xx 50 xx 10^(-6))) = (7 xx 10^(3))/(44) = 159.1` hertz
(c ) (i) At any instant t, change on the capacitor is `q = q_(0) cos omega t = q_(0) "cos" (2 pi)/(T) t`.
At `t = 0, (T)/(2), T, (3T)/(2)`,......`q = q_(0) = `max
`:.` Energy is stored completely in the capacitor as electrical energy.
(ii) Magnetic energy around L will be maximum, when electrical energy in C is zero, i.e., q = 0
This will be at ` i = (T)/(4), (3T)/(4), (5T)/(4)`........
(d) Equal sharing of energy means, energy of capacitor `= (1)/(2)` maximum energy.
`(q^(2))/(2 C) = (1)/(2) (q_(0)^(2))/(2C)` i.e. `q = (q_(0))/(sqrt2)`
From `q = q_(0) cos omega t = q_(0) "cos" (2 pi)/(T) t`
`(q_(0))/(sqrt2) = q_(0) "cos" (2 pi)/(T) t`
`"cos" (2 pi)/(T) t = (1)/(sqrt2) = cos (2n + 1) (pi)/(4) :. (2 pi)/(T) t = (2n + 1) (pi)/(4)`
`t = (2n + 1) (T)/(8)`
Hence, energy will be half on C and half on L at `t = (T)/(8), (3T)/(8), (%T)/(8)`...........
(e) The pressure of resistor in the circuit involves loss of energy. The oscillations become damped and ultimately they disappear, when total energy of 1 joule is dissipated as heat.