An LC circuit contains a 20 mH inductor and a 50 μF 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 LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
(i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e., stored in the inductor)?
(d) At what times 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?
Inductor L = 20 mH
In henry L = 20 × 10-3H
capacitor C = 50 μF = 50 × 10-6F
Charge Q = 10 mC = 10 × 10-3C
(a) Total energy stored in the capacitor can be calculated as follows:
E = 1/2 Q2C
Substituting values in above formula, we get
E = 
On calculating, we get
E = 1 Joule
Yes, the total energy will be conserved during LC oscillations because there is no resistor connected in the circuit.
(b) Natural frequency of the circuit can be calculated using the following expression:
V = 1/(2π√LC)
V = 1/[2 × 3.14 × √ (20 × 10-3H × 50 × 10-6F)]
On calculating, we get
V = 159.24 Hz
Natural angular frequency is given by the following expression:
ωr = 1/√ LC
or ωr = 1/√ (20 × 10-3H × 50 × 10-6F)
ωr = 103 rad/s
(c) (i) Time period T = 1/v
T = 1/159.24 Hz
Or T = 6.29 ms
Therefore, for time period T = 6.29 ms, the total charge on the capacitor can be calculated by the following relation:
Q’ = Q cos 2π/T
Since energy stored is electrical, therefore Q’ = Q
Therefore, energy stored is electrical at time t = 0, T/2, T, 3T/2
(ii) Since when electrical energy is zero, the magnetic energy will be maximum
Therefore, energy stored is electrical at time t = T/4, 3T/4, 5T/4
(d) Q’ is equal to the charge on the capacitor when the total energy is shared between capacitor and inductor
When the energy is shared between the capacitor and inductor then the total energy stored in the capacitor is given as half of the maximum energy
⇒ 
⇒ Q’ = Q/√ 2………………(1)
But as we know, Q’ = Q cos 2π/T
Substituting eqn (1) in above
Thus, the total energy is shared between capacitor and inductor at time t = T/8, 3T/8, 5T/8
(e) If a resistor is inserted in the circuit, then total energy is eventually dissipated as heat.