Question

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?

Answer 1

Given: An LC circuit contains an inductor of 20 mH, its capacitance is 50 μF and the capacitor has an initial charge of 10 mC.

a)

The total energy stored initially in the circuit is given as,

E= 1 2 Q 2 C

Where, the initial charge on the capacitor is Q and the value of capacitor is C.

By substituting the values in the above equation, we get

E= ( 10× 10 −3 ) 2 2×50× 10 −6 =1 J

Thus, the total energy stored initially in the circuit is 1 J.

Since, there is no resistor connected in the circuit therefore, the total energy stored initially in the circuit is conserved.

b)

The natural frequency of the circuit is given as,

ω r = 1 LC

Where, the inductance of the circuit is L and its capacitance is C.

By substituting the values in the above equation, we get

ω r = 1 20× 10 −3 ×50× 10 −6 =1000 rad/s

Thus, the natural frequency of the circuit is 1000 rad/s.

c)

For a time period, T= 2π ω , total charge on the capacitor at time t is given by,

Q ′ =Qcos 2π T t

Hence, it can be inferred that the energy stored in the circuit is completely electrical at times t=0, T 2 ,T, 3T 2 .....

Magnetic energy is maximum when electrical energy is equal to 0.

Hence, it can be inferred that the energy stored in the circuit is completely magnetic at times t= T 4 , 3T 4 , 5T 4 .....

d)

Let Q ′ be the charge on capacitor when the total energy shared equally between the inductor and the capacitor.

When the total energy shared equally between the inductor and the capacitor, the energy stored in capacitor is half the maximum energy.

Hence, the energy stored in capacitor is,

( Q ′ ) 2 2C = 1 2 ( 1 2 Q 2 2C ) = 1 4 Q 2 C Q ′ = Q 2

But,

Q ′ =Qcos 2π T t Q 2 =Qcos 2π T t cos 2π T t= 1 2 =cos( 2n+1 ) π 4

Hence,

t=( 2n+1 ) T 8 .

Thus, total energy is shared equally between the inductor and the capacitor at times t= T 8 , 3T 8 , 5T 8 .

e)

If a resistor is inserted into the circuit, then the total initial energy is dissipated as heat.

Thus, the energy dissipated as heat is 1 J.