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?

Solution

Given, an LC circuit.
Inductance, L = 20 mH = 20

\times 10^{- 3} H

Capacitance, C = 50

μF

= 50

\times 10^{- 6} F

Initial charge, Q = 10 mC = 10

\times 10^{- 3} C

(a) Total initial energy stored in the circuit,

E = \frac{Q_{0}^{2}}{2 C} = \frac{10^{- 2} \times 10^{- 2}}{2 \times 50 \times 10^{- 6}} J = 1 J

This energy stored shall remain conserved in the absence of resistance.

(b) Angular frequency,

ω = \frac{1}{\sqrt{LC}}

= \frac{1}{{(20 \times 10^{- 3} \times 50 \times 10^{- 6})}^{1 / 2}} Hz = 10^{3} rad s^{- 1}

v = \frac{10^{3}}{2 π} Hz = 159 Hz .

Q = Q_{0} \cos ωt

then,

Q = Q_{0} \cos \frac{2 π}{T} t where, T = \frac{1}{v} = \frac{1}{159} s = 6.3 ms

∴

Q is maximum only when,

\cos \frac{2 π}{T} t = \pm 1 = \cos nπ = \frac{nT}{2} where, n = 1, 2, 3, . .

.... (1)

Hence,
Energy stored is completely electrical at t = 0, T/2, T, 3T/2,.. and so on.

Now, let the energy stored be completely magnetic at any instant when electrical charge = 0.
i.e., q = 0.
From equation (1)

\cos \frac{2 π}{T} t = 0 = \frac{nπ}{2} or t = \frac{nT}{4}

where, n =1,2,3,...

Thus, energy stored is completely magnetic at

t = \frac{T}{4}, \frac{3 T}{4}, \frac{5 T}{4}, . . . . . . .

(d) Energy shared between inductor and the capacitor is equal means the energy shared is half times the maximum energy of the circuit.

∴

Electrical energy =

\frac{Q^{2}}{2 C} = \frac{1}{2} \frac{Q_{0}^{2}}{2 C},

which is half of the total energy.

This implies Q=

\frac{Q_{o}}{\sqrt{2}}

Using equation (1) we have,

\frac{Q_{o}}{\sqrt{2}} = Q_{o} . \cos \frac{2 π}{T} t \frac{1}{\sqrt{2}} = \cos \frac{2 π}{T} t

i.e.,

\cos \frac{(2 n + 1) π}{4} = \cos \frac{2 π}{T} t

\frac{(2 n + 1) π}{4} = \frac{2 π}{T} t

t =

\frac{T}{8} (2 n + 1); n = 1, 2,, 3, . . . .

∴ t = \frac{T}{8}, \frac{3 T}{8}, \frac{5 T}{8}, . . . .

During these values of t, total energy will be shared equally between the inductor and the capacitor.

(e) Resistor damps out the LC oscillations. The whole of the initial energy 1.0 J, is eventually dissipated as heat.

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