Ncert Exercises & Solutions

7.12 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?

(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?

Given,

Charge on the capacitor, 𝑄 = 10 mC,

Inductance, 𝐿 = 20 mH

Capacitance, 𝐶 = 50 μF

(a) The total energy stored initially at t = 0,

\begin{array}{l} E = \frac{Q^{2}}{2 C} \\ E = \frac{(10 m C)^{2}}{2 \times 50 u F} = 1 \end{array}

Yes, total energy will remain conserved during the oscillation.

(b) The natural frequency of the circuit,

$\begin{array}{l} ω_{r} = \frac{1}{\sqrt{L C}} \\ ω_{r} = \frac{1}{\sqrt{20 m H \times 50 μ F}} = 10^{3} r a d s \\ f r = \frac{ω_{r}}{2 π} = \frac{10^{3}}{2 π} r a d s^{- 1} = 159 s^{- 1} \end{array}$

$Q^{'} = Q \cos (\frac{2 π t}{T})$

For energy stored in electrical, we can write Q′ = Q.
Hence, it can be inferred that the energy stored in the capacitor is
completely electrical at time, $t = 0, \frac{T}{2}, T, \frac{3 T}{2}, \dots \dots$

(ii) Magnetic energy is maximum when electrical energy is minimum.
Hence, it can be inferred that the energy stored in the circuit is
completely magnetic at time $t = \frac{T}{4}, \frac{3 T}{4}, \frac{5 T}{4}$

(d) The total energy is equally shared between the inductor and the capacitor.
So, the energy stored in the capacitor is-

\begin{array}{l} \Rightarrow \frac{Q^{' 2}}{2 C} = \frac{1}{2} \times \frac{Q^{2}}{2 C} \\ \Rightarrow Q^{'} = \frac{Q}{\sqrt{2}} \\ Also, \\ Q^{'} = Qcos (\frac{2 πt}{T}) = \frac{Q}{\sqrt{2}} \\ \Rightarrow \cos (\frac{2 πt}{T}) = \frac{1}{\sqrt{2}} = \cos (2 n + 1) \frac{π}{4} \\ where n = 0, 1, 2 \dots \dots \\ \Rightarrow t = (2 n + 1) (\frac{π}{8}) \end{array}

Hence, total energy is equally shared between the inductor and the
capacitor at time,

$t = \frac{T}{8}, \frac{3 T}{8}, \frac{5 T}{8}$

(e) Resistor damps out the LC oscillations. The whole of the initial energy 1 J,
is eventually dissipated as heat.

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