Question

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 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

$\left(a\right)$

The energy stored initially in the capacitor is given by:

$E=\frac{Q^2}{2C}$

$=\frac{(10\times {10}^{-3}{)}^2}{2\times 50\times {10}^{-6}}=1J$

Yes. The energy is conserved. It only gets converted from electrical to magnetic and vice versa.

$\left(b\right)$

Natural frequency of oscillation of the circuit is given by,

$\nu =1/2\pi \sqrt{LC}$

$\nu =159.24Hz$

$\left(c\right)$

(i) Time period, $T=1/f=6.28s$

total charge on capacitor at time $t=Q^{{}^{\prime }}=Qcos\left(\frac{2\pi }{T}t\right)$

For energy stored in electrical, we can write, $Q^{{}^{\prime }}=\pm Q$

It can be inferred that the energy stored in capacitor is completely electrical at time, $t=0,T/2,3T/2,...$

(ii) Magnetic energy is maximum when the electrical energy is zero. hence, $t=T/4,3T/4,5T/4,...$

$\left(d\right)$

$Q_1$is the charge when total energy is equally shared.

$\Rightarrow \frac{1}{2}\frac{Q_1^2}{C}=\frac{1}{2}\left(\frac{1}{2}\frac{Q^2}{C}\right)$

$Q_1=\pm Q/\sqrt{2}$

but, $Q_1=Qcos2\pi t/T$

$\Rightarrow cos2\pi t/T=\pm 1/\sqrt{2}\Rightarrow t=(2n+1)T/8$

so, $t=T/8,3T/8,5T/8$

$\left(e\right)$

The energy stored initially is $E=\frac{Q^2}{2C}=\frac{(10\times {10}^{-3}{)}^2}{2\times 50\times {10}^{-6}}=1J$

When the resistor is inserted in the circuit, it will dissipate entire 1J in heat.