Question

A LC-circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. 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 the LC-oscillations?
(b) What is the natural frequency of the circuit?
(c) At what times is the energy stored?
(i) completely electrical (ie., stored in the capacitor)?
(ii) completely magnetic (i.e., stored in the inductor)?
(d) At what time is the total energy shared equally between the inductor and the capacitor

Answer 1

Given LC circuit

Inductance, $L=20mH=20\times {10}^{-3}H$

Capacitance, $C=50\mu F=50\times {10}^{-6}F$

Initial charge $Q=10mC=10\times {10}^{-3}C$

$\left(a\right)$ Total initial energy stored in the circuit

$E=\frac{Q_0^2}{2C}=\frac{{10}^{-2}\times {10}^{-2}}{2\times 50\times {10}^{-6}}=1J$

this energy remain constant in the absence of resistance.

$\left(b\right)$ Natural frequency $f=\frac{1}{2\pi \sqrt{LC}}=\frac{{10}^3}{2\pi }=159H_z$

$\left(c\right)\left(i\right)$ Let, at any instant the energy stored in the circuit is completely the electrical charge on the capacitor

$Q=Q_0\cos \omega t$

Then, $Q=Q_0\cos \frac{2\pi }{T}f$

$T=\frac{1}{f}=\frac{1}{159}=6.3ms$

$\therefore Q$ is maximum only when

$\cos \left(\frac{2\pi }{T}\right)t=\pm 1=\cos n\pi =\frac{nT}{2}$(equation $1$)

where, $n=1,2,3,4,........$(equation$2$)

Hence, energy stored is completely electrical at

$t=0,\frac{T}{2},T,\frac{3T}{2}$ and so on.

$\left(ii\right)$ Now let the energy stored be completely magnetic at instant charge $=0$

From equation $1$

$\cos \frac{2\pi }{T}t=0=\frac{n\pi }{2}$ or $t=\frac{nT}{4}$

Thus, energy stored is completely magnetic at,

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

$\left(d\right)$ Energy shared between inductor and capacitor is equal means the energy shared is half time the maximum energy of the circuit

Electrical energy $=\frac{Q^2}{2C}=\frac{1}{2}\frac{Q_0^2}{2C}$

$Q=\frac{Q_0}{\sqrt{2}}$

Using equation $1$ we have,

$\frac{Q_0}{\sqrt{2}}=Q_0\cos \frac{2\pi }{T}t$

$\Rightarrow \frac{1}{\sqrt{2}}=\cos \frac{2\pi }{T}t$

$\Rightarrow \cos \frac{(2n+1)\pi }{4}=\cos \frac{2\pi }{T}t$

$\Rightarrow \frac{(2n+1)\pi }{4}=\frac{2\pi }{T}t$

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

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