Question

A circuit is shown below.


Currently the switch is in position A and steady current is flowing through circuit. Capacitor is uncharged. At $t=0$, switch is instantly moved to position B.
Find the current in the $LC$ circuit.$\left(\mathrm{Given}\mathrm{that}\omega =\frac{1}{\sqrt{LC}}\right)$

• A. $\frac{V}{R}\sin \left(\omega t\right)$
• B. $\frac{V}{R}\cos \left(\omega t\right)$
• C. $\frac{V}{R}\cos \left(\omega t+\frac{\pi }{5}\right)$
• D. $\frac{V}{R}\cos \left(\omega t+\frac{\pi }{8}\right)$

Answer 1

The correct option is B $\frac{V}{R}\cos \left(\omega t\right)$

At $t=0,$ there is a steady current downward through the inductor i.e.
$i\left(0\right)=\frac{V}{R}$
For $t\ge 0,$ krichhoff's law gave following for the $LC$ circuit
$-L\frac{d^{2}I}{d{t}^2}=\frac{q}{C}.....\left(i\right)$
Here $q$ is the charge on the lower plate of the capacitor and we know
$I=\frac{dq}{dt}$
so differentially equation (i), we get
$\frac{d^{2}I}{d{t}^2}=-\frac{I}{LC}=-{\omega }^{2}I...\left(ii\right)$
The general solution (ii) is given by
$I=I_{max}\cos \left(\omega t+\varphi \right)$
Using initial condition of current $i=\frac{V}{R}$, we get $\varphi =0$
So,
$I=\frac{V}{R}\cos \left(\omega t\right)$