Question

In the figure, if switch $S$ is closed at $t=0$, find the magnitude of charge on the capacitor as a function of time. The inductor is active with initial current $I_0$ in it.

• A. $\sqrt{LC}{I}_0\sin \left(\frac{t}{\sqrt{LC}}\right)$
• B. $\sqrt{LC}{I}_0\cos \left(\frac{t}{\sqrt{LC}}\right)$
• C. $\sqrt{LC}{I}_0\sin \left(\frac{4t}{\sqrt{LC}}\right)$
• D. $\sqrt{LC}{I}_0\cos \left(\frac{t}{\sqrt{2LC}}\right)$

Answer 1

The correct option is A $\sqrt{LC}{I}_0\sin \left(\frac{t}{\sqrt{LC}}\right)$

Let $q_0$ is the maximum charge on the capacitor.

Then, by energy conservation we have

$\frac{1}{2}L{I}_0^2=\frac{q_0^2}{2C}$

$\Rightarrow q_0=\sqrt{LC}{I}_0$

In an $LC$ oscillation circuit, charge on a capacitor as a function of time is given as

$q=q_0\sin (\omega t+\alpha )$

Here $\omega =\frac{1}{\sqrt{LC}}$ and at $t=0,q=0$ we get $\alpha =0$

Thus, the magnitude of charge on the capacitor is given as

$q=\sqrt{LC}{I}_0\sin \left(\frac{t}{\sqrt{LC}}\right)$

Hence, option $\left(\text{A}\right)$ is correct.