A capacitor of capacitance $\text{10μF}$ is charged to a potential difference 60 V and connected to a coil of inductance 4 mH as shown. The resistance in the circuit is negligible. After the switch is closed, the maximum value that the current in the circuit reaches is ____________ A.

 

Let the instantaneous charge on the capacitor be $Q$ and let the instantaneous current be $I$ Then, $I=+\frac{dQ}{dt}$

From KVL: $-L\frac{dI}{dt}-\frac{Q}{C}=0$

Combining the equations, we get $\frac{d^{2}Q}{d{t}^2}=-\frac{1}{LC}Q$

Therefore, $Q\left(t\right)=Q_{\max }\sin (\omega t+\varphi )$

Here, $\omega =\frac{1}{\sqrt{LC}}$ So, $I\left(t\right)=\frac{dQ}{dt}=\omega Q_{\max }\cos (\omega t+\varphi )$

We are given that at $t=0,q=C{V}_0$ and $I=0,$ where $V_0=60 volt$

Therefore, $Q_{\max }=C{V}_0$ and $\varphi =\frac{\pi }{2}$ So, $Q\left(t\right)=C{V}_0\cos \left(\omega t\right)$ and  $I\left(t\right)=-\omega C{V}_0\sin \left(\omega t\right)$

Therefore, $I_{\max }=\omega C{V}_0=\frac{1}{\sqrt{LC}}C{V}_0=\sqrt{\frac{C}{L}}{V}_0=\sqrt{\frac{{10}^{-5}}{4\times {10}^{-3}}}\left(60\right)=3A$

Alternative solution

Mathematically, an LC circuit is exactly like a spring-block oscillator. The charge on the capacitor is analogous to the displacement of the block (and hence the elongation of the spring), and the current in the circuit is analogous to the velocity of the block. Therefore, connecting a charged capacitor to an inductor, and closing the switch with the current initially zero is exactly like releasing a block from rest with the spring elongated by a certain distance. So, just like for a spring-block oscillator, the speed of the block (or its kinetic energy) is maximum when the elongation in the spring is zero, the current in an LC circuit (or the energy stored in the inductor, $\frac{1}{2}L{I}^2$ ) is maximum when the capacitor is completely uncharged.

Therefore, by energy conservation $\frac{1}{2}L{I}_{\max }^2=\frac{1}{2}\frac{Q_{\max }^2}{C}\Rightarrow I_{\max }=\frac{Q_{\max }}{\sqrt{LC}}=\frac{C{V}_0}{\sqrt{LC}}=\sqrt{\frac{C}{L}}{V}_0=3A$