Physics - EMI/AC Question with Solution | TestHub

When the capacitor is fully discharged, the current is maximum and all the energy resides in the inductor. Hence, the energy of the capacitor is equal to the energy of the inductor.

$\frac{L{\left(i^2\right)}_m}{2}=\frac{{\left(Q^2\right)}_m}{2\mathrm{C}}...(i)$

where $i_m$ is the maximum current and $Q_m$ is the maximum charge.

The given values are

$$\begin{gathered} Q_m=2.7\times 10^{-6}\mathrm{C} \\ L=75\times 10^{-3}\mathrm{H} \\ \mathrm{C}=1.2\times 10^{-6}\mathrm{F} \end{gathered}$$

The maximum current is found from equation (i)

$$\begin{gathered} i_m=\sqrt{\frac{1}{L\mathrm{C}}}{Q}_m \\ \Rightarrow i_m=\frac{2.7\times 10^{-6}}{\sqrt{75\times 10^{-3}\times 1.2\times 10^{-6}}}=9\text{mA} \end{gathered}$$