Question

A capacitor of capacity $2\mu F$ is charged to a potential difference of 12 V. It is then connected across an inductor of inductance $6\mu H$. What is the current (in A) in the circuit at a time when the potential difference across the capacitor is 6.0 V?

Answer 1

Initial energy stored in the capacitor,
$\begin{array}{}{U}_i=\frac{1}{2}C{V}_o^2=\frac{1}{2}\times 2\times {10}^{-6}\times (12{)}^2& \text{(1)}\end{array}$
Final energy, $U_f=$ Energy stored in capacitor + energy stored in inductor

$\Rightarrow U_f=\frac{1}{2}C{V}^2+\frac{1}{2}L{I}^2$
$\begin{array}{}{U}_f=\frac{1}{2}\times 2\times {10}^{-6}\times (6{)}^2+\frac{1}{2}\times 6\times {10}^{-6}\times I^2& \text{(2)}\end{array}$
Applying the conservation of energy and equating $\left(1\right)$ and $\left(2\right)$,
$U_i=U_f$
$\frac{1}{2}\times 2\times {10}^{-6}\times (12{)}^2=\frac{1}{2}\times 2\times {10}^{-6}\times (6{)}^2+\frac{1}{2}\times 6\times {10}^{-6}\times I^2$
$\Rightarrow I=6A$