Question

Find the current (in ampere) through the capacitor when the switch $S$ is closed at $t=0$.

Answer 1

Here,

$\tau =R_{eq}C=\frac{RC}{2}$

Where,$R=3\Omega ,C=1\mu F$

$q=q_0(1-e^{\frac{-t}{\tau }})$

$q_0=\frac{CV}{2}=3\mu C$

Current at $t=0$,

$I_0=\frac{q_0}{\tau }=\frac{2\times 3\mu C}{3\Omega \times 1\mu C}$

$I_0=2A$