In the given circuit, the switch is closed in the position $1$ at $t=0$ and then moved to $2$ after $250\mu \text{s}$. Derive an expression for current as a function of time for $t>0$. Also plot the variation of current with time.

Time constant of the circuit is 

${\tau }_C=CR=\left(0.5\times {10}^{-6}\right)\left(500\right)=2.5\times {10}^{-4}$

For $t\le 250\mu s,i=i_0{e}^{-t/{\tau }_C}$

Here, $i_0=\frac{20}{500}=0.04 \mathrm{A}$

$\therefore i=\left(0.04e^{-4000t}\right)\mathrm{amp}$

At $t=250\mu \mathrm{s}=2.5\times {10}^{-4} \mathrm{s}$

$i=0.04e^{-1}=0.015\mathrm{amp}$

At this moment PD across the capacitor

$V_C=20\left(1-e^{-1}\right)=12.64 \mathrm{V}$

So when the switch is shifted to position $2$ the current in the circuit is $0.015 A$ (clockwise) and $PD$ across capacitor is $12.64 V.$

As soon as the switch is shifted to the position $2$ current will reverse its direction with maximum current. 

$i_0^{\text{'}}=\frac{40+12.64}{500}=0.11 \mathrm{A}$

Now it will decrease exponentially to zero.

For $t\ge 250\mu s$

$i=-i_0^{\text{'}}{e}^{-t/\tau }{}^{\text{'}}C=-0.11e^{-4000t}$

The $(i-t)$graph is as shown in figure,