Question

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

Answer 1

Time constant of the circuit is,
${\tau }_C=CR=(0.5\times {10}^{-6})\left(500\right)=2.5\times {10}^{-4}s$
For $t\le 250\mu si=i_o{e}^{-t{\tau }_C}$
Here $i_o=\frac{20}{500}=0.04A$
$\therefore i=\left(0.04e^{-4000t}\right)amp$
At $t=250\mu s=2.5\times {10}^{-4}s$
$i=0.04e^{-1}=0.015amp$
At this moment PD across the capacitor
$V_c=20(1-e^{-1})=12.64V$
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 position 2 current will reverse its direction, with maximum current
$i_o^{\prime }=\frac{40+12.64}{500}=0.11A$
Now it will decrease exponentially to zero.
For $t\ge 250\mu s$
$i=i_o^{\prime }{e}^{-t{\tau }_C}=-0.11e^{-4000t}$
The i-t graph is as shown in figure.