Question

The switch $S$ shown in figure is kept closed for a long time and is then opened at $t=0$. Find the current in the middle $10\Omega$ resistor at $t=1.0s$.

Answer 1

Given,

$t_1=0s,R=10\Omega ,t_2=1.0s$

So in steady state condition current passes through $25\mu f$ Capistor,

$Netresistance=\frac{100}{2}=50\Omega$

Net current is $\frac{12}{6}=2$

So the PD across the capacitor of $10\Omega$ is $\frac{12}{5}\times 10=24V$

$q=Q\left(e^{-\frac{t}{rc}}\right)=v\times \left(e^{-\frac{t}{rc}}\right)$

$24\times 25\times {10}^{-6}\left[\frac{e^{-1\times {10}^{-3}}}{10\times 25\times {10}^4}\right]=24\times 25\times {10}^{-6}\times 0.0183=10.9\times {10}^{-6}$

Now the current in the $10\Omega$ resistance is:

$=\frac{10.9\times {10}^{-6}C}{1\times {10}^{-3}}=11mA$