Question

In the circuit shown in figure $R_1=R_2=6R_3=300M\Omega ,C=0.01\mu F$ and $E=10V$. The switch is closed at $t=0$. Find
(a) Charge on capacitor as a function of time.
(b) Energy of the capacitor at $t=20s$

Answer 1

Charge on capacitor is given by $Q=CV(1-e^{-t/RC})...\left(i\right)$

Here R is the equivalent capacitance which can be calculated from figure

The equivalent resistance is $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...\left(ii\right)$

Substituting values we get $R=37.5\Omega$

Thus charge on capacitor by (i) is

$Q=1\times {10}^{-8}\times 10\times (1-e^{-t\times {10}^8/37.5})$

i.e. $Q=1\times {10}^{-7}\times (1-e^{-2.6t\times {10}^6})$

Energy stored in capacitor is given by $U=0.5QV$

For t = 20 we get charge stored Q = C V = $1\times {10}^{-7}$

Thus energy stored in capacitor is given by $U=0.5\times {10}^{-7}=5\times {10}^{-8}$ Joules