Question

A capacitor of capacitance C is given a charge $q_o$. At time $t=0$, it is connected to a battery of emf E through a resistance R. Find the charge on the capacitor at time t.

Answer 1

Using Kircho$\iint$ Law

$iR+\frac{q}{C}=\epsilon (i=\frac{dq}{dt})\frac{dqR}{at}+\frac{q}{C}=\epsilon \frac{dqR}{dt}=\epsilon -\frac{q}{C}{\int }_{q0}^q\frac{dq}{\epsilon -\frac{q}{C}}={\int }_{t=0}^t\frac{at}{R}-C\ln (\frac{ϵ-\frac{q}{c}}{\epsilon -\frac{q_0}{C}})=+\frac{t}{R}\ln \frac{(\epsilon -\frac{q}{C})}{\epsilon -\frac{q_0}{C}}=-\frac{t}{RC}\frac{\epsilon -\frac{q}{C}}{\epsilon -\frac{q_0}{C}}=e^{-\frac{1}{RC}}\epsilon -\frac{q}{C}=(\epsilon -\frac{q_0}{C})e^{\frac{-t}{RC}}\epsilon -\epsilon e^{\frac{-t}{RC}}+\frac{q_0}{C}{e}^{-\frac{t}{RC}}=\frac{q}{C}C\epsilon (1-e^{\frac{-t}{RC}})+q_0{e}^{\frac{-t}{RC}}=q$