Question

At $t=0$ the switch $S$ is closed. If initially $C_1$ is uncharged and $C_2$ is charged to potential difference $2E$ then the current in circuit as a function of time will be : (Given, $C_1=C_2=C$)

• A. $\frac{3E}{R}{e}^{-2t/RC}$
  
• B. $\frac{E}{R}{e}^{-3t/RC}$
  
• C. $\frac{2E}{R}{e}^{-2t/RC}$
  
• D. $\frac{E}{R}{e}^{-5t/RC}$
  

Answer 1

The correct option is A $\frac{3E}{R}{e}^{-2t/RC}$


Let the charge flowing in circuit at any time be $q$.


The charges on $-ve$ and $+ve$ plates of capacitor ${\text{C}}_2$ will be $(-2C_2\epsilon +q)$ & $(2C_{2}E-q)$ respectively.

Using $\text{KVL}$ in loop:

$E-\left(\frac{q}{C_1}\right)-iR+\left(\frac{2C_{2}E-q}{C_2}\right)=0$

$\Rightarrow E-\frac{q}{C}-iR+\left(\frac{2CE-q}{C}\right)=0$

$(\because C_1=C_2=C)$

$\Rightarrow 3E-\frac{2q}{C}-iR=0$

$\Rightarrow 3EC-iRC=2q$

As, $i=\frac{dq}{dt}$

$\Rightarrow 3CE-\left(\frac{dq}{dt}\right)RC=2q$

$\Rightarrow \frac{dq}{dt}=\frac{3CE-2q}{RC}$

Integrating on both sides:

${\int }_0^q\frac{dq}{3CE-2q}={\int }_0^t\frac{dt}{RC}$

$\Rightarrow \frac{-1}{2}\left[\ln \right(3CE-2q\left)\right]{|}_0^q=\frac{t}{RC}$

$\Rightarrow \frac{-1}{2}\left[\ln \right(3CE-2q)-\ln (3CE\left)\right]=\frac{t}{RC}$

$\Rightarrow \ln \left[\frac{3CE-2q}{3CE}\right]=\frac{-2t}{RC}$

$\Rightarrow \frac{3CE-2q}{3CE}=e^{\frac{-2t}{RC}}$

$\Rightarrow 3CE-2q=\left(3CE\right)e^{-2t/RC}$

$\Rightarrow 2q=3CE(1-e^{-2t/RC})$

$\Rightarrow q=\frac{3CE}{2}(1-e^{-2t/RC})$

Current in the circuit at time $t$ will be,

$i=\frac{dq}{dt}$

$\Rightarrow i=\frac{3CE}{2}\left[\left(\frac{-2}{RC}\right)\right(-e^{-2t/RC}\left)\right]$

$\Rightarrow i=\frac{3CE}{2}\left[\frac{2}{RC}{e}^{-2t/RC}\right]$

$\therefore i=\frac{3E}{R}{e}^{-2t/RC}$

Hence, option (a) is the correct answer.