Question

In the figure shown below, find the current as function of time $t$ on capacitor $C_1$. Initially $C_1$ is uncharged and $C_2$ is charged to a potential of $2E$.

• A. $i=\frac{3E}{R}{e}^{\left(\frac{-2t}{RC}\right)}$
  
• B. $i=\frac{E}{R}{e}^{\left(\frac{-t}{RC}\right)}$
  
  
• C. $i=\frac{E}{R}{e}^{\left(\frac{-2t}{RC}\right)}$
  
• D. $i=\frac{3E}{R}{e}^{\left(\frac{-t}{RC}\right)}$
  

Answer 1

The correct option is A $i=\frac{3E}{R}{e}^{\left(\frac{-2t}{RC}\right)}$


Charge distribution on each plate is shown in the figure below:


Using, $\text{KVL}$ to the given loop we get,

$-\frac{q}{C}+\frac{2CE-q}{C}-iR+E=0$

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

$\Rightarrow 3EC=2q+\frac{dq}{dt}RC$

$\Rightarrow \frac{dq}{3EC-2q}=\frac{dt}{RC}...\left(1\right)$

Integrating eq $\left(1\right)$ with proper limits we get,

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

$\Rightarrow [-\frac{\ln (3EC-2q)}{2}{]}_0^q=\frac{t}{RC}$

$\Rightarrow [\ln (3EC-2q){]}_0^q=-\frac{2t}{RC}$

$\Rightarrow \ln \left(\frac{3EC-2q}{3EC}\right)=-\frac{2t}{RC}$

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

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

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

So, the current at time $t$,

$i=\frac{dq}{dt}=\frac{3E}{R}{e}^{-\left(\frac{2t}{RC}\right)}$

Hence, option (a) is the correct answer.