Question

In the circuit shown, the switch is shifted from position $1\to 2$ at t = 0. The switch was initially in position 1 since a long time. The graph between charge on capacitor C and time 't' is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$V_0+2E-2iR-\frac{q}{C}=V_0$
$i=\frac{E}{R}-\frac{q}{2RC}$
Also, $i=\frac{dq}{dt}$
$\therefore {\int }_{CE}^q\frac{dq}{(\frac{E}{R}-\frac{q}{2RC})}={\int }_0^{t}dt$
$\frac{ln(\frac{E}{R}-\frac{q}{2RC}){|}_{CE}^q}{(-\frac{1}{2RC})}=t$
In $\left(\frac{2EC-q}{EC}\right)=-\frac{t}{2RC}$
$q=2EC-EC{e}^{-\frac{t}{2RC}}$
$q=EC(2-e^{-\frac{t}{2RC}})$

Why this question? Caution: The lower limit of charge is not always zero.