Question

At $t=0,$ capacitor is uncharged. The current through capacitor as function of time would be.

• A. $\frac{2V}{R}{e}^{\frac{-2t}{RC}}$
• B.
  $\frac{V}{R}{e}^{\frac{2t}{3RC}}$
• C. $\frac{V}{R}{e}^{\frac{t}{CR}}$
• D. $\frac{3V}{2R}{e}^{\frac{2t}{3RC}}$

Answer 1

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

In loop 1,

$v-I_{1}R-\frac{Q}{c}=0$

In loop 2,

$v-I_{2}R-\frac{Q}{c}=0$

Adding both equations, $2v-(I_1+I_2)R-\frac{Q}{c}=0$

$I_1+I_2$ is charging the capacitor

$\therefore \frac{dQ}{dt}=I_1+I_2$

$2v+\frac{dQ}{dt}R-\frac{Q}{c}=0$

$2v-\frac{Q}{c}=-\frac{RdQ}{dt}$

$2v-\frac{Q}{Rc}=-\frac{dQ}{dt}$

$2cv-Q=Rc\frac{dQ}{dt}$

${\log }_e\left(\frac{2cv-Q}{2cv}\right)=-\frac{t}{Rc}$

$2cv-Q=2cv{e}^{-t/Rc}$

$Q=2cv[1-e^{-t/Rc}]$

$I=\frac{dQ}{dt}=\frac{2cv}{Rc}{e}^{-t/Rc}=\frac{2v}{R}{e}^{-t/Rc}$