Question

The capacitor in the circuit is initially uncharged. Find out current $I_1,I_2,$ and $I_3$ just after the switch is closed.

• A. $I_1=\frac{2E}{3R},I_2=\frac{E}{3R},I_3=\frac{E}{3R}$
• B. $I_1=\frac{E}{3R},I_2=\frac{E}{3R},I_3=\frac{E}{3R}$
• C. $I_1=\frac{E}{3R},I_2=\frac{2E}{3R},I_3=\frac{E}{3R}$
• D. $I_1=\frac{E}{3R},I_2=\frac{2E}{3R},I_3=\frac{2E}{3R}$

Answer 1

The correct option is A $I_1=\frac{2E}{3R},I_2=\frac{E}{3R},I_3=\frac{E}{3R}$

At $t=0$, the capacitor behaves as short-circuit.
So, the circuit diagram will look like,
Considering the potential at different points applying Kirchhoff's first law at point $X$,

$\frac{E-V_X}{R}=\frac{V_X-0}{R}+\frac{V_X-0}{R}$

$\Rightarrow \frac{3V_X}{R}=\frac{E}{R}$

$\Rightarrow V_X=\frac{E}{3}$

So,

$I_1=\frac{E-\frac{E}{3}}{R}=\frac{2E}{3R}$

$I_2=\frac{\frac{E}{3}-0}{R}=\frac{E}{3R}$,

$I_3=\frac{E}{3R}$

Hence, option $\left(A\right)$ is the correct answer.