Question

In the circuit shown in the figure, the switch $S$ is initially open and the capacitor is initially uncharged. $I_1,I_2$ and $I_3$ represent the current in the resistance $2\Omega ,4\Omega$ and $8\Omega$ respectively.

• A. Just after the switch $S$ is closed, $I_1=3A,I_2=3A$ and $I_3=0$.
• B. Just after the switch $S$ is closed, $I_1=3A,I_2=0$ and $I_3=0$.
• C. Long time after the switch $S$ is closed, $I_1=0.6A,I_2=0$ and $I_3=0$.
• D. Long after the switch $S$ is closed, $I_1=I_2=I_3=0.6A$.

Answer 1

The correct option is B Just after the switch $S$ is closed, $I_1=3A,I_2=0$ and $I_3=0$.

Just after the switch S is closed, the capacitors will start charging through resistor $2\Omega$ and no current will pass through the resistors $8\Omega$ and $4\Omega$.
Thus, $I_1=\frac{6}{2}=3A,I_2=0$ and $I_3=0$.
Long time after the switch S is closed, charging of the capacitors will stop and current will pass through resistors $8\Omega$ and $4\Omega$.
Applying Kirchhoff's law for loop 1 and 2,
$2I_1+8I_2=6\Rightarrow I_1+4I_2=3.....\left(1\right)$
and $-2I_2+4I_3=0...\left(2\right)$
also $I_1=I_2+I_3=I_2+\left(1/2\right)I_2=\left(3/2\right)I_2...\left(3\right)$
From (1) and (3), $\left(2/3\right)I_2+4I_2=3\Rightarrow I_2=0.6A$
Using (3), $I_1=\left(3/2\right)0.6=0.9A$ and $I_3=0.9-0.6=0.3A$