Question

In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the $2 \Omega$ resistor is

• A. 0.9 A
• B. 1.5 A
• C. 0.6 A
• D. 1.2 A

Answer 1

Answer: (A)

Here the capacitor will act as an open circiut so this branch will not have any current in it. The equivalent circuit becomes as shown below
$\therefore $ $R_{eq} = 2.8 + \frac{2 \times 3}{2 + 3} = 4 \Omega$
I = $\frac{6}{4} $ 1.5 A
Current through 2$\Omega$ resistor = $\frac{1.5}{(2 + 3)} \times 3 = 0.9 A$