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

Question

In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the 2 Ω resistor is (A) 0.9 A (B) 1.5 A (C) 0.6 A (D) 1.2 A

Tardigrade Admin · Accepted Answer

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

∴

R_{e q} = 2.8 + \frac{2 \times 3}{2 + 3} = 4Ω

I =

\frac{6}{4}

1.5 A
Current through 2

Ω

resistor =

\frac{1.5}{( 2 + 3 )} \times 3 = 0.9 A

Q. In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the 2Ω resistor is

0.9 A

1.5 A

0.6 A

1.2 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 ∴ Req​=2.8+2+32×3​=4Ω I = 46​ 1.5 A Current through 2Ω resistor = (2+3)1.5​×3=0.9A

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

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
$∴$ $R_{e q} = 2.8 + \frac{2 \times 3}{2 + 3} = 4Ω$
I = $\frac{6}{4}$ 1.5 A
Current through 2 $Ω$ resistor = $\frac{1.5}{( 2 + 3 )} \times 3 = 0.9 A$