Question

Taking the internal resistance of the battery as negligible, the steady state current in the $2\Omega$ resistor shown in the figure will be

• A. 1.8 A
• B. 2.9 A
• C. 0.9 A
• D. 2.8 A

Answer 1

The correct option is C 0.9 A

At steady state capacitor act as open circuit therefore current will not flow in that circuit

Now 2 ohm and 3 ohm are in parallel then equivalent resistance is

$R_{AB}=\frac{2\times 3}{2+3}=\frac{6}{5}\Omega$

Total equivalent circuit is $R_{eq}=\frac{6}{5}+2.8=4\Omega$

$I=\frac{V}{R}=\frac{6}{4}=\frac{3}{2}A$

Voltage across $AB$ is

$V=\frac{3}{2}\times \frac{6}{5}=1.8V$

Current flowing through 2 ohm is,

$I_2=\frac{V}{R}=\frac{1.8}{2}=0.9A$

Option (C) is correct option