Question

In the figure shown, the capacity of the condenser $C$ is $2μF$ . The current in $2\Omega$ resistor is

• A. $9 \, A$
• B. $0.9 \, A$
• C. $\frac{1}{9} \, A$
• D. $\frac{1}{0 .9} \, A$

Answer 1

Answer: (B)

Resistance in uppermost branch $R\text{ = }\frac{2 \times 3}{2 + 3}=\frac{6}{5}=1.2\Omega$
After a long time, branch with the capacitor can be considered open circuited
Hence,
$R_{eq}= 2.8+1.2 = 4\Omega$
$I\text{= }\frac{V}{R}=\frac{6}{4}=\frac{3}{2}A$
$I_{2 \Omega}\text{= }\frac{I R_{2}}{R_{1} + R_{2}}=\frac{\frac{3}{2} \times 3}{5}A$
$I_{2 \Omega}\text{ = }\frac{9}{10}=0.9\text{ A}$