Question

In the circuit shown in figure, the potential difference across $3\Omega$ is

• A. 2 V
• B. 4 V
• C. 8 V
• D. 16 V

Answer 1

Answer: (A)

The resistance of $ABCD$ is $R=\frac{6\times 3}{6+3}+2=4\Omega$
As resistance of arms $ABCD$ and $HG$ is same,
hence current in arm $AB,I=1A$
At junction $B,1=I_1+I_2$...(i)
Potential difference across $A$ and $B=6I_1=3I_2$
or $I_2=2I_2$...(ii)
Solving (i) and (ii), we get, $I_1=\frac{1}{3}A,I_2=\frac{2}{3}A$
Potential difference across $3\Omega =\frac{2}{3}A\times 3\Omega =2V$