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Q.
In the circuit shown in figure, the potential difference across $3\, \Omega$ is
Current Electricity
Solution:
The resistance of $A B C D$ is $R=\frac{6 \times 3}{6+3}+2=4\, \Omega$
As resistance of arms $A B C D$ and $H G$ is same,
hence current in arm $A B, I=1 A$
At junction $B, 1=I_{1}+I_{2}$...(i)
Potential difference across $A$ and $B=6 I_{1}=3 I_{2}$
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 $3 \Omega=\frac{2}{3} A \times 3 \Omega=2\, V$
