Q148 / 200 · WBJEE 2013 · Physics
In the electrical circuit shown in the figure, the current through the $4 \Omega$ resistor is 
- A$1 \mathrm{A}$
- Correct answerB$0.5 \mathrm{A}$
- C$0.25 \mathrm{A}$
- D$0.1 \mathrm{A}$
Solution
The current through the various branches of the circuit will be shown as 
According to Kirchhoff's second law in closed circuit $B C D E B$
$$
\begin{array}{r}
2 I_{1}+4 l_{1}+2 l_{1}-8\left(1-l_{1}\right)=0 \\
\Rightarrow \quad 16 l_{1}-81=0
\end{array}
$$
$\Rightarrow \quad l_{1}=\frac{8 l}{16} \Rightarrow l_{1}=\frac{1}{2} l$
In closed circuit ABEFA
$$
-9+3 l+8\left(1-l_{1}\right)+2 l=0
$$
$\Rightarrow \quad 13l-8 I_{1}=9 \Rightarrow 13 /-8\left(\frac{1}{2} I\right)=9$
$\Rightarrow \quad I=\frac{9}{9}=1 \mathrm{A}$
So current through $4 \Omega$ resistor $l_{1}=\frac{1}{2} \times 1=0.5 \mathrm{A}$
According to Kirchhoff's second law in closed circuit $B C D E B$
$$
\begin{array}{r}
2 I_{1}+4 l_{1}+2 l_{1}-8\left(1-l_{1}\right)=0 \\
\Rightarrow \quad 16 l_{1}-81=0
\end{array}
$$
$\Rightarrow \quad l_{1}=\frac{8 l}{16} \Rightarrow l_{1}=\frac{1}{2} l$
In closed circuit ABEFA
$$
-9+3 l+8\left(1-l_{1}\right)+2 l=0
$$
$\Rightarrow \quad 13l-8 I_{1}=9 \Rightarrow 13 /-8\left(\frac{1}{2} I\right)=9$
$\Rightarrow \quad I=\frac{9}{9}=1 \mathrm{A}$
So current through $4 \Omega$ resistor $l_{1}=\frac{1}{2} \times 1=0.5 \mathrm{A}$Hence, the correct answer is option: 'B'.