Question

In the arrangement shown in figure, the current through $5\Omega$ resistor is

• A. $2A$
• B. Zero
• C. $\frac{12}{7}A$
• D. $1A$

Answer 1

The correct option is A $2A$

Using the Kirchhoff's voltage law

for loop ABCDA, $2i_1+5(i_1+i_2)=12$ or $7i_1+5i_2=12...\left(1\right)$

and for loop EBCFE, $2i_2+5(i_1+i_2)=12$ or $5i_1+7i_2=12...\left(2\right)$

Now, $\left(1\right)\times 7,\left(2\right)\times 5\Rightarrow 49i_1+35i_2=84...\left(3\right)$ and

$25i_1+35i_2=60...\left(4\right)$

$\left(3\right)-\left(4\right)\Rightarrow 24i_1=24$ or $i_1=1A$

Substituting the value of $i_1$ in $\left(1\right)$ we get, $i_2=\frac{12-7}{5}=1A$

Thus the current through $5\Omega =i_1+i_2=1+1=2A$