Question

In the circuit shown in figure $E_1$ = 7V, $E_2=7V$ $R_1=R_2=1\Omega$ and $R_3=3$ respectively. The current through the resistance $R_3$ is:

• A. 2A
• B. 3.5A
• C. 1.75A
• D. none of these

Answer 1

The correct option is A 2A

Using Kirchoff's loop law in loop BCFAB-

$i_1{R}_1+(i_1+i_2)R_3=E_1$

$⟹i_1+3(i_1+i_2)=7$

$⟹4i_1+3i_2=7$

Now, applying Kirchoff's loop law in loop BCDEFAB-

$i_1{R}_1-i_2{R}_2+E_2=E_1$

$⟹i_1-i_2+7=7$

$⟹i_1=i_2$

Hence, in equation-

$4i_1+3i_2=7$

$⟹4i_1+3i_1=7$

$⟹i_1=1A$

and $i_2=i_1=1A$

Current through $R_3$ is $i_1+i_2=2A$

Answer-(A)