In the following circuit, the current $I_3$ is

Question

Aicharya · Accepted Answer

Applying Kirchhoff's voltage law in loop 1 and 2 $\begin{array}{ll} \therefore & 28 \mathrm{I}_1=-6-8=-\frac{1}{2} \mathrm{~A} \\ \therefore & 54 \mathrm{I}_2=-6-12=-\frac{1}{3} \mathrm{~A} \\ \therefore & \mathrm{I}_3=\mathrm{I}_1+\mathrm{I}_2=-\frac{1}{2}+\left(-\frac{1}{3}\right)=-\frac{5}{6} \mathrm{~A} \end{array}$

In the following circuit, the current $I_3$ is

Solution