The value of current through the $5 \Omega$ resistor of the given circuit is
- $\frac{1}{25} \mathrm{~A}$
- $\frac{2}{25} \mathrm{~A}$
- $\frac{2}{23} \mathrm{~A}$
- $\frac{1}{23} \mathrm{~A}$
Solution
By KVL $ \begin{aligned} & 4 \mathrm{i}+8 \mathrm{i}_1-5=0 [In smaller loop]\\ & \Rightarrow 4 \mathrm{i}_1+4 \mathrm{i}_2+8 \mathrm{i}_1-5=0 \\ & \Rightarrow 12 \mathrm{i}_1+4 \mathrm{i}_2=5 ...(i)\\ & \text { and, } 5 \mathrm{i}_2+3-8 \mathrm{i}_1=0 [In bigger loop]\\ & 5 \mathrm{i}_2-8 \mathrm{i}_1=-3 \\ & 5 \mathrm{i}_2-8\left(\frac{5-4 \mathrm{i}_2}{12}\right)=-3 \quad[\text { from }(\mathrm{i})] \\ & \end{aligned} $ $\begin{aligned} & 5 \mathrm{i}_2-\frac{2}{3}\left(5-4 \mathrm{i}_2\right)=-3 \\ & 5 \mathrm{i}_2-\frac{10}{3}+\frac{8}{3} \mathrm{i}_2=-3 \\ & \frac{23}{3} \mathrm{i}_2=\frac{1}{3} \\ & \mathrm{i}_2=\frac{1}{23} \Rightarrow \text { Current through } 5 \Omega=\frac{1}{23} \mathrm{~A}\end{aligned}$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)