In the given figure: $\mathrm{V}_1=\mathrm{V}, \mathrm{V}_2=\alpha \mathrm{V}, \mathrm{R}_1=\beta \mathrm{R}, \mathrm{R}_2=\gamma \mathrm{R}$, where $\alpha, \beta$, and $\gamma$ are positive real numbers. The value of current $I$ is

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Aicharya · Accepted Answer

By Mesh law Loop $1: i_1 R_1+\left(i_1-i_2\right) R_2+V_2+i_1 R_1-V_1=0$ $\mathrm{i}_1 \beta \mathrm{R}+\left(\mathrm{i}_1-\mathrm{i}_2\right) \gamma \mathrm{R}+\alpha \mathrm{V}+\mathrm{i}_1 \beta \mathrm{R}-\mathrm{V}=0...(i)$ Loop 2: $i_2 R_1+V_2+i_2 R_1-V_2+\left(i_2-i_1\right) R_2=0$ $\Rightarrow \mathrm{i}_2 \beta \mathrm{R}+\mathrm{V}_2+\mathrm{i}_2 \beta \mathrm{R}-\mathrm{V}_2+\left(\mathrm{i}_2-\mathrm{i}_1\right) \gamma \mathrm{R}=0$ $\Rightarrow 2 \mathrm{i}_2 \beta \mathrm{R}+\left(\mathrm{i}_2-\mathrm{i}_1\right) \gamma \mathrm{R}=0...(ii)$ Solving (i) \& (ii), we get $ \mathrm{i}_2=\frac{-(\alpha-1)}{4 \beta(\beta-\gamma)} \frac{\gamma_{\mathrm{V}}}{\mathrm{R}} \cdot \text { As } \mathrm{I}=-\mathrm{i}_2 $ So, $I=\frac{(\alpha-1) \gamma}{4 \beta(\beta+\gamma)} \frac{V}{R}$

In the given figure: $\mathrm{V}_1=\mathrm{V}, \mathrm{V}_2=\alpha \mathrm{V}, \mathrm{R}_1=\beta \mathrm{R}…

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