AP EAMCET · PHYSICS · Current Electricity
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
- A \(\frac{(\alpha-1) \gamma}{4 \beta(\beta+\gamma)} \frac{\mathrm{V}}{\mathrm{R}}\)
- B \(\frac{(\alpha-1)}{4 \beta} \frac{\mathrm{V}}{\mathrm{R}}\)
- C \(\frac{(\alpha-1) \beta}{2 \gamma(\beta+\gamma)} \frac{\mathrm{V}}{\mathrm{R}}\)
- D \(\frac{(\alpha-1)(\beta+\gamma)}{2 \beta \gamma} \frac{\mathrm{V}}{\mathrm{R}}\)
Answer & Solution
Correct Answer
(A) \(\frac{(\alpha-1) \gamma}{4 \beta(\beta+\gamma)} \frac{\mathrm{V}}{\mathrm{R}}\)
Step-by-step Solution
Detailed explanation
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:…
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