Two sources of equal emf are connected to an external resistance R. The internal resistance of the two…

Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are $R_1$ and $R_2\left(R_2>R_1\right)$. If the potential difference across the source having internal resistance $\mathrm{R}_2$ is zero, then
  1. $\left.\mathrm{R}=\mathrm{R}_2 \times\left(\mathrm{R}_1+\mathrm{R}_2\right) / \mathrm{R}_2-\mathrm{R}_1\right)$
  2. $\mathrm{R}=\mathrm{R}_2-\mathrm{R}_1$
  3. $\mathrm{R}=\mathrm{R}_1 \mathrm{R}_2 /\left(\mathrm{R}_1+\mathrm{R}_2\right)$
  4. $\mathrm{R}=\mathrm{R}_1 \mathrm{R}_2 /\left(\mathrm{R}_2-\mathrm{R}_1\right)$

Solution

$\mathrm{I}=\frac{2 \mathrm{E}}{\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}}$ $\mathrm{E}-\mathrm{R}_2 \mathrm{I}=0$ $\Rightarrow \mathrm{R}=\mathrm{R}_2-\mathrm{R}_1$

Asked in: JEE Main 2005

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