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(D) Let $E$ be the electromotive force $(EMF)$ of the cell and $r$ be its internal resistance.When resistance $R_1$ is connected,the current $I_1$ is

Vedclass · Accepted Answer

$\frac{I_2 R_2-I_1 R_1}{I_1-I_2}$

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(D) Let $E$ be the electromotive force $(EMF)$ of the cell and $r$ be its internal resistance.When resistance $R_1$ is connected,the current $I_1$ is given by $I_1 = \frac{E}{R_1 + r}$.When resistance $R_2$ is connected,the current $I_2$ is given by $I_2 = \frac{E}{R_2 + r}$.Dividing the two equations: $\frac{I_1}{I_2} = \frac{R_2 + r}{R_1 + r}$.Cross-multiplying gives: $I_1(R_1 + r) = I_2(R_2 + r)$.Expanding the terms: $I_1 R_1 + I_1 r = I_2 R_2 + I_2 r$.Rearranging to solve for $r$: $I_1 r - I_2 r = I_2 R_2 - I_1 R_1$.$r(I_1 - I_2) = I_2 R_2 - I_1 R_1$.Therefore,$r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2}$.

When a resistance $R_1$ is connected across a cell,the current is $I_1$ and if the resistance $R_1$ is replaced by $R_2$,the current is $I_2$. Then the internal resistance of the cell is

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