When a cell is connected to a resistance of 2 | vedclass

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(B) Let the $emf$ of the cell be $E$ and its internal resistance be $r$. The formula for current is $I = \frac{E}{R + r}$.For the first case: $0.5 = \

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$1.5$

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(B) Let the $emf$ of the cell be $E$ and its internal resistance be $r$. The formula for current is $I = \frac{E}{R + r}$.For the first case: $0.5 = \frac{E}{2 + r} \implies E = 0.5(2 + r) = 1 + 0.5r$  --- $(1)$For the second case: $0.25 = \frac{E}{5 + r} \implies E = 0.25(5 + r) = 1.25 + 0.25r$  --- $(2)$Equating $(1)$ and $(2)$:$1 + 0.5r = 1.25 + 0.25r$$0.5r - 0.25r = 1.25 - 1$$0.25r = 0.25 \implies r = 1\,\Omega$Substituting $r = 1$ in equation $(1)$:$E = 1 + 0.5(1) = 1.5\,V$.Thus,the $emf$ of the cell is $1.5\,V$.

When a cell is connected to a resistance of $2\,\Omega$,the current flowing is $0.5\,A$. When it is connected to a resistance of $5\,\Omega$,the current flowing is $0.25\,A$. What is the $emf$ of the cell in $volt$?

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