In an electric circuit,a cell of certain emf | vedclass

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(B) The terminal voltage $V$ across a load resistance $R$ is given by the formula $V = \frac{E \cdot R}{R + r}$,where $E$ is the $emf$ and $r$ is the

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$15$

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(B) The terminal voltage $V$ across a load resistance $R$ is given by the formula $V = \frac{E \cdot R}{R + r}$,where $E$ is the $emf$ and $r$ is the internal resistance of the cell.For the first case: $1.25 = \frac{E \cdot 5}{5 + r} \implies 1.25(5 + r) = 5E \implies 6.25 + 1.25r = 5E \implies 1.25 + 0.25r = E \dots (i)$For the second case: $1 = \frac{E \cdot 2}{2 + r} \implies 2 + r = 2E \implies 1 + 0.5r = E \dots (ii)$Equating $(i)$ and $(ii)$:$1.25 + 0.25r = 1 + 0.5r$$0.25 = 0.25r \implies r = 1\, \Omega$Substituting $r = 1$ into equation $(ii)$:$E = 1 + 0.5(1) = 1.5\, V$Given $E = \frac{x}{10}\, V$,we have:$1.5 = \frac{x}{10} \implies x = 15$.

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