In an electrical circuit, a cell of certain emf provides a potential difference of 1.25V across a load resistance of $5 Ω .$ However it provides a potential difference of 1V across a load resistance of $2 Ω .$ The emf of the cell is given by $\frac{x}{10} V .$ Then the value of ‘ $x$ ’ is

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answer is 15.

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Detailed Solution

$\begin{array}{l} i = \frac{E}{R + r} \\ V = i R \\ V = \frac{E R}{R + r} \\ 1.25 = \frac{E \times 5}{5 + r} .......... (1) \\ 1 = \frac{E (2)}{2 + r} ............ (2) \\ f r o m 1 & 2 \\ E = \frac{3}{2} = \frac{15}{10} \\ x = 15 \end{array}$

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