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In an electric circuit, a call of certain emf provides a potential difference of $1.25 \mathrm{~V}$ across a load resistance of $5 \Omega$ . However, it provides a potential difference of $1 \mathrm{~V}$ across a load resistance of $2 \Omega$ .
The emf of the cell is given by $\frac{x}{10} V$ . Then the value of $x$ is__________.

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$I=\frac{\varepsilon }{\left ( R+r \right )}$

When $R=R_{1}=5\Omega$ $,$ $IR=1.25$

$I=I_{1}=0.25A$

When $R=R_{2}=2 \Omega$, IR $=1 \mathrm{~V}$

$I=I_{2}=0.5A$

$I_{1}=\frac{\varepsilon }{\left ( 5+r \right )}=0.25$

$I_{2}=\frac{\varepsilon}{(2+r)}=0.5$

$\frac{\varepsilon }{\left ( 2+r \right )}=\frac{2\varepsilon }{5+r}$

$5+r=4+2r$

$r=1\Omega$

$\varepsilon =1.5v=\frac{15}{10}\: V$

$\therefore x=15$

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