Question:
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 $\frac{x}{10} \mathrm{~V}$. Then the
value of $x$ is__________
Solution:
Terminal voltage $v=i R=\frac{E R}{R+r}$
$1^{\mathrm{st}} \rightarrow 1.25=\frac{E(5)}{5+r}$ .......(I)
$2^{\text {nd }} \rightarrow 1=\frac{E(2)}{2+r} \ldots$ (ii)
By (i) and (ii)
$\mathrm{r}=1 \Omega, \mathrm{E}=\frac{3}{2} \mathrm{~V}=\frac{15}{10} \mathrm{volt}$
$\Rightarrow x=15$