Q 12-03-204JEE MainJEE Main 2021 (22 Jul, Shift 1)Easy
In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of $5\ \Omega$. However, it provides a potential difference of 1 V across a load resistance of $2\ \Omega$. The emf of the cell is given by $\dfrac{x}{10}$ V. Then the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 15
$V = \dfrac{ER}{R + r}$: $1.25 = \dfrac{5E}{5 + r}$ and $1 = \dfrac{2E}{2 + r}$.
So $E = 0.25(5 + r) = 0.5(2 + r) \Rightarrow r = 1\ \Omega$ and $E = 1.5$ V $= \dfrac{15}{10}$ V. So $x = 15$.
Solution by Sreeraj P, M.Sc Physics