In an electric circuit, a call 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 $$\frac{x}{10}$$ V. Then the value of $$x$$ is ___.

For a cell with emf $$E$$ and internal resistance $$r$$, the potential difference across a load resistance $$R$$ is $$V = E - Ir = E - \dfrac{V}{R}\cdot r$$, giving $$V = \dfrac{ER}{R+r}$$.

From the first observation ($$V_1 = 1.25\,\text{V}$$, $$R_1 = 5\,\Omega$$): current $$I_1 = 1.25/5 = 0.25\,\text{A}$$, so:

$$E = 1.25 + 0.25r \quad \cdots (1)$$

From the second observation ($$V_2 = 1.0\,\text{V}$$, $$R_2 = 2\,\Omega$$): current $$I_2 = 1.0/2 = 0.5\,\text{A}$$, so:

$$E = 1.0 + 0.5r \quad \cdots (2)$$

Subtracting equation (1) from (2):

$$0 = -0.25 + 0.25r \implies r = 1\,\Omega$$

Substituting back: $$E = 1.25 + 0.25 \times 1 = 1.5\,\text{V}$$.

The emf is given as $$\dfrac{x}{10}\,\text{V}$$, so $$\dfrac{x}{10} = 1.5$$, giving $$x = 15$$.