Explanation

(a) When a cell of emf E is connected to a resistance of 3.9$\Omega$, then the emf E of the cell remains constant, while voltage V goes on decreasing on taking more and more current from the cell.
∴ $V = E - ir$
Where, r is internal resistance.
Also, $current\ i = \frac{E}{R + r}$
$\therefore\ \ \ \ \ \ \ \ \ V = E - \left( \frac{E}{R + r} \right)r\ $
Putting the numerical values, we have
$E = 2V,\ r = 0.1\Omega,\ R = 3.9\Omega$
$V = 2 - \left( \frac{2}{3.9 + 0.1} \right) \times 0.1$
$V = 1.95\ Volt$