A cell of emf E and internal resistance r is connected

We know that $I=\frac{E}{R+r}$ and V=IR.

So $V=\frac{E R}{R+r} \ldots($ (I)

$V=\frac{E}{1+\frac{\gamma}{R}} \ldots$ (II)

Here $\mathrm{E}, \mathrm{r}$ are constants. So

$V \propto \frac{1}{1+\frac{\gamma}{R}}($ from II $)$

and $I$ oc $R$ (from I)

With increase in $\mathrm{R}$, P.D. across $\mathrm{R}$ is increased upto maximum value E.