A cell of emf E has an internal resistance r | vedclass

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(D) The potential difference $V$ across the rheostat of resistance $R$ is given by the formula:$V = IR = \left( \frac{E}{R+r} \right) R = \frac{E}{1 +

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(D) The potential difference $V$ across the rheostat of resistance $R$ is given by the formula:$V = IR = \left( \frac{E}{R+r} ight) R = \frac{E}{1 + \frac{r}{R}}$Analyzing the behavior of $V$ with respect to $R$:$1$. When $R = 0$,$V = 0$.$2$. As $R$ increases,the term $\frac{r}{R}$ decreases,so $V$ increases.$3$. As $R 	o \infty$,the term $\frac{r}{R} 	o 0$,so $V 	o E$.The graph starts from the origin $(0,0)$ and asymptotically approaches the value $E$ as $R$ increases. This corresponds to an upward convex curve. Thus,the correct graph is shown in option $D$.

$A$ cell of emf $E$ has an internal resistance $r$ and is connected to a rheostat. When the resistance $R$ of the rheostat is changed,which is the correct graph of the potential difference $V$ across it?

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