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:$V = I \cdot R$Since the circuit consists of a cell wit

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(D) The potential difference $V$ across the rheostat of resistance $R$ is given by the formula:$V = I \cdot R$Since the circuit consists of a cell with $emf$ $E$ and internal resistance $r$ in series with the rheostat $R$,the current $I$ is:$I = \frac{E}{R + r}$Substituting $I$ into the expression for $V$:$V = \left( \frac{E}{R + r} ight) R = E \left( \frac{R}{R + r} ight)$We can rewrite this as:$V = E \left( \frac{R + r - r}{R + r} ight) = E \left( 1 - \frac{r}{R + r} ight)$As $R$ increases from $0$ to $\infty$:$1$. When $R = 0$,$V = 0$.$2$. As $R 	o \infty$,$V 	o E$.$3$. The derivative $\frac{dV}{dR} = E \left( \frac{(R+r) - R}{(R+r)^2} ight) = \frac{Er}{(R+r)^2}$,which is always positive,meaning $V$ is an increasing function.The graph that represents this behavior,where $V$ starts at $0$ and approaches $E$ as $R$ increases,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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