A cell of emf varepsilon and internal resista | vedclass

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(A) The terminal voltage $V$ of a cell is given by the formula:$V = \varepsilon - Ir$Since the current $I = \frac{\varepsilon}{R+r}$,we substitute thi

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(A) The terminal voltage $V$ of a cell is given by the formula:$V = \varepsilon - Ir$Since the current $I = \frac{\varepsilon}{R+r}$,we substitute this into the equation:$V = \varepsilon - \left( \frac{\varepsilon}{R+r} ight) r$$V = \varepsilon \left( 1 - \frac{r}{R+r} ight) = \varepsilon \left( \frac{R+r-r}{R+r} ight) = \frac{\varepsilon R}{R+r}$To analyze the graph of $V$ versus $R$:$1$. When $R = 0$,$V = 0$.$2$. As $R 	o \infty$,$V 	o \varepsilon$.$3$. The derivative $\frac{dV}{dR} = \frac{\varepsilon(R+r) - \varepsilon R}{(R+r)^2} = \frac{\varepsilon r}{(R+r)^2}$,which is always positive,meaning $V$ increases with $R$.$4$. The second derivative $\frac{d^2V}{dR^2} = -\frac{2\varepsilon r}{(R+r)^3}$,which is negative,meaning the graph is concave down.This behavior corresponds to the curve shown in option $A$.

$A$ cell of emf $\varepsilon$ and internal resistance $r$ is connected across a variable load resistance $R$. The graph drawn between its terminal voltage $V$ and resistance $R$ is

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