A cell having an emf varepsilon and internal | vedclass

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(C) The current in the circuit is given by $I = \frac{\varepsilon}{R+r}$.The potential difference $V$ across the external resistance $R$ is $V = IR =

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(C) The current in the circuit is given by $I = \frac{\varepsilon}{R+r}$.The potential difference $V$ across the external resistance $R$ is $V = IR = \left( \frac{\varepsilon}{R+r} ight) R$.This can be rewritten as $V = \frac{\varepsilon}{1 + \frac{r}{R}}$.Analyzing the limits:$1$. When $R = 0$,$V = 0$.$2$. As $R 	o \infty$,the term $\frac{r}{R} 	o 0$,so $V 	o \varepsilon$.As $R$ increases,$V$ increases from $0$ and approaches the value $\varepsilon$ asymptotically. This corresponds to the graph where the curve starts at the origin and levels off at $V = \varepsilon$.

$A$ cell having an emf $\varepsilon$ and internal resistance $r$ is connected across a variable external resistance $R$. As the resistance $R$ is increased,the plot of potential difference $V$ across $R$ is given by:

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