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(C) From the given $V-I$ graph,the electromotive force $(EMF)$ $\varepsilon$ is the intercept on the $V$-axis,so $\varepsilon = 2\,V$.The internal res

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$4$

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(C) From the given $V-I$ graph,the electromotive force $(EMF)$ $\varepsilon$ is the intercept on the $V$-axis,so $\varepsilon = 2\,V$.The internal resistance $r$ is the magnitude of the slope of the graph: $r = \frac{\Delta V}{\Delta I} = \frac{2\,V}{2\,A} = 1\,\Omega$.Let the battery consist of $m$ parallel rows,each containing $n$ cells in series. The total number of cells is $mn = 40$.The current $I$ through an external load $R$ is given by $I = \frac{n\varepsilon}{R + \frac{nr}{m}}$.For maximum current,the external resistance $R$ must equal the equivalent internal resistance of the battery: $R = \frac{nr}{m}$.Given $R = 2.5\,\Omega$ and $r = 1\,\Omega$,we have $2.5 = \frac{n(1)}{m} \Rightarrow n = 2.5m$.Substituting this into $mn = 40$: $(2.5m)m = 40 \Rightarrow 2.5m^2 = 40 \Rightarrow m^2 = 16 \Rightarrow m = 4$.Then $n = 2.5 	imes 4 = 10$.The maximum current is $I = \frac{n\varepsilon}{R + R} = \frac{10 	imes 2}{2.5 + 2.5} = \frac{20}{5} = 4\,A$.

The potential difference across a cell and the current through it are shown in the graph. $A$ battery consists of $40$ such identical cells. The maximum current supplied by the battery through a load of $2.5\,\Omega$ is equal to .............. $A$.

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