Potential difference across a cell and current through a cell is shown in graph. A battery consists of such identical $40$ cells. Max current supplied by the battery through a load of $2.5\,\Omega $ equal to .............. $A$
Medium
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$\varepsilon=2$ $\mathrm{volt},$ $\mathrm{r}=1 \,\Omega$
$m n=40$
Current is maximum
$\mathrm{R}=\frac{\mathrm{nr}}{\mathrm{m}} \Rightarrow 2.5=\frac{\mathrm{n}}{\mathrm{m}} 1$
$\mathrm{n}=\mathrm{m}(2.5)$
$m=4, n=10$
$I=\frac{10 E}{5}=4 \,A$
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