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(C) The total number of cells is given by $m 	imes n = 24$ ... $(i)$.For maximum current in a mixed grouping of cells,the external resistance $R$ mus

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$m = 12, n = 2$

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(C) The total number of cells is given by $m 	imes n = 24$ ... $(i)$.For maximum current in a mixed grouping of cells,the external resistance $R$ must be equal to the equivalent internal resistance of the combination.The equivalent internal resistance is given by $R_{eq} = \frac{mr}{n}$.Given $R = 3 \,\Omega$ and $r = 0.5 \,\Omega$,we have $3 = \frac{m 	imes 0.5}{n}$.This simplifies to $3 = \frac{m}{2n}$,which gives $m = 6n$ ... $(ii)$.Substituting equation $(ii)$ into equation $(i)$: $(6n) 	imes n = 24$.$6n^2 = 24 \implies n^2 = 4 \implies n = 2$.Using $n = 2$ in equation $(ii)$,we get $m = 6 	imes 2 = 12$.Therefore,$m = 12$ and $n = 2$.

The $n$ rows,each containing $m$ cells in series,are joined in parallel. Maximum current is drawn from this combination across an external resistance of $3 \,\Omega$. If the total number of cells used is $24$ and the internal resistance of each cell is $0.5 \,\Omega$,then:

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