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(A) Given: Total number of cells $N = mn = 24$. External resistance $R = 3 \, \Omega$. Internal resistance of each cell $r = 0.5 \, \Omega$.For maximu

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$m = 12, n = 2$

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(A) Given: Total number of cells $N = mn = 24$. External resistance $R = 3 \, \Omega$. Internal resistance of each cell $r = 0.5 \, \Omega$.For maximum current in a mixed grouping of cells,the external resistance must be equal to the total internal resistance of the battery.The total internal resistance of the battery is given by $R_{int} = \frac{mr}{n}$.Setting $R = R_{int}$,we get $3 = \frac{m 	imes 0.5}{n}$,which simplifies to $3 = \frac{m}{2n}$,or $m = 6n$.Substitute $m = 6n$ into the equation $mn = 24$:$(6n) 	imes n = 24 \implies 6n^2 = 24 \implies n^2 = 4 \implies n = 2$.Now,find $m$: $m = 6 	imes 2 = 12$.Thus,$m = 12$ and $n = 2$.

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