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(A) $1$. In one row,there are $n$ cells in series. The total emf of one row is $nE$ and the total internal resistance of one row is $nr$.$2$. There ar

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$\frac{nE}{nr + mR}$

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(A) $1$. In one row,there are $n$ cells in series. The total emf of one row is $nE$ and the total internal resistance of one row is $nr$.$2$. There are $m$ such rows connected in parallel. The total emf of the parallel combination remains $nE$ because all rows are identical and connected in parallel.$3$. The equivalent internal resistance $r_{eq}$ of $m$ rows in parallel,where each row has resistance $nr$,is given by $\frac{1}{r_{eq}} = \frac{1}{nr} + \frac{1}{nr} + ... (m 	ext{ times}) = \frac{m}{nr}$. Thus,$r_{eq} = \frac{nr}{m}$.$4$. The total current $I$ flowing through the external load resistance $R$ is $I = \frac{nE}{R + r_{eq}} = \frac{nE}{R + \frac{nr}{m}} = \frac{mnE}{mR + nr}$.$5$. Since there are $m$ identical rows in parallel,the total current $I$ is divided equally among the $m$ rows. Therefore,the current in each row (and thus in each cell of that row) is $I_{cell} = \frac{I}{m} = \frac{1}{m} 	imes \frac{mnE}{mR + nr} = \frac{nE}{nr + mR}$.

$n$ identical cells,each of emf $E$ and internal resistance $r$,are joined in series to form a row. $m$ such rows are joined in parallel across a load resistance $R$. The current in each cell is:

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