Consider $N = n_{1} n_{2}$ identical cells, each of emf $ϵ$ and internal resistance $r$ . Suppose $n_{1}$ cells are joined in series to form a line and $n_{2}$ such lines are connected in parallel. The combination drives a current in an external resistance $R$ . (a) Find the current in the external resistance. (b) Assuming that $n_{1}$ and $n_{2}$ can be continuously varied, find the relation between $n_{1}$ , $n_{2}$ , $R$ and $r$ for which the current in $R$ is maximum.

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Solution

a) Total emf $= n_{1} E$
in 1 row
Total emf in all news $= n_{1} E$
Total resistance in one row $= n_{1} r$
Total resistance in all rows $= \frac{n_{1} r}{n_{2}}$
Net resistance $= \frac{n_{1} r}{n_{2}} + R$

Current $= \frac{n_{1} E}{n_{1} r / n_{2} + R} = \frac{n_{1} n_{2} E}{n_{1} r + n_{2} R}$

b) $i = \frac{n_{1} n_{2} E}{n_{1} r + n_{2} R}$

for $i = m a x$
$n_{1} r + n_{2} R = m i n$

$\Rightarrow (\sqrt{n_{1} r} - \sqrt{n_{2} R})^{2} + 2 \sqrt{n_{1} m_{2} R} = m i n$

it is min, when
$\sqrt{n_{1} r} = \sqrt{n_{2} R}$

$\to n_{1} r = n_{2} R$
$i$ is max when $n_{1} r = n_{2} R$

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