Question

Consider N = n₁n₂ identical cells, each of emf ε and internal resistance r. Suppose n₁ cells are joined in series to form a line and n₂ 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 n₁ and n₂ can be continuously varied, find the relation between n₁, n₂, R and r for which the current in R is maximum.

Answer 1

(a)

Given:
Emf of one cell = E
∴ Total e.m.f. of n₁ cells in one row = n₁E
Total emf of one row will be equal to the net emf across all the n₂ rows because of parallel connection.
Total resistance in one row = n₁r
Total resistance of n₂ rows in parallel $=\frac{n_{1}r}{n_2}$
Net resistance of the circuit = R + $\frac{n_{1}r}{n_2}$
$\therefore \mathrm{Current},I=\frac{n_{1}E}{R+{\displaystyle \frac{n_{1}r}{n_2}}}=\frac{n_1{n}_2\mathrm{E}}{n_2\mathrm{R}+n_{1}r}$

(b) From (a),
$I=\frac{n_1{n}_2\mathrm{E}}{n_2\mathrm{R}+n_{1}r}$

For I to be maximum, (n₁r + n₂R) should be minimum
$\Rightarrow {\left(\sqrt{n_{1}r}-\sqrt{n_2\mathrm{R}}\right)}^2+2\sqrt{n_1\mathrm{R}{n}_{2}r}=\min$
It is minimum when
$$\begin{gathered} \sqrt{n_{1}r}=\sqrt{n_2\mathrm{R}} \\ {n}_{1}r=n_2\mathrm{R} \end{gathered}$$
∴ I is maximum when n₁r = n₂R .