How will you connect (series and parallel) 24 cells each of internal resistance 1 ohm so as to get maximum power output across a load of $10\mathrm{\Omega }$ ? [a = number of rows that are connected in parallel, b = number of cells in series in each row]

Power delivered to the load is maximum when 

${\mathrm{R}}_{\text{ext }}={\mathrm{r}}_{\text{eff }}\Rightarrow \frac{\mathrm{br}}{\mathrm{a}}=\mathrm{R}\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{R}}{\mathrm{r}}=\frac{10}{1}=10\dots ..\left(1\right)$

total number of cells $\mathrm{ba}=24\Rightarrow \left(10\text{ a) }\right(\mathrm{a})=24$

$\Rightarrow \mathrm{a}=\sqrt{2.4}=1.56$

Now as 1.56 rows are not physically possible. So ‘a’ can be either 1 or 2

(i) If a = 1 then ab = 24$\Rightarrow$b = 24 ; 

$\therefore {\mathrm{P}}_1={\left(\frac{{\mathrm{E}}_{\mathrm{eff}}}{\mathrm{R}+{\mathrm{r}}_{\mathrm{eff}}}\right)}^2\mathrm{R}=\frac{(\mathrm{bE}{)}^2\mathrm{R}}{{\left(\mathrm{R}+\frac{\mathrm{br}}{\mathrm{a}}\right)}^2}=5{\mathrm{E}}^2$

(ii) If a = 2 then ab = 24$\Rightarrow$ b = 12

$P_2=\frac{(bE{)}^{2}R}{{\left(\mathrm{R}+\frac{\mathrm{br}}{\mathrm{a}}\right)}^2}=5.625{\mathrm{E}}^2$

Clearly $P_2>P_1$. To get maximum output across load, a = 2, b = 12.