Question

In a mixed grouping of $24$ identical cells each of internal resistance $2\text{ohm}$ and $n$ cells are connected in a row and likewise $m$ number of rows are connected in parallel. To get the maximum power output across a load of $12\Omega$, the value of $n$ and $m$ are?

• A. $n=12$; $m=2$
• B. $n=24$; $m=1$
• C. $n=3$; $m=8$
• D. $n=4$; $m=6$

Answer 1

The correct option is A $n=12$; $m=2$


Element of each cell is,

$E^{{}^{\prime }}=nE$

Total $R^{{}^{\prime }}=R+\frac{nr}{m}=\frac{mR+nr}{m}$

For maximum power at $I-I_{max}$
when $mR=nr$

$\frac{m}{n}=\frac{r}{R}=\frac{2}{12}=\frac{1}{6}$

$n=6m$

$nm=24$

$6m^2=24$

$m=\sqrt{4}=\pm 2$

Since, negative value of a cell number is not possible. We ignore $-2$.

$m=2$; $n=\frac{24}{m}=\frac{24}{2}=12$

$m=2$;$n=12$