Question

The n rows each containing $m$ cells in series are joined in parallel. Maximum current is taken from this combination across an external resistance of $30$ resistance. If the total number of cells used are $24$ and internal resistance of each cell is $0.5Q$ then

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

Answer 1

Answer: (C)

In mixed grouping the current in the external circuit will be maximum when the internal resistance of the battery is equal to the external resistance,
$R=\frac{mr}{n}$
Given, $R=3\Omega ,r=0.5\Omega$
$\therefore 3=\frac{m}{n}\times 0.5$
$\Rightarrow \frac{m}{n}=6$
$\Rightarrow m=6n$..(i)
Total number of cells $=m\times n=24\dots$ (ii)
From Eqs. (i) and (ii), we get
$6n\times n=24$
$\Rightarrow 6n^2=24$
$\Rightarrow n^2=4$
$\Rightarrow n=2,m=12$