Question

To get maximum current in a resistance of $3 \,ohms$, one can use $n$ rows of $m$ cells (connected in series) connected in parallel. If the total number of cells is $24$ and the internal resistance of a cell is $0.5 \,ohms$ then

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

Answer 1

Answer: (A)

Let, we connect $24 $ cells in n rows of m cells, then if $I$ is the current in external circuit then
$I=\frac{m E}{m r / n+R}$ ....(1)
For I to be maximum, $( mr + nR )$ should be minimum.
It is minimum for $R=\frac{m r}{n}$ ....(2)
So maximum current in external circuit is $I =\frac{ mE }{2 R }$......(3)
here $R=3, r=0.5$ so equation (2) become $\frac{m}{n}=6$
so $n =2, m =12$