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Q. There are $45$ number of cells with internal resistance of each cell is $0.5\Omega$ . To get the maximum current through a resistance of $2.5\Omega$ , one can use $m$ rows of cells, each row having $n$ cells. The values of $m$ and $n$ are

NTA AbhyasNTA Abhyas 2020

Solution:

No. of cells in series $=n$
No. of rows $=m$
Total internal resistance $r^{'}=\frac{n r}{m}$
To have max. current $\frac{n r}{m}=R$
$\frac{n \left(\right. 0 . 5 \left.\right)}{m}=2.5\Rightarrow \frac{n}{m}=5$ 5$
$n\times m=45\Rightarrow n=15,m=3$ 3$