Q. The $n$ rows each containing m cells in series are joined in parallel. Maximum current is taken from this combination across as external resistance of 3$\ \Omega$ resistance. If the total number of cells used are 24 and internal resistance of each cell is 0.5$\ \Omega$, then
A) $m = 8,\ n = 3$
B) $m = 6,n = 4$
C) $m = 12,\ n = 2$
D) $m = 2,n = 12$
Explanation
(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$\Omega,\ r = 0.5\Omega$
$\therefore\ \ \ \ 3 = \frac{m}{n} \times 0.5$
$\Longrightarrow \ \frac{m}{n} = 6\ $
$\Longrightarrow m = 6n$
Total number of cells$= m \times n = 24$ …(i)
From Eqs. (i) and (ii), we get …(ii)
$6n \times n = 24$
$\Longrightarrow 6n^{2} = 24$
$\Longrightarrow n^{2} = 4$
$\Longrightarrow n = 2,\ m = 12$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.