Question

In a series grouping of N cells, current in the external circuit is I. Number of cells to be reversed in polarity such that current becomes $\frac{I}{3}$ is

• A. $\frac{2N}{3}$
• B. $\frac{N}{3}$
• C. $\frac{N}{2}$
• D. $\frac{N}{4}$

Answer 1

The correct option is B $\frac{N}{3}$

Initially, current through external resistance R,
$I=\frac{N_{ϵ}}{R+Nr}...\left(i\right)$
When n cells are reversed in polarity, then current through R,
$\frac{I}{3}=\frac{(N-2n{)}_{ϵ}}{R+Nr}...\left(ii\right)$
From eqn, (i) and (ii)
$3=\frac{N}{(N-2n)}\Rightarrow n=\frac{N}{3}$