Question

A teacher asked a student to connect N cells each of e.m.f. e in series to get a total e.m.f. of Ne. While connecting, the student, by mistake, reversed the polarity of n cells. The total e.m.f. of the resulting series combination is:

• A. $(N-\frac{n}{2})$
• B. $e(N-n)$
• C. $e(N-2n)$
• D. $eN$

Answer 1

The correct option is C $e(N-2n)$

If n cells are connected in -ve polarity then the
$E_{eff}of\text{+ve polarity cells}=(N-n)e$
$E_{eff}of\text{-ve polarity cells}=+ne$
$Total{E}_{eff}=E_{+}-E_{-}$
$=N-ne-ne$
$=e(N-2n)$