Question

$N$ identical cells, each of emf $E$ and internal resistance $r$ are joined in series. Out of $N$ cells, $n$ cells are wrongly connected i.e., their terminals are connected in reverse of that required for series connection $\left(n < \frac{N}{2}\right) .$ Let $E_{0}$ be the emf of resulting battery and $r_{0}$ be its internal resistance. Then

• A. $E_{0}=(N-n) E, r_{0}=(N-n) r$
• B. $E_{0}=(N-2 n) E, r_{0}=(N-2 n) r$
• C. $E_{0}=(N-2 n) E, r_{0}=N r$
• D. $E_{0}=(N-n) E, r_{0}=N r$

Answer 1

Answer: (C)

For every cell that is wrongly connected, the emf decreases by $2 E$. However, internal resistance does not depend on the direction and therefore remains same for all cells.