Q: 12 cells each having same emf , are connected in series and are kept in a closed Box. Some of cells are wrongly connected. This battery is connected is series with an ammeter and two cells identical with the others. The current is 3 A with the cells and battery aid with each other and 2 A when the cells and battery oppose each other. How many cells are wrongly connected.

(a) 1

(b) 2

(c) 3

(d) 4

**Click to See Answer : **

Net emf = (12 – 2 n)E + 2 E and I = 3 A

so , ((12-2n)E +2E )/R = 3 …(i)

When 2 cells oppose the battery, net emf

=(12-2 n)E – 2 E

As per the question

Current = ((12-2n)E -2E )/R = 2 …(ii)

Dividing (i) and (ii),

we get , n = 1