Question

In identical calls are joined in series with its two cells `A` and `B` in loop with reversed polarties. `EMF` of each shell is `E` and internal resistance `r`. Potential difference across cell `A` or `B` is (here `n gt 4`)
A. `(2epsilon(n-2))/(n)`
B. `(2epsilon(n+2))/n`
C. `(4epsilon)/n`
D. `(2epsilon)/n`

Answer 1

Correct Answer - A
a. The two opposite cells A and B will cancel two more cells, so
net emf will be `n-4`. So current is
`I = ((n-4)epsilon)/(nr)`
Now pd across A or B is
`I epsilon + Ir` (as they will be in charging state) ,brgt `= epsilon = ((n-4)epsilon)/n = 2epsilon (1-2/n) = (2epsilon(n-2))/n`.