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In identical calls are joined in series ...

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`

Text Solution

Verified by Experts

The correct Answer is:
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`.
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