n 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. `(2E)/(n)`B. `2E (1 - (1)/(n))`C. `(4E)/(n)`D. `2E (1 - (2)/(n))`

n identical calls are joined in series with its two cells `A` and `B`

1.	n 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. `(2E)/(n)`B. `2E (1 - (1)/(n))`C. `(4E)/(n)`D. `2E (1 - (2)/(n))`
Answer» Correct Answer - D (d) The two opposite cells `A` and `B` will cancel two more other cells, so net emf will be `n - 4`. So current `I = ((n - 4)E)/(nr)`, Now p.d. across `A` or `B` `=E + Ir` (as they will be in charging state) `= E + ((n - 4)E)/(n) = 2 E (1 - (2)/(n))`

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n 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. `(2E)/(n)`B. `2E (1 - (1)/(n))`C. `(4E)/(n)`D. `2E (1 - (2)/(n))`

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