n identical cells each of emf 6 V , connected in series with an external resistor of `5Omega`, carry a current of `10A`. If two cells are connected wrongly in series with the same external resistor, the current flowing through the cells will be 8A. Find the value of n and r the internal resistance of each cell.

Explanation

Let r be the internal resistance of each cell. For series
connection, `epsilon_(eff) =" n"epsilon = 6n`.
`r_(eff) = nr`
or `I = ("n"epsilon)/(nr+R) = (6n)/(nr+5) = 10` ……(i)
For wrong connection of two cells, the effective emf decreases by `2epsilon` because the emfs of two identical cells counteract with each other. Then,
`epsilon'_(eff) = "n"epsilon-epsilon = (n-2)epsilon = (n-2)6`
and `epsilon_(eff) = nr`
or `i' = ((n-2)epsilon)/(nr+R) = ((n-2)6)/(nr +5) = 6.45` ......(ii)
By solving Eqs. (i) and (ii), we have n = 10 and r = 0.1`Omega`.