Question

n identical cells, each of e.m.f. E and internal resistance r, are connected in series. Later on it was found out that two cells ‘X’ and ‘Y’ are connected in reverse polarities. Calculate the potential difference across the cell ‘X’.

Answer 1

Let the current in the series combination be I.

Since two cells X and Y are connected in reverse, the net e.m.f. of the battery is:

E_net​ = (n − 4) E    ...(Two reversed cells reduce the total e.m.f. by 2E each.)

The total internal resistance is:

R_int ​= nr

Hence, if the external resistance is R,

I = `((n - 4) E)/(R + nr)`

For the reversed cell X, current enters its positive terminal, so it is being charged. Therefore, the potential difference across it is:

V_X​ = E + Ir

Substituting I,

V_X = `E + ((n - 4) E r)/(R + n r)`

This is the potential difference across the reversed cell X.

If no external resistance is connected (cells short-circuited), then,

I = `((n - 4) E)/(n r)`

V_X = `E + ((n - 4) E)/n`

= `(2(n - 2) E)/n`

= `((2n - 4) E)/n`