n' cells , each of emf 'e' and internal resistance 'r' are connected in a closed circuit so that the positive terminal of a cell is joined to the negative terminal of the next, as shown in figure. Any two points of the circuit are connected by an external resistance R. Find the current in R.

Explanation

If n be the number of cells connected on one side of resistor R, then the number of cells connected on the other side of Ris (N - n). Current distribution in the equivalent circuit is shown in the adjoining circuit diagram
Here, n cells are shown as equivalent to a single cell of emf (N-n)E nE and internal resistance nr. Similarly, (N-n) cells are(N-n) shown equivalent to a single cell of emf (N-n)E and internal resistance (N-n)r. Applying Kirchoff.s second law to closed abcda `X [N-n]r + yR = (N-n) E`
`X=((N-n)E-yR)/((N-n)r)`
`X=((y)/(N-n))(R)/(r)ldots(1)`
Applying Kirchoff.s second law to closed mesh cdafebc
`X (N-n)r + (x-y)nr = (N-n) E+ nE`
`X [(N-n)r + nr] = NE + ynr`
`xNr = NE + ynr`
`xNE = ynr`
`x=(NE+ynr)/(nr)`
`x=(NE+ynr)/(Nr)`
`x=(E)/(r)+(yn)/(N)`
Combining Eqs (1) and (2), we get
v`(E)/(r)-((y)/(N-n))(R)/(r)=(E)/(r)+(yn)/(N)`
`y[(n)/(N)+(R)/(n-nr)]=0` `thereforey=0`
i.e. current through resistor R is zero. Alternate
Current in the ring I = E//rtherefore`V_(a)+nE - inr = V_(b)` `V_(a)-V-(b) = -nE + nr(E//r) = 0` No current through the resistor R.