`N` sources of current with different emf's are connected as shown in Fig. The emf of the sources are proportional to their internal resistancs, i.e., `E = alpha R`, where `alpha` is an assigned constatant. The lead wire resistance is neglible. Find:
(a) the current in the circuit ,
(b) the potential differences between points `A` and `B` dividing the circuit in `n` and `N- n` links.

Explanation

(a) `i = (E + E)/(R + r_1 + r_2) = (2E)/(R + r_1 +r_2)`
Cell `X` : `p.d., V = E - ir_1 = E - (2 E r_1)/(R + r_1 + r_2) = 0`
`R = r_1 - r_2, (r_2 gt r_1)`
Resistance cannot be negative
Cell `Y` : `p.d. V = E - ir_2 = E - (2 E r_2)/(R + r_1 + r_2) = 0`
`R = r_2 - r_1, O.K`
`p.d` is zero across `Y`
(b) (i) `E_1 = alpha R_1, E_2 = alpha R_2, ...,E_N = alpha R_N`
Net emf `= alpha(R_1 +....+R_N)`
Net resistance `= R_1 + R_2 +...+ R_N`
Current in the circuit
`i = (alpha(R_1 + R_2 +....+R_N))/((R_1 + R_2+ ...+R_N)) = alpha`.
(ii) Between `A` and `B`, `n` cells on one side.
Net emf `= E_1 + E_2 + ...+E_N = alpha(R_1 + R_2 +....+R_n)`
Net resistance `= R_1 + R_2 +...+ R_n`
`p.d. V_A - V_B = (E_1 + E_2 +....+ E_n) - i(R_1 + R_2 +....+ R_n)`
=`alpha(R_1 + R_2 +....+ R_n) - alpha(R_1 + R_2 +....+ R_n)`
`= 0`
( c) Let there are `n` cells in a row and there are such `m` rows.
Total number of cells `mn = 24`...(i)
For current to be maximum in external resistance
`R = (nr)/(m)`
`3 = (n)/(m) xx 0.5`
`(n)/(m) = 6` ...(ii)
`n = 6 m` in (i)
`m xx 6 m = 24 rArr m^2 = 4 rArr m = 2`
`n = 6 m = 12`
(d) `mn = 24`
`R = (nr)/(m)`
`10 = (n)/(m) xx 2`
`(n)/(m) = 5` ...(ii)
`n =5 m` in (i)
`m xx 5 m = 24 rArr m^2 = (24)/(5) = 4.8`
Values of `m` lies between `2` and `3`.
(i) `m = 2, n = 12`,
Current in `R`, `i = (mnE)/(mR + nr) = (24 E)/(2 xx 10 + 12 xx 2) = (24)/(44) E`
=`(6 E)/(11) = 0.54 E`
(ii) If `m = 3, n = 8`
`i' = (24 E)/(3 xx 10 + 8 xx 2) = (24 E)/(46) = (12 E)/(23) = 0.52 E`
i.e. current in arrangement (ii) is higher. eight cells in a row, number of row `= 3`.
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