(a) When a cell is connected in series with some external resistance, a current `alpha` flows through it. When one more identical cell is connected in series with the first one, a current `beta` is found to flow in the circuit. When same cell is connected in parallel with the first one, the current is found to be `delta`. Find the value of `alpha` in terms of `beta` and `delta`.
(b) It is required to send a current of `8 A` through a circuit whose resistance is `5 Omega`. What is the least number of accumulators which must be used for this purpose and how should they be connected ? The emf of each accumulator is `2 V` and the internal resistance is `0.5 Omega`.

Explanation

(a) `alpha = (E)/(R + r) rArr R + r = (E)/(alpha)` …(i)
`beta = (E + E)/(R + r + r) = (2 E)/(R + 2r) rArr R + 2r = (2E)/(beta)` …(ii)
`delta = (E)/(R + (r)/(2)) rArr R + (r)/(2) = (E)/(delta)` …(iii)
(ii)-(i) `rArr r = (2 E)/(beta) - ( E)/(alpha)` ....(iv)
(i)-(iii) `rArr (r)/(2) = (E)/(alpha) -(E)/(delta) rArr r = (2E)/(alpha) - (2 E)/(delta)` ...(v)
(iv) = (v) `rArr (2E)/(beta) -(E)/(alpha) = (2 E)/(alpha) - (2 E)/(delta)`
`(3 E)/(alpha) = (2 E)/(beta) + (2 E)/(delta)`
`(3)/(alpha) = (2)/(beta) +(2)/(delta) rArr alpha = (3 beta delta)/(2(beta + delta))`.
(b) Let `n` be number of cells in a row and `m` the number of rows
`R = (nr)/(m) rArr 5 = (n xx 0.5)/(m) rArr n = 10 m`
`i = (nE)/(R + (nr)/(m)) rArr 8 = (10 m xx 2)/(5 + (10 m xx 0.5)/(m)) = (20 m)/(10)`
`m = 4`
`n = 40`
Total cells `= m n = 160`.
(a) ,