8 cells, each of internal resistance `0.5 Omega` and emf 1.5V are used to send a current through an external ressistor of (a) `200 Omega` (b) `0.002 Omega` (c ) `1.0 Omega`. How would you arrange them to get the maximum current in each case? Find the value of current in each case.

Explanation

Here, total number of cells= 8,
`r=0.5 Omega , epsilon=1.5V`.
(a) When `R=200 Omega` , then R gtgt r, so for maximum current, the cells are to be connected in series in circuit.
Total internal resistance of 8 cells = 8r
Current in circuit,
`I= (8 epsilon)/(R +8r)= (8xx1.5)/(200+8 xx 0.5)=12.0/204 = 0.059A`
(b) When `R=0.002 Omega` , then R ltlt r, so for maximum current, the cells are to be connected in parallel in circuit.
Total internal resistance of 8 cells = `r//8`
Total resistance of circuit `= R + r//8`
`0.002 +0.5 //8 = 0.0645 Omega`
Effective emf of all cells = emf of each cell
1.5V
Current in circuit, `I=1.5/0.0645=23.26A`
(c ) When `R= 1.0 Omega`, then R is comparable to r. For a mximum current, the cells are to be connected in mixed grouping. Let there be m rows of cells in parallel with n cells in sereis, in each row. Then,
`m n=8`
For maximum current, `R= (nr)/(m)`
or `1.0 = nxx0.5/m or n=2 m`
From (i), `m xx 2 m = 8 or m^(2) = 4 or m=2`.
So `n = 2xx2=4`
Thus, 4 cells in series in a row and 2 such rows of cells in parallel.
Max. current, `I=( n epsilon)/(R + nr//m) = ( m n epsilon)/(mR + nr)`
`= ( 8 xx 1.5)/(2 xx 1 + 4 xx 0.5)=3 A`