It is required to send a current of 10 A through a resistance `R = 3 Omega`. (a) What is the minimum number of cells required if each cell has an emf of 10 V and internal resistance `1 Omega`. (b) Find the power dissipated in R.

Explanation

(a) Let N be the total number of cells required. Let there be m rows of cells and n cells in each row. Then N = mn.
The total internal resistance of all the cells
`= (nr)/m = (n xx 1)/m" "( :' r = 1 Omega)`
Total emf of all the cells `= n epsilon = n xx 10`
Total resistance of the whole circuit
`=(nr//m) + R + ( n xx 1//m) + 3`
Current through external resistance will be
`I = n epsilon/((nr//m) + R)`
or `10 = (n xx 10)/(( n xx 1// m )+ 3) = (10 nm)/(n+3m)`
or `n + 3 m = nm or n + 3 (N//n) = N`
or `n^(2) + 3N = nN`
Differentiating it w.r.t. n, we get
`2 n + 3 (dN)/(dn) = n (dN)/(dn) + N`
When N is minimum, `(dN)/(dn) =0`.
`:. 2 n = N or n = N//2`
Putting this value in (i), we get
`N^(2)/4 + 3N = N/2 xx N or N/4 + 3 = N/2`
or ` N = 12`
(b) Power dissipated in R
`=I^(2)R = (10)^(2) xx 3 = 300` watt