MIXED GROUPING OF BATTERY CELLS (m+n=constant)
* In electricity, when there is a mixed group of cells of emf E each, internal resistance r each such that there are n cells in series in a row, and there are m equal rows arranged in parallel and the system is linked to an external resistance of R , current in the resistor R is given by I = nE / (R + nr/m ) = mnE / (mR+nr). If mn = k=constant, the arrangement of m and n which results in I being maximum is derived below--
We can get the result by calculus method by putting m+n=k=constant and putting f(m)= E/ [R/n +r/m] = E/ [R/(k-m) +r/m] to be extremum. df(m)/dm = 0 or -r/m2 + R/(k-m)2 = 0 or m2(R-r)+2krm-rk2=0
or m =[k/(R-r)]*[r ± √(Rr)] and Imax =E / [ r(R-r) /k( rą√(rR)) +R/n ]
* For mn=24, m=4,n=6, r=2, mR + nr will be minimum when R= 6*2/4=3; In other words, when R=3, R/r=3/2. thus n/m =6/4=3/2 will be the combination at which maximum current will flow through the resistance. Similarly, when R=2, R/r=1 and so m/n=1 or m*m=24 or m =4.8 , n=5. And at this combination, maximum current will flow. Rounding up, m=4,n=6 and R=2.
|since m,n,r, R i.e. 4 numbers are involved, we have tried to find out whether the numbers can form a quadrilateral . For example, taking mn= 24, the combination can be (m,n)=(24,1) , (1,24) , (12,2), (2,12), (8,3), (3,8), (6,4), (4,6)|
|taking (24,1) -> and r=2, permissible k=2 only and permissible R=23,24,25. The minimum R which will result in maximum current is not permissible since triangle ACD cannot be constructed with that value of R.|