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# 12 cells, each of EMF 1.5 volt and internal resistance of 1.5 ohm are arranged in M rows each containing n cells connected in series such that this combination would send Maximum current through an external resistance of 0.5 ohm when m and n are:1)m=3,n=42)m=6,n=23)m=4,n=34)m=2,n=6

Vikas TU
14149 Points
3 years ago
If n cells are associated in lines, at that point the aggregate resistance will be (R1) = n*r where r is the inside resistance of every cell.
For m lines, at that point the aggregate resistance will be
    1/R = 1/R1+1/R1+… … .m
= m/R1
As R1 = n*r
1/R = m/(n*r)
R= (n*r)/m
Current and power will be most extreme when the inward resistance is equivalent to the outside resistance
(n*r)/m = 1.5
(n*0.5)/m = 1.5
n/m = 3
n=3m
mn = 12
3m^2 = 12
m^2 = 4
m=2 and n=6