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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

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

Grade:12th pass

1 Answers

Vikas TU
14149 Points
6 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

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