Question

42) To get a maximum current through a resistance
of 2.52 one can use m rows of cells each row
having n cells. The internal resistance of each
cell is 0.502 Jf the total number of cells are 20,
then the values of m and n are
A) m=3,n=4
B) m=2,n=10
C) m-4, n=5
D) m=6,n=3
​

Answer 1

Answer:

m=3;n=15.

Explanation:

Given,

Resistance=2.5Ω Internal resistance =0.5Ω,mn=45

Let n be the cell in the series in one row and m rows in parallel let all the cells are identical. Let each cell be of emf E and internal resistance is r,

In each row of cell in parallel so, total internal resistance (r_p) of all the cell given by,

rp1 = nr

1+ nr1+.....n terms

nrm⇒rp= mnr

Total distance in the circuit R+ mnr

Effective emf of the cell is nE

I= R+ mnr

nE= mR+nr

=mnE

The current will maximum if, MR=nr ,⇒R=nr.mn

=45...1

m×2.5=n×0.5⇒n=5m...........2

From equation 1 and 2 we get,

m 2 =45⇒m2 =9⇒m=3

Therefore, n=15;m=3

............I hope it will help you............