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Grouping of Identical Cells

Series grouping

E.m.f. of the cell is and internal resistance is r.

 

791_series grouping.jpg

                                             Alt txt: series grouping

Applying Kirchhoff’s Law

                e– ir + e– ir +… (to n times) – iR = 0

       Þ    i = nz / R + nr = ε / (R/n) + r

 Illustration:

In a series grouping of N cells, current in the external circuit is l. The polarity of how many cells should be reversed so that the current becomes (1/3)rd of the earlier value?

Solution:

Before reversing the polarity of the cells, the current is

                l = NE / R + Nr

Let n cells be reversed in their polarity.

\Net emf = (N – n)E – nE = (N – 2n) E

Total resistance NR + R

   Þ i’ = (N – 2n)E / Nr + R

        But, 1' / 1 = (N–2n)E / (R + Nr) / (NE / Nr + R) = N – 2n / N 

   Þ n = N/3,

This is valid only when N is a multiple of 3, otherwise it cannot be done.

Parallel grouping

Let there be n identical rows each containing a cell of emf e

and a resistance r arranged as shown in the figure.

1735_parallel grouping.jpg

Alt txt : parallel grouping

Applying Kirchhoff’s law

        e–1/ n r – lR = 0
Þ

     l = nz / r + nR

* To get maximum current, cells must be connected in series if effective internal resistance is less than external resistance and in parallel if effective internal resistance is greater than external resistance.

Mixed grouping

Let emf of each cell is e

and internal resistance is r.

Number of rows is m and number of cells in each row is n

2366_mixed-grouping.jpg

Alt txt : mixed-grouping

Applying Kirchhoff’s law,

        ne– n1/m r – lR = 0

Þ    l = mnz / mR + nr

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