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  • Complete Physics Course - Class 11
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Grouping of Identical Cells





Series Grouping



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



 



Series Grouping



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.



parallel grouping



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



 Mixed Grouping



Alt txt : mixed-grouping



Applying Kirchhoff’s law,



ne– n1/m r – lR = 0



Þ  l = mnz / mR + nr



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