# Grouping of Identical Cells

## Series Grouping

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

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.

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

Alt txt : mixed-grouping

Applying Kirchhoff’s law,

ne– n1/m r – lR = 0

Þ  l = mnz / mR + nr

To read more, Buy study materials of Current Electricity comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Physics here.

### Course Features

• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

r