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Grouping of Identical Cells Series Grouping E.m.f. of the cell is e 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.
E.m.f. of the cell is e 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.
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.
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.
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