Series, Parallel Grouping
The two cells are said to be connected in series between two points A and C if one terminal of each cell is joined together and the other terminals of each cell is free
Let ε1, ε2 be the emf of the two cells and r1, r2 be their internal resistance respectively. Ket the cells be sending the current in a circuit not shown in fig. let VA VB and Vc be the flowing through them.
Potential difference between positive and negative terminals of the first cell is V AB = VA – VB = ε1 – I r1
Potential difference between positive and negative terminals of the first cell VBC = VB – VC = ε2 – I r2
Potential difference between A and C of the series combination of the two cells is
VAC = VA – VC = (VA – VB) + (VB – VC) = (ε1 – I r1) + (ε2 – I r2) = (ε1+ ε2) – I (r1 + r2)
Rules from the series combinations of cells are as follows:
The equivalent emf of a series combination of cells is equal to the sum of their individual emf.
The equivalent internal resistance of a series combination of cells is equal to sum of their individual internal resistances.
Two cells in parallel
The two cells are said to be connected in parallel between two points A and C if positive terminal of each cell is connected to one point and negative terminal of each cell at the other point.
Let VB1VB2 be the potentials at B1 and B2 respectively and V be the potential difference between B1 andB2 here the potential difference across the terminals of first cell is equal to the potential difference across the terminals of the second cell. So for the first cell
V = VB1 – VB2 = ε1 – I1 R1 or I1 = ε1 – V / r1
For the second cell
V = VB1 – VB2 = ε2 – I2 r2 or I2 = ε2 – V / r2
Putting values in (32) we have
I = (ε1 – V / r1) + (ε2 – V/ r2) = (ε1/r1 + ε2/r2) – V (1 / r1 + 1 / r2) = ε1 r2 + ε2 r1 / r1 r2 – V (r1 + r2 / r1 r2)
Or V = ε1 r2 + ε2 r1 / r1 + r2 – Ir1 r2 / r1 + r2
If the parallel combination of cells is replaced by a single cell between B1 and B2 of emf ε and internal resistance r them
V = εeq – I req
Comparing (33) and (34) we have
εeq = ε1 r2 + ε2 r1 / r1 + r2
And req = r1 r2 / r1 + r2
Or 1 / req = r1 + r2 / r1 + r2 = 1 /rt + 1/ r2
Dividing (x) by (xi) we have
εeq / req = ε1 r2 + ε2 r1 / r1 + r2 = 1 / r1 + 1 / r2
Dividing (x) by (xi) we have
εeq / req = ε1 r2 + ε2 r1 / r1 r2 = εi / r1 + ε2 / r2
If the two cells connected in parallel are of the same emf ε and same internal resistance r, then
From (35) εeq = εr + εr / r + r = ε
From (37) 1 / req = 1 / r + 1 / r = 2 / r
If identical cells are connected in parallel then the equivalent emf of all the cells is equal to the emf of one cell. It is so because in parallel combination of cells, there is only increase in sizes of the electrodes and not the emf of the combination of cells.
Equivalent internal resistance of parallel combination of n cells is
1 / req – 1 / r + 1 / r +….+ n terms = n / r
Or req = r/n.
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