Electrostatics - Series, Parallel Grouping

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 

ε_eq = 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 = ε_i / 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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