Series Combination of Cells

The cell is the smallest unit of a battery, and it has four main components: anode, cathode, electrolyte and separator. More than one cell forms the battery, it can be used in powering electrical appliances. When there are multiple cells in a given circuit, they are either wired in parallel or in series.

Cells Connected in a Series

There is only a single path between the terminals of the cell in a series combination of cells. The cells are said to be connected in series if the positive terminal of the first cell is connected to the negative terminal of the second cell, and the negative terminal of the second cell is connected to the positive terminal of the third cell. The same current flows through each cell.

Let us consider that ‘n’ identical cells are connected in series with the same polarity. The EMF of individual cells is E₁, E₂, E₃ —– E_n. Similarly, the internal resistance of each cell is r₁, r₂, r₃ ——–r_n.

The equivalent EMF is the terminal voltage across the cell when the cell is not in use.

The equivalent EMF of the cell is given by

E_eq = E₁ + E₂ +E₃ ——-E_n = nE

The equivalent internal resistance is given by

r_eq = r₁ + r₂ + r₃ ——-r_n = nr

The combination can be replaced with a single cell of equivalent EMF ‘nE’ and equivalent internal resistance ‘nr’.

The external resistance ‘R’ is connected between the free terminals of the first and the last cells. Now, the equivalent resistance of the circuit is

R_eq = nr + R

The current flowing through the load will be I = E_eq/R_eq

I = nE/(R+nr)

Case 1: If nr ＜＜R then I = nE/R

If the value of the internal resistance is much lesser than the external resistance, then the current in the circuit will be n times the circuit current due to the single cell.

Case 2: If nr ＞＞ R then I = E/r

If the value of the internal resistance is much greater than the external resistance, then the current in the circuit will be equal to the short-circuited current obtained from a single cell.

Solved Example

1. From the given circuit,

find

i) Equivalent emf of the combination

ii) Equivalent internal resistance

iii) Total current

iv) Potential difference across the external resistance

Solution:

i) Equivalent emf of the combination

E_eq = nE = 4 x 9 = 36 V

ii) r_eq = nr = 4 x 0.1 = 0.4 Ω

iii) Total current I = nE/(R + nr)

= (4 x 9)/[10 + (4 x 0.1)]

= 36/10.4

I = 3.46 A

(iv) Potential difference across the external resistance

V = IR = 3.46 x 10 = 34.6 V

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