Combinations of Cells

A single electrochemical cell often cannot supply sufficient EMF (voltage) or current for practical circuits. To overcome this limitation, multiple cells are combined into a battery, arranged in series, parallel, or mixed (series-parallel) groupings depending on whether higher voltage or higher current is required.

Real-life Analogy: Think of cells like water pumps. Connecting pumps in series (one after another) increases pressure (like voltage), while connecting pumps in parallel increases the total flow rate (like current).

When n identical cells, each of EMF E and internal resistance r, are connected in series with an external resistance R:

Key Insight: Series grouping is advantageous when the external resistance R is much greater than the total internal resistance nr — i.e., R ≫ nr. In this case, current is nearly independent of the number of cells and voltage output is maximised.

For cells of different EMFs and internal resistances connected in series:

For n identical cells (EMF E, internal resistance r each) connected in parallel:

Key Insight: Parallel grouping is advantageous when the external resistance R is much smaller than the internal resistance r — i.e., R ≪ r. This configuration delivers a stronger current through low-resistance circuits.

For two cells with EMFs E₁, E₂​ and internal resistances r₁, r₂​:

Consider nnn cells in series per row, and mmm such rows connected in parallel — a total of mn identical cells.

Condition for Maximum Current

Maximum current flows through the external resistance R when:

Under this condition, maximum current is given by:

Problem: Six cells, each of EMF 2.0 V and internal resistance 0.015 Ω, are connected in series with an external resistance of 8.5 Ω. Find the current and terminal voltage.

Solution:

Answer: Current = 1.4 A, Terminal Voltage = 11.9 V

This problem asks: at what external resistance does connecting the two cells in series give the same current as connecting them in parallel?

• In series: Total EMF = 6 + 12 = 18 V, total internal resistance = 2 + 1 = 3 Ω. Current = \[\frac {18}{R+3}\].
• In parallel: Equivalent EMF = \[\frac {6(1)+12(2)}{1+2}\] = 10 V, equivalent internal resistance = \[\frac {2×1}{2+1}\] = 0.67 Ω. Current = \[\frac {10}{R+0.67}\].
• Setting both currents equal and solving for R gives R = 2.25 Ω.

Simple takeaway: At low external resistance, parallel gives more current; at high external resistance, series wins. 2.25 Ω is the "crossover point" where both arrangements perform identically.

Here, 12 identical cells (1.5 V, 0.5 Ω each) must be arranged to push the maximum possible current through a 1.5 Ω external resistor.

• Rule for max current in mixed grouping: current is maximised when external resistance = equivalent internal resistance, i.e., R = \[\frac {nr}{m}\], where n = cells in series per row, mmm = number of parallel rows.

• With 12 cells total, possible splits are (n = 12, m = 1), (n = 6, m = 2), (n = 4, m = 3), (n = 3, m = 4), (n = 2, m = 6), (n = 1, m = 12).

• Testing n = 6, m = 2: r_eq = \[\frac {6×0.5}{2}\] = 1.5 Ω, which exactly matches R = 1.5 Ω.

Simple takeaway: Arrange the 12 cells as 2 parallel rows, each with 6 cells in series — this makes the battery's internal resistance equal to the external resistance, which is exactly the condition for drawing the largest possible current.