Combinations of Cells

Cells one after another

A battery is a group of cells joined so that they act as one. Whatever the grouping, the circuit outside sees only two terminals. So every combination can be replaced by one equivalent cell, with an emf \(\mathcal E_\text{eq}\) and an internal resistance \(r_\text{eq}\). The task of this lesson is to find those two numbers.

In series the same current passes through every cell. Walk from one end of the chain to the other and add every rise and fall. Each cell adds its emf and each internal resistance adds its drop. So the emfs add and the internal resistances add: \(\mathcal E_\text{eq} = \mathcal E_1 + \mathcal E_2 + \dots\) and \(r_\text{eq} = r_1 + r_2 + \dots\).

Four cells of \(2\,\text{V}\) and \(1\,\Omega\) in series make \(8\,\text{V}\) and \(4\,\Omega\). On a \(4\,\Omega\) load they drive \(8/(4 + 4) = 1\,\text{A}\). A torch with two cells of \(1.5\,\text{V}\) and \(0.5\,\Omega\) has \(3\,\text{V}\) and \(1\,\Omega\). With a \(2\,\Omega\) bulb it draws \(1\,\text{A}\), gives the bulb \(2\,\text{W}\) and heats its cells by \(1\,\text{W}\).

A cell put in backwards

The emfs add with their signs. A cell turned round, with its positive terminal facing its neighbour's positive terminal, pushes against the others. Its emf enters as \(-\mathcal E\). Its internal resistance still enters as \(+r\), because a resistance opposes current either way. Of \(n\) equal cells with one reversed, the reversed cell cancels one good cell: \((n - 2)\mathcal E\) remains, with \(nr\).

Six cells with one reversed give \(4\mathcal E\) and \(6r\). Four \(1.5\,\text{V}\) cells with one reversed give \(3\,\text{V}\), not \(6\,\text{V}\). A child who puts one cell of a toy in backwards finds the toy barely runs.

The weaker of two opposed cells is charged by the stronger, so its terminal voltage is above its emf. Values used in this lesson: the cells and loads of the workbook's first sheet and a lead-acid cell of about \(2\,\text{V}\). The inverter battery is \(12\,\text{V}\) and \(150\,\text{Ah}\), a common size.

Cells side by side

Join equal cells in parallel, positive to positive. Every cell has the same terminal voltage, and the current divides equally among them. The emf does not grow; each cell carries only its share of the current. So \(n\) equal cells in parallel act as one cell of emf \(\mathcal E\) and internal resistance \(r/n\).

Four cells of \(2\,\text{V}\) and \(1\,\Omega\) in parallel give \(2\,\text{V}\) and \(0.25\,\Omega\). On a \(0.75\,\Omega\) load they drive \(2/(0.75 + 0.25) = 2\,\text{A}\), half an ampere from each cell. A parallel bank is chosen when a large current is wanted at a modest voltage.

Unequal cells in parallel

Now let the cells differ. The chapter Circuits: Kirchhoff's Laws, Networks and RC Transients found the common voltage of two cells in parallel by the junction rule. Let the pair share a terminal voltage \(V\) and supply a load current \(I\). Cell 1 then gives \((\mathcal E_1 - V)/r_1\) and cell 2 gives \((\mathcal E_2 - V)/r_2\), and the two add to \(I\).

The equivalent emf is an average of the emfs, each weighted by \(1/r\). A cell with a small internal resistance has a loud voice in it. Cells of \(6\,\text{V}\) and \(3\,\text{V}\), \(1\,\Omega\) each, give \(\mathcal E_\text{eq} = (6 + 3)/2 = 4.5\,\text{V}\) and \(r_\text{eq} = 0.5\,\Omega\). On a \(4\,\Omega\) load they drive \(4.5/4.5 = 1\,\text{A}\).

Look inside that pair. The terminals stand at \(4 \times 1 = 4\,\text{V}\). The \(6\,\text{V}\) cell gives \((6 - 4)/1 = 2\,\text{A}\). The \(3\,\text{V}\) cell gives \((3 - 4)/1 = -1\,\text{A}\): one ampere flows into its positive terminal. The weak cell is being charged by the strong one, even while the load runs. That is why old and new cells should never share a parallel bank.

Rows of cells

A mixed grouping uses both ideas. Put \(n\) equal cells in series to make a row of \(n\mathcal E\) and \(nr\). Put \(m\) such rows in parallel. The rows are equal, so the emf stays \(n\mathcal E\) and the internal resistance falls to \(nr/m\). The current in a load \(R\) is \(I = n\mathcal E/(R + nr/m)\).

In Fig 2.2 each row is \(8\,\text{V}\) and \(4\,\Omega\). Three rows make \(8\,\text{V}\) and \(4/3\,\Omega\). With the load of \(4/3\,\Omega\) the current is \(8/(8/3) = 3\,\text{A}\). Try the other shapes of twelve cells on this load. Twelve in series give \(1.8\,\text{A}\); two rows of six give \(2.77\,\text{A}\). Four rows of three give \(2.88\,\text{A}\), and all twelve in parallel only \(1.41\,\text{A}\).

The winning shape is the one whose internal resistance equals the load, \(nr/m = R\). It is the matching condition of the last lesson again. With \(R\) fixed, the largest current also means the largest power in the load.

Series or parallel?

With \(n\) equal cells, series gives \(I_s = n\mathcal E/(R + nr)\) and parallel gives \(I_p = \mathcal E/(R + r/n)\). When \(R \gg r\), the internal resistances hardly matter, and series wins because its emf is \(n\) times larger. When \(R \ll r\), the internal resistance controls the current, and parallel wins because it is \(n\) times smaller.

In the figure above the curves cross at \(R = 1\,\Omega\), the internal resistance of one cell. There both groupings give \(1.6\,\text{A}\). At \(4\,\Omega\) series gives \(1\,\text{A}\) and parallel only \(0.47\,\text{A}\). At \(0.75\,\Omega\) parallel gives \(2\,\text{A}\) and series \(1.68\,\text{A}\). The tie is exact for any \(n\): both currents equal \(n\mathcal E/((n + 1)r)\) when \(R = r\).