Cells in combination

Why join cells at all?

A torch cell gives $1.5\ \text{V}$. A car starter wants $12\ \text{V}$, and a pocket radio is happy with $3\ \text{V}$. To get what we need, we join cells.

In Electric Circuits you met the rule for one cell, $V = E - Ir$. You saw that cells in series add. Now we look at every way of joining cells. We also ask what goes wrong when two cells pull in different directions.

Cells in series

In series the same current $I$ passes through every cell.

• Aiding. The positive end of one cell is joined to the negative end of the next. The emfs add: $E_{\text{eq}} = E_1 + E_2$.
• Opposing. Two positive ends are joined, or two negative ends. The emfs subtract: $E_{\text{eq}} = E_1 - E_2$, where $E_1$ is the larger.
• Resistance. In both cases $r_{\text{eq}} = r_1 + r_2$. A resistance never subtracts, because current meets the chemicals of the cell whichever way it flows.

In the opposing case the stronger cell wins. It drives the current round the circuit. That current goes through the weaker cell against its own emf, from its positive terminal into the cell. The weaker cell is being charged. Its terminal voltage is larger than its emf:

A worked example. Cell A has $E_1 = 4.5\ \text{V}$ and $r_1 = 0.60\ \Omega$. Cell B has $E_2 = 1.5\ \text{V}$ and $r_2 = 0.90\ \Omega$. They are joined in series, opposing, across a $2.5\ \Omega$ resistor.

1. The current. $I = \dfrac{E_1 - E_2}{R + r_1 + r_2} = \dfrac{3.0\ \text{V}}{4.0\ \Omega} = 0.75\ \text{A}$.
2. Cell A, discharging. $V_A = 4.5\ \text{V} - 0.75\ \text{A} \times 0.60\ \Omega = 4.05\ \text{V}$.
3. Cell B, charging. $V_B = 1.5\ \text{V} + 0.75\ \text{A} \times 0.90\ \Omega = 2.175\ \text{V}$.
4. Check. $V_A - V_B = 1.875\ \text{V}$, which is $IR = 0.75\ \text{A} \times 2.5\ \Omega$.

Cells in parallel

In parallel all the positive terminals are joined to one wire and all the negative terminals to another. Every cell has the same terminal voltage $V$, and they share the current.

You know the case of $n$ identical cells: the emf stays $E$ and the resistance becomes $r/n$. What if the cells are different?

Take two cells, $(E_1, r_1)$ and $(E_2, r_2)$, joined in parallel across a load. Let them send out currents $I_1$ and $I_2$, with $I = I_1 + I_2$ through the load. Each cell has the same terminal voltage $V$:

So $I_1 = \dfrac{E_1 - V}{r_1}$ and $I_2 = \dfrac{E_2 - V}{r_2}$. Add them, and solve for $V$:

This has the shape $V = E_{\text{eq}} - Ir_{\text{eq}}$. So the pair acts as one cell with

Look at $E_{\text{eq}}$. It is an average of $E_1$ and $E_2$, but a weighted one. The cell with the smaller internal resistance counts for more.

A warning. With no load, a current $i = \dfrac{E_1 - E_2}{r_1 + r_2}$ circulates round the two cells. It heats them and charges the weaker one. That is why we put in parallel only cells with the same emf.

Why use parallel cells? They give the same emf as one cell, but a much smaller internal resistance. That means a bigger current into a small load, and each cell works less hard, so the set lasts longer.

Mixed grouping

Join $N$ identical cells in $m$ rows, with $k$ cells in series in every row. Then $N = mk$.

Each row is a battery with emf $kE$ and internal resistance $kr$. The $m$ rows are in parallel, so the group keeps the emf and divides the resistance by $m$:

The current through a load $R$ is

The second form comes from multiplying the top and the bottom by $m$, and using $mk = N$.

Which arrangement gives the most current? With $N$ fixed, only the bottom line $mR + kr$ changes. The two terms $mR$ and $kr$ multiply to $(mk)(Rr) = NRr$. That product does not depend on the arrangement. Two positive numbers with a fixed product have the smallest sum when they are equal. So the current is greatest when

Match the load to the internal resistance of the group. This is the same idea as the maximum-power rule, $R = r$, that you met in Electric Circuits. It gives $m = \sqrt{Nr/R}$. In practice $m$ and $k$ must be whole numbers with $mk = N$, so we pick the arrangement closest to it.

A worked example. $24$ cells of $1.5\ \text{V}$, each with $r = 0.40\ \Omega$, drive a load of $0.60\ \Omega$.

1. The best rows. $m = \sqrt{\dfrac{24 \times 0.40}{0.60}} = \sqrt{16} = 4$, so $k = 6$.
2. The group. $E = 6 \times 1.5\ \text{V} = 9.0\ \text{V}$ and $r_{\text{group}} = \dfrac{6 \times 0.40\ \Omega}{4} = 0.60\ \Omega$. It matches the load.
3. The current. $I = \dfrac{9.0\ \text{V}}{0.60\ \Omega + 0.60\ \Omega} = 7.5\ \text{A}$.
4. Compare. All $24$ in series would give $36\ \text{V}$ and $9.6\ \Omega$, so $I = 3.5\ \text{A}$. All in parallel would give only $2.4\ \text{A}$.

Start with the defaults. The red point sits at $4$ rows. Make the load bigger, $R = 3\ \Omega$. Does the best arrangement move towards more rows, or fewer? Why does a big load favour long rows of cells?

Practise with new numbers

Try it in Python

The program tries every way to arrange $24$ cells into equal rows and finds the best.

Common mistakes

• Subtracting resistances for opposing cells. The emfs subtract. The resistances still add.
• Using the emf as the terminal voltage of a charging cell. A cell that is being charged has $V = E + Ir$, which is more than $E$.
• Averaging the emfs of two parallel cells. The average is weighted by the other cell's resistance.
• Using $kr$ for the group. The rows are in parallel, so the group has $kr/m$.

Remember these

• Series, aiding: $E = E_1 + E_2$. Opposing: $E = E_1 - E_2$. Always $r = r_1 + r_2$.
• Charging cell: $V = E + Ir$. Discharging cell: $V = E - Ir$.
• Two cells in parallel: $E = \dfrac{E_1r_2 + E_2r_1}{r_1 + r_2}$ and $r = \dfrac{r_1r_2}{r_1 + r_2}$.
• Mixed group: $E = kE_0$ and $r = kr_0/m$, with $N = mk$.
• Greatest current: $R = kr/m$, so $m = \sqrt{Nr/R}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Two cells, $6.0\ \text{V}$ and $2.0\ \text{V}$, each with $0.50\ \Omega$ inside, are joined in series, opposing. What are the emf and the internal resistance of the pair?

2. A battery of $12\ \text{V}$ and $0.20\ \Omega$ is charged with a current of $5.0\ \text{A}$. What is its terminal voltage?

3. Twelve cells are arranged as $3$ rows of $4$. Each cell has $E = 1.5\ \text{V}$ and $r = 0.30\ \Omega$. What are the emf and the internal resistance of the group?

4. The internal resistance of the group in question 3 is $0.40\ \Omega$. Is $3$ rows of $4$ the best choice for a $0.40\ \Omega$ load?