Internal resistance

Internal resistance, r, is the resistance inside a cell or battery itself. When current flows, some of the energy the cell gives each coulomb is used up inside the cell, so the voltage at its terminals is less than its emf. That drop, Ir, is called the lost volts. The key equations are ε = I(R + r) and V = ε − Ir, where V is the terminal pd and R is the external resistance.

Emf and terminal pd

The electromotive force (emf), ε, is the energy a source transfers to each coulomb of charge that passes through it. AQA writes it as ε = E ÷ Q, so 1 V means 1 joule per coulomb. Despite its name, emf isn't a force.

The terminal pd, V, is the potential difference across the cell's terminals: the energy per coulomb actually delivered to the outside circuit. If no current flows, nothing is lost inside the cell and V equals ε. As soon as current flows, V is smaller. More on potential difference itself is in what is potential difference.

Check the link between energy, charge and pd with a question from the electricity course, then press Mark.

What is internal resistance?

A real cell behaves like a perfect source of emf ε in series with a small resistor r, both hidden inside the case. The current passes through r on its way round, so r has a pd of Ir across it, inside the cell, where you can't use it.

Energy is conserved round the loop: the emf supplies energy to each coulomb, and it is shared between the external resistor and the internal resistance.

Internal resistance equations

ε = I(R + r) is the one AQA's specification gives. The others follow from it: IR is the terminal pd V, and Ir is the lost volts, so ε = V + Ir. Rearranged, the current is I = ε ÷ (R + r).

Worked examples

Example 1: current, terminal pd and lost volts

A battery of emf 6.0 V and internal resistance 1.5 Ω is connected to a 10.5 Ω resistor. Find the current, the terminal pd and the lost volts.

1. I = ε ÷ (R + r) = 6.0 ÷ (10.5 + 1.5) = 6.0 ÷ 12.0 = 0.50 A.
2. Terminal pd: V = IR = 0.50 × 10.5 = 5.25 V.
3. Lost volts: Ir = 0.50 × 1.5 = 0.75 V. Check: 5.25 + 0.75 = 6.0 V.

The energy lost inside the battery heats it: the power wasted is I²r = 0.50² × 1.5 = 0.375 W, while the resistor receives I²R = 0.50² × 10.5 = 2.6 W.

Example 2: finding r from one reading

A cell has an emf of 1.50 V. When it supplies a current of 0.40 A, its terminal pd is 1.20 V. Find its internal resistance.

Lost volts = ε − V = 1.50 − 1.20 = 0.30 V. Then r = 0.30 ÷ 0.40 = 0.75 Ω.

Example 3: finding ε and r from two readings

A battery gives a terminal pd of 2.6 V at a current of 0.20 A, and 2.0 V at 0.50 A. Find its emf and internal resistance.

1. Write ε = V + Ir for each: ε = 2.6 + 0.20r and ε = 2.0 + 0.50r.
2. Set them equal: 2.6 + 0.20r = 2.0 + 0.50r, so 0.6 = 0.30r and r = 2.0 Ω.
3. Then ε = 2.6 + 0.20 × 2.0 = 3.0 V.

Build any of these in the Circuit Builder with a resistor in series with the cell to stand for r, and watch the voltmeter reading fall as the current rises.

Required practical: measuring emf and internal resistance

AQA's required practical 6 investigates the emf and internal resistance of cells and batteries by measuring how the terminal pd changes with current. AQA's practical handbook lists the apparatus: the cell in a holder, a variable resistor (such as a rheostat), a digital ammeter, a digital voltmeter, a switch and leads.

1. Connect the cell, ammeter, switch and variable resistor in series, with the voltmeter across the cell's terminals.
2. Close the switch, record V and I, then open it again.
3. Change the variable resistor and repeat to get a range of currents.
4. Plot terminal pd V (up) against current I (across) and draw the line of best fit.

AQA advises switching off between readings so the cell doesn't run down, checking the variable resistor's power rating is high enough for the largest current, and avoiding rechargeable cells, which have very low internal resistance and can deliver high currents.

Reading the graph

Rearranging ε = V + Ir gives V = −rI + ε, which is y = mx + c. So the graph of V against I is a straight line falling to the right. These results are for a battery with ε = 4.0 V and r = 1.0 Ω:

• y-intercept = ε. At I = 0 there are no lost volts, so V = ε = 4.0 V.
• Gradient = −r. Between (0.5, 3.5) and (2.5, 1.5) the gradient is (1.5 − 3.5) ÷ (2.5 − 0.5) = −1.0, so r = 1.0 Ω.
• x-intercept = ε ÷ r, the short-circuit current, when the terminals are joined and R = 0.

Cells in series and why internal resistance matters

Identical cells in series add both their emfs and their internal resistances, since every resistance in series adds. Internal resistance is why a battery's voltage sags when it delivers a large current, and why batteries warm up in use: the I²r power is dissipated inside them. It also limits the largest current a cell can give, ε ÷ r.

Common mistakes

• Forgetting r in the current. I = ε ÷ (R + r), not ε ÷ R.
• Calling the terminal pd the emf. They are only equal when no current flows.
• Getting the sign of the gradient wrong. On a V against I graph the gradient is −r; the internal resistance is its size.
• Reading the emf off the first data point. Extend the line to I = 0; the intercept is the emf.
• Thinking lost volts are lost energy from nowhere. The energy heats the cell.

Emf and internal resistance are in the first year of A-level Physics, so practise them on today's A-level Physics: Year 1 short timed paper. For combining the external resistors first, see resistors in series and parallel.

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