LessonPower sources and safety · 15 min

Internal resistance: why the voltage sags under load

Measure a motorbike battery with a meter: 12.5 V, looks fine. Press the starter: nothing but a click. The battery is not lying - the meter draws almost no current, and a battery's weakness only shows when current flows. Every real source has an internal resistance r, and the current through it costs voltage.

The model: an ideal 9 V source E and r = 5 Ω inside the case (an old 9 V battery). With the 40 Ω load on, I = 9 / 45 = 0.2 A and the terminals show 9 − 0.2 × 5 = 8.0 V.

V_terminal = E − I × r I_short = E / r

E: the open-circuit voltage (no current): the "EMF"
I: the current being drawn
r: the internal resistance
I_short: the current if the terminals are shorted

Low r is good for delivering power - and makes short circuits violent.
Source	Typical r	Short-circuit current
9 V PP3 alkaline	1 - 2 Ω	a few amps
AA alkaline (new)	0.1 - 0.3 Ω	about 10 A
18650 Li-ion cell	0.02 - 0.05 Ω	100 A or more
Car / motorbike battery	0.005 - 0.02 Ω	hundreds to thousands of amps

Worked example: the starter motor

A 12.6 V battery feeds a starter motor that draws 150 A. What reaches the motor if r = 0.01 Ω (healthy) and r = 0.05 Ω (old, sulphated)?

Healthy: 12.6 − 150 × 0.01 = 11.1 V.
Old: 12.6 − 150 × 0.05 = 5.1 V - too little to turn the engine.

Answer: 11.1 V and 5.1 V. Both batteries read 12.6 V on a meter with no load. Garages use a load tester.

Terminal voltage of a 12.6 V battery against current

r = 0.01 Ω (healthy)r = 0.05 Ω (old)

Measuring r

Measure the voltage with no load (that is E), then with a known load current: r = (E − V_loaded) / I. You will do exactly this in the lab.

InteractiveTry it: a battery and a load

Battery EMFVInternal resistanceΩLoadΩ

Current-A

Terminal voltage-V

Power into the load-W

Efficiency-%

The load gets the most power when RL = r - but then half the energy heats the battery. Real designs keep RL much bigger than r.

Circuit for: Try it: a battery and a load — The circuit at the starting values

Where you meet it

Battery testers

A load test (a big current for a few seconds) shows r; a plain voltmeter cannot.

Design

Put a big capacitor near a GSM module or a motor so short current bursts do not pull the supply down through the battery's r.

Safety

A low-r battery shorted by a spanner or a ring delivers enormous current.

Practice 1A battery reads 9.0 V with no load and 8.4 V while 0.3 A flows. What is its internal resistance?

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Answer: 2 Ω

r = (9.0 − 8.4) / 0.3 = 2 Ω.

Practice 2An 18650 cell (4.1 V, r = 0.04 Ω) supplies 5 A. What is its terminal voltage?

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Answer: 3.9 V

4.1 − 5 × 0.04 = 3.9 V.

Practice 3What current flows if a 12.6 V battery with r = 0.01 Ω is short-circuited?

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Answer: 1260 A

12.6 / 0.01 = 1260 A - enough to weld metal.

Challenge 4A battery gives 11.8 V into a 10 Ω load and 11.0 V into a 2 Ω load. Find its internal resistance.

Hint

Two equations: E = V + I r with I = V / R for each load.

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Answer: 0.1852 Ω

I1 = 11.8 / 10 = 1.18 A, I2 = 11.0 / 2 = 5.5 A. E = 11.8 + 1.18 r = 11.0 + 5.5 r, so 0.8 = 4.32 r and r = 0.185 Ω (and E = 12.0 V).