What if a battery were old and tired?

Olly has a battery and four little bulbs. The bulbs sit side by side.

With a new battery, they all glow brightly. What if the battery were old and tired?

You've probably seen it happen: a flashlight that used to shine brightly starts to glow weakly, and you know the battery is getting old. But what is actually changing inside the battery?

In this lab, a 3-volt battery lights four bulbs wired side by side, each on its own path. First the battery is fresh. Then you make it old and tired by turning up its internal resistance, the resistance inside the battery itself. What do you think will happen to the bulbs?

A battery is often drawn as a perfect source of voltage, but OpenStax explains that every voltage source has two fundamental parts: a source of electrical energy (its emf) and an internal resistance, in series. Because the chemicals inside have resistance, a battery cannot make its emf without one.

This lab asks what happens to a heavily loaded circuit, four 3 Ω bulbs in parallel on a 3 V battery, when the battery's internal resistance rises from 0 to 2 Ω, as it might in an aging battery.

The new battery gives each bulb its full push.

All four bulbs glow brightly. The battery sends out lots of electricity to light them all.

With a fresh battery that has no internal resistance, each of the four bulbs gets the full 3 volts, because each sits on its own path straight across the battery. Each bulb carries 1 ampere of current and uses 3 watts, just as if it were the only bulb.

The battery has to supply the current for all four paths at once: 4 amperes in total, and 12 watts of power. That's a big job for a small battery.

With r = 0, each bulb sees the full 3.00 V, so each carries I = V/R = 1.00 A and dissipates P = V²/R = 3 W. The four branch currents add, so the battery supplies 4.00 A and 12.0 W; the equivalent resistance of the four bulbs is 0.75 Ω.

That low load resistance is the key to this lab. OpenStax's rule is that the smaller the internal resistance for a given emf, the more current and power the source can supply, so a load this hungry depends heavily on r being small.

Now Olly makes the battery old and tired.

All four bulbs go very dim! The battery even gets warm inside.

Turn the battery's internal resistance up to 2 ohms. The battery's push is the same, but now some of it is used up just getting the current through the battery itself.

The current falls from 4 amperes to about 1.09, and the voltage reaching the bulbs drops from 3 volts to about 0.82. Each bulb now uses only 0.223 watts, so on the model's brightness meter it glows at about 7% of what one bulb gets from a fresh battery. Most of the energy now warms the battery's insides instead of lighting the bulbs.

With r = 2 Ω, the total resistance is 2 + 0.75 = 2.75 Ω, so I = 3 V ÷ 2.75 Ω ≈ 1.09 A. The terminal voltage is V = emf − Ir = 3 − 1.09 × 2 ≈ 0.82 V, each bulb carries 0.27 A and dissipates 0.223 W (7% on the brightness readout, which compares each bulb with one bulb alone on a fresh 3 V battery).

The energy budget shifts dramatically: the bulbs share 0.893 W, while I²r ≈ 2.38 W is dissipated inside the battery. Over a 12-second run the model counts 39.3 J in total, of which only about 10.7 J reaches the bulbs.

The tired battery has to push the electricity through itself first. That uses up lots of its push.

So the bulbs get only a little. Big jobs make an old battery struggle most.

Think of the internal resistance as a narrow gate inside the battery. Every bit of current has to squeeze through it before it reaches the bulbs, and the bigger the current, the more push is lost at the gate.

That's why an old battery struggles most with big jobs. In the model, the same tired battery lights just one bulb at about 36% brightness, but four bulbs at only 7% each. Our model is simple, though: its bulbs keep a fixed resistance, and its battery never runs down during a run.

The terminal voltage falls by Ir, so the loss grows with the current. A light load barely notices r; a heavy load is starved. In the model with r = 2 Ω, one bulb (R = 3 Ω) still gets 1.80 V and 36% of full power, while four bulbs in parallel (0.75 Ω) get 0.82 V and 7% each. In series the same four bulbs draw so little current that r costs them little: 6% each with a fresh battery, 5% with a tired one.

Model limits: r is a fixed number you choose, while OpenStax notes that real internal resistance can depend on the current, temperature and history of the battery; the emf stays at 3 V, bulbs are ohmic, and brightness is compared by power, not light output.

When car lights are left on, they fade slowly as the battery runs down.

Some battery testers check a battery by making it work a little.

OpenStax, a free physics textbook, points out that car lights left on fade slowly as the battery runs down, instead of blinking off. That gradual dimming happens because a battery's output voltage drops as it is used up, and its internal resistance rises.

Battery testers use this idea. Some draw a current from the battery and check whether its voltage drops too far. A battery with high internal resistance is weak, even if it still seems to have its full voltage when nothing is connected.

OpenStax describes several consequences. Car lights dim gradually as the battery is depleted, because internal resistance rises. A big 12 V truck battery has a smaller internal resistance than a 12 V motorcycle battery with the same emf, so it can deliver a larger current. And if a device measures a battery's voltage without drawing current, it reads the emf, even for a very depleted battery.

That is why battery testers use small load resistors to draw current: they really test the internal resistance. OpenStax also notes that the internal resistance of rechargeable nickel-cadmium cells depends on how many times and how deeply they have been depleted.