What if you kept adding bulbs side by side?

Olly lights one bulb with a battery. It glows brightly.

Now he adds another bulb side by side. Then another, and another! Will the bulbs get dimmer?

If you add more bulbs to a circuit, do they have to share, and glow more dimly? It depends on how they are wired.

In this lab, every new bulb goes side by side with the others, on its own path straight across a 3-volt battery. That's called a parallel circuit. You start with one bulb, then keep adding them, up to four. Watch two things: how brightly each bulb glows, and how much current the battery has to supply.

In a parallel network every branch is connected across the same two points, so each branch sees the same voltage. Adding a branch adds a new path for current rather than lengthening an existing one.

This lab starts with one 3 Ω bulb on an ideal 3 V battery and adds identical bulbs in parallel, up to four. It tracks the brightness of each bulb, the battery's total current and the total power, then asks what a real battery, with its own internal resistance, would do.

With one bulb, the battery gives it the full push.

The bulb glows brightly. The battery sends out a small amount of electricity.

With one bulb, the bulb gets the battery's full 3 volts. A current of 1 ampere flows through it, and it uses 3 watts of power. The model calls that full brightness: 100%.

The battery supplies exactly what that one bulb needs, 1 ampere. Over a 12-second run in the model, the circuit uses 36 joules of energy.

One 3 Ω bulb on 3 V carries I = V/R = 1.00 A and dissipates P = V²/R = 3 W, the model's 100% brightness reference. The battery's current equals the bulb current, and in a 12-second run the model counts 3 W × 12 s = 36 J.

This is the baseline: one path, one current, the battery's full voltage across the only load.

Now Olly adds bulbs side by side, up to four.

Every bulb still glows brightly! But the battery has to send out four times as much electricity.

Add bulbs one at a time. Each new bulb gets its own path and the full 3 volts, so each one carries 1 ampere and glows at 100%, just like the first.

But the battery must feed every path at once. With two bulbs it supplies 2 amperes, with three 3 amperes, and with four 4 amperes. The total power climbs from 3 watts to 12 watts, and a 12-second run uses 144 joules instead of 36. The bulbs don't pay for the extra bulbs: the battery does.

Each added branch draws its own 1.00 A, so the battery current is the sum: 1, 2, 3, then 4.00 A. The equivalent resistance falls from 3 Ω to 1.5, 1 and finally 0.75 Ω, which OpenStax describes as always less than the smallest single resistance in a parallel connection.

Total power grows in step, from 3 W to 12 W, and the energy counted over a 12-second run from 36 J to 144 J. With an ideal battery each bulb's brightness readout stays at 100%, because nothing between the battery and the bulbs uses up any voltage.

Each bulb has its own path. So each bulb gets the whole push.

But every path needs its own electricity. More paths means more electricity from the battery.

In a parallel circuit, each path connects straight across the battery, so each bulb gets the battery's whole push no matter how many other bulbs there are. That's why the bulbs stay bright.

But each path also carries its own current, and all those currents come from the same battery. So every bulb you add means more current and more power from the battery. Our model's battery is perfect, though: it never runs down, and it has no resistance inside unless you add some. A real battery does.

Kirchhoff's junction rule: the battery current equals the sum of the branch currents, I = I₁ + I₂ + …, while each branch voltage equals the source voltage. With an ideal source, each bulb's power, V²/R, is independent of how many bulbs are present.

Add internal resistance and that independence breaks. With r = 0.2 Ω, one bulb draws 0.94 A and glows at 88%; four bulbs draw 3.16 A in total, the terminal voltage sags to 2.37 V and each bulb falls to 62%. The model's other limits: ohmic bulbs, wires with no resistance, a battery that never runs down, and brightness compared by power.

The lights and plugs in a house are wired side by side.

If too much electricity flows in the wires, a safety switch turns it off.

Homes are wired in parallel, so every lamp and appliance gets the same voltage and works on its own. But each one you switch on adds its own current to the wires that feed them.

OpenStax, a free physics textbook, explains that wires carrying too great a current can overheat. That's why homes have fuses and circuit breakers: they switch the circuit off automatically when the current stays above a safe limit. And it notes that too many lights in parallel on a car battery glow dimly, even when the battery is fresh.

OpenStax explains that power distribution mostly uses parallel connections so devices share one voltage and operate independently, but the supply wires carry the sum of all the branch currents. The power dissipated in those wires is P = I²R_w, so an excessive current makes them overheat; fuses and circuit breakers open the circuit automatically when a sustained current exceeds safe limits.

The battery itself limits the load too. OpenStax notes that connecting an excessive number of 12 V lights in parallel to a car battery leaves them dim even when the battery is fresh: its output voltage falls under the overload because of its internal resistance.