The Brightness Lab

What you need

If it doesn't work

The boring ways this circuit actually breaks — check these first before you unwire everything.

Both LEDs look equally bright. Same-value resistor in both lanes. That's a fair test showing they match — now swap one for the 1 kΩ or 10 kΩ to see Ohm's law kick in.

One LED dark. Flip it — long leg toward +. LEDs are one-way doors; reversing them is harmless.

10 kΩ lane looks completely off. That's the lesson: bigger resistance, smaller current. Look straight down at the LED in a shadow — you should see a tiny glow. If nothing at all, check the resistor colour bands.

Both LEDs dark. Battery pack switch, then that each lane has its own path from + through resistor through LED back to −.

1 kΩ lane also totally dark. Check the bands. 1 kΩ is brown-black-red, NOT brown-black-orange (that's 10 kΩ).

What you're building

Not a gadget this time — an experiment. Two identical LED lanes, side by side, so your eyes become a measuring instrument. Scientists call this a controlled comparison, and it's how you'll make Ohm's law something you've seen, not just read.

Set up the lab

Build two First Light lanes next to each other (no switches needed): battery + → resistor → LED long leg → LED short leg → battery −.
Lane A always keeps its 220 Ω resistor. It's your control — the thing that never changes, so everything else can be compared against it.
Lane B is where you experiment.

Run it: predict → swap → observe

For each round, write your prediction down first. (A prediction you didn't write down is very easy to quietly "always have been right" about.)

Round 1 — 220 Ω vs 220 Ω. Both lanes identical. They should match. This proves your lab is fair.

Round 2 — 220 Ω vs 1 kΩ. The math says lane B drops to 2.5 V ÷ 1000 Ω = 2.5 mA — about a quarter of lane A's 11 mA. Prediction: dimmer, but by how much? Now look. Is brightness proportional to current, or does your eye play tricks? (Eyes compress brightness — a quarter of the current looks like more than a quarter of the light. Your eye is a measuring tool with a built-in bias. Real scientists learn their instruments' biases.)

Round 3 — 220 Ω vs 10 kΩ. Now lane B gets about 2.5 V ÷ 10 000 Ω ≈ 0.25 mA (a touch more in practice: at tiny currents a red LED only takes about 1.9 V, leaving 2.6 V for the resistor — the "entry fee" is not a fixed 2.0 V, it shrinks a little as the current drops). A quarter of a milliamp! Will it glow at all? Look with the room lights on, then off. LEDs still glow at surprisingly tiny currents.

Round 4 — two LEDs in series. Put both LEDs in one lane, in a chain, with the 220 Ω. Each LED needs its "entry fee" first, and at this low current a red LED takes about 1.95 V, so: 4.5 − 1.95 − 1.95 ≈ 0.6 V left, and 0.6 ÷ 220 ≈ 2.8 mA. Dim — and now you know why dim, before you ever see it.

What you proved

One equation predicted every round before your eyes confirmed it: current = leftover voltage ÷ resistance. Engineers use it the same way: predict first, build, then check.

🧠 Your challenge

No single right answer. That's the point.

Design a lane that's as dim as you can make it while still visibly glowing in a dark room. Which resistor combo wins? What current is that?
Round 4 left only about 0.6 V spare. Predict: will three LEDs in series light at all on 4.5 V? Work out the leftover voltage, then test.
Resistors in series add up: what current do 220 Ω + 1 kΩ chained give? Calculate it, then compare that lane's glow to your Round 2 result.

For grown-ups: safety notes

Battery-safe voltages — experiment freely, nothing here can shock you.
Keep the control lane's 220 Ω in place. The one lane you never want to test is "no resistor at all" — that's the burned-out-LED experiment.
Staring directly into a bright LED up close is unpleasant; compare the lanes from the side.

The Brightness Lab

What you need

Circuit map