tinyΩhms

Level 1 · Light · 30 min · ●●●○○

The Brightness Lab

Run a real experiment: predict, measure with your eyes, and prove Ohm's law.

What you need

  • 1× 3×AA battery pack (4.5 V) L1
  • 1× Solderless breadboard L1
  • 6× Jumper wires L1
  • 2× Red LED (5 mm) two matching LEDs make a fair comparison L1
  • 2× 220 Ω resistor L1
  • 1× 1 kΩ resistor L1
  • 1× 10 kΩ resistor L1

you'll learn: Ohm's law · Series & parallel · LED polarity

4.5 V + lane A · 220 Ω control — never changes ? lane B · swap me 220 → 1k → 10k Ω
a fair experiment: identical lanes, change one thing, compare by eye

🔧 Open in Workbench — test this exact circuit virtually, then rewire it however you like.

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

  1. Build two First Light lanes next to each other (no switches needed): battery + → resistor → LED long leg → LED short leg → battery −.
  2. 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.
  3. 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 2.5 V ÷ 10 000 Ω = 0.25 mA. A quarter of a milliamp! Will it glow at all? Look with the room lights on, then off. LEDs are astonishingly efficient at tiny currents — this is why sensor gadgets can run for a year on a coin cell.

Round 4 — two LEDs in series. Put both LEDs in one lane, in a chain, with the 220 Ω. Each LED needs its ~2.0 V "entry fee" first, so: 4.5 − 2.0 − 2.0 = 0.5 V left, and 0.5 ÷ 220 ≈ 2.3 mA. Dim — and now you know why dim, before you ever see it. That's the whole superpower.

What you proved

One equation predicted every round before your eyes confirmed it: current = leftover voltage ÷ resistance. Engineers use it exactly the way you just did — predict first, then 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 0.5 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.

Checked against