Level 1 · Light · project 14 of 24 · 15 min · ●●○○○

Bright and Dim

Two red LEDs on one battery — one through 220 Ω (bright), one through 1 kΩ (candle). Same LEDs, two currents. The resistor sets each lane's brightness.

new here? Start with First Light — one battery, one resistor, one LED. It teaches the loop, LED direction, and why the 220 Ω resistor matters.

You'll make a bright + dim LED pair

You'll learn resistor sets LED brightness

Watch for 220 Ω vs 1 kΩ bands — red-red-brown vs brown-black-red

What you need

🔧 Try before wiring →

1× 3×AA battery pack (4.5 V)
1× Solderless breadboard
6× Jumper wires
1× Slide switch one master slide feeds both lanes together — the "arm" switch
1× 220 Ω resistor red-red-brown — the BRIGHT lane, First-Light bump
1× 1 kΩ resistor brown-black-red — the DIM lane, candle-sized bump
2× Red LED (5 mm) two identical reds — same bag, same batch; the brightness difference is PURELY the resistors

you'll learn: Circuits are loops · Ohm's law · Series & parallel · Switches · LED polarity

Circuit map

The circuit you'll build, running on the Workbench right now.

simulating…

🔧 Open in Workbench 📱 Build this in the app Test this exact circuit virtually, then rewire it however you like. On an iPhone or iPad with the app installed, this page opens the circuit straight on your bench.

If it doesn't work

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

Both LEDs are the same brightness. You either have the same resistor on both lanes, or the LEDs are swapped with the resistors (two resistors on one lane, zero on the other → the resistorless lane would pop immediately). Read the colour bands: lane A should be red-red-brown (220 Ω), lane B brown-black-red (1 kΩ). If both are the same colour, you're missing one of the two values.
The bright lane LED is DIM and the dim lane LED is BRIGHT. You swapped the two resistors between lanes. Pull them and put the 220 Ω in the lane you call bright, 1 kΩ in the lane you call dim. (Or decide this is the layout you wanted — either way, the brightness follows the resistor.)
One LED is bright, the other is completely dark. The dark lane's LED is flipped (long leg must be toward the resistor), or the dark lane's resistor is 10 kΩ (brown-black-ORANGE — would give 0.27 mA, below the visible glow threshold). Check both: pull the LED first and re-seat long-leg-up, then verify resistor bands.
Both LEDs are dark. The master slide is open, or the + rail isn't jumpered to both lanes, or battery pack switch is off. Flip the slide ON (slider visible past the detent), confirm the battery pack switch is on if your pack has one, and trace the + rail from the slide's output to each lane's top.
The bright lane LED is MUCH brighter than First Light (uncomfortable to look at). Your 'bright lane' resistor is probably 100 Ω or less, or it's missing. 100 Ω gives (4.5 − 2.0) ÷ 100 = 25 mA — above the LED's 20 mA pop limit; the LED will overheat within minutes. Pull power, check the resistor bands (220 Ω = red-red-brown), replace if wrong.
LEDs light briefly, then go dark; the switch feels warm. Not expected on this circuit — 14 mA total is nothing to a slide switch. If this happens you probably have a short across the battery (a stray jumper from + rail directly to − rail), which the slide is passing. Pull power, remove any jumper that touches both rails, re-check.

What you're building

Two red LEDs sitting side by side on your breadboard, both wired to one battery through one slide switch. Flip the slide on and both LEDs light at the same instant — but one is clearly First-Light bright while the other is candle dim right next to it. Flip the slide off: both go dark together.

The brightness mismatch isn't a dud LED or a bad connection. Both LEDs are identical, from the same bag. The only difference is which resistor sits in each lane: a 220 Ω for the bright one, a 1 kΩ for the dim one. Every lane runs its own Ohm's law, so every lane gets its own current, so every lane gets its own brightness.

The big idea: lanes are independent — each one does its own math

In Three Sisters, three LED lanes each had the same 220 Ω and lit at the same brightness. In Step Up, three lanes with the same 1 kΩ each delivered the same 2.5 mA to the LED. The pattern there was identical lanes = identical outcomes.

This project flips that. Two lanes, different resistors, same LED colour. The lanes still don't talk to each other — one 220 Ω knows nothing about the 1 kΩ next to it — but because they carry different speed bumps, they carry different currents. Side-by-side proof that the resistor is what sets LED brightness, not the LED itself.

Do the math — two lanes, two Ohm's laws

Bright lane (220 Ω, same as First Light):

Battery push: 4.5 V
LED eats: ~1.98 V (red forward voltage at ~11 mA)
Leftover for the 220 Ω: 4.5 − 1.98 = 2.52 V
Lane current: I = V ÷ R = 2.52 ÷ 220 ≈ 11.4 mA — First-Light bright

Dim lane (1 kΩ, same as Step Up with one slide):

Battery push: 4.5 V
LED eats: ~1.88 V (same LED, but at lower current the forward voltage relaxes a bit — fewer electrons means less squeeze)
Leftover for the 1 kΩ: 4.5 − 1.88 = 2.62 V
Lane current: I = V ÷ R = 2.62 ÷ 1 000 ≈ 2.6 mA — candle dim

Battery total:

Both lanes together: 11.4 + 2.6 ≈ 14 mA
The battery sees the sum of all independent lane currents. The 3×AA pack handles cool.

So the LED on the 220 Ω lane runs at about four times the current of the LED on the 1 kΩ lane. Your eye reads this as "distinctly brighter" — not quite 4× brighter in perceived light (the eye follows a log curve), but very clearly a lamp vs. a glow.

Why the LED's forward voltage changes between lanes

The two LEDs are identical. They "eat" slightly different amounts of voltage (1.98 V vs. 1.88 V) because they're conducting different currents, and a diode's forward voltage follows the Shockley equation:

Vf = N × Vt × ln(I ÷ IS)

where Vt ≈ 25.85 mV and the natural log ln grows slowly. Doubling the current only adds about 36 mV to Vf; dropping it by 4× only subtracts ~70 mV. That's the whole reason resistor sizing is a resistor problem, not a diode problem — the LED's Vf is nearly constant across the useful range, and the current is set by (V_battery − Vf) ÷ R, dominated by R.

Keep this in your head: sizing an LED current is a resistor sizing problem. The LED just does what it always does — eat ~2 V and conduct whatever current the resistor allows.

Where you'll see this pattern

Dashboard indicator LEDs vs. headlights. A car headlight runs at hundreds of mA; a dashboard indicator LED runs at a few. Same 12 V battery, different series resistors — exactly this circuit, scaled up.
"Power" LED next to a "status" LED on an appliance. The "power is on" LED is usually sized for a very low current (1-3 mA, like your dim lane) because it only needs to be visible, not blinding. The "warning" or "error" LED next to it may run at 10-15 mA so your eye picks it out immediately.
A multi-colour LED strip with varying brightnesses in the same product. Any time you see "ambient" vs. "task" lighting in a single device, there are two different resistors behind the scenes doing exactly this split.
Debugging a dim LED. If a lamp is dimmer than you expected, the first question is "what's the resistance in that lane?" Compare it against another lane you know is bright. The resistor is the first thing to check — the LED is almost never the problem.

Build it

Power column. Battery + at the top of column 1, − at the bottom. A jumper from column 1 bottom to the breadboard − rail.
Master slide (column 1 top). Insert a slide switch vertically at the very top of column 1, flipping ON. The slide's output sits at the top rail — this is where both lanes tap in.
Top rail (row 0). Run jumpers along the top row from column 1 to column 4 so both lanes can share the slide's output.
Lane A — BRIGHT (column 2). From the top rail at (2,0), drop a 220 Ω (red-red-brown) into column 2, then a red LED below it with its long leg at the resistor's bottom and short leg two rows further down. The LED's cathode row becomes the lane's return.
Lane B — DIM (column 4). Same shape, column over: top rail at (4,0) → 1 kΩ (brown-black-red) → red LED (long leg up, short leg down).
Bottom rail (row 3). Jumpers along row 3 from column 1 (where battery − waits) across to column 4, so both LED cathodes return to −.
Flip the master slide ON. Lane A's LED fires up at full First-Light brightness. Lane B's LED glows candle-soft right next to it. Side by side, same colour, same bag — the brightness difference is the resistor and only the resistor.

In a well-lit room the brightness ratio looks like ~2–3× to the eye (the eye logs the current). In a dark room the two LEDs look much closer in brightness because both are above threshold and the "fully on" LED can't look infinitely bright. Try it both ways to feel the eye's response curve.

Troubleshoot by swapping lanes

If one lane is dark or one is popped, swap the LEDs between lanes:

If the brightness pattern follows the resistor (whichever lane has 220 Ω is brighter), your LEDs are both fine and the circuit works — the brightness lives in the resistor.
If the brightness pattern follows the LED (one specific LED is always brighter whichever lane it's in), you grabbed two LEDs with different forward voltages (manufacturing spread) — rare but possible.
If one LED stays dark no matter which lane it's in, that LED is dead, flipped, or in a bad hole. Pull and reinsert long-leg-up.

🧠 Your challenge

No single right answer. That's the point.

Make the dim lane TWICE as bright without changing resistors. Replace the 1 kΩ with… another 1 kΩ in parallel with it. R_parallel = 500 Ω → I = 2.6 ÷ 500 ≈ 5.2 mA, about double. (You've just rediscovered the parallel-resistor rule from Step Up. The dim lane is now halfway between candle and First Light.)
Make the bright lane HALF as bright by adding ONE part. Add a 220 Ω in series with the existing 220 Ω → R = 440 Ω → I ≈ 5.7 mA (close to the modified dim lane). Series stacks resistance; parallel shrinks it. Both arrangements give you ~half current from different directions.
Match the two lanes. Replace the bright lane's 220 Ω with a 1 kΩ. Now both lanes have 1 kΩ, so both LEDs light at ~2.6 mA — a Twin Candle. Or replace the dim lane's 1 kΩ with a 220 Ω and get a Signal Pair with no buttons: both at 11 mA.
Measure the lane currents with a multimeter in series. Place the meter in series with the 220 Ω — expect 11 mA. Then in series with the 1 kΩ — expect 2.6 mA. Then in series with the battery pack as a whole — expect the sum: 14 mA. Three measurements, one Kirchhoff's current law.
What if you used a THIRD colour? Replace the dim lane's red LED with a green one. Green has Vf ≈ 2.1 V (0.1 V higher than red). The 1 kΩ now sees (4.5 − 2.0) ÷ 1 000 = 2.5 mA instead of 2.6 mA — barely different. The lane doesn't care about colour when there's plenty of voltage left over. (One for All only fails because the LEDs share the resistor and compete; here each lane has its own 1 kΩ bottleneck.)
Reverse the lanes. Swap the 220 Ω and the 1 kΩ. The physical left/right of the breadboard doesn't matter — the bright lamp is always the one with 220 Ω. Nothing on the breadboard knows what "left" means.
Design a four-stop brightness scale. Using what you now know, sketch four lanes using 220 Ω, 470 Ω, 1 kΩ, and 2.2 kΩ. What currents do each give? (Approximate: 11.5, 5.3, 2.6, 1.2 mA — a visual "staircase" of brightnesses. Add a slide to each and you have a four-way lamp selector.)

For grown-ups: safety notes

Battery-safe at every state. Both lanes together draw ~14 mA from the 3×AA pack — well within what AAs supply cool. The 220 Ω burns V² ÷ R = 2.52² ÷ 220 ≈ 29 mW; the 1 kΩ burns 2.62² ÷ 1 000 ≈ 7 mW. Both are tiny; neither resistor gets warm.
Each lane needs its OWN resistor. The whole point of this build is that each lane sets its own brightness. If you share one resistor between both LEDs (either lane's resistor serving both), you've built One for All — two LEDs fighting over one bottleneck, mismatched brightnesses that have nothing to do with the design you intended. Keep one resistor firmly per lane.
Red LED polarity. Long leg (anode) toward the resistor (and the + rail), flat side (cathode) toward − and the bottom rail. A flipped LED makes that lane completely dark no matter what resistor it has. Classic L1 "half the LEDs are dark" fault.
220 Ω is red-red-brown, 1 kΩ is brown-black-red — these are easy to confuse because both use brown and red bands. The giveaway: 220 Ω has red FIRST, 1 kΩ has brown FIRST. If you swap them accidentally, the brightness pattern reverses (the "dim" lane is now bright, and vice versa). Lesson still works — it's just inverted — but check the bands twice if you expected a specific lane to be the bright one.
Two identical red LEDs — try to use two from the same bag. If you grab one red and one yellow (both have Vf ≈ 2.0 V), the colours differ but the math is unchanged — the lane currents are the same. Different colours work fine as long as each lane has its own resistor.
Small parts stay away from pets and toddlers. Resistor leads and LED legs are the same swallow-hazard shape as every other breadboard piece. Never wire this to mains.

Checked against

Keep building

shares ideas with this one

Signal Pair Two push buttons, two LEDs, one battery — no sharing. Each button lights its own lamp with its own current, independent of the other. L1 15 min Three Sisters Three red LEDs side by side from one battery — each with its own 220 Ω — proving parallel lanes do their own Ohm's law and the battery adds the bill. L1 15 min One for All Three LEDs on ONE shared 220 Ω — the shortcut that fails. Red and yellow dim-split the current; green sits dark. Lesson: each LED needs its own resistor. L1 15 min