Level 1 · Light · project 10 of 24 · 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.
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.
What you need
- 1× 3×AA battery pack (4.5 V)
- 1× Solderless breadboard
- 7× Jumper wires
- 1× Slide switch one master slide feeds all three lanes — the "arm" switch
- 3× 220 Ω resistor ONE per LED — no sharing. Three resistors is non-negotiable; see the sibling project *One for All* for why.
- 3× Red LED (5 mm) three identical reds, long leg of each toward the + rail
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…
If it doesn't work
The boring ways this circuit actually breaks — check these first before you unwire everything.
- All three LEDs are dark. Master slide is off, or the battery pack switch is off, or the top rail jumper doesn't reach past column 1. Check each in that order.
- One LED is dark, the other two glow. That one LED is flipped. Pull it, rotate 180°, push back in. The parallel topology means a flipped LED only kills its own lane — a free debug clue.
- Two LEDs lit, the third is very bright and the others dimmer. Lane C skipped its 220 Ω — current from the + rail is going straight through the LED. Pull the LED FAST (before it pops) and insert the resistor.
- All three LEDs are visibly dimmer than First Light. You wired them in parallel through ONE shared 220 Ω instead of giving each its own. That's the *One for All* mistake. Rewire so every lane has its own resistor between the top rail and the LED.
- Flipping the master slide does nothing — LEDs stay on. A rail jumper bypasses the master. Pull the master slide out of the breadboard; if LEDs stay lit, there's a sneak path. Trace from battery + to the top rail — every path should go THROUGH the slide, not around it.
- LEDs flicker when you press on the breadboard. Loose leads in a lane — usually the resistor isn't fully seated. Press each lane's resistor and LED firmly; the flicker will go away as the loose one gets contact.
What you're building
A row of three red lamps across your breadboard, all lit at full First-Light brightness the moment you flip one slide switch. Flip it off: all three go dark together. Hold a finger over any one lamp — the other two keep glowing exactly as before. The lanes are side by side, but they don't talk to each other.
This is what a bank of indicator lights looks like inside anything you own: dashboards, charger status rows, equaliser bars on a stereo. The topology is: one master switch, three independent lanes, each lane complete in itself.
The big idea: parallel lanes each do their own math
Twin Reds taught you that LEDs in series share the loop current — one LED's physics changes the other's. This project is the opposite move: three LEDs in parallel, each sitting on its own branch, each wired straight back to the battery through its own resistor.
Because every lane is a complete loop from + to − on its own, every lane runs the First Light calculation independently:
I = (V_battery − V_LED) ÷ R = (4.5 − 2.0) ÷ 220 ≈ 0.011 A = 11 mA
Lane A doesn't know lane B exists. Lane B doesn't know lane C exists. All three carry 11 mA when the master is on, same as a single First Light.
Do the math — three lanes, one battery
Per-lane current (identical for all three):
- Battery push: 4.5 V
- LED eats: ~2.0 V (red forward voltage)
- Leftover for the 220 Ω: 4.5 − 2.0 = 2.5 V
- Lane current: I = V ÷ R = 2.5 ÷ 220 ≈ 11 mA
Battery total (what the pack has to supply):
- Three identical lanes, each pulling 11 mA
- I_total = 3 × 11 = 33 mA
Think of the battery like a water tap and the three lanes like three garden hoses off a splitter. Each hose runs at its own flow rate, set by its own nozzle (the 220 Ω). The tap just has to supply the sum of all three flows. Add a fourth hose — the tap supplies four flows. Add ten — ten.
Your 3×AA pack is rated for roughly 1 500 mAh at low loads. At 33 mA this bank runs for ≈ 45 hours of continuous light. A single AA has that much energy in it; three of them in the pack has three times. Nothing is close to stressed.
Why "independence" matters
Three real-world things only work because of this trick:
- One lamp failing doesn't take the others down. Pull any one LED out of the breadboard mid-run. The other two keep glowing at 11 mA, unchanged. The failed lane just goes dark; the healthy lanes don't notice. Compare with Twin Reds: pull one LED from that series chain and both go dark, because they share the current.
- You can mix brightnesses lane-by-lane without affecting the others. Swap one 220 Ω for a 1 kΩ and that lane goes dim (2.5 mA). The other two still carry 11 mA. The resistor on a parallel lane controls only its own lane — nothing bleeds across.
- Lanes can be different colours or output types. Replace the red at lane C with a yellow — it still runs at 11 mA because yellow's forward voltage is also about 2.0 V. The math is lane-local. (A green LED needs ~2.1 V, so its lane runs at ~11.3 mA — tiny difference, same formula.)
Independent-parallel lanes are the fundamental building block of anything that drives multiple outputs at once. Every project from here that lights more than one thing uses this pattern.
Build it
- Power column. Battery + at the top of column 1 of your breadboard, − at the bottom. Drop a slide switch into column 1 at the very top — this is your master slide.
- Top rail. A jumper fans out along the top row from the master switch's output, reaching across six columns. This row carries + to every lane. (On a real breadboard, the top power rail runs the whole length — you just need a wire from the master to that rail.)
- Lane A (col 2). From the top rail, drop a 220 Ω straight down. Below the resistor, insert a red LED, long leg at the top (toward +, matching the resistor's output). Short leg down to the bottom rail.
- Lane B (col 4). Repeat exactly: top rail → 220 Ω → red LED (long leg up) → bottom rail.
- Lane C (col 6). Repeat again: top rail → 220 Ω → red LED (long leg up) → bottom rail.
- Bottom rail. A jumper from each LED's short-leg hole back to the breadboard's − power rail, and from that rail to the battery − terminal.
- Flip the master slide ON. All three LEDs light at full brightness, same as First Light. Flip OFF: all three dark.
Troubleshoot by lane: if one LED is dark and the others are fine, the fault is in that lane only — almost always a flipped LED. If all three are dark, the master slide is off, the battery pack switch is off, or the + rail isn't fanned out past the first lane.
Where you'll see this
- LED strings in series/parallel combinations. Decorative string lights are a long parallel chain of short series groups — each 2-or-3-LED group has its own tiny resistor, and the groups hang in parallel across the mains supply. If one group fails open, the others stay on; if your string is "when one goes out they ALL go out", it's a straight series string, not a parallel one.
- Charge-level bar graphs. Battery meters, VU bars on an audio mixer, mode indicators on an amp — a horizontal row of 5–10 LEDs, each driven from its own transistor and its own resistor. Lane independence is why one meter LED can go bright while its neighbour is still dark.
- Pilot lights on power strips with per-outlet switches. Each outlet gets its own switch and its own lamp. Flipping one does nothing to the others. Same topology as this build, scaled to mains voltage with a different lamp.
Learn parallel independence once, see it everywhere.
🧠 Your challenge
No single right answer. That's the point.
- Add a fourth lane. Daisy-chain a fourth 220 Ω + red LED off the + rail. Predict the new battery current before you measure. (Answer: 4 × 11 = 44 mA. The per-lane current doesn't change at all — each lane still does its own 11 mA. Only the battery total goes up.)
- Swap one 220 Ω for a 1 kΩ. Replace the resistor in lane C only. What should happen? (Lane C current = 2.5 ÷ 1 000 = 2.5 mA — about a fifth of the brightness. Lanes A and B still carry 11 mA each. Battery total: 11 + 11 + 2.5 = 24.5 mA. Lane-local change, lane-local effect — the hallmark of parallel independence.)
- Mix colours. Replace lane C's red LED with a green. Green LED Vf ≈ 2.1 V. Redo the math: lane C current ≈ (4.5 − 2.1) ÷ 220 = 10.9 mA. Nearly identical to the red lanes. Lesson: a 0.1 V difference in forward voltage barely budges the current because the 220 Ω dominates the Ohm's-law equation.
- Add a per-lane kill switch. Insert a push button between the + rail and each lane's resistor. Now you have Signal Pair with three colours. Press any combination — each lane fires only when its own button is pressed, and the battery supplies exactly the sum. Which real device uses this? (Game-show buzzer panels, DJ pad lamps, the mode indicator row on a synth.)
- Pull one LED out while the circuit is running. Confirm with your own eyes that the other two stay at full brightness. Now pull another. The last one keeps running, same brightness, until you pull it too. A parallel bank fails gracefully. Compare: pulling one LED from a series chain (Twin Reds) kills the whole chain. The two topologies have opposite failure modes.
- Series-vs-parallel brightness race. Rebuild Twin Reds (two reds in series through one 220 Ω) next to a two-sister version of this project (two reds parallel, each with their own 220 Ω). Which is brighter per LED? (The parallel version: each LED carries 11 mA, the series version carries about 2 mA per LED since the two LEDs together eat 4 V, leaving only 0.5 V for the resistor. Four to five times brighter per lamp. The cost: parallel uses two resistors instead of one, and twice the battery current.)
For grown-ups: safety notes
- Battery-safe at every lane count. Three lanes = 33 mA, four = 44 mA, ten would be 110 mA — all well within what a 3×AA pack supplies cool. Nothing warms in this build; if any part feels warm, you have a short somewhere, usually a lane that skipped its 220 Ω.
- Each LED gets its own 220 Ω — no sharing. The temptation when you have three LEDs is "I only need one resistor, right? The battery is just 4.5 V." Wrong, and we built a whole second project (One for All) specifically to show you why. One shared resistor means the three LEDs fight over the current and the one with the lowest forward voltage hogs it. Give every LED its own 220 Ω.
- Three LEDs, three polarities to get right. Long leg = anode = toward +, flat side = cathode = toward −. If one lane is dark while the others light, the FIRST thing to flip is that lane's LED (it is almost always flipped). The parallel topology makes this debug trivial: a flipped LED only breaks its own lane, so working siblings prove the rail is fine.
- 220 Ω colour code is red-red-brown. Easy to mix up with 220 kΩ (red-red-yellow) or 1 kΩ (brown-black-red) if you grabbed from the wrong bag. In this circuit a wildly-wrong resistor doesn't fry anything, but you'll chase a brightness mystery (dim LEDs with 2.2 kΩ, almost-dark with 220 kΩ).
- Small parts stay away from pets and toddlers. Resistors, LED leads and the slide are the same swallow-hazard shape as every other breadboard piece.
- Never wire this to mains. The 3×AA pack is the whole supply; higher voltages blow past the 220 Ω's safe current window even on a healthy LED.
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