Level 1 · Light · project 13 of 24 · 15 min · ●●○○○
Longer Loop
Three 1 kΩ speed bumps stacked end-to-end on one lane feed a single red LED. Resistors in series ADD up — the LED is throttled to a near-dark 0.87 mA.
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
- 6× Jumper wires
- 1× Slide switch the master gate — one switch for the whole series chain
- 3× 1 kΩ resistor brown-black-red × 3 — three identical speed bumps stacked end-to-end, same lane
- 1× Red LED (5 mm) long leg toward the resistor chain, short leg toward −
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 — shown with a resistor swapped.
simulating…
If it doesn't work
The boring ways this circuit actually breaks — check these first before you unwire everything.
- LED is completely dark even in a dim room. Either the LED is flipped, the slide is open, or the chain has a gap. Pull the LED and reinsert long-leg-toward-the-chain. If still dark, bridge the three resistors with a jumper (temporarily) and look for a 2.5 mA candle glow — that confirms the LED and the slide are fine, so the fault is somewhere in the three-resistor chain.
- LED is bright — far brighter than a candle. Two of your 1 kΩs are accidentally parallel instead of series. Look at where their leads land: each resistor's right lead must be the ONLY wire going into the next resistor's left lead. Row-sharing between two resistors turns 2 kΩ series into 500 Ω parallel — a 4× current jump.
- LED glows noticeably but it looks orange, not red. The LED is probably running at ~2 mA (one resistor missing from the chain, so R = 2 kΩ). Count the resistors — you should have exactly three in series. If one looks very short, it may be a bent leg forming a tiny loop that shorts around one of them. Pull and re-seat.
- Loop current reads 1 mA instead of ~0.87 mA. One of your '1 kΩ' resistors is actually a 470 Ω (yellow-violet-brown) or similar. Pull each one and read the bands: brown-black-red is 1 kΩ. If one is wrong, swap it for the right value — the chain adds, so a 470 Ω drops R_total to 2.47 kΩ and nudges current up by ~15 %.
- LED stayed dark, then suddenly flashed when I touched a wire. You bypassed one of the resistors by touching your finger (humans are moist and conduct a bit). Not a bug — your body made a parallel lane. Clean break: keep fingers off the leads while powered. If the flash is bright and painful enough to notice, the "parallel finger" was effectively shorting out a chunk of the 3 kΩ.
What you're building
A single red LED, a single slide switch, and three 1 kΩ resistors in a row between them. Flip the slide on and the LED barely glows — a hint of red that you can only see in a dark room, cupping your hand around it. Flip it off and the hint vanishes.
That hint is the lesson. The LED isn't broken; the current is simply tiny, because three resistors in a chain each take their bite out of the voltage budget. This is the sibling project to Step Up, which built the same three 1 kΩs in parallel and finished at 7.5 mA. Here they sit in series and finish at under 1 mA. The parts are identical. The arrangement is the whole story.
The big idea: series resistors ADD
In Twin Reds, you learned that two LEDs stacked in series share one lane and share one current — each LED added its own 2 V bite to the voltage budget until the loop ran out of headroom. The same rule applies to resistors in series:
R_total = R1 + R2 + R3
Three 1 kΩs stacked on one lane:
R_total = 1 000 + 1 000 + 1 000 = 3 000 Ω
That's three times the resistance of a single 1 kΩ lane, in the same loop as a single LED. Ohm's law then does what Ohm's law always does: divides the leftover voltage by the resistance.
Do the math
- Battery push: 4.5 V
- LED eats: ~1.88 V (red LEDs drop less than their 2.0 V book value at low current — fewer electrons, less squeeze across the junction)
- Leftover for the resistor chain: 4.5 − 1.88 = 2.62 V
- Resistor total: 3 000 Ω
- Current: I = V ÷ R = 2.62 ÷ 3 000 ≈ 0.00087 A = 0.87 mA
Under one milliamp. The 5 mm red LED's data-sheet "visible glow" threshold is right around 1 mA. You've landed just below it. The LED is on in the sense that current is flowing in the right direction through its junction — but there's so little of it that the eye barely picks up the red.
Compare the whole family:
| Build | Loop total R | LED current | How it looks |
|---|---|---|---|
| First Light | 220 Ω | 11 mA | Fully lit (bright) |
| Signal Pair / Three Sisters per lane | 220 Ω | 11 mA | Fully lit |
| Step Up (3 slides × 1 kΩ parallel) | 333 Ω | 7.5 mA | Full lamp |
| Step Up (1 slide × 1 kΩ) | 1 000 Ω | 2.5 mA | Candle |
| Longer Loop (3 × 1 kΩ series) | 3 000 Ω | 0.87 mA | Just-visible pinprick |
Same battery, same LED, same resistor value. Resistance climbs — current falls — brightness falls with it.
Series vs. parallel, side by side
The rules are symmetric, and the reason they're symmetric is deep:
- SERIES means one loop, one current. Add resistors to the loop and each one takes its own bite from the voltage budget. Resistances ADD: 1k + 1k + 1k = 3 kΩ.
- PARALLEL means separate loops, independent currents. Add resistors as new lanes and each one carries its own share. Resistances COMBINE SMALLER by the reciprocal rule — for identical resistors, R/n: 1k || 1k || 1k = 333 Ω.
A quick memory trick: series stacks the same way parts stack physically (end-to-end, lots of them in a line = lots of resistance to push through). Parallel is the "many doors" case — more lanes mean easier passage, so resistance drops. If this feels backwards, run the math on both arrangements with the same parts: Step Up for parallel (7.5 mA), this build for series (0.87 mA). The arithmetic burns the pattern in.
Where you'll see this pattern
- Resistor substitution upward. You need a "roughly 3 kΩ" resistor to tame a buzzer or set a sensor divider, and your bag only has 1 kΩs. Three in series = 3 kΩ. Buy two values of resistor, make six values with a jumper and some clever stacking. (Step Up does this trick in the other direction — three parallel 1 kΩs = 333 Ω.)
- Night-light trim. When you want a soft indicator LED that's barely visible — "the power is on" — you don't need a 220 Ω burning 25 mW on an 11 mA glow. A single 10 kΩ, or three 1 kΩs in series, gives a quarter of a milliamp and a clean dim glow for essentially zero power.
- Reading a resistor divider by ear. Audio engineers chain resistors to make precise attenuations ("drop the signal by 6 dB"). The chain adds. The math is the same math you just did.
- Figuring out why your LED is dim. The first question when a lit LED surprises you with how dim it is: how much series resistance is actually in the loop? Add up every resistor in the single loop — if the sum is over 1 kΩ for a 5 mm LED on 4.5 V, you've landed in dim-territory and the fix is less resistance, not a bigger battery.
Build it
- Power column. Battery + at the top of column 1, − at the bottom. A jumper to the − rail.
- Master slide (column 2). Slide switch sitting horizontally in the top row, left lead on the + rail, right lead on the next column over. Flip it on to arm the chain.
- The three resistors. Three 1 kΩs (brown-black-red) stacked end-to-end along the top row: slide output → 1 kΩ A → 1 kΩ B → 1 kΩ C → open hole in column 6. Each resistor's right lead becomes the left lead of the next one. One continuous electrical line, three resistors long.
- The LED (column 6). Insert the red LED vertically, long leg (anode) in the row that holds resistor C's right lead, short leg (cathode) two rows down.
- Bottom rail. A jumper from the LED's cathode row back to the breadboard's − rail, and from there to the battery − terminal.
- Flip the slide on. In a lit room, you'll see nothing. Cup your hand over the LED and lean in — a soft red hint should emerge. Switch the room lights off and the LED is clearly on, just dim.
- Pull one of the three 1 kΩs out. The chain is now 2 kΩ — current rises to 2.65 ÷ 2 000 ≈ 1.33 mA, right at the "clearly visible" threshold. Pull two out (just one 1 kΩ left in the chain), and the LED jumps to ~2.5 mA (the Step Up "candle" step). Pull all three and short the gap with a jumper — bare 220 Ω would make this First Light, but 0 Ω with no resistor at all would pop the LED. Don't do that.
Troubleshoot by counting bumps
If the LED is darker than expected (or completely dark), count the resistors in the loop:
| Resistors in the loop | Expected current | How it looks |
|---|---|---|
| 0 (bare wire — DO NOT BUILD) | 25+ mA | LED pops |
| 1 × 1 kΩ | 2.5 mA | Candle glow |
| 2 × 1 kΩ | 1.33 mA | Clearly dim but visibly lit |
| 3 × 1 kΩ | 0.87 mA | Pinprick in a dark room |
| 4 × 1 kΩ | 0.65 mA | Barely visible, even in the dark |
The dimmer the LED, the more resistance in the loop. The brightest fault to look for is a lane with less than you think — maybe one of the "1 kΩ"s is a 100 Ω (brown-black-brown). Read the colour bands twice.
🧠 Your challenge
No single right answer. That's the point.
- Predict a four-bump chain. Add one more 1 kΩ in the series chain. What's the new current? (R_total = 4 000 Ω, I = 2.62 ÷ 4 000 ≈ 0.66 mA — about 25 % dimmer than the three-bump version. The brightness drop between 3 and 4 bumps is much less visible than between 1 and 2, because the human eye follows a log curve.)
- Mix two different values. Replace one 1 kΩ with a 220 Ω. New total: 220 + 1 000 + 1 000 = 2 220 Ω. Current: 2.62 ÷ 2 220 ≈ 1.18 mA. The series chain adds whatever is in it — identical resistors just make the math easier. (This is why dim-candle works: 1 kΩ + 10 kΩ in series = 11 kΩ for a 0.25 mA candle; closing the bypass drops the chain to just 1 kΩ for a 2.6 mA glow.)
- Chain two LEDs onto the same loop. Add a second red LED in series after the three resistors. Now the voltage budget has to feed TWO LEDs (~2 V × 2 = 4 V) plus the 3 kΩ chain. Does any current flow? (Just barely — 4.5 − 4.0 = 0.5 V across the chain → 0.17 mA. The LEDs are essentially dark; you've landed below the "visible glow" threshold entirely. The same reason Twin Reds is dim-but-lit, pushed further by the extra 3 kΩ.)
- Rebuild in parallel. Take the same three 1 kΩs and arrange them as three parallel lanes under one master slide (one lane each). That's Step Up with the master gating all three — current rises from 0.87 mA to 7.5 mA, roughly nine times brighter. Build both back-to-back if you have the parts; the brightness difference is the whole series-vs-parallel lesson in one evening.
- Design a 50 % dimmer. Starting from First Light (220 Ω, 11 mA), how many ohms do you need to add in series to drop the LED to about 5 mA? (Set up Ohm's law backwards: (4.5 − 2.0) ÷ 0.005 = 500 Ω total. So you need to add 280 Ω in series with the 220 Ω → closest standard values are a 330 Ω, or 220 Ω + 100 Ω stacked. Hold a thumb over half the LED and compare brightness — the eye says "a bit dimmer", the math says "half the current".)
- Walk the chain with a multimeter. If you had a meter, what voltage would you read across each individual 1 kΩ in the three-bump chain? (Each one burns I × R = 0.87 mA × 1 000 = 0.87 V. Three of them: 0.87 × 3 = 2.62 V, which equals the budget leftover after the LED. KVL closes.)
For grown-ups: safety notes
- Battery-safe at every length of chain. Three 1 kΩs in series means 0.87 mA — one eleventh of First Light's current, one twenty-third of the LED's 20 mA pop threshold. Nothing in this build gets warm. If any resistor feels warm you've built a short that bypasses at least two of them.
- Each resistor must sit END-TO-END with the next. The whole point of a series chain is that current has only one path and must squeeze through every resistor. If two resistors share a row (parallel instead of series), the pair becomes 500 Ω instead of 2 000 Ω, and the chain's current triples. If all three share a row, you've accidentally built Step Up with a candle step. Watch where the resistor leads land: each resistor's right lead must be the ONLY wire going into the next resistor's left lead.
- Red LED polarity. Long leg (anode) toward the resistor chain, flat side (cathode) toward −. A flipped LED with a 3 kΩ throttle will stay dark forever — unlike First Light where you'd see the dark LED immediately, here the LED should look dim-near-dark, so a flipped LED is harder to spot. If you can't see any glow even in a dark room with your hand cupped, flip the LED first before suspecting anything else.
- 1 kΩ colour code is brown-black-red (three identical resistors). It's easy to confuse with 10 kΩ (brown-black-orange) or 100 Ω (brown-black-brown). A 10 kΩ snuck into the chain gives R_total ≈ 12 kΩ → 0.22 mA (invisible). A 100 Ω snuck in gives R_total ≈ 2.1 kΩ → 1.25 mA (noticeably brighter). Read the colours twice and keep the three resistors oriented the same way so the bands are easy to compare.
- Switch orientation doesn't matter, but switch POSITION does: the slide must be the FIRST element after the + rail, so you can arm the chain without pulling a jumper. A slide buried inside the chain still works but makes "arm" less obvious.
- 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
You might also like
shares ideas with this one