Level 1 · Light · project 17 of 24 · 15 min · ●●○○○
Short Cut
First Light with a slide switch wired straight across the LED. Flip the slide and the LED goes dark even though the loop stays complete.
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 the SHORT — wires in parallel with the LED, not in series with the loop
- 1× 220 Ω resistor red-red-brown — the current limiter that keeps the LED (and the slide) safe in BOTH states
- 1× Red LED (5 mm)
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.
- The LED is dark in both slide positions. Either the LED is flipped (dark in the open-slide state too) or the loop is broken past the 220 Ω. Pull the LED and reinsert long-leg-toward-the-220. If still dark in both states, remove the slide completely and verify it's First Light — one slide at a time, add complexity after you've got the basics.
- The LED is bright in BOTH slide positions. The slide isn't actually shorting the LED. Its two jumpers aren't landing on the LED's anode and cathode rows. Trace: anode row ↔ col 5 row 1 ↔ slide ↔ col 5 row 2 ↔ cathode row. If one of those jumpers is on the wrong row (very easy on a breadboard, where rows look identical), the slide is a dead-end and the LED just lights normally.
- I close the slide and smell something burning. You've shorted the 220 Ω instead of the LED. Pull the slide IMMEDIATELY. The LED is probably already dead from the overcurrent spike. The 220 Ω is one of your two protection parts; it must always stay in the loop. Rewire so the slide straddles the LED's two leg rows only.
- LED is dim (not dark) when the slide is closed. Your 'slide' has substantial contact resistance — maybe a dirty old slide switch, or you're using a push button that doesn't fully close. Replace the slide with a known-good one. Flaky contact resistance across the LED gives the characteristic 'weirdly dim' look that mimics a real circuit fault — great for teaching but confusing in a real build.
- LED works but the 220 Ω gets noticeably warm after a minute in slide-closed mode. You've picked a lower-value resistor than 220 Ω — maybe a 22 Ω (red-red-black) or 47 Ω (yellow-violet-black). At those values the loop current in the shorted state exceeds 50 mA and the resistor approaches its power rating. Swap for a real 220 Ω (red-red-brown).
What you're building
First Light with a stealth switch. The loop is identical to First Light — battery, 220 Ω, red LED, back to battery — plus one extra piece: a slide switch wired as a direct wire across the LED. Flip the slide one way: the LED is bright, exactly like First Light. Flip it the other way: the LED goes dark even though every jumper is still in place. The circuit is still a loop. Current is still flowing. The LED is just no longer part of the lit path.
This is the single most-confusing-at-first fault pattern in electronics: the loop is complete but the LED doesn't light. Beginners instinctively believe "complete loop = LED on", and the first time a loose jumper lands in the wrong hole and silently shorts around their LED, they spend an hour chasing a problem that isn't there. Building this circuit on purpose, with a slide switch making the fault repeatable, teaches your eye to spot it in a messier project later.
The big idea: current takes the easiest path
When current has a choice between two paths from the same + node to the same − node, it doesn't split equally. It splits inversely by resistance — the easier path gets most of the current, and the harder path gets almost none.
With the slide open (a gap), the LED is the only path from the 220 Ω's output back to the − rail. All the current runs through the LED.
With the slide closed (a wire, ~0 Ω), the current has two paths:
- Path A: through the LED — a diode has an effective "resistance" of around 150 Ω when conducting at ~10 mA (really it's a non-linear junction, but 150 Ω is a useful approximation).
- Path B: through the slide — a wire, essentially 0 Ω.
The current splits as 1 / R: Path B gets ~100 % of the current, Path A gets essentially nothing. More precisely, the slide clamps the voltage across the LED to ~0 V, and a diode with 0 V across it conducts zero current (that's what being a diode means — it needs its forward voltage to turn on). So the LED current drops not just "a little", but to zero.
Meanwhile the loop still carries current — just through the slide instead. The battery still feels the full 220 Ω speed bump, so I_loop = 4.5 V ÷ 220 Ω = 20.5 mA. The LED is dark; the 220 Ω is doing its job (keeping the slide and battery safe); the slide passes 20.5 mA like a wire does. Nobody's unhappy except the LED, which went from bright to dark without the loop ever opening.
Shorts starve the parts around them. A low-resistance path in parallel with a higher-resistance one (an LED, a motor, a resistor you wanted to drop voltage across) steals the current.
Do the math — two states
State 1 — slide OPEN (First Light):
- Loop: + → 220 Ω → LED → −
- Battery push: 4.5 V
- LED eats: ~2.0 V
- Leftover for 220 Ω: 2.5 V
- Current through LED and 220 Ω: I = 2.5 ÷ 220 ≈ 11.4 mA
State 2 — slide CLOSED (short across LED):
- Loop: + → 220 Ω → slide (0 Ω) → −
- LED is a parallel branch, but with 0 V across it, so I_LED = 0
- Current through slide and 220 Ω: I = 4.5 ÷ 220 ≈ 20.5 mA
- Current through LED: 0 mA — fully dark
The 220 Ω carries more current when the LED is dark (20.5 mA) than when it's lit (11.4 mA). That's unintuitive — shouldn't current drop when "less is happening"? No: the LED's 2.0 V bite was eating into the voltage budget. Remove the LED from the lit path, and the 220 Ω now sees the whole 4.5 V minus the slide's near-zero drop, so the current goes up.
Why doesn't the LED pop?
This is the critical safety point. In the shorted state, the loop current is 20.5 mA — right near the LED's 20 mA rating. If the LED were still in the lit path, 20.5 mA through it would be at the edge. But the LED is not in the lit path. It has 0 V across it. 0 V on a diode junction means 0 mA through it, no matter how much current is swirling around the rest of the loop. So the LED is safe, cool, and dark.
The 220 Ω sees V × I = 4.5 × 0.0205 = 92 mW, well under its 250 mW (quarter-watt) rating. The slide switch handles 20 mA like a wire. The battery pack supplies 20 mA like it does any day. Nothing is stressed.
The real danger would be if you removed the 220 Ω. Then the slide would short the battery directly — ~1 A, warm batteries, dead batteries in minutes, a hot slide contact. The 220 Ω keeps the shorted state safe.
Where you'll see this pattern (as a FAULT)
- A loose jumper landing in the wrong hole. This is the "my LED went dark but all my parts look right" story. Trace the LED's anode and cathode rows — if any jumper connects them directly, the LED is shorted. Pull every "extra" jumper until the LED comes back.
- A soldered joint bridging to the hole next door. On a PCB, two solder blobs touching can create exactly this fault. The LED is fine, the trace is fine, there's just a tiny tin-coloured whisker tying anode to cathode. Diagnose with a magnifier.
- A multimeter probe dropped across the LED. Shorting an LED by accident with a probe tip briefly extinguishes it; the LED relights the moment the probe lifts. Spooky if you don't know the trick.
- A transistor's collector shorted to its emitter by heat damage. Advanced, but same pattern: the "load" (LED, motor) in the collector lane goes dark/stops because a toasted transistor is now a wire.
Where you'll see this pattern (as a FEATURE)
- Flash-off buttons. Press to momentarily extinguish an LED without unpowering the whole circuit. Used in car dashboards (press to acknowledge an alarm — alarm LED dims, system stays on).
- "Blink the hard way". Rapidly toggle the short across an LED and you get an on-off blink without opening the whole loop. Not useful at 50 Hz (buzzer land), but at 1 Hz it's a hand-driven blinker.
- Testing an LED without unpowering. If you suspect an LED is dead, briefly short across it while watching the rest of the circuit. If the rest keeps running, the LED is a passive consumer that can be lifted out. If something else changes, the LED was coupling to it.
Build it
Power column. Battery + at the top of column 1, − at the bottom. Jumper from + to the row where the 220 Ω starts.
220 Ω (col 2). Lay the 220 Ω (red-red-brown) horizontally on row 1 from col 1's rail into col 2's working node.
LED (col 4). Red 5 mm LED standing vertical: long leg (anode) in row 1 at col 4, short leg (cathode) in row 2 at col 4. The anode row connects back to the 220 Ω's output via a jumper on row 1 from col 2 to col 4.
Bottom rail (row 3). Jumper from the LED's cathode row (col 4, row 2) DOWN to row 3, then along row 3 back to the battery − terminal.
The short across the LED (col 5). Here's the trick. The LED connects anode-row (1) to cathode-row (2) through its junction. We're going to make a second path from anode-row to cathode-row through a slide switch at col 5. Jumper from col 4 row 1 (anode-row) → col 5 row 1 (horizontal jumper). Slide switch vertical at col 5, from row 1 to row 2. Jumper from col 5 row 2 (cathode-row) → col 4 row 2 (horizontal jumper).
Now there are two paths from anode-row to cathode-row: the LED (through the diode junction), and the slide (through col 5). Both are in parallel with each other.
Flip the slide OPEN. The LED should light at First-Light brightness — ~11 mA. The slide is a gap, so all current goes through the LED.
Flip the slide CLOSED. The LED goes dark. Not off-the-loop dark — in-the-loop-but-starved dark. The slide is now a wire sitting across the LED, carrying 20 mA while the LED carries 0 mA. The 220 Ω is a bit warmer than before (92 mW vs. 29 mW) but not noticeably hot.
Pull a cathode-row jumper while the slide is closed. The slide's return path is broken. Current drops to 0. LED and slide are both dark. Reinsert — the slide's short reappears.
What to notice
The slide switch is not controlling the LED in the way "a switch controls an LED" usually means. It's not opening the loop. The loop is always closed as long as the battery, 220 Ω, and bottom rail are intact. The slide is only choosing where the current flows within the loop — through the LED (slide open) or around it (slide closed).
This is why a careful search of a dim-but-should-be-bright circuit always traces the LED's two terminals: anode row to cathode row. If anything bridges them directly, that something is stealing the current.
🧠 Your challenge
No single right answer. That's the point.
- What if the "short" has 10 Ω of resistance? Imagine the slide is a slightly corroded contact — not 0 Ω, but 10 Ω. Does the LED still go completely dark? (Approximately — but not exactly. With 10 Ω across a 2 V LED, there's still 2 V trying to drive 10 Ω → 200 mA ideally, but the LED's forward drop means the diode will conduct SOME current as the voltage across it rises above ~1.5 V. Rough analysis: at the lowest LED Vf that still conducts significantly (~1.7 V), current through the "short" branch is 1.7/10 = 170 mA, and through the diode is a few mA — but the 220 Ω is now in a near-short, drawing 4.5 ÷ (220 + 10/parallel-stuff) ≈ 20 mA total, split mostly through the 10 Ω. In short: a 10 Ω "near short" still mostly kills the LED but doesn't fully extinguish it. Real-world "flaky shorts" show up as "LED is weirdly dim but not off" — a huge debugging hint.)
- Short the 220 Ω instead of the LED. Rewire so the slide is parallel to the 220 Ω instead of the LED. What's the current when the slide is closed? (No 220 Ω in the loop → 4.5 − 2.0 = 2.5 V across effectively 0 Ω → 100s of mA, LED pops in a few seconds. The 220 Ω is the LED's seatbelt in First Light; shorting it is unsafe. The slide-across-LED is safe because the 220 Ω stays in the loop. Where you put the short matters a lot.)
- Add a second LED in series with the first. Where would the slide short go, and what happens? (If the slide is across just the FIRST LED, the second LED is still in the lit path — it'll still try to light at the series-rules current, but with only one LED's Vf instead of two, so the current goes UP from Twin Reds's 2.3 mA to First Light's 11.4 mA. The slide "chooses" one or both LEDs. If the slide is across BOTH LEDs together, both go dark exactly like the single-LED case here. Mind-bender: the slide works on groups, not individual parts.)
- Replace the slide with a push button. Now the short is only active while the button is held. What's the behavior? (LED default-on; pressing the button kills it momentarily. "Hush" button for a status LED — common in alarm panels where you want to acknowledge-but-keep-running.)
- Can you build a 2-LED alternator? Two red LEDs, each with its own slide-across. When slide A is closed and slide B is open, which LED lights? (The LED whose slide is OPEN. The other is shorted. Build this and you have a 2-way manual selector — the slides act like inverse-switches, one shorting its LED while the other lets its LED light. The lesson: a slide across a load is a reverse-logic switch for that load.)
- Measure the current through the 220 Ω in both states with a multimeter. How does it change? (Open-slide: 11.4 mA. Closed-slide: 20.5 mA. The 220 Ω works harder when the LED is dark — the LED's 2 V bite is gone from the voltage budget, so more leftover for the 220 Ω. This is the "paradox" every beginner has to internalise: pushing MORE current through the circuit can mean LESS light.)
For grown-ups: safety notes
- The 220 Ω must stay in the loop in both states. This project is safe BECAUSE the slide shorts the LED, not the 220 Ω. In the shorted state the 220 Ω still carries ~20 mA (92 mW across its 250 mW rating). If you accidentally wire the slide across the 220 Ω instead, the LED is unprotected and pops within seconds. Trace the slide's two ends: both must land on the two LEG-ROWS of the LED (anode row and cathode row), not the 220 Ω's two ends.
- The 220 Ω runs a touch warmer in the shorted state. 92 mW is still cool-to-the-touch; a 250 mW resistor can handle this all day. If your 220 Ω feels obviously warm after a minute, you've probably shorted the 220 Ω instead of the LED and are drawing 100+ mA — pull the slide open and recheck wiring.
- LED polarity still matters even though you're going to short it. The LED must be oriented correctly (long leg toward +, short leg toward −) because the lit state is First Light, which needs correct polarity. A flipped LED makes the open-slide state look identical to the closed-slide state (both dark), which defeats the "flip the slide to see the magic" demo. If flipping the slide doesn't change anything visible, flip the LED first.
- The slide switch's rated current. Hobby slide switches are typically rated for 300 mA to 500 mA — far above our 20 mA. The slide is doing nothing stressful here. No slide-switch model you can buy at a hobby shop will struggle with 20 mA on 4.5 V.
- Don't short directly across the battery. The whole safety story depends on the 220 Ω being present. If a wire ever runs directly from + rail to − rail with nothing in between, the battery pack sees a dead short and sinks ~1 A continuously, warming up and dying in minutes. This is a general rule, not Short-Cut-specific: never make a + → − path without a load. The 220 Ω is a load; the LED is a load; wires are not loads.
- Don't leave the slide closed indefinitely. 20 mA for an hour is fine; 20 mA for a week drains the AAs noticeably faster than usual. Switch the main power off (unplug the battery pack) when you're done playing.
- Small parts stay away from pets and toddlers. Never wire this to mains.
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