🕯️ Dim Candle

What you're building

A single red LED on your breadboard, running so faintly you can barely see it in daylight — about a tenth of First Light's brightness. In a dark room it's a soft red pinprick, like a cheap LED candle. Then you flip a slide switch and click: the LED pops to normal brightness.

The pop is the lesson. The two states aren't magic — they're two different resistor totals in the same loop, and Ohm's law tells you exactly how dim versus how bright.

The big idea: series resistors add

Up to now your circuits have had one resistor at a time. This one has two in a row in the LED's loop: a 1 kΩ and a 10 kΩ, end to end. When current has to squeeze through both, their resistances add:

R_total = 1 000 Ω + 10 000 Ω = 11 000 Ω

That's the same "parts in series share the loop current" rule you learned with Twin Reds, now applied to resistors instead of LEDs. Series = add.

Do the math — both states, side by side

Dim state (bypass slide open — both resistors in the loop):

• Battery push: 4.5 V
• LED eats: ~1.8 V (red LEDs drop a bit less than their 2.0 V book value at low current — fewer electrons means less squeeze across the junction)
• Leftover for the resistors: 4.5 − 1.8 = 2.7 V
• Resistor total: 11 000 Ω
• Current: I = V ÷ R = 2.7 ÷ 11 000 ≈ 0.00025 A = 0.25 mA

A quarter of a milliamp. In a dark room: visible pinprick. In daylight: squint-to-see.

Bright state (bypass slide closed — 10 kΩ shorted out):

The closed bypass slide is now a near-zero-resistance wire right across the 10 kΩ. Current ignores the 10 kΩ (why go through 10 000 Ω of pain when 0.01 Ω of wire is right there?) and takes the easy path.

• Resistor the current actually sees: just the 1 kΩ
• LED now eats: ~1.9 V (closer to book value, more current = more squeeze)
• Leftover: 4.5 − 1.9 = 2.6 V
• Current: I = 2.6 ÷ 1 000 ≈ 0.0026 A = 2.6 mA

Ten times brighter. Still a quarter of First Light (which runs at 11 mA with a 220 Ω), but very clearly on.

Flip the slide back open: the 10 kΩ is back in the chain, back to a candle. One slide, two orders of magnitude of brightness.

Why the shorted resistor doesn't fry anything

A beginner reflex when they hear "short out the 10 kΩ" is to worry something will overheat. Nothing does, and the math tells you why.

A short across a part means the two ends of that part are now at the same voltage. If both ends are at the same voltage, the voltage across the 10 kΩ is zero. If the voltage across a resistor is zero, so is the current through it (I = V ÷ R = 0 ÷ 10 000 = 0). A resistor with no current through it dissipates no power. The 10 kΩ sits there, electrically ignored, perfectly comfortable.

The bypass wire has to carry whatever the 1 kΩ is passing: 2.6 mA — nothing. The slide switch inside is rated for a hundred times that.

Why ten times, not a thousand

Here's the subtle thing. You dropped the resistor total from 11 000 Ω to 1 000 Ω — an 11× cut. But the current went up by only about 10× (0.25 → 2.6 mA). Where did the missing factor go?

Into the LED. At 0.25 mA the LED only had to push ~1.8 V aside. At 2.6 mA it pushes ~1.9 V aside — 0.1 V more. That 0.1 V is 0.1 V the resistor doesn't get. The LED is slightly nonlinear: forward voltage climbs a bit with current. Resistors are perfectly linear; LEDs aren't. Real circuits are the two curves meeting in the middle.

You just saw it. You can see it on every project from here on.

Build it

1. Power rails. Battery + to a top row, battery − to a bottom row.
2. Main slide: + rail → first slide switch (the master on/off). Start it ON.
3. The series stack: out of the main slide → 1 kΩ → 10 kΩ — one after the other on the same row. One hole must sit between the two resistors; call it the mid-point.
4. The LED lane: out of the 10 kΩ → red LED long leg → short leg down to the − rail.
5. The bypass detour (the whole lesson): run a jumper from the mid-point up to a free row above, then a second slide switch, then another jumper down to the far end of the 10 kΩ. Start this slide OFF.
6. Power on. With the bypass slide open, the LED glows faintly. In daylight you may need to shade it to see at all.
7. Flip the bypass slide ON. The LED jumps to a clear, obvious glow. Flip it off: back to a pinprick.

If you can't see any difference, check that the bypass detour's two ends are across the 10 kΩ (one end on each side of it) — not across the whole stack or across the LED. If the detour accidentally crosses the LED, the LED will go dark instead of bright when you close the slide, which is a different (very educational) mistake.

Where you'll see this

• Dimmer switches: real dimmers use fancier parts (triacs), but the idea is the same — insert resistance to drop current, bypass it to raise current.
• Preamp pads: audio preamps often have a "−10 dB" switch that drops a loud signal before the input stage. Underneath, the switch adds or removes a series resistor.
• Multi-brightness LED torches: cheap three-mode flashlights ("low / medium / high") usually just swap different resistors in series with the LED array. You just built the "two of three" version.

Series resistors are the simplest form of "turn it down". Learn it here; it reappears every time a product needs more than one brightness or volume setting.