tinyΩhms

Level 1 · Light · 15 min · ●○○○○

First Light

Build a real flashlight circuit and learn why every circuit is a loop.

What you need

  • 1× 3×AA battery pack (4.5 V) L1
  • 1× Solderless breadboard L1
  • 3× Jumper wires L1
  • 1× Red LED (5 mm) long leg toward plus L1
  • 1× 220 Ω resistor red-red-brown — see the math below L1
  • 1× Slide switch L1

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

4.5 V + slide switch (on) 220 Ω long leg (+) LED
one loop: + → switch → 220 Ω → LED → − · (4.5 − 2.0) V ÷ 220 Ω ≈ 11 mA

🔧 Open in Workbench — test this exact circuit virtually, then rewire it however you like.

What you're building

A flashlight. Slide the switch, light comes on. Slide it back, light goes off. It sounds simple — and it teaches the single most important idea in all of electronics.

The big idea: circuits are loops

Electricity only flows if it can travel in a complete circle: out of the battery's + side, through your parts, and back into the battery's − side. Break the circle anywhere and everything stops. That's all a switch is — a gap in the loop that you control.

Build it

  1. Push the slide switch into the breadboard so its three legs sit in three different rows.
  2. Push the LED in nearby. Look closely: one leg is longer. The long leg is the + side (its fancy name is the anode).
  3. Connect the battery pack's red (+) wire to the switch's middle leg.
  4. Wire one outer leg of the switch to the LED's long leg... but wait — put the 220 Ω resistor in between. (Why? Math below. Never skip it.)
  5. Connect the LED's short leg to the battery pack's black (−) wire.
  6. Slide the switch. Light!

If nothing happens: is the loop really complete? Is the LED in backwards? (Flipping it won't hurt it — it's a one-way door, not a fuse.)

Why the resistor? Do the math

An LED is greedy. Connected straight to the battery it gulps far too much current and burns out — sometimes instantly. The resistor is a speed bump that sets how much current flows, and Ohm's law tells us how big a bump we need:

current = voltage ÷ resistance

  • Your battery pack pushes with 4.5 V.
  • A red LED uses up about 2.0 V of that just being on (its forward voltage).
  • That leaves 4.5 − 2.0 = 2.5 V pushing current through the resistor.
  • Through 220 Ω: 2.5 V ÷ 220 Ω ≈ 0.011 A, which is 11 mA.

A 5 mm LED is happiest around 10 mA and must stay under about 20 mA. Our 11 mA sits right in the comfy zone. You didn't just build a flashlight — you engineered one.

🧠 Your challenge

No single right answer. That's the point.

  • Swap the 220 Ω resistor for the 1 kΩ one. Predict first: brighter or dimmer? Then check. Can you calculate the new current before you look? (Hint: 2.5 V ÷ 1000 Ω.)
  • Can you move the switch to the other side of the LED — between the short leg and the battery? Does it still work? What does that tell you about loops?
  • Design a flashlight that uses the push button instead. When would a button be better than a slide switch? When would it be worse?

For grown-ups: safety notes

  • 4.5 V from AA batteries is safe to touch — this whole level is skin-safe.
  • Never connect the battery's red wire directly to its black wire (a short circuit). The wires and batteries get hot fast.
  • An LED without its resistor can pop. It won't hurt you, but it's a dead LED.
  • Batteries stay out of mouths and away from younger siblings.

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