troubleshooting
Why is my resistor getting hot?
Short answer: Resistors turn electricity into heat — P = V²÷R. A 100 Ω across 9 V burns 0.81 W, more than three times a ¼ W resistor can shed. Pick a bigger R, or add an LED.
See it happen
This is the broken circuit, simulated live. The readout under the board tells you what the parts are doing.
simulating…
What's going on
Resistors are the one part on the breadboard whose job is to turn electrical energy into heat. How much heat they turn it into is their power dissipation, measured in watts (W):
P = V × I — and because I = V ÷ R on a resistor, that's also P = V² ÷ R or P = I² × R.
The ¼ W resistors in every starter kit can shed about 0.25 W of heat before they start to char. That's not a lot. A single resistor stretched across a 9 V battery — nothing else in the loop — pays for the whole voltage:
- 9 V, 100 Ω: I = 9 ÷ 100 = 90 mA, P = 9² ÷ 100 = 0.81 W — more than three times the ¼ W rating. Within seconds the resistor browns, smells, and may crack.
- 9 V, 1 kΩ: I = 9 ÷ 1 000 = 9 mA, P = 9² ÷ 1 000 = 0.081 W = 81 mW — fine, but there's nothing in the loop to look at.
- 9 V, 680 Ω with a red LED: the LED takes 2 V, so the resistor only has 7 V across it. I = 7 ÷ 680 ≈ 10 mA, P = 7 × 0.01 = 70 mW — safely cool, and the LED lights.
The LED matters because it drops voltage you don't have to burn as heat. A resistor across a bare battery is doing nothing useful — all 9 V end up as warmth in a tiny ceramic cylinder.
The fix
Two moves, usually both:
- Put something in the loop that uses the energy — an LED, a buzzer, a motor. That part drops voltage the resistor would otherwise have to swallow.
- Size the resistor to the leftover voltage, keeping its power under its rating. For a ¼ W part at 9 V, that means R ≥ 9² ÷ 0.25 = 324 Ω across the whole battery, or less once an LED is in there.
Standard choices:
- 9 V + red LED: 680 Ω → 10 mA, R dissipates 70 mW.
- 4.5 V + red LED: 220 Ω → 11 mA, R dissipates 28 mW.
- 3 V + red LED: 100 Ω → 10 mA, R dissipates 10 mW.
For an exact number, use the LED resistor calculator or the general power calculator.
simulating…
Also check
- Just warm, not hot? Up to about 100 mW is normal for a ¼ W resistor and nothing to worry about — it'll be noticeably warm but safe. Only when P > 0.25 W does it start to fail.
- Resistor turning black. It's past the point of return. Replace it and pick a new value — the old one failed because of the maths, not bad luck.
- Two resistors in series share the heat. 9 V across two 100 Ω in series is still 45 mA total and 0.4 W total — but 0.2 W in each resistor, which both ¼ W parts can handle. Series resistors are a quiet way to raise the power rating.
- Need to burn more than ¼ W on purpose (bigger LED, small motor)? Buy ½ W or 1 W resistors — they look chunkier but behave the same. Datasheets call this the power rating.