Try it
Resistor drops the leftover voltage so the LED sees a safe current
Understand the result
What's going on
An LED has no built-in resistance to speak of - wire one straight to a battery and it draws as much current as the supply can give, which is usually enough to destroy it in about a second. The series resistor’s only job is to eat the extra voltage the LED doesn’t need, so the current stays at a safe level.
The formula
R = (Vsupply − Vforward) / Iforward
Show the derivation, mnemonic, and worked example
Build it up
An LED has a fixed forward voltage - the voltage it drops once it’s lit, regardless of the supply. Whatever voltage is left over from the supply has to be absorbed by something else in the loop, or it forces too much current through the LED. That something else is the resistor: it drops the leftover voltage while limiting current to whatever the LED is rated for.
Worked example
Worked example
Vsupply = 5 V, Vforward = 2 V, I = 20 mA
R = (5 − 2) / 0.02
R = 150 Ω
