Swap a 21 W indicator bulb for an LED and the flasher sees almost no current: it thinks the bulb has blown and blinks twice as fast (hyper flash), or doesn't blink at all. A load resistor in parallel with the LED draws the current the old bulb used to draw.
Enter the system voltage, the wattage of the original bulb and the power of the LED bulb. You get the resistance, the current and how much heat the resistor has to handle with the engine running.
How to use
- Pick 12 V for cars and motorcycles, 24 V for trucks and 6 V for old motorcycles.
- Pick the original bulb: 21 W (P21W, 1156, 7440) for front and rear indicators, 10 W or 5 W for side repeaters and some motorcycles.
- Enter the LED bulb power printed on its box (usually 0.5 to 3 W).
- Fit one resistor per LED bulb, from the indicator wire to ground, close to the bulb.
How the load resistor is calculated
R = V² ÷ (bulb watts − LED watts). For a 21 W bulb at 12 V: 144 ÷ 21 = 6.9 Ω, so a 6.8 Ω resistor (or the common 6 Ω) does the job. The calculator picks the standard value just below, so the flasher always sees at least the current of the original bulb.
With the engine running the voltage rises to about 14.4 V, and the same 6.8 Ω resistor turns about 30 W into heat while it is on. That's why load resistors for 21 W bulbs are sold as 50 W aluminium resistors.
Load resistor or LED flasher relay?
If your car has a separate plug-in flasher relay, an electronic LED flasher relay fixes hyper flash for all four corners with no heat and no wasted power. Load resistors are the answer when the flasher is built into a control module (most cars since about 2005) or when you only changed some of the bulbs.
On cars that check the bulbs, the resistor also clears the bulb failure warning on the dashboard.