Sky_Render wrote:jstand wrote: If I recall correctly: Watts = volts * amps Therefore if your led is lower wattage it has higher resistance. So Sky and Ross ( in his first post) had the right area about putting it in parallel with the bulb to lower the resistance. That makes the calculation a little more complicated, but readily available online ( iPhone doesn't do well with math formulas). But is your time to modify the harness really worth less that the $20 for the flasher? Especially after buying the bulbs?No. Resistors are rated in ohms (the resistance) and a wattage they can safely handle. So, a 5 ohm resistor rated at 5 watts can safely handle 5 watts of heat dissipation. A 5 ohm resistor rated at 25 watts can safely handle 25 watts of heat dissipation. You must select a resistor of the correct resistance and a high enough wattage to handle the current you're putting through it. As for wiring the resistor in series, this is correct in some applications, but not this one. The LED bulbs are designed for use in 12-Volt systems. You're not trying to limit the current going to the LED, which is what you're doing if wiring in series. Instead, you're trying to match the impedance that the blinker circuit "sees." That's why you wire the resistor in parallel--to make the blinker circuit "think" there's still a bulb there.
I understand you are trying to match the impedance in the circuit as seen by the flasher.
What i was getting at it is if the LED consumes a lower wattage ( and I think the source of confusion, not the rating as you mentioned) than the incandescent, then the impedance is higher since the voltage is 12 volts in both cases.
If you know the wattage of the bulb that was taken out, then substituting current*resistance for volts in the formula in my post can be used to determine the impedance normally there with an incandescent bulb.
Then if you need more impedance ( and/ or current limiting) the resistor goes in series, if you need less it goes in parallel.
