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The electrical engineer in me can't stand having resistors sitting around for no reason generating heat lol.

But yes, centerpunch is correct, the resistors needed are going to be a high wattage, low resistance type so the ones you have leftover probably will not work for this. The bright filament of a 1157 is about 27 watts, so for the sake of making math easy lets call battery voltage 13.5, meaning 2 amps (P = V * I). This is roughly 10x the current of most 1157 LED bulbs, so our resistor needs to draw 1.8 amps to mimic a 1157. Pulling 1.8 amps at 13.5 volts means a resistance value of around 7.5 ohms. (V = I * R). There's probably some range where the factory flasher relay is happy so you could start off with something like a 10 ohm, 40+ watt resistor and see if that works. That would pull 1.5 amps combined with the LED and would probably make the flasher relay happy.

That's not the sort of thing I would buy on Amazon for cheap TBH. I'd get something like this and make a custom harness for it: https://www.digikey.com/en/products/detail/yageo/AHA50AFB-10R/2169297

<blockquote data-quote="MotorMouth93" data-source="post: 2078761" data-attributes="member: 33247">The electrical engineer in me can&#039;t stand having resistors sitting around for no reason generating heat lol.But yes, centerpunch is correct, the resistors needed are going to be a high wattage, low resistance type so the ones you have leftover probably will not work for this. The bright filament of a 1157 is about 27 watts, so for the sake of making math easy lets call battery voltage 13.5, meaning 2 amps (P = V * I). This is roughly 10x the current of most 1157 LED bulbs, so our resistor needs to draw 1.8 amps to mimic a 1157. Pulling 1.8 amps at 13.5 volts means a resistance value of around 7.5 ohms. (V = I * R). There&#039;s probably some range where the factory flasher relay is happy so you could start off with something like a 10 ohm, 40+ watt resistor and see if that works. That would pull 1.5 amps combined with the LED and would probably make the flasher relay happy.That&#039;s not the sort of thing I would buy on Amazon for cheap TBH. I&#039;d get something like this and make a custom harness for it: <a href="https://www.digikey.com/en/products/detail/yageo/AHA50AFB-10R/2169297" target="_blank">https://www.digikey.com/en/products/detail/yageo/AHA50AFB-10R/2169297</a></blockquote>

[QUOTE="MotorMouth93, post: 2078761, member: 33247"] The electrical engineer in me can't stand having resistors sitting around for no reason generating heat lol. But yes, centerpunch is correct, the resistors needed are going to be a high wattage, low resistance type so the ones you have leftover probably will not work for this. The bright filament of a 1157 is about 27 watts, so for the sake of making math easy lets call battery voltage 13.5, meaning 2 amps (P = V * I). This is roughly 10x the current of most 1157 LED bulbs, so our resistor needs to draw 1.8 amps to mimic a 1157. Pulling 1.8 amps at 13.5 volts means a resistance value of around 7.5 ohms. (V = I * R). There's probably some range where the factory flasher relay is happy so you could start off with something like a 10 ohm, 40+ watt resistor and see if that works. That would pull 1.5 amps combined with the LED and would probably make the flasher relay happy. That's not the sort of thing I would buy on Amazon for cheap TBH. I'd get something like this and make a custom harness for it: [URL]https://www.digikey.com/en/products/detail/yageo/AHA50AFB-10R/2169297[/URL] [/QUOTE]

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