| Author | Message | ||
     Zoolander |
I'm replacing the front turn signals on my CR. The mirror/signal combo is gone. I am installing an LED turn signal with six little bulbs. I have a resistor that states "25W10RJ" on the side of it. I understand that it has to be placed "in parallel". I understand that to be along one line (if that makes sense to anyone?). When I hold the resistor up into position along the line, nothing happens to the flashing rate. It just continues as a fast(er) rate. Is the resistor the wrong type? | ||
| Kinder |
I just replaced my rear signals and used a 48ohm 10 watt resistor(one on each side). That seemed to do the trick. I extended the wires so I could tuck the resistors behind the tail light. They did not get hot after 1min of a signal being on. (Message edited by kinder on June 02, 2011) | ||
| Marcodesade |
It sounds like you're wiring it in series rather than parallel. For parallel, you would splice one end into the signal's positive wire, and the other end into the negative wire. Series would be cutting one of the wires and splicing the resistor in there. | ||
| Zoolander |
It didn't seem to work in parallel or in series, however, I did have it in series. Didn't know, now I do, thanks! I'll try it again...something has to work. | ||
| Kinder |
FWIW my kit came with some resistors that didn't do the job. That's why I had to grab some different ones. | ||
| Arcticcr |
This question can’t be answered properly without additional information. You need to know the current of the stock turn signals and the current of the new turn signals. The stock current is around 180 milliamps. The new turn signals voltage is specific to the kind that you’ve purchased, therefore, you’ll need to measure with an ohmmeter. Once you’ve gotten both numbers, confirmed the 180 and the new turn signals current, you can then calculate what resistance you need. Here’s the math from when I did this change on my bike. Know: 13 volt system Measured: Stock amps - .180 amps New amps - .0156 amps Calculated: V = IR 13 = (.180-.0156)*R 13/.1644 = 79.075 ohm (close enough to an 80 ohm, that’s what I got) Let me know if you have any more questions. Jay |
