Electrical light question

[jetmechmarty]

Steve,
I looked up the 1156 lamp at Advance Auto Parts and it's 26.9 watts. Is there a reason to go someplace else?

Parts n' More has OEM style replacement signal lights.

P=I*V (Power = Current * Voltage)

[crazy steve]

The motorcycle lamps have heavier-duty filaments; I particularly like the Harley lamps, as theirs feature a mid-point filament support to prevent failure from vibration. They were about $3 each last time I bought some...

The math to find the right resistance is:
P = I x I x R (P is watts, I is current squared, and R is resistance). In this case, you need to add 4 watts and .33 amps load, so...
4 = .109 x R, 4/.109 = R, or 36.67 ohms. 40 is about as close as you'll get...

[nzemke]

right now I have an additional 12V .33 amp in there and its not seeming to pull enough load to trigger the flashing unit. Thats why I tried to use the 1200 ohm resistor to make up for an additional .009 amps in hopes of getting it to trigger which it sadly did not on the way home. Thinking I will have to try out this 552 blinker unit at O'Reilys.

[MaximPhil]

right now I have an additional 12V .33 amp in there and its not seeming to pull enough load to trigger the flashing unit. Thats why I tried to use the 1200 ohm resistor to make up for an additional .009 amps in hopes of getting it to trigger which it sadly did not on the way home. Thinking I will have to try out this 552 blinker unit at O'Reilys.

Hi nzemke,
Or clean the grounds. My left flasher has stopped flashing again so I know it must be time to clean it once again. I will check mine to make sure I have the correct bulb in that side too. Doxit and a toothbrush with no Math calculations Electrickery is still a mystery to me
Phil