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Yes. Volts/R1 + R2 = Amperage rating of bulb. R1 would be the light and R2 either another bulb or a resistor. But don't forget that Amps x Amps x R2 value (in ohms) = watts. Resistor has to be wattage rated more than working wattage; bulb is already compensated correctly. But there is a much simpler solution that doesn't wast those "watts". Simply wire your lights in series rather than parallel. In the above equation, if R1 and R2 were your L and R 6 Volt light, you would have 12 V of capability and not need an additionally resistor. Lights are usually wired in parallel. A feed wire goes to one side of both lights. The other side of both lights grounds to the chassis usually inside the light housing. All you do is come from the feed to one side of a light (R1), from the other side (of the light R1)you only go to one side of the other light R2 (the wire leaving R1 light doesn't wire to ground at all), and only leave the ground in place on the other side of the second light. Presto you are there. So your voltage setup is 12v to R1, 6v out of R1 and into R2, and chassis ground on the output side of R2. Put your light switch in your initial feed line. If you have more than 2 lights, just wire them in pairs as you did R1 & R2. Good luck txblu
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