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  • please check for me

    I would like to use a 12 volt / 3watt bulb fed by a 43 volt ac power supply.
    Is a 3k2 / .3watt the resistor I have to put in series?

    Thanks a lot.

  • #2
    I'd like to see how you got there.

    I get 124 ohms dissipating 7.75 watts

    Using 12v and 3w I get the following:

    To make 3 watts, 12v across a bulb must be flowing at 0.25 amp. That implies a hot resistance of 48 ohms.

    The supply is 43v, so we need to drop 31v across a series resistor. Current through the resistor wil be the same as through the bulb - 0.25 amp.

    E=IxR, so 31v across R with 0.25A means R must be 31/0.25 = 124

    Power? ExE/R = 31x31/124 = 7.75W So a 15 watt resistor would be about right.

    ANyone spot a glaring error in my figures? My day is just starting.

    Is your 43v supply up to providing 11 watts over and above whatever else it is doing?

    Might a 31v zener work? Or a pair of them for AC.

    What are you actually doing here?
    Education is what you're left with after you have forgotten what you have learned.

    Comment


    • #3
      Originally posted by Enzo View Post
      I'd like to see how you got there.
      Me too
      I failed here: I=W/E, and for some reason I did 3/31=.0097 instead of 3/12=.25
      I found the calculator just before I read your post, and of course you're right.
      It is 124 ohms, 7.75 watt
      Thank you.

      What I try to use is a 12 volt/3W automotive bulb replacing the 48 volt/4W bulb in an ebs td600

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      • #4
        I decided to order a replacement bulb. A lot easier, but I have to wait a few days.

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