Watts triangle

Discussion in 'General Electronics Chat' started by Webby, Sep 6, 2008.

  1. Webby

    Thread Starter Active Member

    Jun 15, 2008
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    Hi all, with a 12v circuit with two 85 watt bulbs in parralel do I add the bulbs values together before calculating the current flowing or just divide by one bulb value only? Trying to work out the current flowing to use a suitable fuse rating.

    Power divided Volts = current

    85/12= 7.08amps

    or

    170/12=14.1amps
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Your second equation is correct.

    Is your 12V source capable of delivering that much current?

    hgmjr
     
  3. Webby

    Thread Starter Active Member

    Jun 15, 2008
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    Hi, see what you mean this question was on a test paper I had to compleat thats all what fuse rating would be suitable?

    If this was a series circuit i presume i use the same equation adding the bulb wattage together?
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    If the bulbs were is series you would not use the same equation. Current in both bulbs would be the same but the voltage would be cut in half.

    hgmjr
     
  5. Webby

    Thread Starter Active Member

    Jun 15, 2008
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    Thanks if in the paralel circuit we have 14.1amps currnt flow what fuse would be suitable 15amps or less?
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    A fuse less than 15 amps might prove too small for the parallel case. There are no absolute standards for fuse selection but typically you are safe to use a fuse large enough to insure that the heat rise in the interconnecting wires remains below 10 to 20 degrees celcius above ambient in the event of a short circuit.

    hgmjr
     
  7. hgmjr

    Moderator

    Jan 28, 2005
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    The fuse manufacturer Littelfuse has a pretty good Whitepaper on the subject of fuses at this link.

    hgmjr
     
  8. iamspook

    Member

    Aug 6, 2008
    27
    0
    Assuming this is (say) a tungsten globe, then there are two characteristics to consider.
    Steady-state - which is governed by the resistance once the bulb has reached steady temperature, and the non-linear surge that happens when you first apply voltage. For steady state - consider the resistances as lumped components. For considering surge, then
    take the peak current drawn just after voltage is applied. The resistance of the filament falls as it heats up.
     
  9. Webby

    Thread Starter Active Member

    Jun 15, 2008
    66
    0
    Hi, was trying to find the answer in the "eebook of all about circuits" why would you have to half the system voltage to work out the current flow if two bulbs are in series.

    Do I add the bulb power i,e 85watts togther =170 watts divide by 6v= 28.3amps cuurent flow.
     
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