the wire in a house circuit is rated at 15.0 A and has a resistance of 0.15. What is the power rating? What is the voltage?

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  • Anonymous
    1 month ago

    You are off on a tangent! 

    15 amp RATING means 14 gauge in the USA. 

    You shouldn't load a 14 gauge wire continuously (there is a time rating) with more than 80%, so 12 amps. 

    Is this circuit a 120 volt circuit? 

    Your resistance number means nothing. 

    12 amps at 120 volts is 1440 watts.

  • 2 months ago

    E= I*R=15*.15=2.25V

    P=E*I=2.25*15=33.75W

  • 2 months ago

    The question as asked does not really make sense, but..

    If you have a wire carrying 15 amps, and it has a resistance of 0.15 ohms, it will be dissipating:

    P = I²R = 15² * 0.15 = 33.75 watts.  This will be distributed along the length of the wire. 

    The voltage dropped across this segment of wire will be:

    E = I*R = 15 * 0.15 = 2.25 volts. 

  • 2 months ago

    @billrussell42. watch out, there is another "billrussell42" prowling Y!A level1.

    "https://in.answers.yahoo.com/activity/questions?sh...

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  • 2 months ago

    NO, it has a resistance of 2.58 Ω per 1000 feet for AWG #14 copper.

    resistance and voltage drop of the wire depend on the length.

    It has a max voltage rating that varies with the type of insulation, typically 600 volts. That is the max voltage the cable (not the wire) is rated to carry.

    voltage drop and power in the wire depends on length. If you have 100 ft, R is 0.258 Ω, voltage drop is E = IR = 15•0.258 = 3.9 volts, which brings up the point that long lengths of wire have a lower current rating. a 100 ft length would have a lower rating, see below. But that comes to a power of 58 watts.

    for long wires, the spec is: voltage drop must be less than 3%. For 120 volt circuits that is 3.6 volts. For a 2 wire system, that is 1.8 volts per wire. for the above example, E = IR, I = E/R = 1.8/0.258 = 7 amps.

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