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Finding wire with the right resistance

  1. Oct 16, 2016 #1
    1. The problem statement, all variables and given/known data
    So we are doing an experiment where we are trying to place a 60W light bulb in a room. It is connected to the mains power supply, but we are imagining that it is DC instead of AC. Now we have to make sure that we only loose 0.1% of power in the wires.

    2. Relevant equations
    P=VI
    V=IR
    R=αL/A
    3. The attempt at a solution
    So if the bulb draws 99.9% of the power I calculated that the remaining one percent is about 0.06W. This is how much we can afford to loose in the wire. Now we can use derived form of the Power equation P=I^2R or P=V^2/R to find R. But I noticed you'd need I or V and I'm not sure what value to use pleases help.
     
  2. jcsd
  3. Oct 16, 2016 #2

    berkeman

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    Staff: Mentor

    You know the Mains Voltage and the power, so that gives you the current in the wire. That gives you the resistance of the wire for the 0.1% power level.

    You need to know the length of wire (and double that for the two pieces of wire in the cable), and from that you can calculate what gauge wire you need to use to achieve the target resistance. Makes sense?
     
  4. Oct 16, 2016 #3
    The power you area referring to, is that the total power drawn by the wire and the bulb?
     
  5. Oct 16, 2016 #4

    berkeman

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    Staff: Mentor

    Well, the wire is dissipating so little, I was just referring to the 60W light bulb. What AC Mains voltage (or equivalent DC voltage) is given?
     
  6. Oct 16, 2016 #5
    120V , I plugged in this and the 60W into one of the power equations and got a resistance of 240 Ω, which I suppose is how much power was decapitated in the circuit. I wasn't clear before but what we have to do is find the properties of a wire that would allow us to lose only this much power through it .
     
  7. Oct 16, 2016 #6

    berkeman

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    Staff: Mentor

    Not quite right.

    Use the bulb power and voltage to give you the wire current. Then you want about 0.1% of 60W to be dissipated in the wire, so figure out what resistance dissipates 0.1% of 60W given that current. Then use wire tables to tell you what AWG (or mm square) wire has that resistance for 2xL, where L is the length of the power cord to the bulb.

    Try that, and post your work so we can check it please. :smile:
     
  8. Oct 16, 2016 #7
    WIN_20161016_20_54_06_Pro.jpg
    I definitely did something horribly wrong
     
  9. Oct 16, 2016 #8

    berkeman

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    Staff: Mentor

    No, you're fine up through R = 0.24 Ohms. Then you are only left with figuring out what AWG or mm squared wire to use to get that resistance for the length of wire you are given. What length of wire are you given in the problem? How far is it from the power feed to the bulb?
     
  10. Oct 16, 2016 #9
    we were given 2m.
    WIN_20161016_21_09_22_Pro.jpg
    looks like it worked out, thank you so much :D
     
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