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Homework Help: DC Theory

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    What Voltage is delivered to a 120-volt/5000-watt load which is fed with #10 AWG wire (1.24 ohms/1000') and located 750 feet from the 120-volt source?

    2. Relevant equations

    Ohm's Law E= I * R

    3. The attempt at a solution

    I am in an apprenticeship program for Electrical. I was wondering if I could get some help on this problem. This is what I understand...I need to find Resistivity(K) since they do not tell me the type of wire being used. the Area in circular mils for #10 wire is 10380 circular mils. I have the length of 750 feet. I also need to find the current throughout the circuit. Since I am given the Power and Voltage of the load I calculated the Current from that:

    I=41.7 Amps

    Now in order to find K resistivity I know that the resistance is 1.24 ohms/1000'. I understand the K is resistance per mil-foot, but this is where I come up with a problem. Any help would be appreciated...the apprenticeship program I am in is not that intense, so to work out these problems is not necessary. I am trying to learn this material because I intend to go for my bachelors degree afterwards. Thanks again
  2. jcsd
  3. Nov 23, 2008 #2


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    Welcome to PF, enzo.

    Don't worry about resistivity. What can you figure out about the wire from this information:

    1. The wire is 124 ohms per 1000 ft.
    2. The load is 750 feet from the source.

    Extra hint: how many wires would run from the source to the load?
  4. Nov 23, 2008 #3
    i can get .00124 ohms per foot and get .93 ohms per 750 feet. if E=I*R and multiply the current of 41.7 amps and multiply that by the resistance of .93 ohms per 750 feet than I get a voltage drop of 38.8 Volts. That value doesn't seem right to me, but is that the correct way? I believe 1 wire would run from the source to the load in this problem, is that correct?
  5. Nov 24, 2008 #4


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    You're getting there. Yes, 750 feet of wire would be 0.93 ohms.

    It might help to treat the load as a simple resistor. We know that it draws 41.7A at 120V. The actual current will be different than the 41.7A number, since the voltage at the load is different than 120V.

    That's what many people will assume, but if you think about it just 1 wire is not enough to make a complete circuit ...

    So I would draw a circuit diagram, using resistors for the load and for the wire(s) also. From there, it's Ohm's Law to figure it all out.
  6. Nov 24, 2008 #5
    so drawing it out it makes sense that there are 2 wires and the resistance for the wires 1.86 but with the load if the current is different how would I calculate the resostance for the load to treat it as a simple resistor?
  7. Nov 24, 2008 #6


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    Good question. Since we know that it draws 41.7A at 120V, we can use Ohm's Law to figure out R for the load.
  8. Nov 27, 2008 #7
    Yes I agree with red, find the load Resistance, you should then be able to get the voltage drop across the load R from the total resistance at 120V
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