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Current = Voltage/Resistance or Power/Voltage?

  1. Oct 28, 2013 #1
    I was doing (ii) in the following question.

    The ESB supplies electrical energy at a rate of 2 MW to an industrial park from a local power station, whose
    output voltage is 10 kV. The total length of the cables connecting the industrial park to the power station is
    15 km. The cables have a diameter of 10 mm and are made from a material of resistivity 5.0 × 10–8 Ω m.
    Calculate
    (i) the total resistance of the cables; (15)
    (ii) the current flowing in the cables; (6)

    (iii) the rate at which energy is “lost” in the cables. (6)
    Suggest a method of reducing the energy “lost” in the cables. (5)


    When trying to work out current I used V/R but from the marking scheme I found out that the right (and different) answer is obtained by using P/V. Are these two formula not equivalent? If not, what is different about them?
     
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  3. Oct 28, 2013 #2

    phinds

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    The two formulas are the same, but perhaps the difference is in how you base your calculations since the power supplied by the source is not the same as the power received by the load, due to the loss in the lines.
     
  4. Oct 28, 2013 #3
    If you want to calculate the current in the cables, you can use I=V/R but you need the voltage drop on the cables. Which is not given. The 10kV is not the voltage drop on the cables but the total voltage at the output of the local power station.
     
  5. Oct 28, 2013 #4

    russ_watters

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    Similarly, 2MW is the power used by the industrial park, not what is lost in the wires.
     
  6. Oct 28, 2013 #5
    I thin that 2MW is the total power delivered (supplied). And 10 kV is the voltage of the power station, as it says in the text.
    Part of this voltage drops on the wires, resulting in some power being dissipated on the same wires.
    But to find the current you don't need to know how is the power distributed, do you?
     
  7. Oct 28, 2013 #6

    phinds

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    One of us is reading the problem incorrectly. The way I read it, the problem specifically says the 2MW is delivered to the load. It does NOT say that the 2MW is the power produced by the station although I certainly see how you can read it that way.

    The problem is badly stated and the OP has to go with whatever interpretation gives him the right answer, since he seems to know the answer.
     
  8. Oct 28, 2013 #7
    So you would take it that the 10kV is the output of the power station (before cables) - this is explicitly given. The 10kV is the output of the power station.
    But the 2 MW power is after the cables, at the input of the industrial park.

    It makes sense if "supplied" means the net power entering the industrial park rather than what "leaves" the power station.
    However I don't see how can you say that one interpretation is "incorrect", based just on the text
    Unless you have some inside info about what the author have in mind.:smile:
     
  9. Oct 28, 2013 #8

    phinds

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    I said that ONE OF US is reading the problem incorrectly. I do not presume to say which one. They are both interpretations. Was that not clear from my post?
     
  10. Oct 28, 2013 #9
    Yes, I understand this. I did not try to defend my "reading". Actually, thinking about real world, we pay directly the energy delivered to the house. The energy lost in cables comes into fees and surcharges (which we pay as well). So your reading would be more likely.

    It's just the way you said that one of us is reading it incorrectly. I don't think that the fact we read it in two different ways would necessarily makes one "incorrect". As long as the text is formulated such that multiple interpretations are possible, they may all be "correct". Only extra information can provide a criterion to decide otherwise. They may be both wrong, after all.:smile:
     
  11. Oct 29, 2013 #10

    CWatters

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    Best show your working so we can see where you went wrong.

    It's surprising how many people use the "line voltage" instead of the "voltage drop down the line" to calculate the power loss.
     
  12. Oct 29, 2013 #11

    CWatters

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    That would be my reading but I agree the question isn't clear.

    In most cases it's the load (the industrial park) that determines how much power is consumed. The power station just delivers whatever is required to maintain a 10kV output.
     
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