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Math for why voltage is stepped-up in power lines?

  1. Nov 22, 2009 #1
    I've long understood that voltage is stepped-up in power t-lines to decrease losses, however, I've never really understood the math reasoning behind it.

    Isn't P = V^2/R the same as P = I^2*R? If so, it seems that either increasing V or increasing I would invoke the same losses... Could someone explain?

    Thanks
     
  2. jcsd
  3. Nov 22, 2009 #2
    When you are talking about the power lost over the transmission line, it's the voltage difference between the start and the end of the line, not between operating voltage and earth. This voltage is dependent on current and impedance of the line only.
     
  4. Nov 22, 2009 #3
    OK, thanks but that isn't really answering my question. Why does increasing the voltage decrease the power loss in the line?
     
  5. Nov 22, 2009 #4

    mgb_phys

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    Good answer - this is the thing most students misunderstand when first seeing this
     
  6. Nov 22, 2009 #5

    dlgoff

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    Another good reason for stepping up the voltage is that you need less current to get the same amount of energy transfer. Hence you can use a smaller gauge wire to carry the load saving wire cost.
     
  7. Nov 22, 2009 #6

    mgb_phys

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    Yes that's the point - but it confuses people who think of power = V^2/r and v as the powerline voltage rather than the voltage drop.
     
  8. Nov 22, 2009 #7

    dlgoff

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    There are losses due to corona discharge also when the line to ground voltage gets very high. Not trying to confuse the issue; just noting.
     
  9. Nov 22, 2009 #8

    jambaugh

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    It helps if you draw a circuit with a load and consider that you have two voltages, the voltage at the generator and a lowered voltage at the load due to line resistance.

    Generator:============Load
    (= is wire pair across which we measure voltage).

    You have V at the power source and V' at the load with voltage drop: V-V' = RI.
    Where R is the line resistance (...for both legs. You can put the load anywhere in the current loop and get the same numbers.)

    The load power is V'I and of course current is constant.
    (We're doing this in DC for simplicity but you can see the reasoning will apply to AC as well.)

    Efficiency is power received at load over power leaving the source:
    P at source = VI
    P at load = V'I
    Efficiency is V'I/VI =V'/V= (V-RI)/V = 1-RI/V. So the less current and more voltage, the close to 100% efficiency.
     
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