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How does high voltage decrease energy loss in a wire

  1. Feb 2, 2009 #1
    If you raise the voltage you must raise either the current or the resistance since V = IR.

    So how does a wire with higher voltage and lower current have less resistance?

    It should have more resistance to make up for the lower current.
     
  2. jcsd
  3. Feb 2, 2009 #2

    mgb_phys

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    If you raise the voltage you can reduce the current to give the same power since P=IV
    Higher voltage and lower current is less power lost to I2R
     
  4. Feb 2, 2009 #3
    If you lower the Resistance the voltage will go DOWN, not up, the R in your equation is only a component of V, not of the current.
     
  5. Feb 2, 2009 #4

    mgb_phys

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    The R is generally fixed - unless you want to buy more wire.
    The V in your equation is the voltage DROP along the wire, the V in my equation is the voltage on the wire (ie relative to ground)

    Everybody must have reached the same page in the book because this is coming up a lot
    https://www.physicsforums.com/showthread.php?t=287174
     
  6. Feb 2, 2009 #5
    Ok so lets assume that R is a constant and make it 1. Now we have Power = IV and V = I, so power = II and we can increase I or I, whats the difference?
     
  7. Feb 2, 2009 #6

    mgb_phys

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    Be careful about which V, there is V=the voltage you are running the system at, and V = the voltage drop along the wire.

    eg. Suppose you need to supply 1MW (enough for a city block) and have a 1 ohm wire.

    At V=4Kv (a typical trash can size pole transformer) current from I=P/V = 250A
    Vdrop= I R = 250V so you get 3.75Kv out at the other end and waste I2R = 62KW as heat.

    At V=25Kv (typical sub station) I = 1,000,000/25,000 = 40A, so you only lose V=IR=40volts, and only 1.6KW is wasted (about a tea-kettle)

    At 1MV only 1Watt would be wasted - however running million volt cables is difficult since they have to be kept a long way from the ground to prevent arcing and the switching gear is expensive.
     
  8. Feb 2, 2009 #7
    So Vdrop = IR, not V = IR?

    Otherwise they couldn't increase V to 1MV without increasing I
     
  9. Feb 2, 2009 #8

    mgb_phys

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    Correct, the voltage in a wire you can make as anything you like using transformers (and so change the current) the voltage drop along the wire is what causes the problems, this depends only on the current and resistance.
     
  10. Feb 2, 2009 #9
    So the Voltage in a wire between two specified points is not IR? I seem to remember this being the case in an LRC circuit when taking the integral over the closed loop.
     
  11. Feb 2, 2009 #10

    russ_watters

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    Just to make sure it is said succinctly: P=IV describes the power lost by a resistive load, with V being the drop across the load. The voltage of power lines is not a voltage drop, it is a voltage supplied. It isn't where the loss comes from. Different equations for different situations.
     
  12. Feb 2, 2009 #11

    mgb_phys

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    Yes the voltage difference between two points on a wire is IR, but it doesn't matter what the voltage is (above ground). If you put one amp through a one ohm wire with one end at 1V then the voltage drop is 1V and the other end is at zero. If you put one end of the wire at 1000V there is the same 1V drop and the other end is at 999V.

    But the power you can deliver down the wire is the voltage above ground (ie the 1000V) times the current. And you want to maximize power while minimizing current.
     
  13. Feb 2, 2009 #12
    Ok, got it thanks.
     
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