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Why Dissipation is Minimized in Power Line Transmission through High Voltage

  1. Apr 11, 2012 #1
    So, power lines are designed to deliver power at extremely high voltage to minimize energy dissipation. This was presented in lecture by the fact that I=PV, so for a given power high V => low I, and since (I^2)R represents the energy dissipation, this term will be minimized.

    That makes sense to me, however, when I think of it from a different perspective, I am tangling ideas and getting something incorrect. If you have power lines, they will be made out of a given material and should have a (relatively) fixed resistance (assuming Ohm's Law applies). Thus, raising the voltage should raise the current. So in trying to minimize energy dissipation, wouldn't increasing the voltage also increase the current? (of course this would also, increase the power at which it is delivered)

    So to me it seems like if you pick a power level and have a given material (with fixed resistance), you really don't have much choice in varying the voltage-- you just have to pick the one where V*I equals that power level. Otherwise, you can change the geometry of your power lines, but this is now minimizing R, not I.

    Could someone please clarify? Thank you.
  2. jcsd
  3. Apr 11, 2012 #2


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    Higher voltage means lower current FOR A GIVEN POWER LEVEL. The important thing is that energy is conserved, but voltage is not, so we can use transformers to step up the voltage while keeping the power the same. This, of course, reduces the current since P = IV. If transformers didn't exist then we wouldn't easily be able to use high voltage lines for power.

    The power load will change depending on how many appliances the end users have plugged in. So, the resistance at the house level changes with time, but the voltage is more or less fixed, so the current varies going into the house. This means some amount of power is moving into the house. The house is drawing less current from the high voltage line to draw the same amount of power. Less current means less I^2*R in the high voltage lines. We pretty much want the lowest resistance practical in the power lines, which means fairly large cables.
    Last edited: Apr 11, 2012
  4. Apr 11, 2012 #3
    I got some new useful things from here, thanks very much
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