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I'm currently learning about the transportation of electrical energy through power lines. I now know that transporting current at a very high voltage cause minimal power loss compared to transporting it at a very high amperage.

However, one of the formulas I've been given to calculate power loss is P=V^2/R, which in this context produces ridiculously high numbers. For example,

P=40,000^2/1 (assuming the cable has a resistance of 1 ohm) = 1,600MW. This obviously isn't what happens in reality. So, is the formula only applicable in certain circumstances? Should I only use it when working with closed circuits?

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    \$\begingroup\$ Where do you get 40kV dropping through 1\$\Omega\$? Nothing is wrong with the formula! \$\endgroup\$ – Daniel Jun 18 '16 at 13:33
  • \$\begingroup\$ Simply put that voltage isn't being dropped across the cable otherwise there would be zero volts for the load. \$\endgroup\$ – JonRB Nov 4 '18 at 18:08
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The way P=V^2/R works is very much different than they way you applied it.

The voltage in that formula must be the one across the load or subject of which you are interested to calculate the power, and the R is the resistance of that subject.

This raise a question. How did you come to the state the 40KV is across the end to end of the power line? The power line which is having only mere one ohm of resistance?

In order to calculate power loss in the transmission lines, you need to know the power received by the load end and the voltage at the load end. Use these two to calculate the current drawn from the line. Here you use P=VI. Then use the this current value and the total resistance of the transmission line to calculate the power loss. Here you use P=I^2.R.

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    \$\begingroup\$ Ok, I think I see where I'm going wrong. The "V" in this formula is not actually referring to the voltage, but to the voltage drop across a load? So although the current is charged with 40kV, only a fraction of this will be dropped across the distance of the power line? \$\endgroup\$ – Jake Davies Jun 18 '16 at 13:52
  • \$\begingroup\$ Yes... You got it. \$\endgroup\$ – soosai steven Jun 18 '16 at 13:53
  • \$\begingroup\$ "The "V" in this formula is not actually referring to the voltage, but to the voltage drop across a load?" No, the voltage drop across the line. (I think you may have meant to write that.) \$\endgroup\$ – Transistor Jun 18 '16 at 18:18
  • \$\begingroup\$ Yes.. Voltage across the line at very close proximity to the load. \$\endgroup\$ – soosai steven Jun 18 '16 at 18:22
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It is helpful to think of it this way, where

$$P = {V^2 \over R}$$

is actually a rearrangement of this, more direct relationship:

$$P = I^2 \times R$$

This more clearly shows that heating in resistive elements is directly related to the CURRENT and not some abstract combination of things.

It also helps illustrate that power loss is distributed along a distributed resistance, ie power transmission lines.

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    \$\begingroup\$ They're both restatements of \$P=VI\$, which applies to any branch of a circuit, not just to a resistor. \$\endgroup\$ – The Photon Jun 18 '16 at 14:20
  • \$\begingroup\$ Is that truly the more fundamental statement? (metaphysics, ugh) . This form just better illustrates why wires and fuses are rated at particular currents. \$\endgroup\$ – Daniel Jun 18 '16 at 14:26
  • \$\begingroup\$ Which works well because they rarely drop much voltage. \$\endgroup\$ – Daniel Jun 18 '16 at 14:27
  • \$\begingroup\$ And also helps illustrate how you get more transmission line capability 'for free' by jacking up the voltage, for a given wire (and a fixed R) \$\endgroup\$ – Daniel Jun 18 '16 at 14:28
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    \$\begingroup\$ I don't consider \$P=I^2R\$ to be the fundamental statement because it doesn't apply to, for example, a capacitor or inductor. Whereas \$P=IV\$, taken instantaneously, does also tell you how power flows in and out of those elements. \$\endgroup\$ – The Photon Jun 18 '16 at 14:30

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