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During the transmission, the generated electric power is delivered after stepping up to hundreds of thousands or even more voltages by transformers.

In that case since P = I*V, increasing V reduces I in the secondary of the transformer.

The reason given is to reduce Ploss = I^2*R losses. Here I decreases so the power loss.

Is that the real reason to step up?

I'm asking because we can write the power loss equation as:

Ploss = V^2/R

Or if we use both I and V in the power equation:

Ploss = (V/R)*I

It seems like if we step up the voltage I decreases but V increases. How about the power loss?

Does the power loss decrease? Or the real reason to step-up the voltage is to reduce the cross section area of the transmission lines significantly?

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    \$\begingroup\$ You're thinking V is the voltage between Wire A and Wire B. But it's not - here V is the voltage between one end of Wire A and the other end of Wire A (and the same for Wire B). \$\endgroup\$ – user253751 Jun 9 '16 at 6:58
  • \$\begingroup\$ This is a "why" question. Physics simply is what it is. Nobody can explain why our creator chose to make things the way they are. This topic is discussed in this Q&A, in any case: electronics.stackexchange.com/questions/333574/… \$\endgroup\$ – SDsolar Oct 10 '17 at 19:42
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For these transmission lines, not the voltage will result in a power loss, but the voltage drop on these lines.

I thinks its easiest explained on an example: Lets say your transmission line has a resistance of R=100Ohm and you want to transfer P=1kW.

With "P=U*I" you get:

@1000V, you need to transfer 1A

@100kV you need to transfer 0.01A

By "dU=R*I" the voltage drop across your transmission line will be:

100V @1000V

1V @100kV

As the voltage drop in your transmission line is the power lost you can now calculate the Power loss by "Ploss = dU * I" which results in:

100W @1000V which is 10% of your original 1kW

10mW @100kV which is 0.001% of your original 1kW

Thus: The higher the voltage, the lower the power loss in your transmission line (which of course also has it's upper limits due to for example isolation of these lines and isolation of the transformers, but thats another topic).

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  • \$\begingroup\$ I think I got the point which confused me. In a power supply with a series resistor dV = V - 0(GND) thats whay we use V^2/R because V = dV there. But in this case dV is not V. And explaining it with examples makes it more easier to see as you did. \$\endgroup\$ – user16307 Jun 9 '16 at 0:32
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I'm asking because we can write the power loss equation as:

Ploss = V^2/R

Well... no. A more accurate representation would be:

\$P_{loss} = {{{\Delta V}^2} \over R}\$

That is to say, you only lose power when you have a voltage (i.e. energy) drop in the resistance. Increasing the voltage decreases the voltage drop due to the now lower current passing through the same resistance.

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  • \$\begingroup\$ Alright so the Ohms's law is the starting point here: dV = I*R. I think what creates confusion is it is not a power supply where current increases with voltage so it is not convenient to use V^2/R. \$\endgroup\$ – user16307 Jun 8 '16 at 23:44
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V^2/R holds but you need to be careful about what you mean by V, in particular V is the voltage drop due to cable resistance not total supply voltage....

Lets say we have 1MVA @ 1,000V, so 1,000A flowing, and our line is say 0.01 Ohms so our I^R losses calculate to be 1000^2 * 0.01 = 10KW.

Lets see what V^2/R makes it, well the V in this case is the voltage drop due to the line resistance = 1000A * 0.01 ohms = 10V, 10^2/0.01 = 10KW, exactly the same as the calculation done the other way.

Now lets raise the line voltage to 10kV, current is now 100A for the same power delivered, so if the cable is still 0.01 ohms, we get I^2R losses as 100^2 * 0.01 = 100W, doing the same calculation for V^2/R, we get voltage drop due to cable resistance as 100A * 0.01 = 1V, and 1^2/0.01 = 100W.

It should be no surprise that both calculations come out the same, as to find the voltage term for the cable losses we do V = IR, the we can substitute that into P = V^2/R = (IR^2)/R = IR*IR/R = I^2R.

As to reasons to do it, it is an optimisation problem, you trade off insulation hassle and cost for less weight in the wires and possibly less line loss (or some combination of both).

Regards, Dan.

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When using such equations we must be careful which voltage, which current and which resistance we are talking about.

Lets assume a simple case, the supply is DC so there are no capacitive or inductive affects and the cable has perfect insulation but the conductors have some resistance. Now lets write some equations.

$$V_{load}=V_{source}-V_{drop}$$

Where \$V_{load}\$ is the voltage delivered to the load V_{source} is the voltage supplied by the source and V_{drop} is the voltage dropped in the cable.

By ohms law we can write

$$V_{drop} = I * R_{cable}$$

Where \$I\$ is the current flow in the circuit and\$R_{cable}\$ is the total resistance of the conductors (both positive and negative) in the cable supplying the load. We can now write an equation for the power loss in the cable.

$$P_{loss} = V_{drop} * I = I^2 * R_{cable} = V_{drop}^2 / R_{cable}$$

So to reduce \$P_{loss}\$ we need to either reduce \$R_{cable}\$ or reduce \$I\$. To reduce \$I\$ while keeping the power delivered to the load the same we have to increase \$V_{load}\$ and hence \$V_{supply}\$

Now this example isn't a perfect reflection of the real world. In reality insulators aren't perfect and systems are usually AC so capacitive and inductive affects have to be considered. The result is that increasing voltage helps up to a point but eventualy you reach a point where further voltage increases are not helpful.

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  • \$\begingroup\$ I was thinking the motivation is to reduce the cross section of transmission lines hence the costs. One might need a very thick line to pass the current if voltage were not stepped up. \$\endgroup\$ – user16307 Jun 9 '16 at 0:19
  • \$\begingroup\$ Well you can look at it either way. Increasing the voltage reduces the loss for a given size of cable or reduces the size of cable needed for a given loss. \$\endgroup\$ – Peter Green Jun 9 '16 at 0:45

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