high voltage transmission

2012-04-13 4:34 am

In high voltage power transmission, electricity is transmitted at a high voltage so that. according to P=VI, the current will decrease greatly. Then according to P=I^2R, the power loss can be greatly reduced. However, if we use P=V^2/R instead, then the use of high voltage will lead to great power loss.

Is this correct? Or is that the power in the two cases actually representing different things? Why it seems that the application of these two equations are contradicting?

I have read the reply in some other similar questions, but I am still confused about it. So, please give me answers different from those reply=]

THZ=]

回答 (1)

[匿名]

2012-04-19 7:05 am

✔ 最佳答案

Of course, using high voltage to transit electricity will reduse its current by
P=VI.
But the power loss = V (the volage drop because of resistance) I(its current) = (I^2)R .
P=V^2/R, P represents the power input, V represents the input voltage.

For example:
A 10MW electricity is transitted from a power station to a town through two transmission lines. Each transmission line has a resistance of 1Ω per kilometre.

If the power is transitted at 10kV,
By P=VI,
10^7=(10^4)I
I=1000A
The voltage drop per kilometre=IR=1000(1)=1000V
Power loss=1000(1000)=1MW

If the power is transitted at 400kV,
By P=VI,
10^7=(400000)I
I=25A
The voltage drop per kilometre=IR=25(1)=25V
Power loss=25(25)=625W

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