Why electric power plants transmit electricty at high voltage?

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Offline Vorador

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I searched around google for this and I got the answer that: Since P=I^2*R, so high-voltage and low current would result in less power losses.

What confuses me is that P=I^2*R can also be written as P=V^2/R; which tells a different story. According to this equation, the lower the voltage, the lower the power losses. In fact, this second equations seems to suggest that transmission at lower voltages should be more efficient, since the factor V is being divided by R, whereas in the previous equation I was being multiplied by the same factor..

So, yeah, I'm confused.


Offline syhprum

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The equation  P=I^2*R refers to the power transmitted along the line to the load R whereas P=V^2/R refers to the power dissipated in the power line V being the voltage drop along the line which equals I*R  hence when you feed a higher voltage into the line the current I is less for a given amount of power transmitted hence the power dissipated in the line is reduced.
« Last Edit: 21/06/2013 21:21:28 by syhprum »


Offline highvoltpower

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Power rule of transmission link is defined the ratio between transferring and getting end voltage to receiving end voltage of a transmission link between conditions of load and full load.
P=V2R is derived from V=IR and P=IV You get P=V2R because P=IV=VRV solved the equation V=IR for I and plugged it in to that Power equation P=IV Do the same for P=I2R
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