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"In theory, theory and practice are the same. In practice, they are not." ..."Albert Einstein (1879~1955, Mathematical physicist, Nobel Prize 1921-Physics)"
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Author Topic: Power Loss in electric power transmission network .. V^2/R OR I^2*R?/board:37-100-  (Read 85061 times)
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on: March 08, 2010, 05:53:28 pm »

I understand that in an electric power transmission network, the power loss in the transmission line is given by Ploss = I^2R.Hence if we transmit electricity at high voltage, current I is reduced and hence Ploss is reduced.

But Ohm's law states that V = IR.
So if I were to substitute this into the Ploss equation above, I end up with Ploss = V^2/R.  Which means if the voltage is higher, the Ploss would be higher.

Do these seem conflicting?

I appreciate your help to reconcile this.

Thanks in advance for your assistance.
Lim Heng Leng
Fu-Kwun Hwang
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Reply #1 on: March 08, 2010, 06:29:49 pm » posted from:Taipei,T\'ai-pei,Taiwan

To delieve electric power to the user, a transmission line is needed.
The power loss is due to the resistor(e.g. r=1.0Ω in the above picture) from the transmission line.

P\equiv IV , It can be re-write as P=I V =I_R (I_R R)=I_R^2 R orP=IV=\frac{V_R}{R}V=\frac{V_R^2}{R}
However, you need to use the voltage across the resistor V_R or the current flow through the resistor I_R.

If you want to calculate the power loss of the transmission line with P=V_t^2/R,
then the voltage should be the voltage from the transmission line (not the total voltage).
It is not the same as the voltage from the power plant.

From the above picture:
The source voltage is V, the resistor of the transmission line is r=1.0Ω, and the resistor of the transformer is 99Ω. So the voltage on the transmission line is only V_t=V \frac{r}{r+R}=V\frac{1}{1+99} =\frac{V}{100}.
For the transmission line, the resistor from the transform will change according to voltage ratio between transformer. It is not a constant so a simulation was created to help you understand it.
The power loss can be calculated from P_{loss}=I_r V_r= I_r^2 r and The power delivered P=I_r V
P_{loss}=I_r^2 r=(\frac{P}{V})^2 r= \frac{P^2}{V^2}r
And the percentage of power loss can be calculated as \frac{P_{loss}}{P}= \frac{P }{V^2}r
which is inverse proportional to V^2

Please check out Why we need High Voltage Transmission Line

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Reply #2 on: October 14, 2016, 06:41:10 pm » posted from:Neubiberg,Bayern,Germany


Could you please tell me from which software you took that picture. and where I can get it ?

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"In theory, theory and practice are the same. In practice, they are not." ..."Albert Einstein (1879~1955, Mathematical physicist, Nobel Prize 1921-Physics)"
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