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 Author Topic: Power Loss in electric power transmission network .. V^2/R OR I^2*R?/board:37-100-  (Read 88094 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
hengleng
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 « Embed this message 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.

Lim Heng Leng
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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: March 08, 2010, 06:29:49 pm »

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 $P=I V =I_R (I_R R)=I_R^2 R$ or$P=IV=\frac{V_R}{R}V=\frac{V_R^2}{R}$
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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Shahin1990
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 « Embed this message Reply #2 on: October 14, 2016, 06:41:10 pm »

Hi,

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

thanks
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himanshuswaraj
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Posts: 2

 « Embed this message Reply #3 on: January 03, 2021, 08:23:42 pm »

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 $P=I V =I_R (I_R R)=I_R^2 R$ or$P=IV=\frac{V_R}{R}V=\frac{V_R^2}{R}$
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

Great, you rock.. this helped..
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