# NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/

## Information about this web site => Question related to Physics or physics related simulation => Topic started by: hengleng on March 08, 2010, 05:53:28 pm

 Title: Power Loss in electric power transmission network .. V^2/R OR I^2*R?/board:37-100- Post by: hengleng 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 Title: Re: Power Loss in electric power transmission network .. V^2/R OR I^2*R?/board:37-100- Post by: Fu-Kwun Hwang on March 08, 2010, 06:29:49 pm (http://www.phy.ntnu.edu.tw/ntnujava/snapshotejs/941_smf_highpowertransmissionline_20090208192238.gif)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$$P_{loss}=I_r^2 r=(\frac{P}{V})^2 r= \frac{P^2}{V^2}rAnd the percentage of power loss can be calculated as \frac{P_{loss}}{P}= \frac{P }{V^2}r which is inverse proportional to V^2Please check out Why we need High Voltage Transmission Line (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=941.0) Title: Re: Power Loss in electric power transmission network .. V^2/R OR I^2*R?/board:37-100- Post by: Shahin1990 on October 14, 2016, 06:41:10 pm Hi,Could you please tell me from which software you took that picture. and where I can get it ?thanks Title: Re: Power Loss in electric power transmission network .. V^2/R OR I^2*R?/board:37-100- Post by: himanshuswaraj on January 03, 2021, 08:23:42 pm Quote from: Fu-Kwun Hwang on March 08, 2010, 06:29:49 pm(http://www.phy.ntnu.edu.tw/ntnujava/snapshotejs/941_smf_highpowertransmissionline_20090208192238.gif)To delieve electric power to the user, a transmission line is needed.The power loss is due to the resistor(e.g. r=1.0Ω in the above picture) from the transmission line.P\equiv IV , It can be re-write as P=I V =I_R (I_R R)=I_R^2 R orP=IV=\frac{V_R}{R}V=\frac{V_R^2}{R}However, you need to use the voltage across the resistor V_R or the current flow through the resistor I_R.If you want to calculate the power loss of the transmission line with P=V_t^2/R, then the voltage should be the voltage from the transmission line (not the total voltage).It is not the same as the voltage from the power plant.From the above picture:The source voltage is V, the resistor of the transmission line is r=1.0Ω, and the resistor of the transformer is 99Ω. So the voltage on the transmission line is only V_t=V \frac{r}{r+R}=V\frac{1}{1+99} =\frac{V}{100}. For the transmission line, the resistor from the transform will change according to voltage ratio between transformer. It is not a constant so a simulation was created to help you understand it.The power loss can be calculated from P_{loss}=I_r V_r= I_r^2 r and The power delivered P=I_r V$$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 (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=941.0)Great, you rock.. this helped..