[img]http://www.phy.ntnu.edu.tw/ntnujava/snapshotejs/941_smf_highpowertransmissionline_20090208192238.gif[/img]

To delieve electric power to the user, a transmission line is needed.
The power loss is due to the resistor(e.g. r=1.0? 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=1.0?, and the resistor of the transformer is 99?. 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 [url=http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=941.0]Why we need High Voltage Transmission Line[/url]