Why Is Power Loss in Transmission Lines Calculated as I²R and Not V²/R?

ı asked the similar question once more but ı couldn't understand with the answer please somebody explain it clearly for me. The question was that while we are transmitting the power across a transmission line, we use high voltage and low current to make lower the power loss. power loss is equal to i^2*r and we can also write it like v^2/r but if we write it like the second equation it gets a huge power loss. so what is the point that ı couldn't see.

Do an equivalent circuit, assume the transformers are perfect for simplicities sake.
Voltage Generator on the primary of a voltage step up transformer, secondary connected to an equivalent transmission line resistance. Connect the other end of the transmission line to the primary of a voltage step down transformer, the secondary to the load.
Now look at what is happening, all the power is transformed to the load, apart from the power loss in the transmission line. If the turns ratio on the input transformer is adjusted to give a higher secondary voltage, and on the output transformer to give the same load voltage, for the same load power, the current in the transmission line has dropped(this is the same current in secondary of the source transformer and the primary of the output transformer). Since this is the primary source of loss in the whole system, then the power loss must drop.
Best I can do, does that help at all, put the whole circuit into spice if you must, but sketching the whole thing up and analyzing it on a bit of paper should do it just as well,
Cheers,
Richard

It is the power which is transmitted and it is constant i.e., P=VI. For the constant power, if we increase voltage(V) then current reduces and vice versa. If we stepdown voltage then what happens is that current increases due to which we have to increase the size of conductor which makes conductor very bulky and almost impossible to erect the line. So if we increase voltage only thing we have to do is to increase the insulation level i.e., usually height of tower is increased for high voltages.

I^2*r would give you the power losses over the power line. v2^/r would give the power loss through the line if the other end were connected to ground (ouch). (v1-v2)^2/r would be the correct equation for find power losses if the voltage leaving the power plant (v1) and the voltage at the end of the power line (v2) is known. As resistance of the power line drops, the voltage drop over the power line would also drop so (v1-v2)^2/r would go to zero as r goes to zero.

High voltage (V) is used for transmission because the same power (P) can be moved from one point to another using less current (I) as we know from Ohm’s Law (P=IV). Power losses across the line can be calculated by either i^2*r or (v1-v2)^2/r, but i^2*r is typically used so that you don’t have to measure the voltage drop at the end of the line in order to calculate power loss and current at one end of the line is the same as current at the other end (+ or -).

Yes, P=V^2/R is a basic law, this equation can't be wrong and can absolutely be applied to any situation.

The simple explanation is that the voltage drop over a long transmission line is very small because of the low current, V=IR, so that the power can be reduced at last.