# High voltage power lines - clarification of energy loss

I've been having a bit of trouble understanding the high-voltage power lines. If I was sending power from $$A \rightarrow B$$, we have:

Ohm's law $$V = IR$$

Power lost in the form of heat $$P = I^2 R$$

Power delivered to $$B$$ is $$P = VI$$

But using Ohm's law on the power lost formula, we get $$P = V I$$ also.

Does this mean that the total amount of power $$A$$ loses is $$P_A = 2 VI$$, and the total amount of power delivered to $$B$$ is just $$P_B = VI$$? Will transferring energy always result in half of it being lost?

The difficulty is that there are three voltages involved.
The voltage at the power station end $V_S$, the total voltage drop across the cables $V_L$ and the voltage at the consumer end $V_C$.

The voltages are related as follows.

$V_S=V_L+V_C$

So you have power supplied by power station is equal to the power lost in the transmission cables plus the power used by the customer.

$V_SI = V_L I+ V_CI$

So the power loss in the cables is $V_LI = RI\cdot I = I^2R$ where $R$ is the total resistance of the transmitting cables.

The reason for transmitting the power at high voltage is that then the current through the cables $I$ is less and so the ohmic loss $I^2R$ in the cables is considerably smaller.

• I completely forgot about the voltage drop from the power loss. That's perfect - thank you very much. Apr 8, 2016 at 12:33
• It is important to remember that the current is the same everywhere and that we cannot think of using V^2/R since it does not make sense . There is no single V but 3 Vs as stated above. May 29, 2021 at 10:54