2016-04-08 12:21:13 8 Comments

I've been having a bit of trouble understanding the High-Voltage powerlines. 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?

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## 1 comments

## @Farcher 2016-04-08 12:30:41

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.

## @Tweej 2016-04-08 12:33:33

I completely forgot about the voltage drop from the power loss. That's perfect - thank you very much.