By Tweej

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?


@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.


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.

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