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I'm trying to understand why high voltage is used in transmission but I seem really confused by the explanations I read.

Here's what I could make out:

$$P_\text{loss} = \dfrac{ΔV^2}{R_t} = \dfrac{(V_0-V'_h)^2}{R_t}$$ So increasing $V'_h$ must reduce the power loss in the transmission line. ($R_t$ is the resistance of the transmission line.)

Is my understanding flawed or correct?

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  • $\begingroup$ What does your question have to do with a transmission line. A coaxial cable is an example of a transmission line. What role plays the transformer you have drawn in the circuit diagram? $\endgroup$
    – oliver
    Commented Mar 29, 2022 at 16:16
  • $\begingroup$ I understand the OP to ask about transmission over the power grid, not signal transmission. $\endgroup$
    – Ralf Kleberhoff
    Commented Mar 29, 2022 at 16:18
  • $\begingroup$ I think the $\Delta{}V$ may be confusing you. It refers to the voltage loss from one end of the power line to the other. But the voltage loss depends on the resistance of the wire, $R_t$, and on the current that the load is consuming. If you know the current, $I$, then a more direct way to calculate the lost power is $P=I^2R_t$ $\endgroup$
    – Solomon Slow
    Commented Mar 29, 2022 at 19:36
  • $\begingroup$ This should have been better posted on EE.SE $\endgroup$
    – Miss Mulan
    Commented Apr 5, 2022 at 13:15
  • $\begingroup$ There's another reason besides power loss. Because the current transmitted is lower, thinner wires can be used which costs WAY less money!!!! Not everything is about physics. Lol $\endgroup$
    – Kyle B
    Commented Apr 6, 2022 at 1:26

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Your understanding is correct. To be a bit more precise:

You want some power $P_h$ to arrive at the consumer. And, as (ideal) transformers transform voltage and current, but keep the power value constant, the same power is also consumed at the ${V'}_h$ side of the transformer.

The current at the transmission line then must be

$$I_t = \frac{P_h}{{V'}_h}$$

The power loss at the $R_t$ resistance is

$$P_{loss} = {I_t}^2 \cdot R_t$$

$$P_{loss} = \left(\frac{P_h}{{V'}_h}\right)^2 \cdot R_t$$

You see, keeping power consumption and line resistance constant, the power loss decreases with the square of the transmission-line voltage. E.g. raising the transmission-line voltage by a factor of 10 decreases the power loss to 1/100.

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    $\begingroup$ In electrical engineering, we actually have a name for this effect --- We call it the "I-squared-R loss". The term obviously coming from your 2nd equation above. $\endgroup$
    – Kyle B
    Commented Mar 29, 2022 at 18:02
  • $\begingroup$ @KyleB In the last place I lived, the power company paid a discounted price for my excess solar power. The discount was calculated based on one ohm's worth of loss. $\endgroup$
    – John Doty
    Commented Apr 5, 2022 at 12:40

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