# Is Ohm's law obeyed in power transmission?

We learnt in high school that according to Ohm's law $V/I=R$. We also learnt that during power transmission in an electric line $P=VI$ and that in order to minimize loss voltage is raised. As a consequence current is reduced. This contradicts with the Ohm's law which states that $V$ is proportional to $I$. What is the difference?

The usable voltage out of a battery or a generator is $V = E - i R_{wire}$ where $E$ is the rated voltage (being 12V or 768kV), $i$ is the current and $R_{wire}$ is the resistance of the wire. The current is usually found by the load on the circuit as $i=\frac{V}{R_{load}}$.

So alltogether with power $P=i V$ we have

$$V = \frac{E R_{load}}{R_{load}+R_{wire}} \\ P = \frac{E^2 R_{load}}{(R_{load}+R_{wire})^2}$$

Only when the wire resitance is negligible you get $$V=E \\ P=\frac{V^2}{R_{load}}$$

The relevant voltage here in $P = V\, I$ is NOT the whole voltage across the power generator (in which case your reasoning would hold and there would be a contradiction), but rather the voltage drop across the transmission lines. The transmission lines look like a resistor $R_{line}$ in series with the load $R_{load}$ (the power users) at the other end. The following applies to resistive loads only, which power companies like to force their customers to use anyway, but the concepts are general.

Power users need power. We can send it to them either as a huge current through a low voltage load, or as a small current through a very high voltage load. The product of these two: $V_{load} I$ is constant: the rate at which the power users want their machines to do work in watts. The transmission line resistance is constant: we can use thick power lines and good materials, but we can't lower the resistance to nought. So the power losses are $I^2 R_{line}$. If we send the power to the users as a high current through a low voltage load, the losses $I^2 R_{line}$ will be higher than if we send a small current through a higher voltage load to achieve the same transmitted power. Hence, near where the power is used, transformers transform the impedance of the user loads so that a higher load impedance is presented to the the transmission lines, which then supply a low current through the accompanying high voltage.

I would draw the series circuit and do the following analysis: suppose we wish the power dissipation through $R_{load}$ to be some constant $P$. We can use a transformer to change the value of $R_{load}$ by the phenomenon of reflected impedance: see my answers here and here. So we our reflected $R_{load}$ as big as we technologically (i.e. limited by the accompanying high voltages and the need to insulate them) can. The current is then $\sqrt{P/R_{load}}$, yielding transmission losses$P\,R_{line}/R_{load}$.

• Actually because the power lines are long relative to the wavelength they need to be modeled as transmission lines. In addition you need to include reactive power as well as real power in the calculation. But that's probably TMI. – user6972 Oct 4 '13 at 4:51
• TMI indeed. But if you match the load the source/load doesn't experience any delay (phase shifts) they just see real resistances. – user6972 Oct 4 '13 at 5:16
• @user6972 Of course, but one principle at a time. As you say, TMI. Also, I understand that alot of engineering goes into adding reactances so that loads are matched: this reduces reflexions and makes the transmission line look like a resistor with delay. In any case, the full calcs show the same thing. Incidentally, this kind of calc was exactly what this thing (flip over) and also here, with its Sinh and Cosh scales, was developed to do in 1929. – Selene Routley Oct 4 '13 at 5:17

If you use high voltage for long distances, you'll need a transformer to connect your household load. A transformer not only changes the voltage, but also the current, hence the R (impedance) with the square of the voltage ratio. On either side of the transformer Ohm's law will hold.