# AC versus DC Power? **URL:** \ **Category:** Factual Questions\ **Created:** [November 10, 2003, 4:57am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349 "2003-11-10T04:57:42Z")\ **Posts on this page:** 12\ **Page:** 2
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [November 10, 2003, 11:17pm UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/21 "2003-11-10T23:17:08Z")
> [@](#): > > \*Originally posted by Crafter\_Man \* > \*\*As stated in my first post, a DC power transmission line must be more efficient than an AC power transmission line. This is because there’s no such thing as an inductor with an infinite Q factor, and there’s no such thing as a capacitor with an infinite Q factor. A capacitor will dissipate energy due to loss tangent and ESR, and an inductor will dissipate energy due to magnetic hysteresis losses. It’s probably not very significant, but real none-the-less. \*\* I left out **skin effect**. Which may or may not be significant. ---
**Author:** ![bughunter](https://avatars.discourse-cdn.com/v4/letter/b/aeb1de/32.png) [@bughunter](https://boards.straightdope.com/u/bughunter)\ **Post date:** [November 11, 2003, 12:20am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/22 "2003-11-11T00:20:01Z")
Sorry - off on the other boards for a bit. **gazpacho** No, I am not implying that there are no resistive losses in an AC transmission line, but that they appear as _current_ losses, not voltage losses. Current is not lost in DC systems because there are no shunt components in a point-to-point transmission line. But the parallel capacitance of the tranmission line allows AC current to return short of the terminating element. **Crafter\_man** , still looking for cite. In my attempt to explain transmission lines without using jargon like resonance and phase margins, I may have oversimplified to the point of confusion to an EE. If you read to the end of the link I provided, it describes how most models omit completely the resistive element of the transmission line, because for a properly tuned and terminated line, the reactive components dominate completely. It’s a suitable approximation for all but the longest transmission lines. And even in those, the inclusion of a resistance element is not the same value as the linear resistance of a DC transmission wire. I recall this being glossed over as the “simple answer” in my undergrad transmission theory course. This hasn’t been so easy to find a cite for, but the web is a big place. I didn’t mean to imply that there was any overall gains using AC, just that at the time the power distribution systems were invented, the voltage losses were much less than using DC, allowing much further transmission over practical distances using practical voltages. This isn’t the only tradeoff point, but it’s a very important one, and in the histories I’ve read, the one that shot DC down in the early battle of AC vs DC. ([cite](http://www.nextekpower.com/acrobat/ACDC2.pdf)) A lot of the cites I’m finding point to multiple point generation and distribution more difficult with DC due to the lack of a DC equivalent to a transformer, which was basically **gazpacho’s** ’s point, and some cite both. But Edison was confident he could overcome this scheme by building local generators to serve neighborhood-scale markets. The efficiency that makes a practical difference, in the end, is cost efficiency, and the ability to centrally generate power and distribute it cost effectively to rural and suburban customers won out in the end. ---
**Author:** ![bughunter](https://avatars.discourse-cdn.com/v4/letter/b/aeb1de/32.png) [@bughunter](https://boards.straightdope.com/u/bughunter)\ **Post date:** [November 11, 2003, 12:45am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/23 "2003-11-11T00:45:49Z")
> [@](#): > > \*Originally posted by CurtC \* > \*\*And if energy is lost, the voltage is reduced at the load end. \*\* Not necessarily. If energy is lost, the _available power_ is reduced at the load end. This will appear as a voltage loss in a single-mesh DC loop, due to Kirchoff’s Current Law. But in an AC circuit, the loss can appear as a phase shift between the current and voltage, or as a reduction in voltage, or in current, depending on the specifics of the source, load, and transmission line. You cannot apply the simple rules of DC circuits to AC transmission lines. It just don’t work… and it’s not nearly as intuitive. If you imagine DC electricity as water thru a pipe you gain some useful insight, but AC is more like sound transmission thru your plumbing. It’s nothing like the flow of water in your tap, and more like the weird noises your pipes make when the taps are running. ---
**Author:** ![gazpacho](https://avatars.discourse-cdn.com/v4/letter/g/6f9a4e/32.png) [@gazpacho](https://boards.straightdope.com/u/gazpacho)\ **Post date:** [November 11, 2003, 1:47am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/24 "2003-11-11T01:47:11Z")
> [@](#): > > gazpacho No, I am not implying that there are no resistive losses in an AC transmission line, but that they appear as current losses, not voltage losses. Current is not lost in DC systems because there are no shunt components in a point-to-point transmission line. But the parallel capacitance of the transmission line allows AC current to return short of the terminating element. A current loss? If there is current flowing in the conductors there will be a voltage drop across the line. There is a current flowing in AC power lines. Your explanation of losses due to the capacitance only make AC power lines worse than DC power lines. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [November 11, 2003, 1:47am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/25 "2003-11-11T01:47:11Z")
**bughunter** , I’m having difficulty accepting your explanation of voltage relationships as having any impact on the choice of AC over DC. The Ferranti effect wasn’t even _known_ about until after a fire at Ferranti’s Deptford power station. The Ferranti effect doesn’t rely on using a distributed parameter model, either. It works out just fine using a lumped parameter model. Say you have a transmission line of length _x_. It has a series inductance of L per unit length, and a shunt capacitance of C per unit length. When the line is not supplying any load current, the receiving end voltage is given by: V[sub]R[/sub] = V[sub]S[/sub] / (1 - 0.5 [symbol]w[/symbol][sup]2[/sup] L C _x_[sup]2[/sup]) Note that V[sub]R[/sub] is necessarily greater than V[sub]S[/sub]. As the load current is increased above zero, the receiving end voltage falls because of the voltage drop across R and [symbol]w[/symbol]L. For high voltage systems, [symbol]w[/symbol]L is about ten times larger than R, meaning that voltage regulation is much worse on AC systems than on DC systems. At some particular level of load current, the receiving end voltage will equal the sending end voltage, and this is the situation you are describing. However, transmission lines are operated over a wide range of loading levels, and it would be pure coincidence if I[sup]2[/sup][symbol]w[/symbol]L equalled V[sup]2[/sup]/[symbol]w[/symbol]C at any particular time. What really happens is that as the loading level on the line changes, shunt inductors and shunt capacitors are switched in and out of service to counteract the piss-poor voltage regulation of the AC line. ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [November 11, 2003, 2:12am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/26 "2003-11-11T02:12:50Z")
All I know is that the biggest west coast electric grid backbone uses several DC lines. The 845 mile[Inter-Tie](http://www.usbr.gov/dataweb/html/pninter.html) lCelilo (OR)-Sylmar (CA) line, uses 800-kV DC. I have a slight tendency to believe that the EEs who figured out that sending all that juice was best done with DC weren’t idiots. (I was good friends with two sons of the then BPA head in college. They were smart as could-be. Not so sure about the father.) In short: DC is better if you have the tech but it costs more so the longer the link the better. ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [November 11, 2003, 2:37am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/27 "2003-11-11T02:37:13Z")
The **Ferranti Effect**? Never heard of it. You’re the man, **Desmostylus**. 🙂 So it would appear that, if certain conditions are met, the load voltage can actually be _higher_ than the source voltage. Fascinating. But this does not necessarily mean more power will be available to the load. All else being equal, is it still be true that a DC transmission line is less lossy (i.e. more efficient) than an AC transmission line? I would think that this would be true even with the Ferranti Effect in play… ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [November 11, 2003, 3:00am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/28 "2003-11-11T03:00:37Z")
> [@](#): > > is it still be true that a DC transmission line is less lossy (i.e. more efficient) than an AC transmission line? Yes. There’s no charging current, so more _load_ current can be carried for a given conductor size. No charging current also means less I[sup]2[/sup]R losses for a given load current. There’s no losses cause by induced currents in parallel conductors (e.g. earth wires, fences, telcoms cables). No EM-radiation losses, either. You also have smaller towers and narrower easements (right-of-ways), because the peak voltage is the same as the rms voltage. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [November 11, 2003, 4:26am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/29 "2003-11-11T04:26:24Z")
> [@](#): > > _Originally posted by bughunter_ > **If you read to the end of the link I provided, it describes how most models omit completely the resistive element of the transmission line…** It doesn’t say that at all. > [@](#): > > _Originally posted by bughunter_ > **And even in those, the inclusion of a resistance element is not the same value as the linear resistance of a DC transmission wire. I recall this being glossed over as the “simple answer” in my undergrad transmission theory course. This hasn’t been so easy to find a cite for, but the web is a big place.** Don’t worry about cite for this. **Crafter\_Man** has already mentioned skin effect, which serves to increase R. There’s also another factor. To accurately model a long transmission line using lumped equivalents, it’s not sufficient to simply take the R, L and C per unit length, and multiply by the total length _x_. The R and L also have to be multiplied by a correction factor: sinh([symbol]w[/symbol]_x_√(LC))/([symbol]w[/symbol]_x_√(LC)), whilst C has to be multiplied by a correction factor of tanh(0.5[symbol]w[/symbol]_x_√(LC))/(0.5[symbol]w[/symbol]_x_√(LC)). But, unless _x_ is very large, those correction factors are very, very close to 1, meaning you can safely ignore them. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [November 11, 2003, 4:30am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/30 "2003-11-11T04:30:04Z")
What’s “very large” in this instance? Transmission lines can run into the hundreds of miles… ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [November 11, 2003, 6:11am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/31 "2003-11-11T06:11:13Z")
In this case, “very large” happens to be hundreds of miles. The wavelength in air at 60 Hz is 5,000 km. 500 km is 1/10 of a wavelength, and at this point, the effect becomes noticeable. ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [November 11, 2003, 7:03am UTC](https://boards.straightdope.com/t/ac-versus-dc-power/212349/32 "2003-11-11T07:03:26Z")
Let’s try an example. [symbol]w[/symbol]L = 0.3 [symbol]W[/symbol]km[sup]-1[/sup] [symbol]w[/symbol]C = 3 x 10[sup]-6[/sup] Skm[sup]-1[/sup] The factor sinh([symbol]w[/symbol]_x_√(LC))/([symbol]w[/symbol]_x_√(LC)) for different values of _x_ is tabulated below: ```auto x (km) factor 50 1.000 100 1.002 150 1.003 200 1.006 250 1.009 300 1.014 350 1.018 400 1.024 450 1.031 500 1.038 550 1.046 600 1.055 650 1.065 700 1.075 750 1.087 800 1.099 850 1.112 900 1.126 950 1.141 1000 1.157 ``` [Previous page](https://boards.straightdope.com/t/ac-versus-dc-power/212349.md?page=1)