# Voltage/Current Questions **URL:** \ **Category:** Factual Questions\ **Created:** [November 27, 2002, 3:12pm UTC](https://boards.straightdope.com/t/voltage-current-questions/139108 "2002-11-27T15:12:18Z")\ **Posts on this page:** 6\ **Page:** 2
**Author:** ![andy\_fl](https://avatars.discourse-cdn.com/v4/letter/a/49beb7/32.png) [@andy\_fl](https://boards.straightdope.com/u/andy_fl)\ **Post date:** [November 28, 2002, 12:56pm UTC](https://boards.straightdope.com/t/voltage-current-questions/139108/21 "2002-11-28T12:56:55Z")
> [@](#): > > \*Originally posted by David Simmons \* > \*\*The generators that provide the input to transmission lines are 3-phase machines. A full-wave, three-phase rectifier has 6 half cycles per cycle of the basic 60 Hz. wave. This gives an rms ripple of less 9% without any filtering, so the alternating component of the output is small. \*\* So Dave. The ripple frequency I think would be 6 times of the supply. Also can you please explain what kind of rectifiers are these. And also how the inverter works. From, what I remember, its difficult to make a inverter which gives pur sinusoidal output. ---
**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 28, 2002, 2:16pm UTC](https://boards.straightdope.com/t/voltage-current-questions/139108/22 "2002-11-28T14:16:24Z")
One scenario where DC might be transmitted: Let’s say a power plant sells 60 Hz power to all of its customers. Along comes a customer who wants **50 Hz** power. In this situation the 60 Hz is converted to DC then reconverted to 50 Hz AC. But _where_ should the DC-to-50 Hz conversion be done? It is more efficient to perform the 60 Hz-to-DC conversion at the _source_ plant, transfer the power at _DC_, then perform the DC-to-50 Hz at the _customer’s_ location. As I mentioned in a previous post, this is because DC has a slight efficiency advantage over AC when you’re considering just the conductors… ---
**Author:** ![SandWriter](https://avatars.discourse-cdn.com/v4/letter/s/8491ac/32.png) [@SandWriter](https://boards.straightdope.com/u/SandWriter)\ **Post date:** [November 28, 2002, 4:22pm UTC](https://boards.straightdope.com/t/voltage-current-questions/139108/23 "2002-11-28T16:22:38Z")
So based on Chronos’s post, you’d have to put in a 17.7(ish) ohm 3k watt resistor to get maximum power draw without blowing a fuze. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [November 28, 2002, 5:41pm UTC](https://boards.straightdope.com/t/voltage-current-questions/139108/24 "2002-11-28T17:41:49Z")
> [@](#): > > \*Originally posted by andy\_fl \* > \*\*So Dave. The ripple frequency I think would be 6 times of the supply. \*\* That’s right, but the frequency is immaterial. This power is lost to the system anyway irrespective of frequency. Even with that loss, the DC is still competitive provided, as was said, if the line is long enough. > [@](#): > > \*\*Also can you please explain what kind of rectifiers are these. And also how the inverter works. From, what I remember, its difficult to make a inverter which gives pur sinusoidal output. \*\* I think they are solid state diodes. At one time, Phillips Industries was the only manufacturer of diodes that would do the job. Even then, several of their diodes had to be stacked in series with auxilliary circuitry to make sure they shared the inverse voltage equally. There were no, and probably still aren’t any, diodes that will take the inverse voltage for such a DC power line. The line was 1.5 million volts between the lines, + and - 750 K to ground for the two lines. I hope someone who has more detailed knowledge will come up with better answers. Mine are sort of generic based on my experience and without any exact data on the details of the installation. The inverters were silicon controlled rectifiers triggered from the local AC power to make sure they were in frequency and phase with the local power system. The SCR’s also had to be stacked in series because of device inverse voltage limitations. The raw output of the inverters isn’t sinusoidal, but filtering will correct that, and besides the power lines themselves are filters of a sort. It was interesting to watch the evolution of the Sylmar converter station. It was first erected as an open-air structure with screen shielding on the walls to hold down radio interference from the switching noise. As time passed, more and more shield was added until the structure’s walls became almost solid. Apparently the noise was a lot worse than they had originally anticipated. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [November 28, 2002, 8:04pm UTC](https://boards.straightdope.com/t/voltage-current-questions/139108/25 "2002-11-28T20:04:31Z")
Well, apparently the original DC system was quite different that it is now. Here are a couple of references about it. [Pacific Intertie.](http://www.transmission.bpa.gov/oasis/bpat/outagecoord/pdci_public_process.htm) And [Modifications to converter sations.](http://www.pgd.siemens.ca/news/trans/news43.htm) Apparently the original rectifiers and switches were mercury arc tubes, still used as of 2000 at one station, and the voltage wasn’t as high as it is now. And Siemens, not Phillips Industries, is the supplier. Other than those few minor details you really have to admit that my information was right on the button. Sure it was. ---
**Author:** ![mrcrow](https://avatars.discourse-cdn.com/v4/letter/m/838e76/32.png) [@mrcrow](https://boards.straightdope.com/u/mrcrow)\ **Post date:** [November 28, 2002, 8:27pm UTC](https://boards.straightdope.com/t/voltage-current-questions/139108/26 "2002-11-28T20:27:38Z")
voltage is a differential made up at the power station current is what will flow through any resistance put up against the voltage current is the voltage divided by the resistance numerically smaller the resistance greater the current till you get none and then bang I=V/R the watts is the current x the voltage for power calcs.=VxI or I^2 x R this is my elementary understanding of it all 🙂 [Previous page](https://boards.straightdope.com/t/voltage-current-questions/139108.md?page=1)