OT Electrical question

For mains power via a transformer (ie; a sine wave @ 24VAC RMS), it's the same. Either way, you'd also need to know the resistance/foot (or metre) of the wire he's using.

Ain't that the dyin' truth!

Esotic theory is fun for some (including me at times) but KISS gets 'er done when it works.

In all fairness, Bruce, you sorta started the argument with an assertion --but who cares. In this sit, KISS works nuff said. Hook 'em up and make light, do the job.

On Tue, 20 Mar 2007 04:38:18 GMT, with neither quill nor qualm, "Leo Lichtman" quickly quoth:

I thought he discovered keys, and had to split it with his sound effects man.

-- Losing faith in humanity, one person at a time.

Yeah, and I said right up front that it really doesn't matter one itty bitty teeny tiny /infinitesimal/ bit in the big picture.

But I should know better. Bringing up an esoteric techie question here - even though the answer is pedantic, pointless, and nobody should waste the mental effort and/or CPU cycles needed to actually answer it - is like waving a T-Bone steak in front of a Rottweiler. You are GOING to get a reaction...

"If I do it, I'm gonna get a whippin'... I dood it." ;-D

-->--

Amd the dissipation of the diode stack likely outweighs the gain, all by itself.

Not hardly, but your post most certainly is.

Where as you find a need to jump in the "defense" of some one who doesn't need it. There was an error, and I pointed it out. If you don't care for it, get over it.

What a nonsensical, over general and uninformed statement. No one is so well informed, in every aspect, as to be the equal of all experts in a field, any field.

Hardly "curt", but conversational, learn the difference. I am sure if HE has a problem with what I said, HE would chime in.

Perhaps you should go back and read the post I was responding to, which wasn't his any way. In fact it was yours, which perhaps explains, but not excuses your tone. All I did was clarify a statement of yours, in summary. Unless of course you are referring to some other post, but you fail to communicate that effectively.

jk

Small yes, negligible, no. Apparently you come to this party late. jk

Hi jk

You may be confused about my "defending" anyone. I did want to point out that you deal with minute, unimportant details that tend to confuse the discussion. You never did establish how much the loss will increase when AC is used instead of DC. Maybe you can tell me more about the amount of loss will be included as the result of using AC rather than DC in the circuit with a purely resistive load such as the Malibu lights.

Jerry

The distances between the three phases of the t-line and a point on the ground beneath them are not identical, so the vector sum of the H-fields will be nonzero at that point. (Ampere's law, I think, and superposition) It won't be much, but it won't be zero. If one has a long enough wire under the lines, usable voltage will be induced in it.

Over a domestic lighting cable run? Of course it's negligible. Considering the inductive losses over such short runs at 60Hz makes about as much sense as considering blue-shift of a cars headlights at

Apparently you came to it without a sense of humour or a sense of proportion.

Yeah, jk seems to be a whiny, pedantic idiot.

In partial answer to your questions; while I'm too lazy to dig out the EE texts to compute the answer rigorously, I can assure you that the difference between DC & 60Hz AC is so low that jk is too embarrassed to state it.

And in fact, in the OP's situation, even the I^2R losses on good quality cable to a bunch of 12V halogens will be small enough that you'd need a good multimeter to measure it. In his case, the best reason to run AC would be to prevent electrolytic corrosion of the cables & fittings from condensation, etc, which can be a major problem with DC.

"Don Foreman" wrote: (clip) If one has a *long enough*wire under the lines, usable voltage will be induced in it. ^^^^^^^^^^^^^^^^ Yeah, and if you give me *enough* sand, I will build you a golden castle.

Leo sez:

"> Yeah, and if you give me *enough* sand, I will build you a golden castle."

Tsk, tsk, Leo ! Only a golden castle ? And all the while I thought you were smart enough to make transistors from sand.

Bob (stiffling a tear) Swinney

Hi Lionel

You and I agree. You state it better than I do. I did want jk to have an opportunity to re-consider his original statement when we are comparing DC vs RMS in lawn lights. But, he seems to be fixed on demonstrating his depth of knowledge of electricity.

Jerry

I recall some guy in a rural setting that stole enough power to get stuck with a bill of many thousands when the power company took him to court. I don't recall if it was revealed how much wire he used.

The coupling coefficient would indeed be miniscule -- but miniscule ain't zero and the numbers associated with highline power transmission are enormous. 0.01% of 10 megawatts is a kilowatt which is certainly a usable amount of power. Nameplat capacity of Hoover Dam is about

Gosh, there may be significant opportunity for you in Iraq!

A broad, incorrect assertion was made, and I corrected it.

Actually I had already answered that, all be it indirectly, and did not figure that it needed repeating.

jk

I recall hearing of one such case,(the truth of which I have no idea) where a rancher used this to light a gate light, and fought the power company in court, and won. THis because these "stray" power losses are already accounted for in their tarrifs, and to collect from him, the would have had to rebate to everyone, the amount of those stray losses. jk

Nice weasel-words, Mr Expert - tell us all what the inductive voltage loss will be on 200' of cable carrying 12V AC vs 12V DC for a bunch of small lamps, then compare that to the I^2R losses on the same setup.