OT Electrical question

I got a Malibu 600 watt transformer. Model ML600TW. When I rotate the on/off knob, it makes the sound of smacking the outer housing with a stick. A loud audible clank. Then it hums.
When I apply the tester to the bottom connectors, I can't get it to read anything. Double checked tester with various voltage alkaline batteries on DC, and they tested fine, so tester is good.
Is this transformer fried? Is the humming normal? Is that clanking noise normal?
Steve

Maybe it is a.c. output.
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Nope. Sticker says 12vDC output.
Steve

On Thu, 15 Mar 2007 12:41:08 -0700, "Steve B"
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Trust me, the sticker is almost certainly wrong. Set your meter to AC Volts and check it again - hook up a scope and you'll see nice sine wave AC. And the SMACK-Hummmm... is normal as the transformer energizes and settles down to work.
I work with these every day, the Malibu Model ML600TW is a standard unit. I haven't seen a lighting transformer yet with a diode bridge and any sort of filter capacitors or chokes. They are built to a price, and AC goes further than DC on the same size and gauge wire.
If a device wants DC (like a 12V fluorescent floodlight) they have a rectifier inside the device.
--<< Bruce >>--

On Fri, 16 Mar 2007 05:32:22 GMT, Bruce L. Bergman
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Makes sense. Low voltage lightbulbs run on AC just fine, no benefit in rectifying it.
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It does? I'd think that the drop in given wires would be the same for same RMS current whether AC or DC, RMS current for DC being the DC current.
I know and respect that you are a professional electrician and know what you're doing. What am I missing?

Naw ! Don. It's that pesky sine wave thing again. If you hook on at the right time you may not read anything, don't you know.
Bob (wet fingers) Swinney
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Uh, no it doesn't. For the exact same piece of wire in the exact same installation, with the same RMS current. There will be less voltage drop for the DC, than the AC. This difference will get bigger as the frequency goes up. jk

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Well, if there is any series inductance in the cable the voltage drop could go down with frequency. But it seems to me that the capacitance or inductance in a typical power cable will be so low as to make no significant difference at 60 Hz.

snipped-for-privacy@kcwc.com (Curt Welch) wrote:
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To clarify, with series inductance, the voltage drop increases with the frequency. (As 2*Pi* f*L)
And it all depends on what you consider "significant". At 60 Hz, it is small, but not insignificant, and I deal with it almost every working day. A reference for calculating the difference is included in the back of the NEC.
jk

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Quite right. I had it backwards. If there were an inductance effect across the line, the drop would go down with frequency. (too many double negatives). There wouldn't be an inductance effect across the line however as far as I understand. So any inductance effect in series with the line would have the same effect as any capacitance effect across the lines - which both cause the power loss to increase with frequency.
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Cool.

wrote:
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Hi jk
If this guy has 100 feet of # 12 wire drawing 10 amps at 24 volts, how much difference would there be for AC compared with DC?
Jerry

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It is irrelevant what the distance is, whatthe current is, or what the voltage is.
relevant data is Conductor material WIre Size Frequency WHat type of wireway it is in, and configuration of circuit(s)
Assuming a circuit that returns along the same path, magnetic wire way, #12 uncoated copper wire, STRANDED DC resistances ohms/kilometer = 6.50, Impedance Ohms/km .223 AC resistance Ohm/km at 60 Hz 6.6
jk

wrote:
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Hi jk
Well, I didnt actually think there would be even 1/10th of an ohm difference between 60 Hz and DC. But, the glibness of your post to correct Don Foreman where you begin with " Uh, no it doesn't." caught my attention. I am sure Don is well aware of the methods that can be used to account for loss due to frequency, but he probably figured that would have confused what could be a practical answer, like the one he gave.
I do realize that *I* learned something.
Jerry

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OK, but low-voltage yard lighting wire does not typically run for kilometers thru magnetic raceway.
In Jerry's example, resistance is 0.374 ohms, inductive reactance (for insulated wires spaced 0.2" apart) is .0063 ohms so Z = ( | R+jX | ) = .37405 ohms. I'd call that difference negligable.
References: http://members.aol.com/marctt/CV/Abstracts/inductance.htm and wire tables on page 114-115 of Glover's "Pocket Ref".

On Sat, 17 Mar 2007 04:28:22 GMT, "Jerry Martes"
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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.

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You're confusing 1^2R losses with inductance. The inductance of power cables is negligible at the 50-60Hz frequencies used by power companies.

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

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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 60MPH. In both cases, those effects - while real - are tiny, & swamped by far more mundane components.
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Apparently you came to it without a sense of humour or a sense of proportion.

On Fri, 16 Mar 2007 04:16:28 -0500, Don Foreman
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Voltage losses due to wiring resistance rise with the square of the load current (I^2 x R, or current squared, times resistance), so the smaller the current, the lower the losses. For a given power consumption (ie; Watts or VA), you can lower the current by raising the voltage down the wires, then lower it again at the other end. (ie; in your gizmo). This is expensive & difficult to do with DC, but can be done cheaply & easily on AC with just a transformer.

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I saw someone else mention this idea recently but I've never heard it before and don't understand how it could possibly be true.
Care to explain why this is true that AC "goes further" than DC?
Anyone else know about this?