Dumb question

Light bulb

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    Boris Mohar

Could you just put 20 watts worth of lights in parallel with the relay to make the photo cell unit happy?

M

pmoseley wrote:

There's been an ugly rumor going around that Ohm's Law might be somewhat useful here.

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Many thanks,

Don Lancaster                          voice phone: (928)428-4073

OK the way I understand it is as follows: 20W/V=1.429I. And using R=V/I (be gentle with me guys. It's been a long time since I've done this) R=14/1.429=9.8 Ohms. Radio Shack has a 10 Ohm, 10W wire-wound resistor. So, I guess two of these in parallel would do it. Right?

TIA Peter Moseley

Sounds like you're using what is usually called an "electronic transformer" intended for low voltage lighting applications. That would then be a 12 or 24 volt output selection and not an oddball 14 volts. The minimum loading requirement of 20W seems to be a standard and is something they call a "demand circuit design." If your "unit" is in fact an "electronic transformer" of this type, then it is most likely a high frequency inverter type, producing a low voltage AC at several tens of thousands of kilohertz. This would make sense because high frequency operation makes for much smaller size (volume, "footprint," or what have you) which is what most people are after when they use the "electronic transformer." If your limited concentration span has lasted this far then you need to know that an ordinary low voltage relay will not operate at this of a high frequency, and this aside from the fact that a

14V relay is rather rare.

On Mon, 3 Dec 2007 15:42:18 -0800 (PST) in sci.electronics.design, pmoseley wrote,

Just use 20 watts of your lighting, then put the rest on the relay. Don't waste the 20 watts.

Plus, if the original poster uses resistors, depending on how small (physically) they are, they can get pretty hot, dissipating 20 watts. That's a lot of heat, especially if placed in an enclosed space - could possibly start a fire.

Just my 2 cents, and fwiw, I'm not an electronics/electrical expert by any stretch of the imagination

M

nope. two 10 ohm rs in || would be 5 ohms, consuming 40 W total or 20 W per, and goodby resistors. unless you need a space heater also, forget using resistors, and load it down with some automotive bulbs which handle 13.8 V or so nicely - if, in fact, that thing really has a 14 V option - otherwise, 12 V isn't such a bad choice. I suppose 24 snipped-for-privacy@gmail.com auto bulbs are available since, IIRC military trucks use 2 batts in series for 24 V - can't remember why.

mike

nope. two 10 ohm rs in || would be 5 ohms, consuming 40 W total or 20 W per, and goodby resistors. unless you need a space heater also, forget using resistors, and load it down with some automotive bulbs which handle 13.8 V or so nicely - if, in fact, that thing really has a 14 V option - otherwise, 12 V isn't such a bad choice. I suppose 24 V auto bulbs are available since, IIRC military trucks use 2 batts in series for 24 V - can't remember why.

mike

OK. Thanks Mike. How about a 5 ohm 50 watt resistor with the 12V output. That would pull 2.4 amps for 28.8 watts. I'm worried that a

12V lamp would burn out. If the resistor doesn't put out too much heat I can just plug it in and forget it.

peter

Now you get more heat. At one time some people might not care, but with rising energy costs these days, you might put them in a box for a bun warmer ;) whatever works. Do it if there's nothing better staring you in the face.

Mike

28.8 watts will make that resistor hot. Use 5 25 ohm 25 watt resistors in parallel instead to spread the heat out over a much larger area.

Ed