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In an 12v outdoor lighting circuit, is our electrical consumption (our electrical bill) determined by the 50W 12V bulb that is connected to the transformer, or the 300W rated output of the transformer? I have equally received two answers; i.e.....(1) our electrical usage is based upon the 50W bulb, (2) our usage is determined by the transformer making available 300W, regardless how much of the 300W we use.

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  • $\begingroup$ Ethan - Thanks so very much. We have been trying to determine where our electrical is being consumed. I thought it might be the Malibu 300W transformer running +/-12hrs per day X 30 days, or 108000W per/mo. 108Kw X 13 cents = $14/mo. I'll absorb that for the security. That leaves $160 per month unaccounted for. Yes, it does appear that my household is the root cause of brown outs. $\endgroup$ – Steven M Sep 5 '15 at 21:48
  • $\begingroup$ As Olin notes - the no load wattage of a power supply MAY be around 10% of the rated power - so a 50W load on a 300W transformer MAY draw somewhat more than expected. Using a transformer more closely matched to the load MAY save some money. $\endgroup$ – Russell McMahon Sep 7 '15 at 10:24
  • $\begingroup$ Worth noting: actual transformers have relatively low losses. OTOH, there exist cheap, crappy alternative PSU devices that simply waste most of surplus power, dissipating it as heat. Luckily you usually won't find these in 300W+ range. $\endgroup$ – SF. Sep 18 '17 at 7:28
  • $\begingroup$ Just to be clear, the lights-off consumption is 0 because the thing that turns the lights off switches the primaries, dumping the transformer. The lamps are always connected as a load, whether powered or not. So you are only left to worry about the draw in the on position. Some of the previous discussion may have confused some readers on this point. $\endgroup$ – Phil Sweet Sep 18 '17 at 21:00
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The answer is #1, with a bit of an asterisk.

If the bulb pulls 50 watts as it's rated, the current (power divided by voltage) will be 4.17 amps. We know that anywhere in a closed circuit, the current is the same, so the power supply is outputting 4.17 amps as well (assuming it is attached to only one light bulb.)

So, if your power supply is outputting 12 volts (which it is because it's designed to regulate its output voltage) and it's only outputting 4.17 amps (which it is because of the resistance of the light bulb) we can calculate the power it is outputting as 4.17 amps times 12 volts and we end up back at 50 watts. The 300 watt rating on the power supply just means that that's what it can output, not that it always outputs that. If you attached 5 more light bulbs, all in parallel, then it would produce the whole 300 watts.

The only trick is that we have checked how much power the power supply is outputting. Your electric bill is based on how much power you consume at line voltage which is the input to the power supply. Power supplies are not perfectly efficient (some energy gets lost as heat,) so your consumption will be a bit higher, but hopefully not a whole 300 watts. If we assume, conservatively, that your power supply is 50% efficient, you'd still only be drawing 100W at the input side, not 300. Also of note is the fact that if your power supply were outputting a whole 300 watts, the input would be even more than 300W.

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A perfect ideal transformer converts some combination of Volts and Amps going into it, to some other combination of Volts and Amps coming out but a the same power. Put another ways (Volts in)(Amps in) = (Volts out)(Amps out).

By this logic, it is the power used by the bulb that matters. Just because the transformer could have supplied 300 W (and then drawn 300 W from the power line) is irrelevant. All that means is that you could connect a brighter bulb or more bulbs that add up to 300 W.

However, back in reality, transformers aren't perfect. OutPower = InPower x efficiency, where efficiency will always be somewhat less than 1. As you might imagine, higher efficiency requires tricks that add to the cost of the transformer, so some compromise was inevitably made. The efficiency might be 90%, for example, or maybe just 80%, or even lower.

One way to get some idea of this is to feel the transformer after it's been on a while. If it's barely warmer than when off, its efficiency is good. If it's considerably warmer than when off, then clearly it's dissipating power that is not getting delivered to the light. Both the power to drive the light (50 W) and to heat the transformer are delivered by the power line.

When the light is off, the transformer is really just a inductor connected to the power line since its secondary is open-circuit and therefore has no current thru it and no relevance to the magnetic field in the transformer. There will be current in the primary, which is ideally 90° out of phase with the voltage, so no net power is delivered to the transformer and you are not charged for this "reactive power", as apposed to "real power".

However, the wires the primary coil is made from don't have perfect 0 resistance. Electrically, this looks like a resistor in series with the inductor. The inductor still doesn't dissipate power, but the resistor does, and you do get billed for that.

So overall, the power drawn from the line by the transformer is largely a function of how much power is drawn from its secondary, but there is always some loss which is made up for by drawing more power, and even with the light off there will be some power drawn. The higher the transformer rating, the higher this no-load power will likely be. Up to around 10% of the full power rating wouldn't be much of a surprise for such a transformer that was likely optimized for cost much more than efficiency or no-load power.

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  • $\begingroup$ 10% is a good estimate. I have a friend who replaced tungsten lighting with LEDS throughout a 'ski chalet' and was annoyed to find that the no load power was indeed about 10% - so power with LEDs was about double what may have been expected. Downsizing the transformers would be a major task in the context. Ppower consumption mattered more there than in most cases as the chalet is fed by long circuits and the power company charges per unit are based on the peak draw over at any time over a short period during any year. Keeping the peak load down saves money all year. (Non electric heating). $\endgroup$ – Russell McMahon Sep 7 '15 at 10:22

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