I just purchased a 600w 12v transformer for outdoor lighting. Is the DC load going to directly determine the current draw (similar to AC)? Or is this baby going to suck down 600w no matter what the DC load is? It has an eerie hum with just the AC plugged in with no DC load, and it’s warm, and it’s huge (at least 20#).

Reason I ask: I only have about 200w worth of lighting that I need to power, but I opted for the larger power supply for the possibility of expanding, and it was only $20 more than a unit with half the capacity. Just wondering if the smaller unit would have been notably more efficient (300w unit driving 200w worth of lights versus a 600w unit driving 200w worth of lights)?

Thanks

 

awww. I got all excited.

 

Current draw will vary with load.

More lights = more current on the AC side.

The wattage rating on the transformer indicates the maximum that it can supply before it blows up.


Yup. Transformers are not 100% efficient. They draw a small amount of current even with no load.

If you were to measure a transformer's primary winding with a regular DC ohmmeter, it would appear to be a dead short. It's just a big coil of wire connected directly across both legs of the AC supply. The reason it doesn't immediately catch on fire and blow a fuse is because it has a very large AC impedance, and impedance is basically a resistance to change in the flow of current (it's sort of like resistance, but applicable to AC only.) If you were to feed 120v DC into it, it would melt and catch on fire. This is the big reason why Westinghouse beat Edison in the War of the Currents- you can run AC it into transformers, and the electromagnetic properties of the transformer cause it to magically load-follow. With DC, you can't do passive voltage conversion.

Those poor elephants...



Very slightly, but you're not likely to notice it on your power bill.

 

P = I * I * R
Where P = power (measured in watts)
I = current (measured in amps)
R = resistance (measured in ohms)

So power varies with load.