I have a pool light whose bulb physically broke, but I've found a way to open the water-tight bulb enclosure to get at the actual bulb. I'm trying to either hunt down a replacement bulb (rather than the whole expensive assembly) OR make some LED magic.

Either way, the transformer's somewhat permanently mounted and I'd like to use it as it's already there, water-tight, and has the specific fitting to match the pool light flange.

I just can't figure out what 35va-12v means, and google didn't really help. I assume 35VA means 35VAC, but 12V is a common DC voltage and it's on there too!

Of course, I tested it and I actually get 14 VAC. How would I, in the future, know that based on the label? Is there a spec for this kind of stuff?


If it is a plain transformer, without any additional electronics, then the 12V mentioned means 12V(AC). This is the voltage the transformer outputs at its rated load.

Now what is this rated load I speak of? It is about how much power the transformer can potentially output. The 35VA indication is a measure of how much power the transformer can deliver. This is simply the product of voltage and current. The voltage is 12V, so maximum current can easily be calculated I(max) = 35VA / 12V = 2.9A When experimenting with this transformer, you might want to add a 3A fuse at the 12V(AC) side, just to prevent overloading.

You wrote your lightbulb is broken, so your transformer has no load. This is a bit like you driving your car down hill. Your engine doesn't have to work very hard and you probably start accelerating a bit. An unloaded transformer increases its output voltage. 14V output voltage is perfectly normal with this transformer when unloaded. When you attach a 25W lightbulb to it, you'll notice the voltage drops to around 12V. With a conventional light bulb, you can use bulbs up to 35W with this transformer. When using other devices the maximum number of watts drops, depending on the exact device.

120V and 60Hz is related to your mains supply. It is what you get from the wall socket.

  • 2
    \$\begingroup\$ A 3A fuse isn't much headroom unless it's a slow-blow. An incandescent bulb is a very low resistance before it's lit, and will draw way more than 3A getting up to temperature. \$\endgroup\$ – Bryan Boettcher Jun 29 '12 at 15:36
  • \$\begingroup\$ So I could build a rectifier inside the bulb and get 12vdc? \$\endgroup\$ – kavisiegel Dec 22 '13 at 18:46
  • \$\begingroup\$ More or less, and only when not using a buffer capacitor. The peak voltage is more like \$12V× \sqrt(2) - ( 2 × 0.7V ) = 15.6V\$, which you will see with a buffer capacitor. Rectifying a transformer's output can be a bit tricky when it comes to the exact numbers. Probably worth more than a comment here. \$\endgroup\$ – jippie Dec 22 '13 at 20:22

Here 35 VA means 35 volt-amperes.

If a system is purely resistive, that would be equivalent to 35 W, but in cases where it isn't, we need to use complex power which consists of real part measured in watts and imaginary part measured in vars. Since the ratios of those parts can be different, the transformers are specified in VA, which takes both parts into account.

The voltages would be 120 V AC nominal input at 60 Hz and 12 V AC nominal output. Do note that in cases where the load is light, the output voltage will be higher than 12 V.


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