OK... VA (not V/A...) is Volt-amps. It's called the 'apparent power' rating of the load. Remember that in a DC circuit you calculate the power by multiplying

voltage (in volts) by the current in amps. So the VA unit does have the right dimensions to be power. In an AC circuit, things are not so simple. I'm going to stick to sinusoidal waveforms here, BTW and let someone else handle the nasty cases. It turns out that if the voltage and current are not in phase

each other (and they won't be unless the load is perfectly resistive), then the useful power you can get out of the load - the so-called true power - is given by V*I*cos(phi) where phi is the phase angle between

voltage (V) and current (I) waveforms. V and I are the rms values of

voltage and current, as usual. cos(phi) is called the 'power factor' of the system. But, the size of the transformer you need is really determined by the

values of V and I, no matter what the phase shift is. So that, for example, if you have a load giving out a true power of 50W, with a

factor of 0.5, then you need to use a 100W transformer to supply it. What does this mean to you? I think it's fair to say that 0.5 is an excessively low power factor

any micro. 0.66 would be pretty low as well. So if you (say) see that your machine takes 100W, then a 150W or 200W transformer would be

A 100W transformer _might_ be OK, but it might be marginal.

I'd probably pick smalllest transformer with a VA rating in excess of

wattage of the computer, and if it doesn't get excessively hot in use, it'll be fine. -tony