MTFCA Forum

Look on surplussales.com. They have one for about $15.

What is VA? Another name for watts?

VA is Volt Amperes. Equivalent to Watts only when the load is purely resistive.
if its the kind of tester I think it is, the transformer won't be critical; anything 12 to 16V with at least 1.5A current rating would do.

What I learned the first year of my trade(Electrical) 1954.
R=e divided by i, P=e times i.
Where V or E is voltage,
R is the resistance, ohms.
i is the current, amps.
P is the power in Watts.
This will act as my first new forum "Dummy Question" trial reply. Neil.

I have been following along and John has it nailed and I'd dare guess you do to Steve.
I had to use an online calculator but 1.666666667 is the amps. Once that was done I remembered how to do it.
25 watts divided by 15 volts gives us the 1.666666667 amps.
I had been thinking a wall wart-ya know one of those warts you plug into the wall and it gives you dc'ish voltage back.
But it would take, let's say, a GOOD external hard drive power unit.
John , may I ask, what is purely restive?

The above explanations are correct but do not address the difference between the names: watts versus volt-amps.

I was taught that watts are a measure of pure DC voltage and amperage. While volt-amps are a measure of AC voltage and amperage.

Often, you will not have a pure DC voltage. It will be unregulated having spikes of higher and lower voltages. This tends to act somewhat like AC voltage without going into a negative voltage shift. In this case, it is the average voltage or the root means squared (RMS) voltage, that is important. Again, this is measured in volt-amps due to it not being regulated or pure DC voltage.

Both terms are still measurements of power but are different terms used to identify the type of voltage being used/measured.

I hope this makes it all clear as mud.

Good Luck,
Terry

Duey, a purely resistive load is one that has no inductance or capacitance. A good example is an incandescent light bulb or a heating element.
It's hard to avoid getting too technical here, but if one measures the current and voltage fed into a resistive load on AC, using an oscilloscope, the current and voltage waveforms will seen to be in phase. In this case, the power is 'real power' and is simply current multiplied by voltage.

Volt Amps are the unit of 'apparent power'. If the load is reactive; i.e. it is inductive or capacitive, the current will not be in phase with the voltage. The apparent power is then greater than the real power. The best way I can put this into simple terminology is that the current drawn by the load is higher than what it would be if the load was resistive. This is why power companies do not like large reactive loads on their mains - it causes extra losses in the transmission network. This then leads into the equally complex subject of power factor correction.
The important thing to know is that power in an AC circuit is not always simply current multiplied by volts, as it is with DC.

Steve, I suspected the tester had a Ford coil in it - which is why the transformer is not very critical. A Ford coil presents a largely inductive load.