🚀 go-pugleaf

In article <o4RwXB1w165w@brewhau.bison.mb.ca>, magus@brewhau.bison.mb.ca (James Copland) writes:
> cs000rjp@selway.umt.edu (Russell J. Pagenkopf) writes:
> 
>> Hi! I have a sealed power supply that I need to test for voltage and
>> amperage under load and this is my problem, how do I create a 2 amp load?
>> The power supply uses 120VAC 0.40A grounded and outputs (supposedly) 20V,
>> 2.0A, 40VA Class 2  with four pins ( I have no idea what pins do what yet). I
>> understand that if I place my meter in series between the load and the
>> power supply, I should be able to read 20V (dc right?) and 2.0 amps. But
>> what does the 40va mean? 40VAC? I guess what this comes down to is, how
>> should I test this thing to make sure it is giving me what it should?
>>
>> Thanks for any help and sorry for wandering ;-).
>>
>> --
>>   Russ Pagenkopf                                cs000rjp@selway.umt.edu
>>  School of Journalism, University of Montana         cs__rjp@lewis.umt.edu
>> <There is no other wisdom, and no other hope for us but that we grow wise.>
> Salut!
> 
> Have you tried lamps in series and parallel for a load?  40VA means
> 40VoltAmps, just what you would expect from 20V * 2A.  If you place
> your meter in series with the load and source, you should measure
> Amps.  If your meter is parallel the load, you should measure Volts.
> All readings on the output should be DC (only a scope knows for sure
> 8-).
> 
> Regards,
> Magus.
> -- If I already posted this, file this one under _old timers_ disease. --
> --              magus@brewhau.bison.mb.ca (James Copland)              --

If you want a fixed load, buy a 10 ohm 50 Watt resistor; it will draw 2.0 Amps
at 20V, and dissipate 40Watts (or 40VA). If you want a variable load, the
easiest way to make one is to use a power transistor as a load, with a pot
controlling the bias current. I gather from your question, however, that you
haven't much electronics experience, so the resistor is probably the easiest
solution.