Hi all,

i have a confusion about measuring the power in the presence of a transformer. Looking at the attached pic, a 3 phase R=10 Ohms load is supplied by a 3 phase 110Vrms line to line voltage through a 2:1 delta-wye 3 phase non-ideal transformer. As you can see there are 2 wattmeters one at the primary side and one on the secondary side.
my question is what would be the reading of these wattmeters? The wattmeter on the secondary side should be lower than the one at the primary because of transformer resistance? The wattmeter reading on the secondary side is equivalent to only the power consumed by the load and nothing else?
Please advice.Thanks

 

It's been many many years since I did 3-phase calculations, so I'd have to look it up to refresh the exact details. I guess just multiply the rms values of current that are in phase with voltage.

As for the transformer in the calculation, assuming it's ideal (lossless), you don't have to worry about it with regard to power. In this case just remember that power into the transformer equals power out. If voltage is stepped down by a factor of 3 across the xformer then the current drawn on the primary (input) side is 1/3 the amount drawn on the secondary. Power in and out will work out to be the same.

Probably another more complicated way you could do the calculation is to reflect the load impedance to the primary side of the xformer by dividing (or multiplying) the load by the square of the turns ratio.

 

Hi bearing01,
Thanks for your explanations. The power in and out of the transformer is the same in ideal transformer case.How about non-ideal with R,L,Lm? How's the power going to be on both sides? Another question would be in the case of a three phase delta to wye ideal transformer,how to calculate the power on the delta side? Assuming its supplying a resistive load,is P=3VphIph? where Vph=VLL and Iph =IL/sqrt(3) or P=3VphIL?. Thanks