If a typical class 2 transformer is rated at 40 va.what is its true power rating(purely resistive load)and how is it calculated? Thx
If the load is purely resistive then the power factor is 1. This is because the true power is equal to the apparent power. The reason the true power is not always the same as the apparent power is because the load may be either capacitive or inductive. This causes the phase of the current waveform to be either leading or lagging compared to the voltage waveform. Power is still the voltage times the current but since they are out of sync if you slice up the waveform into small segments and do the calculation for each slice and the average them all together the real power is less then the apparent power. The apparent power is the the product of the rms voltage and the rms current. The answer to your question is that the true power rating is the same as the apparent power rating. It is not calculated because by definition if the load is purely resistive the true power rating is the same as the apparent power rating. I hope this helps.
For resistive loads, 40VA can be taken as 40 watts.
Lots of people don't understand this rating, but it is not that difficult. The power rating of a transformer is limited by the losses in the transformer; these are core losses and wire losses. Core losses are a function of voltage and more or less fixed. Wire losses are I^2 R where I is the RMS current. Obviously, these are a function of the current squared. Now the bad news is that the VA rating of the transformer is not the same as the power rating except when the current is purely sinusoidal. This means that in case where there are rectifiers, the RMS current will be much greater than the average current, so the output power will be much less that the VA rating.
