Watt (W) vs Volt-Ampere (VA): What’s the Difference in AC Circuits?

Question

This topic could be raised more than once ... but I did not find anything specific (because there were no specific answers)
(....)
So 1W = 1VA

I hope I'm thinking well, if not, please correct me and stick the topic in order to avoid the next posts of this type.
(...)

gregor10 · Accepted Answer

K4BE - YOU WERE RIGHT, the luck will not be equal, and if you look at VA and W, then if you have a choice of 500VA and 500W transformers, take 500W.

k4be · Accepted Answer

Rather, they will be the same, but a manufacturer writing 500W will assume that the resistive load will be powered.

avatar · Accepted Answer

There is a mistake in this reasoning. No COS Fi because 1VA can sometimes equal 0.3W and sometimes 0.82W If you have an audio amplifier that is able to give 400 watts to the speakers and is in class AB, then you have to assume that the luck must be with an adequate margin I would look for 620VA luck

krzyjak · Answer

P = U * I = S only holds for an ideal resistor. It can be said that the active power (unit W) across the resistor is equal to the apparent power (unit VA). I don't know what exactly you mean. Make your statement more specific. greetings

Added after 3 [minutes]:

In general, active power is not the product of the RMS current and voltage, this only occurs for an ideal resistor.

mateusz885 · Answer

it is a transformer to drive 2x 200W 4r mosfets (such with 0.5 kW but usually in stores the power is given in VA)

@Edit
however, the amplifier is not a perfect resistor and will not consume electricity like a welder ... but most sources say that you should mount a transformer a little more power than the sum of the power of all channels at a given loudspeaker load

@Edit
in my case it is 4 ohms

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I don't know anyway, I don't know anything about it - that's why I'm asking

k4be · Answer

if the current is phase-shifted in relation to the voltage (cos phi) or is non-linearly dependent on it, the number VA will be greater than the number W.

mateusz885 · Answer

I don't have to look around because for 99% of the order; pi will probably be 600 or 650 VA 2x 40V (after straightening it will give 2x 57V) - 50V is recommended and 65 is the max; p
I think you should be ok and you? (I will find out how much the transformer will cost, because 600VA 2x 35V is 157 PLN gross - toroidy.pl)

gruniu · Answer

I did not know for a long time what Ampere is. Only when I was about to buy a power supply and calculated how much Watt the LEDs would consume, I found out how many Amps the power supply must have.
Now I ask someone to explain to me what VA is in a simple way.
If my plant's transformer says 30MVA, does it mean that devices or machines can draw 30MW of electricity in an instant?

krzysssztof · Answer

Not. Search for power triangle, active power, reactive power, apparent power. W is a unit of active power, and VA is apparent power.

Anonymous · Answer

To select a transformer that powers the amplifier, you do not have to delve into "what is the difference between W and VA" - read about active, reactive and apparent power.
For your needs, it is enough to assume that W = VA, and then how many VA should the transformer have if the amplifier has, for example, 400W?
Generally - no matter what the power supply is to supply, you should know that a lot depends on the type of transformer load. If you load the 12V transformer itself with a 12V / 50W bulb, i.e. that the current consumed by the bulb is I = 50W / 12V = 4.17A and winding. the secondary of the transformer must provide this current I-4.17A, i.e. a 12V transformer with a power of 50VA at least is enough. This is a resistive load.
It will be similar when you rectify this voltage, but then the transformer must give a slightly higher voltage of 12V + Udiod. For a Graetz bridge Udiod = approx. 2V, so the transformer must give approx. 14V, transformer power P = 14 x 4.17 = approx. 60W at least.
But if you want a light bulb (or other resistive receiver) or an amplifier to be powered with a DC filtered (without ripple) using capacitors, then: the 12VAC voltage from the transformer will drop by Udiod, the capacitors will "increase" them by about 40% (because the value of the peak sinusoidal voltage is 1.41 Usk), but with such a capacitive load (capacitors), the way the transformer is loaded changes and the current should not be taken from it greater than approx. 60% of the nominal current (i.e. for a resistive load).
For example, for a transformer with a capacity of 500VA 2x40VAC you will get Uwy = approx.2x53VDC [1.4 x (40-2)] - for full load, but since the max current of the transformer is 500/80 = 6.25A, the maximum current that can be taken from the power supply will be 6.25 / 1.6 = 3.9A and the max power taken from the power supply is (2x53V) x 3.9 = approx. 413W.
A lot depends on what the amplifier will be used for - if it is a guitar amplifier, then the power of the power supply should be at least the power of the Pwy amplifier increased by the power of Pstr losses in the amplifier. For class AB amplifiers, the efficiency is approx. 50-60%, so the total power consumed by such an amplifier is approx. (1.7-2) x Pwy, and only from this power one should calculate (select) a transformer with appropriate voltage and power P = U x I.
For an acoustic amplifier for home (music), a transformer with the same power as the power of the amplifier is completely sufficient - here for a 400W amplifier, a 400VA transformer is enough, although a 500VA, 600VA or even 1kVA transformer will not hurt it.
For such an amplifier, from the considerations "1W or 1VA, transformer 400VA or 500VA", it is much more important to ensure good voltage filtration and adequate cooling.

Watt (W) vs Volt-Ampere (VA): What’s the Difference in AC Circuits?

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