VoltAmpere

Thread Starter

donskiter

Joined Nov 17, 2006
29
Good Day...Guys can you give me idea about VoltAmpere ( VA ), I know that the Power = Volt x Ampere = Watts but I noticed that some equipment have different amount of power and VoltAmpere (VA ) for example i saw last week a medela suction machine have a rating of ( 1500Watts and 80VA)..I cant remember the exact value of VA but im sure the value was to far on 1500 watts rating..Are they different:confused:
 

Dave

Joined Nov 17, 2003
6,970
The difference is that Real Power is measured in Watts, whereas Apparent Power is measured in VA.

You may be aware that power in circuits other than purely-resistive circuits is comprised of a real component and an imaginary component. The real component is the Real Power (R) measured in Watts and the imaginary component is the Reactive Power (Q) measured in Volts-Ampere-Reactive (VAr).

The Complex Power (S) is comprised of both the real and imaginary components:

S = R + jQ

Where the Apparent Power is the modulus of S, i.e. the magnitude of the complex power vector |S|:

|S| = √(R^2) + (Q^2)

I hope that explains a little bit.

Dave
 

allied

Joined Mar 11, 2007
33
I plan to buy a UPS...I notice in the specification that the watts and voltampere is not the same. ( ex. 1000va , 650 watts). I call the RS component techsupport about this they not able to explain it to me they told me the will call me back to explain it but they not call me. Im just want to know the relation of watts and va on the UPS.
 

Dave

Joined Nov 17, 2003
6,970
The relationship is given in my previous post:

The difference is that Real Power is measured in Watts, whereas Apparent Power is measured in VA.

You may be aware that power in circuits other than purely-resistive circuits is comprised of a real component and an imaginary component. The real component is the Real Power (R) measured in Watts and the imaginary component is the Reactive Power (Q) measured in Volts-Ampere-Reactive (VAr).

The Complex Power (S) is comprised of both the real and imaginary components:

S = R + jQ

Where the Apparent Power is the modulus of S, i.e. the magnitude of the complex power vector |S|:

|S| = √(R^2) + (Q^2)
Dave
 

recca02

Joined Apr 2, 2007
1,214
consider a load of some power factor cos(phi)
then the active power is given by W = VA.cos(phi)
where VA is the apparent power(as already explained) obtained by multiplication of voltage and current at full load.
sometimes ratings are specified in terms of VA since it remains constant irrespective of load power factor as in case of transformers.
 
Thread starter Similar threads Forum Replies Date
A General Electronics Chat 7
Similar threads
Voltampere
Top