It is like saying 'I just drove 4 miles. How many miles west did I go?' The question has meaning if you supply additional information.

In an AC circuit, both the voltage and the current continuously cycle from 0 to + to 0 to - and back again. The current will change at the same frequency as the voltage, but does not need to be perfectly in step with the voltage. However the _real power_ delivered to the load depends upon that portion of the current flowing that is exactly in step with the applied voltage. kVA is the raw product of voltage and current; kW is the real power delivered to the load. kW and kVA are equal if the voltage and current are perfectly in step.

kVA and kW are related by the 'power factor' of the load. kVA = kW/pf

When pf=1.0, then kVA = kW.

pf can never be greater than 1.0 .

Most (but not all) loads have pf > 0.7. So as a reasonable _guess_ you can say 'kVA will certainly be greater than or equal to kW, and probably be less than kW * 1.4 '

Some inductive loads can have pf that approaches 0, so without knowing the load pf, you can never be sure.

If power factor is unknown or cannot be determined, it is common to assume pf=0.8.

So, for your case kw = 4.0 kva = kw/pf = 4.0/0.8 = 5.0

Note, this assumes the circuit your dealing with is predominatly resistive & inductive. For a resistive & inductive circuit the power factor is termed a "lagging" power factor. On the other hand, if the circuit is predominately capacitive, power factor is termed a "leading" power factor.

Hope this sheds some electrons on the situation for you ;-)

If we were using DC, or AC with onlt resistance loads (IE: simple light bulbs), a watt and a volt-amp would be the same thing.

Unfortunately, we don't live in that kind of world. Large motors, lots of motors, variable spped controls, electronics, all sorts of things we use distort the wave form...and can fool our meters into thinking we're using more electricity than we actually are.

For the most part, this difference is of interest only to engineers. Most of us can get by equating one watt to one volt-amp.

Now there's a trick question if there ever was one!

For fluorescent lighting calculations, you use the figure on the ballasts, rather than adding up light bulbs. For traditional ballasts and when using T-12 lamps, yes, volt-amps can be treated as watts. The same applies for any fluorescents on single-phase systems.

However- on some fixtures you will see a sticker warning you about sharing the neutral wire between different circuits. This sticker is a major clue that not only are the ballasts electronic- but that "harmonics" may be an issue. And, since with three phase systems no single leg can completely "cancel out" another, you will want to up-size the neutral, or run separate neutrals.

All fixtures using the narrower T-8 and T-5 lamps use electronic ballasts.

The explanation here as to "why" volt-amps and watts are not the same is that the distorted waveform can fool an amp meter into reading lower than the actual current being used.