Sorry, this might be a stupid question as I'm not an electrical person. I'm analyzing an electricity consumption dataset and the consumption is measured by apparent energy (VAh).

How do I convert this VAh to kWh as normally seen on a power bill? Or are they the same thing, just different ways to call?


  • 2
    \$\begingroup\$ See @Christian answer below. In most parts of the world, you pay only for actual energy but big industries pay penalties for apparent energy too. Assume a PF of 1 and it will get you very close to the real value. Then 1000 VAh = 1 kWh or 1 kVAh = 1 kWh. \$\endgroup\$
    – winny
    Commented Mar 29, 2017 at 8:54
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    \$\begingroup\$ Related, but probably not a duplicate: What is the practical difference between watts and VA (volt-amps)? \$\endgroup\$
    – user
    Commented Mar 29, 2017 at 15:23

3 Answers 3


You need to know the power factor, which is highly depedant on your application. Then you simply use this formula: $$ P = S * PF $$ where

P = actual energy
S = apparent energy
PF= Power Factor

This answer tells us, that power factor in general public households can be expected to be greater than 0.9.

  • \$\begingroup\$ Thanks a lot. Also, I have 2 rows of data in chronical order (60 sec interval) like this: [PF=0.98; VAh=30925] and [PF=0.97; VAh=30926]. If convert to the actual energy kWh, the kWh values would be decreasing (30306 -> 29998), which doesn't make sense for me. So should I analyze the electricity consumption using apparent energy instead of actual energy? \$\endgroup\$
    – foo
    Commented Mar 29, 2017 at 12:14
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    \$\begingroup\$ 30306 kWh in 60 seconds is 1.8GW, are you working at a nuclear power plant? Why would you think that the power is constant? \$\endgroup\$
    – Christian
    Commented Mar 29, 2017 at 12:19
  • \$\begingroup\$ The data is cumulative so it looks that big. I meant the power went from 30306 to 29998 kWh in 60 seconds. This is data of a household appliance. \$\endgroup\$
    – foo
    Commented Mar 29, 2017 at 12:23
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    \$\begingroup\$ My main concern is why the kWh can ever decrease. \$\endgroup\$
    – foo
    Commented Mar 29, 2017 at 12:50
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    \$\begingroup\$ Either the PF is a average since the beginning of the dataset, which would make sense, because your power consumption is a sum. Or it's the PF in the last 60 seconds, in that case you should apply the new PF value only to the delta since the last measurment, not to the whole set. \$\endgroup\$
    – Christian
    Commented Mar 29, 2017 at 12:56

You can’t convert them.

All you know is that the actual power(’s absolute value) is ≤ the apparent power. The actual power might be 0 (purely inductive or capacitive load) or negative (if you are analyzing a generator). It might also be = the apparent power (purely resistive load).

As for the units, VA and W are equivalent.

As for the energy, give than the power is the derivative of the energy, you have the same result that actual energy(’s absolute value) is ≤ the apparent energy, and that VAh and Wh are equivalent units.

  • 3
    \$\begingroup\$ VA and W are only equivalent when the phase angle between voltage and current is zero. \$\endgroup\$
    – Steve
    Commented Mar 29, 2017 at 15:00
  • 1
    \$\begingroup\$ @Steve Did you read the whole answer before commenting? I wrote that VA and W are equivalent units. I know that apparent power is different from actual power. I also know that VA is generally used for apparent power while W is generally used for actual power. \$\endgroup\$ Commented Mar 29, 2017 at 18:52
  • \$\begingroup\$ Yes I did, I just felt it was worth pointing out. \$\endgroup\$
    – Steve
    Commented Mar 29, 2017 at 19:26

Power: W=V*I

Effective power: VA=VIcos(Φ) where Φ is the phase angle between the current and the power.

For a pure capacitive load the current will lead the voltage by 90° or Π/2 Radians


Therefore for a purely capacitive load, although the effective power would be 0, the actual power drawn is V*I

The same goes for a purely inductive load, where the current lags the voltage by 90° and cos(-90) = cos(270) = 0

This would mean that under those conditions, although the system draws V*I watts out of the system, the work done is 0, which is what the meter on the incoming mains, will also read.

This is not a good state for either the consumer or the supplier, therefore power factor correction is applied. As most loads are inductive, the method used to correct the power factor, is by using a capacitor of the appropriate size, for the load.

Domestically, washing machines, drills, etc. have in built capacitors, and in anycase, their effect is minimal.

Industrially, where they have 3 phase power feeding large transformers, large motor driven machines, ones which require 10kVA or more, will also require the incoming mains to have power factor correction capacitor connected. These however tend to be dynamic, where the power factor is constantly monitored and capacitors are automatically switched in to keep the PF as close to 1 as possible.





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