Why is a load always rated in kW and not in kVA or kVAr? Isn't it better to know how much kVA or kVAr an appliance or device has? Isn't it more useful?

  • \$\begingroup\$ It depends on the appliance and how your electricity is billed. What kind of appliance you are talking about, home appliances such as hair dryers or microwaves, or some industrial appliances such as motors? \$\endgroup\$
    – Justme
    Sep 21, 2022 at 5:43
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    \$\begingroup\$ The question you ask is which is better. And I don't think there's a bright-line answer to that. Seems to me, context matters. Assuming you are attached to a mains power system at the consumer level, where the metering and billing is based upon real power, then the watts are what you'd like to know. If you have a true RMS meter and measure volts and amps separately and multiply them, then you might like to know VA (so you can match up.) If you are sizing copper wiring as a contractor, you'd like to know the vars over and above the watts to be supported, I suppose. Define better. \$\endgroup\$
    – jonk
    Sep 21, 2022 at 5:45

2 Answers 2


kW and kVA have different purposes.

As a domestic consumer, I'm billed for my real consumption, kW, not my kVA. So I don't care about kVA, and neither does the industry selling me consumer equipment, so they badge stuff in kW.

If I was an industrial customer, I'd be billed for kVA, or some combination of kW and an additional penalty for excessive kVA.

Quite why the domestic consumer is forgiven bad power factor (excess of VA over W) I'm not sure. It would be an incentive to buy loads with a good power factor. Maybe the powers that be thought it would be too complicated for the unwashed masses to handle two numbers and the consequences of their ratio? So domestic customers are charged for the real power, just like delivered fuel. Good power factor is handled through legislation - anything above X watts has to have a power factor better than Y. For some long time now, PCs above 300 watts have needed power factor corrected supplies instead of the bad old rectifier capacitor supply.

When an electricity supply company has to put in a multi-MVA feeder to an industrial user, they will want to use as little copper as possible. High VA means more copper is needed than low VA for the same real power, so they charge for it. At least the industrial customer understands the figures.

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    \$\begingroup\$ I suspect it's down to the history of domestic electricity. When measured energy billing was first developed, most domestic loads would've been resistive - electric lights, heaters, and clothes irons. Even when electric motors became cheap and small enough to include in domestic appliances, they represented transient loads rather than continuous loads. By the time nonlinear electronics represented a significant portion of domestic energy use, we had already standardised on kWh billing and measurement, and it was easier to regulate PFC into products. \$\endgroup\$
    – Polynomial
    Sep 21, 2022 at 10:26
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    \$\begingroup\$ Comparatively, large inductive loads have always been a staple of electrified industry, so it would've made sense to bill for kVA there instead. \$\endgroup\$
    – Polynomial
    Sep 21, 2022 at 10:28
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    \$\begingroup\$ The story I was told in college (never confirmed) was that residential loads are typically many in number, and power factor issues can be solved with statistics, with power factors changing relatively slowly. An industrial load is capable of a very high power shift in power factor, which is hard to to correct for. \$\endgroup\$
    – Cort Ammon
    Sep 21, 2022 at 19:29
  • \$\begingroup\$ @CortAmmon power factor in domestic households is less about capacitive vs inductive loads and more about a current spike at the top of the voltage sine. (Diode + capacitor) So statistics doesn't solve anything. \$\endgroup\$
    – Christian
    Sep 21, 2022 at 19:38
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    \$\begingroup\$ @Christian: It would generally be rare for all households in a an apartment building or neighborhood to simultaneously present resistive loads, or simultaneously present capacitive loads. If one house served by a transformer present an inductive load while another presents a capacitive load, then during the parts of each cycle where one house would be returning power back to the utility mains, the other house will be demanding extra power. Power would thus flow from one house to the other, without having to flow all the way back to the power plant. \$\endgroup\$
    – supercat
    Sep 21, 2022 at 20:17

Why the load is always rated in kW ...?

It isn't. Many industrial loads and generators are rated in kVAr.

Isn't it better to know how much kVA or kVAr an appliance or device has? Isn't it more useful?

In general, kVA will allow us to calculated expected current and from that we can correctly size circuit breakers and cables. kW will determine what it costs to run the machine (provided the user doesn't suffer power-factor penalties).

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    \$\begingroup\$ If we’re going to be pedantic about spelling: “the correct symbol is lower-case ‘var’, although the spellings ’Var’ and ‘VAr’ are commonly seen, and ‘VAR’ is widely used throughout the power industry.” en.wikipedia.org/wiki/Volt-ampere \$\endgroup\$
    – Reid
    Sep 21, 2022 at 15:01
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    \$\begingroup\$ @Reid, I'd never read that before and didn't know it was considered a word. I've always treated as the unit of voltage, V, times the unit of current, A. Thanks. \$\endgroup\$
    – Transistor
    Sep 21, 2022 at 15:12
  • \$\begingroup\$ Re, "kW will determine what it costs to run the machine." That's only true for the customer if the customer's bill is determined entirely by how many kWh they use. As the other answer points out. The billing for industrial customers takes more than just kWh into account. \$\endgroup\$ Sep 21, 2022 at 16:03
  • \$\begingroup\$ @SolomonSlow. I covered that in the last sentence of my answer, I think. \$\endgroup\$
    – Transistor
    Sep 21, 2022 at 16:07

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