I am trying to calculate how long I can run a given load (a sound system and laptop charger) from a 55 Ah 12 V battery using a pure sine wave inverter (specifically, this one). Though, at the end of the day, I'm mostly interested in figuring out "can this battery run the system for the duration of an event".

I used a Kill-a-Watt to measure the power used to run the system for a 4-hour event, and at the end it read 0.28 kWh.

I've calculated the capacity of my battery in watt-hours, including a factor for inverter inefficiencies (which admittedly, I just pulled out of the internet, if this looks wrong feel free to point me in the right direction!)

12 V * 55 Ah * 90% efficiency = 594 watt-hours. 594 is less than 280, so yes, the battery has the capacity to run my system for the duration I'm interested in (with plenty of wiggle room).

However, I noticed that the Kill-a-Watt reported a power factor fluctuating around 0.65-0.67.

I have heard (somewhere online, though I couldn't find it when I tried searching) that there's a difference between "apparent power" and "actual power". I can't find the reference for this claim, but I have heard that power factor is a concern in large industrial plants because "the power plant has to do the work to generate the whole sine wave, even if the load only uses part of it", and the power company measures the power factor for these industrial plants and charges them for the "apparent power", or "the entire sine wave".

Question #1: Is that understanding of power factor correct?

Question #2: Does this mean that I have to take power factor into consideration when calculating how long my battery can run my system? If my inverter needs to generate the "entire sine wave" but my load only uses 60% of it, where does that other 40% of the power go? Dissipated somewhere as heat?

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    \$\begingroup\$ I am not sure about the exact calculations, that is actually one of my concerns as well, which makes me very irritated when some solar equipment stores also sell LED bulbs which contain a capacitor voltage dropper circuit (VERY poor power factor) and advertise them as "power saving bulbs". What I do know is that a low power factor means you need to provide more power than used by the load, but can't tell you exactly how much more. I actually intend to do an experiment measuring EXACTLY such inverter loss due to poor power factor of a load. \$\endgroup\$ May 28, 2023 at 15:21
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    \$\begingroup\$ If the inverter design is good, you should see very little difference in current consumed from you battery with X W of AC power consumed with a PF of 1 and X W AC power with a PF of 0.65. \$\endgroup\$
    – winny
    May 28, 2023 at 18:20

1 Answer 1


Apparent power is what you get when you multiply Vrms and Irms. You get units of power, but you don't necessarily get real power flow.

The easiest way to see this is probably the fact that, suppose a load draws current as voltage crosses zero. The power is therefore near zero during that time, but Irms includes the current during this time, raising Irms and therefore Vrms*Irms as a result.

What this implies for the inverter, depends on its design. The power dissipation and power output (apparent or real) of an amplifier are in general unrelated.

Most likely, it's a class D amplifier, which will dissipate some idle power while operating, and additional power under load. Mostly depending on load current. But exactly how much, who knows.

So, as a basic assumption, consider the increase in load current. If the load is 280Wh and efficiency is 90%, then 311Wh would be drawn with no apparent-power penalty, or 31.1Wh dissipated by the inverter. For a power factor of 0.6, suppose the losses increase by the reciprocal, i.e. 51.8Wh. For a total 332Wh.

But this is at best an assumption, and either consult your inverter's manual -- if it gives load and efficiency curves at all -- or just try it yourself and see. A four hour test isn't that big a deal in the grand scheme of things, but it's understandable if this isn't practicable.

It certainly seems unlikely that it would be so bad as to draw full apparent power (or worse), so having some extra capacity should be fine.

Also beware of depth of charge, depending on what type of battery you're using, and expected life / cycle time. Or time between discharge and charge (when that matters e.g. lead acid).

  • \$\begingroup\$ I hadn't thought about checking with the manufacturer - the manual doesn't have any curves, but I can try emailing them and asking! \$\endgroup\$
    – maples
    Jun 1, 2023 at 15:58
  • \$\begingroup\$ Regarding your last statement about "time between discharge and charge" - I am using an AGM lead acid battery. I've heard it's generally considered "good" to only discharge it to about half its capacity before recharging, but this is the first I've heard about timing between charge and discharge. Can you point me to more information about that? \$\endgroup\$
    – maples
    Jun 1, 2023 at 16:11
  • \$\begingroup\$ e.g. louwrentius.com/… I'm not sure if anyone gives lifetime vs. depth of discharge vs. duration in discharged state, but it stands to reason that there is a continuum between these, which gives rise to the "shouldn't be left in discharged state for very long" advice. AFAIK, the mechanism is amorphous PbSO4 slowly crystallizing, and/or the crystals growing in size. Crystals are slowly reactive, or inactive, contributing to grid corrosion and reduced active area ("sulfation"). I imagine a solid state diffusion mechanism is in play. \$\endgroup\$ Jun 1, 2023 at 22:01

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