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I have a ballast that's rated to run two 32 watt bulbs, which means (I think) that the working power is 64 watts.

The line rating of the ballast is 0.5A at 120V, which means the apparent power (the total amount of power drawn, including reactive losses) is 60VA.

The power factor PF is the working power divided by the apparent power (PF = W/VA), which comes out to 1.06, which is impossible. What am I doing wrong here?

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    \$\begingroup\$ nitpick: reactive power isn't "losses". It causes losses, since it increases line current, but what it really is is just energy being stored and then returned to the source. \$\endgroup\$
    – Hearth
    Commented May 20, 2023 at 3:17
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    \$\begingroup\$ When you say "64 watts", how was that measured? And 0.5A and 120V, were those measured? If you're just assuming from the nameplate rating, I have bad news for you -- products aren't always what the label says. (Or, less nefariously: the ballast might simply not be delivering full rated power to them, for a variety of reasons.) \$\endgroup\$
    – Tim Williams
    Commented May 20, 2023 at 4:05
  • \$\begingroup\$ @Hearth right, my bad \$\endgroup\$
    – Robert M.
    Commented May 21, 2023 at 4:50
  • \$\begingroup\$ @TimWilliams In other words, the tubes might not actually run at 32W and the ballast might not actually draw half an amp during normal operation? What confuses me though is, even if the manufacturer is lying, if you're labelling a ballast as rated for two 32W tubes, why would you put a VA that's less than 64VA? \$\endgroup\$
    – Robert M.
    Commented May 21, 2023 at 4:54
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    \$\begingroup\$ Ampere ratings are usually maximum, for fusing. V*A should be an overestimate then. I'm not sure offhand what VA should represent, whether typical or maximum. \$\endgroup\$
    – Tim Williams
    Commented May 21, 2023 at 7:50

2 Answers 2

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What am I doing wrong here?

You are trying to do sums with ratings, as if they are measurements.

At best, ratings are usually some maximum, for setting fuse and cable sizes. At worst, especially when buying from dubious online marketplaces, they are some hallucination from the advertising department, and their only relation to reality is a factor between 5 and 10.

That said, even if you do want to compute results from measurements, you have to be very careful how you make those measurements. If you have an AC reading meter, you have to know whether it's reading true RMS, or averaged rectified scaled as sinewave. If it's a cheap meter, it's usually the second one. The difference on some waveforms can be very significant.

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  • \$\begingroup\$ That makes sense. In other words, it's questionable that the ballast draws half an amp during normal operation, or that the tubes actually run at 32 watts? \$\endgroup\$
    – Robert M.
    Commented May 21, 2023 at 4:43
  • \$\begingroup\$ The thing that confuses me is, even if the manufacturer is lying, why would they put a VA on the nameplate that is less than the wattage of the tubes it claims to be rated to drive? Why not lie and put 0.66A so that at least the PF appears to be 80%? \$\endgroup\$
    – Robert M.
    Commented May 21, 2023 at 4:56
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Fluorescent ballasts have a b*gger-factor called “ballast factor”, which can be something like 0.78 to 1.2. A lower ballast factor means less light and less power consumption. A typical ballast factor is said to be 0.88 so a 64W fixture might consume 56W.

If your ballast has power factor correction the PF may be very good. Or it’s possible it’s all lies.

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  • \$\begingroup\$ That's interesting, how can the ballast-factor be greater than 1? I'm assuming the two tubes draw 64 W together, since they're rated as 32 W tubes. Wouldn't that mean that the ballast would be consuming more than 64W i.e. 64W / 0.8 = 80W? Or does the ballast factor mean that it's intentionally running the tubes at a lower wattage than rated? \$\endgroup\$
    – Robert M.
    Commented May 21, 2023 at 4:48
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    \$\begingroup\$ @RobertM. A ballast factor greater than 1 means that it runs the tubes at a slightly higher power than their rated power. It makes the lights brighter, but shortens the lifespan a bit compared to nominal power. A "32 W" tube just means it draws 32 watts on a reference ballast; due to how the ballast works, the power used (and heat and light produced) isn't just a function of the line voltage like it is for incandescent lights. \$\endgroup\$
    – Hearth
    Commented May 21, 2023 at 5:34
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    \$\begingroup\$ What's "b*gger-factor"? \$\endgroup\$
    – Oskar Skog
    Commented May 21, 2023 at 6:05
  • \$\begingroup\$ @OskarSkog generally speaking a number which exists primarily to make the calculations work out. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 21, 2023 at 15:11
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    \$\begingroup\$ @RobertM. If greater than one then the tubes are running at a higher than rated wattage, so more light, more power (and presumably less life). \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 21, 2023 at 15:12

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