One of those annoying electrical questions

OMG

I suggest you look up single phase supply.

The OP also missed out that we are working at 50Hz, though less surprised you didn't spot that.

Reply to
Fredxxx
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Eh?? I stated 50Hz in the original question - you wouldn't have got the right answer otherwise!

Reply to
Cursitor Doom

The aim of a pf 0.9 is probably an arbitrary trade-off between the cost of the correction capacitor and the benefit it brings. As motors go through their power curve / startup the reactive current will change so I tend to consider the pf a finger in the air, or at a known running load (measured or from the motor spec).

I am not aware of any instability issues with this potential parallel circuit. If a smaller capacitor is used then any resonant frequency will be correspondingly higher. The system is damped, so while a "AC" waveform might persist, it will decay away very quickly. If you could get oscillation then you would have discovered perpetual motion!

One thing that isn't generally considered is the current spike when switching a correction capacitor across the supply!

Reply to
Fredxxx

Generally a PF correction capacitor is there to bring the LC resonant frequency to 50Hz. That's how you get the zero phase shift.

without amplification, it can't be unstable.

Reply to
The Natural Philosopher

For a typical induction motor (if there is such a thing) the field current/reactive current remains pretty static. It's the *load* current component of total motor current that varies according to the amount of work the motor's doing. The PF improves as the motor becomes fully loaded.

True, but the 50Hz is fixed, so you're basically reinforcing my original point here.

AIUI, "perpetual motion" requires no external energy input which is not the case here. Oscillators require an in-phase energy pulse to keep them running.

Reply to
Cursitor Doom

MM. That's true, but only on account of the way PF is defined. What yiou have, with a motor, amounts to a fat inductor in parallel with a variable resistor, representing the variable mechanical load.

What te PF corrector does, is make a 50Hz tuned circuit out of it to offset the inductance and bring the phases back in line.

So the PF correction capacitor size is constant as this the inductance.. Its the Q that varies as the 'resistance' part varies.

In electrical terms, you need amplification. And te thing would ring at near enough 50hz, which is not an issue

Reply to
The Natural Philosopher

You're not going to get resonance when it's connected to the mains, which has close to zero impedance. Aim for 1.

NT

Reply to
tabbypurr

On this occasion I am going to have to disagree with you.

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might give the impression that reactive current will be heavily dependant on slip.

I accept slip could be close to zero, but there has to be some slip to get an output torque.

The oscillator drive is simply the mains. It will keep "running" as long as mains is applied. Any apparent oscillatory waveform after the supply is removed will decay very quickly.

Reply to
Fredxxx

Oooh - you did. It's normally a given!

Apologies.

Reply to
Fredxxx

I have an actual button on my keyboard between F23 and PrtScn that pops up the calculator thing. Use it all the time at home but at work I have those calculators that have a Tax+ and Tax- button for VAT which are more useful for the daily requirements.

Reply to
0345.86.86.888

I thought the only people, other than design engineers, who buy one of those over-complicated calculators are schoolkids who hope it will magically do their hard maths for them. :-)

"There's probably more computing power in one of those calculators than in the Apollo moon landing module" etc etc etc.

Reply to
pamela

FFS google it and you will get the user guide. support.casio.com/en/manual/manualfile.php?cid=004009082

Reply to
bert

***Watch out, guys! Troll alert!!!
Reply to
Julian Barnes

If you have some actual content with some relevance, feel free to provide it. Otherwise you are merely trolling.

NT

Reply to
tabbypurr

Urgh! NYE!! I have to confess I was "very, very drunk" :-D when I came home this morning and read your rejoinder, so you'll have to excuse me as I'm still not even 50% normal yet, but until I recover completely, please take a butcher's at this:

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Reply to
Cursitor Doom

Well it says in rather more detail exactly what I was trying to explain.

Reply to
The Natural Philosopher

Perhaps you too were "very, very drunk" at the time? BTW, here's the cultural reference for those who may have missed it:

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Reply to
Cursitor Doom

You seem to have an association in your mind between 440V and 3-phase since all the 440Vs you've ever seen have probably been 3-phase. But that doesn't mean that you can't have 440V single phase. You can also have

230V 3-phase, too, which I believe is the standard in the US.
Reply to
Cursitor Doom

I doubt anyone with any sense would waste their time with you. You have already demonstrated that:

A) electronics is not your field (to put it politely)

B) when things are patiently explained to you at great length by several well-meaning people you persist in clinging to your erroneous notions of how things work.

If you're not a troll you certainly have a sub-standard IQ.

Reply to
Cursitor Doom

The article is a very simple model of a motor of unknown type and the article seemingly sole purpose is to demonstrate a (real) load current and out of phase magnetising current. For the average electrician it is probably all they need to know.

Most motors in the kw region tend to be squirrel case induction motors, which exhibit a reactive component that is dependant on load.

The article uses the term "magnetising current" which is more commonly a term associated with transformers and not induction motors. It's a shame the article doesn't inform us for what type of motor this would be constant. I can't tell from the photo.

Reply to
Fredxxx

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