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Re: [Phys-L] AC power calculations



there's a gentle-voiced YouTuber named Jeremy Fielding who will discuss
practicalities of appliance motor capacitors all day long, kind of Mr Rogers of
the washing machine. Jeremy is good watching.

https://www.youtube.com/c/JeremyFieldingSr/playlists

Dan M


On Feb 22, 2021, at 23:01, John Denker via Phys-l <phys-l@mail.phys-l.org> wrote:

Some emendations and elaborations:

2) In particular, what about the motor-start capacitors

I have a friend who is not a rocket scientist, but does know
about motor-start capacitors. The big fan on her A/C condenser
stopped working, but she noticed that if she spun it by hand
it would keep going and accelerate to full speed. So she saved
herself a $200 service call by replacing the $10 capacitor.
Quick and easy. Requires a screwdriver and maybe pliers.

I don't know how people are supposed to learn about this sort
of thing. Feynman mentions shaded-pole motors but not motor-start
capacitors. And most people don't read Feynman as closely as I
do. I learned about capacitors when I was about 8 years old, by
watching my father replace one.


On 2/22/21 1:02 PM, I wrote:

Forsooth, in any context whatsoever, the physics depends on ΔV, never
directly on V. The fancy name for this is /gauge invariance/. It is
a fundamental property of the EM field. It is built into the Maxwell
equations.

Also, it's why voltmeters have two leads, not one.
This is a point you can make in the introductory class, as a way
of getting students to think about reeeealy fundamental issues.


And (!!) your typical dimmer circuit draws current during some
small part of the circuit ...

Should have said:

] small part of the cycle ...

This is a really big deal because many of their costs (including
dissipation in the distribution network) depend on ⟨I⟩² and/or ⟨ΔV⟩

Should have said:

] ⟨I⟩² and/or ⟨ΔV⟩²

Let [P] represent the algebraic average of P.

Should have said:

] Let [P] represent the arithmetic average of P, i.e. the plain old mean.

And similarly in other places: algebraic -> arithmetic

Since Ohm's law is linear (!) we can average both sides, so:
ΔV = I R
becomes:
⟨ΔV⟩ = ⟨I⟩⟨R⟩

Should have said:

] ⟨ΔV⟩ = ⟨I⟩ R

since R is constant, which is a key point here.


This is a really big deal because many of their costs

... where "they" are the power grid operators.

===========
Sorry for the screwups.
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