# Residence Power Meter And Reactive Loads Question ?

B

#### Bob

Jan 1, 1970
0
Hi,

Think I have a very basic understanding of VA, Power Factor, etc. but

Even though most of the load in a typical residence is probably all
resistive, how does the power meter outside "decide" what's reactive and
what's resistive ?

It measures the total current going into the house, doesn't it ?

But you should, I think, be charged Only for the current
consumed/required by all the resistive loads. Is this true ?

Does it just measure total VA, and assume a Power Factor ?

If someone a lot better at this stuff could explain this for me, would
be most appreciative.

Thanks, and Happy Holidays,
Bob

A

#### Archimedes' Lever

Jan 1, 1970
0
Hi,

Think I have a very basic understanding of VA, Power Factor, etc. but

Even though most of the load in a typical residence is probably all
resistive, how does the power meter outside "decide" what's reactive and
what's resistive ?

What makes you think that it needs to?
It measures the total current going into the house, doesn't it ?

No, it measures power.
But you should, I think, be charged Only for the current
consumed/required by all the resistive loads. Is this true ?

No. You get charged for your power consumption. What makes you think
Does it just measure total VA, and assume a Power Factor ?
No.

If someone a lot better at this stuff could explain this for me, would
be most appreciative.

quite capable of determining just how much you use, regardless of what
type of load is presented to the load side of the meter.

D

#### Dean Hoffman

Jan 1, 1970
0
Bob said:
Hi,

Think I have a very basic understanding of VA, Power Factor, etc. but

Even though most of the load in a typical residence is probably all
resistive, how does the power meter outside "decide" what's reactive and
what's resistive ?

It measures the total current going into the house, doesn't it ?

But you should, I think, be charged Only for the current
consumed/required by all the resistive loads. Is this true ?

Does it just measure total VA, and assume a Power Factor ?

If someone a lot better at this stuff could explain this for me, would
be most appreciative.

Thanks, and Happy Holidays,
Bob

There was a long thread here "Balancing the Breaker Box"

J

#### James Sweet

Jan 1, 1970
0
Bob said:
Hi,

Think I have a very basic understanding of VA, Power Factor, etc. but

Even though most of the load in a typical residence is probably all
resistive, how does the power meter outside "decide" what's reactive and
what's resistive ?

It measures the total current going into the house, doesn't it ?

But you should, I think, be charged Only for the current
consumed/required by all the resistive loads. Is this true ?

Does it just measure total VA, and assume a Power Factor ?

If someone a lot better at this stuff could explain this for me, would
be most appreciative.

Thanks, and Happy Holidays,
Bob

The meter measures true power, so the power factor of the load has no
effect on the charges.

A

#### Archimedes' Lever

Jan 1, 1970
0
They did?

The basic mechanism generated a torque proportional to the voltage X current
average. (That's power!) I don't think any "funny" load could spoof it.

If anything they slight cheated the customer who had a very unbalanced load
because the basic meter was usually 2-wire.

There was a time when the number of capacitors hanging on the line
determined how well aluminum disc power meters would "read" inductive
consumption.

A

#### Archimedes' Lever

Jan 1, 1970
0
There was a time when the number of capacitors hanging on the line
determined how well aluminum disc power meters would "read" inductive
consumption.

Note that I am referring to the utility provider side, up on the local
HV sub for the cap banks.

A

#### Archimedes' Lever

Jan 1, 1970
0
They did?

The basic mechanism generated a torque proportional to the voltage X current
average. (That's power!) I don't think any "funny" load could spoof it.

No, the basic mechanism APPLIED a torque to an Alunimnum plate, and it
had error when the current and the voltage were not in phase.

The torque applied is less when the two are out of phase, and the
subsequent turns per watt/hour will be less as well.

The power company puts the cap banks on the line to offset their losses
on those meters.

They do not need to any more as the meters no longer present the
problem.
If anything they slight cheated the customer who had a very unbalanced load
because the basic meter was usually 2-wire.

It only needed to be 2 wire, and even if all use was on one side, it
STILL reads resistive accurately, and inductive less so if the phase
correction is not just right.

K

#### krw

Jan 1, 1970
0
There was a time when the number of capacitors hanging on the line
determined how well aluminum disc power meters would "read" inductive
consumption.

You surely are an idiot, DinBulb. AlwaysWrong too.

A

Jan 1, 1970
0
A

#### Archimedes' Lever

Jan 1, 1970
0
DimBulb, the torque is lower because the W are lower. The meter reads
WATTS.

No, the meter reads the number of turns the disc gets put to it. When
usual), when it is responding to an inductive load, it reads LESS
Watt/hours than it should. That is one reason why that type of meter has
been phased out of service, idiot.

If a load turns it less than usual, that means that the METER fails to
make a valid claim of what was actually consumed.

I know it is hard for you to understand, K-Tard, but less is less, and
the number of functioning brain cells between your ears is no exception.

K

#### krw

Jan 1, 1970
0

Like I said, folks, AlwaysWrong is *ALWAYS* wrong. He didn't get the
name because he knows anything.

K

#### krw

Jan 1, 1970
0
No, the meter reads the number of turns the disc gets put to it. When
usual), when it is responding to an inductive load, it reads LESS
Watt/hours than it should. That is one reason why that type of meter has
been phased out of service, idiot.

You really don't have a clue. The torque (thus power) is proportional
to the watts. The integral of that (the accumulator gears) is energy
(watt-hours).
If a load turns it less than usual, that means that the METER fails to
make a valid claim of what was actually consumed.

Wrong again, AlwaysWrong. The meter reads *exactly* the power
consumed. It doesn't care about inductance, capacitance, or anything
else, but WATTs.
I know it is hard for you to understand, K-Tard, but less is less, and
the number of functioning brain cells between your ears is no exception.

You've been proven wrong again, AlwaysWrong, though you will never
understand.

A

#### Archimedes' Lever

Jan 1, 1970
0
You really don't have a clue. The torque (thus power) is proportional
to the watts. The integral of that (the accumulator gears) is energy
(watt-hours).
You really do not get it. The torque is LESS when the two coils that
drive it are out of phase. That means that the same POWER will yield
LESS metered reporting of it. It does not get much more simple than
that, K-Tard.

See if you can wrap your single neuron around that, dumbfuck.

K

#### krw

Jan 1, 1970
0
You really do not get it. The torque is LESS when the two coils that
drive it are out of phase.

No, AlwaysWrong, it is *you* who doesn't get it (or anything). The
torque is LESS because the POWER is lower when the current in the
coils is out of phase.
That means that the same POWER will yield
LESS metered reporting of it.

Now, DimBulb, the POWER is NOT the same when the current in the coils
is out of phase as it is when they are. The meter reads the POWER, no
matter what the phase relationship.
It does not get much more simple than that, K-Tard.

Evidently it does, because *you* are too simple to understand simple
electricity, AlwaysWrong.
See if you can wrap your single neuron around that, dumbfuck.

Once again, AlwaysWrong proves how wrong he always is.

Keep it up DimBulb, you're making a real fool out of yourself. Still.

A

#### Archimedes' Lever

Jan 1, 1970
0
No, AlwaysWrong, it is *you* who doesn't get it (or anything). The
torque is LESS because the POWER is lower when the current in the
coils is out of phase.

You really do not understand the relationship between torque and total
number of turns over time.

Less torque will mean slower rotation. It really is THAT simple, you
moronic sub-human chump.

K

#### krw

Jan 1, 1970
0
You really do not understand the relationship between torque and total
number of turns over time.

You're a dumbass, DimBulb. Do you have any clue what in integral is?
No, I suppose not.
Less torque will mean slower rotation. It really is THAT simple, you
moronic sub-human chump.

Integrated over time is fewer turns, so? The TORQUE is a function of
POWER, moron.

Don't try anything as technical as brushing your teeth, AlwaysWrong.
You're not up to it. Oh, that's right. You don't.

Have fun in mommy's hamper, DimBulb.

G

#### Guest

Jan 1, 1970
0
Archimedes' Lever said:
There was a time when the number of capacitors hanging on the line
determined how well aluminum disc power meters would "read" inductive
consumption.
---------------------------------
Puzzlement??
a) The induction disk meter "reads" the real power only.
b)whatever is on the line "upstream" of the meter has no bearing on the
meter.
Capacitors "downstream" will reduce the inductive load (which is being
ignored by the meter except for a decrease in losses and possible harmonic
effects).

G

#### Guest

Jan 1, 1970
0
Archimedes' Lever said:
No, the basic mechanism APPLIED a torque to an Alunimnum plate, and it
had error when the current and the voltage were not in phase.
---
Errr, Not true. If it were, the meters would be useless as the load pf
is not constant. The meter inherently reads the <real power>, ignoring the
reactive component.
Are you thinking of the "lag coil"? This is not a power factor correction
device.

Nor is the torque "applied". A rotating magnetic field is applied and the
reaction of induced eddy currents in the disc with this field produces (or
generates) torque.

------------
The torque applied is less when the two are out of phase, and the
subsequent turns per watt/hour will be less as well.
-----------
When the two are out of phase, the torque at a given current and voltage
will be lower, as you indicate, because the torque is proportional to the
component of current in phase with the voltage- ie -the real power which
will be lower than the VI product. Voltage*current is, contrary to what you
appear to believe, not power (watts) for AC.
The turns per watt-hour will not change.
.. The speed of rotation will decrease because the real power and hence the
torque will decrease.
----------------------
The power company puts the cap banks on the line to offset their losses
on those meters.
----------------
No. They put capacitors on the line to provide vars locally to compensate
for inductive loads, reducing line losses "upstream" of the capacitors, or
to improve the voltage. Note that these capacitors will have absolutely no
effect on the meter and meter losses will not be corrected by such
capacitors.

A

#### Archimedes' Lever

Jan 1, 1970
0
Whether you like it or not, krw is right.

-

The retarded **** has not ever once said a goddamned thing about it,
you stupid ****.

Now, IF he comes back with some brains, instead of his stupid, peanut
gallery dumbfuck crap, MAYBE your remarks would have some merit.

Since all he does is parrot the same old, tired, retarded horseshit,
both he, and his retarded butt buddy (you) can shove it up each others
asses for all I care.

So, his CRAP is meaningless without stating what his refutation is.

And since you do not even know what fucking torque is, I doubt
seriously that you have any real depth of knowledge as it relates to ANY
mechanical device in existence. This meter is a mechanical device,
dipshit.

A

#### Archimedes' Lever

Jan 1, 1970
0
Nor is the torque "applied". A rotating magnetic field is applied and the
reaction of induced eddy currents in the disc with this field produces (or
generates) torque.

Dumbass. The field IS APPLYing the torque force.

Yes, it IS applied. Whenever a shaft is rotated by a force, torque has
been applied. And whenever a shaft is turned, torque IS the force that
was applied to the shaft, by whatever means, to turn it. Go back to
basic physics classes, dimwit.

So, if the shaft turns at one speed under an in phase resistive load,
and it does... and IF the shaft gets slightly less torque applied to it
by the eddy currents when "looking at" an inductive load, then ANY IDIOT
can declare that the number of turns will be less, despite the fact that
the same power was consumed, if the torque is less due to the two coils
being out of phase, and as it happens, that IS the case with inductive

So, same power consumed in both cases, and the meter will read the
inductive load as having been less. Period. The difference was not
enough for them to be concerned about, and it was made up for ON the FEED
side (utility) by putting cap banks up on the poles to shift the phase so
that an inductive load would make the metering a bit more accurate on

It is not just for reducing line losses, it is also for reducing the

You all need to go back to school... very basic school.

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