Ohms law

[Sahib]

But if you look at the time lot of the waveforms, they are, in at least one sense.
What does that tell you?

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Circuit theory has given us precise language to talk about that in terms of phase differences. Why not employ it?

 

[GoldDigger]

Circuit theory has given us precise language to talk about that in terms of phase differences. Why not employ it?

You are free to do so. But some people learn better with other analogies or descriptions.
I agree that the "opposite" description can be misinterpreted without additional information, but that does not mean that it has no value.
But we have gone pretty far from the original subject of instantaneous values versus various representative single value measures.

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[Sahib]

I agree that the "opposite" description can be misinterpreted without additional information, but that does not mean that it has no value.

It should not be misleading as does the last sentence of mivey in post#55.

But we have gone pretty far from the original subject of instantaneous values versus various representative single value measures.

Yes. As regards instantaneous power, it is sum of active power and reactive power at any instant. So simply designating its unit as watt is not correct, IMO.

 

[GoldDigger]

...
Yes. As regards instantaneous power, it is sum of active power and reactive power at any instant. So simply designating its unit as watt is not correct, IMO.

Reactive power is only meaningful when considered in terms of its total effect over the course of a period of the periodic waveform.

As far as the physics of energy transfer are considered, actual energy is moving from one place to another in the circuit in an amount which corresponds to the product of the instantaneous voltage and the instantaneous current. That movement of energy does not depend in any way on whether the power is being dissipated in a load or will be returned at a later time in the cycle. The proper unit for that rate of transfer of energy is the watt.

It is the fictitious measure of energy transfer obtained by multiplying the RMS voltage times the RMS current that has the dimensions of energy by does not have any real value in terms of energy transferred across an interface and so is, for convenience, assigned the units of VA instead of Watts.

JMO, of course.

 

[Sahib]

Reactive power is only meaningful when considered in terms of its total effect over the course of a period of the periodic waveform.

Hmm. Over a period of the waveform, the reactive power is zero and so its total effect is nil.

As far as the physics of energy transfer are considered, actual energy is moving from one place to another in the circuit in an amount which corresponds to the product of the instantaneous voltage and the instantaneous current. That movement of energy does not depend in any way on whether the power is being dissipated in a load or will be returned at a later time in the cycle. The proper unit for that rate of transfer of energy is the watt.

Even though the average value of reactive power is zero, its instantaneous value is not zero. That means its instantaneous energy content is not zero and so it should not disregarded when evaluating the total energy/power of an electrical system.

It is the fictitious measure of energy transfer obtained by multiplying the RMS voltage times the RMS current that has the dimensions of energy by does not have any real value in terms of energy transferred across an interface and so is, for convenience, assigned the units of VA instead of Watts.

The units VA, W and VAR are really assigned for a different reason. There exist different types of powers: active, reactive and apparent. To avoid confusion in dealing with them, they are assigned different units. Otherwise they have the same nature in that they can be converted to one another under favorable conditions in consistent with energy conservation law.

 

[Besoeker]

If you remove pf from the equations, you get "apparent power" in units of VA

True but still quite a useful measure for sizing things like conductots, transformers, switckgea, etc.
Of course you need Watts to specfy the prime nover.

 

[mivey]

Are you saying there is a 180 degree phase difference between voltage drops of resistance and inductance in a circuit, which is really wrong? It is really 90 degree.

You still don't understand instantaneous values.

Loop around the series RL circuit I was talking about at the instant I was talking about. The input is zero volts, the current in the loop is non-zero. There are only two elements left under consideration: the resistor and inductor. The resistor has current through it so it also has a voltage. Think about the inductor voltage and what it must be. Maybe you will figure it out but at this point I have my doubts.

Why don't you try some brain power and work out the math rather than just sputtering around making nonsense statements? It seems at this point the best way for you to understand is to actually crank the math so you will understand since you are not listening. Figure it out for yourself and perhaps it will clear it up for you.

 

[mivey]

It is "opposite" in the sense that the peaks of one are at the zero crossings of the other.
Not the most common usage, but supportable.

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Opposite as in positive and negative. The loop voltages must sum to zero. For instantaneous values you have plus and minus, that's it.

 

[mivey]

If you look at the phasor diagram, the two are not opposite in any sense.

Instantaneous values: they are positive or negative, nothing else.

 

[mivey]

Circuit theory has given us precise language to talk about that in terms of phase differences. Why not employ it?

Circuit theory? Why not try some?

Run the numbers instead of talking all around about them. Perhaps you will understand then.

 

[mivey]

It should not be misleading as does the last sentence of mivey in post#55.

If you understood the instantaneous values, you would see that they are simply positive or negative.

Follow any sinudoid. At any phase location the non-zero instantaneous value is either positive or negative. Nothing misleading. Nothing complicated. Only as complicated and misleading as you make it. This is as simple as you can get so if you feel mislead it is your own doing.

Think about it, don't complicate it. Positive, negative, positive, negative, ... back and forth over time above and below the zero axis. Tracing that out over time traces a sinusoidal wave but at each instant we have a single value.

Yes. As regards instantaneous power, it is sum of active power and reactive power at any instant. So simply designating its unit as watt is not correct, IMO.

At an instant there is only power and it has the unit of watt.

 

[Sahib]

Opposite as in positive and negative. The loop voltages must sum to zero. For instantaneous values you have plus and minus, that's it.

The voltage across the resistor is exactly opposite of the voltage across the inductor.

You are not taking into account the source voltage to complete the loop. Your imagination is at play again.

 

[mivey]

You are not taking into account the source voltage to complete the loop. Your imagination is at play again.

One must listen in order to understand what is being said:

...at the instant source voltage is zero...

You Loop around the series RL circuit I was talking about at the instant I was talking about. The input is zero volts, the current in the loop is non-zero.

Please pay attention. Calc the numbers. Learn.

 

[Sahib]

mivey:
You are describing an impossible circuit. I bet you can't find one in reality.

 

[Sahib]

In a RL circuit, the below condition never occurs under steady state condition, when the source voltage is zero:

The voltage across the resistor is exactly opposite of the voltage across the inductor.

because it violates the the phase difference of 90 degree between the voltage drops across resistance and inductance under steady state condition.

 

[Besoeker]

Stop arguing
This is how is for a restive load,
Twice the frequency and all positive..

 

[mivey]

In a RL circuit, the below condition never occurs under steady state condition, when the source voltage is zero:



because it violates the the phase difference of 90 degree between the voltage drops across resistance and inductance under steady state condition.

Again, follow the loop around the circuit at the instant given. We start at zero and end at zero. If there is a voltage rise in the resistor, there must be a voltage fall in the inductor. If there is a voltage fall in the resistor, there must be a voltage rise in the inductor.

Opposites. Sum to zero around the loop. Basic circuit analysis.

Set aside the inductor for a moment. Take a charged capacitor and put it across a resistor. Do you maintain there will be a 90 degree phase difference there? Think about what is wrong with your premise.

 

[mivey]

Stop arguing
This is how it is for a resistive load,
Twice the frequency and all positive..

No arguing with that.

 

[Sahib]

mivey: After some deep meditation on your words about voltage of resistor is exactly opposite to that of inductor when source voltage is zero in RL circuit, I found that it contradicts KVL when values of inductive reactance and resistance are not same.

 

[K8MHZ]

mivey: After some deep meditation on your words about voltage of resistor is exactly opposite to that of inductor when source voltage is zero in RL circuit, I found that it contradicts KVL when values of inductive reactance and resistance are not same.

Were you using 'Ohm' for a mantra?