Ohms law

[GoldDigger]

Got ya


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The other way to get it, which I find easier to remember, works as follows for a 208Y/120 source:
If each line current is I, and is in phase with the line to neutral voltage, then the power is I x 120 x 3, since there are three line conductors.
But since 208 is equal to 120 x sqrt(3), the above is also equal to I x 208 x sqrt(3). (Since by definition sqrt(3) x sqrt(3) equals 3.)

The final step is to note that the power into a three phase load is not dependent on whether the source is delta or wye.


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[Sahib]

It was just like this, with a ohms law circle for help...


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One basic assumption of your test is the circuit is linear for application of ohm law.

 

[K8MHZ]

Correct me if I am wrong.

Vrms x Irms = apparent power. Without a power factor, we don't know the watts.

V x I instantaneous would = instantaneous watts, without the need of a power factor, but only at that instant.

 

[Sahib]

V x I instantaneous would = instantaneous watts, without the need of a power factor, but only at that instant.

You are correct for a purely resistive circuit.

 

[mivey]

You are correct for a purely resistive circuit.

For any circuit.

 

[Sahib]

For any circuit.

How? Watt is for active power only and for a circuit with mixed elements, the unit is VA and not W.

 

[gar]

170708-1357 EDT

Sahib:

mivey is correct.

Further the instantaneous product of voltage and current producing instantaneous power applies to any waveform and any load, linear or non-linear. This product may vary from + to 0 to - from one instant to another.

If you integrate this product over time T, and divide that by T, then you have the average power over T. What would be read on a wattmeter. And that is how a wattmeter works.

If you have a cyclic process, then you might like to make T equal to one cycle, or an integer number of cycles.

If you have a random process, such as white noise, then you want to make T long enough so that your error is less than some desired or tolerated value.

When your process is non-stationary, stationary is a statistical term, then pick a value of T that produces some adequate result. For power line monitor this might be 1/60 second, but that produces a huge amount of data. The original TED system (The Energy Detective) used one second, and even that is a lot of data. Power line data is not stationary.

.

.

 

[K8MHZ]

For any circuit.

I was asking about any circuit, thanks.

 

[K8MHZ]

170708-1357 EDT

Sahib:

mivey is correct.

Further the instantaneous product of voltage and current producing instantaneous power applies to any waveform and any load, linear or non-linear. This product may vary from + to 0 to - from one instant to another.

If you integrate this product over time T, and divide that by T, then you have the average power over T. What would be read on a wattmeter. And that is how a wattmeter works.

If you have a cyclic process, then you might like to make T equal to one cycle, or an integer number of cycles.

If you have a random process, such as white noise, then you want to make T long enough so that your error is less than some desired or tolerated value.

When your process is non-stationary, stationary is a statistical term, then pick a value of T that produces some adequate result. For power line monitor this might be 1/60 second, but that produces a huge amount of data. The original TED system (The Energy Detective) used one second, and even that is a lot of data. Power line data is not stationary.

.

.

I was thinking more along the lines of visualizing the voltage and current waveforms on an oscilloscope. A phase shift would change the result of V x I at any instant and time, which I presumed would be equal to real power.

That is to say, I would think that with a resistive load, Vinst x Iinst = Vrms x Irms, wheras that would not be true with a reactive component.

 

[Carultch]

I was thinking more along the lines of visualizing the voltage and current waveforms on an oscilloscope. A phase shift would change the result of V x I at any instant and time, which I presumed would be equal to real power.

That is to say, I would think that with a resistive load, Vinst x Iinst = Vrms x Irms, wheras that would not be true with a reactive component.

When one uses RMS values for voltage and current, they are using a specialized average calculation that acts as a representative value for the waveform as a whole. The special case of a resistive load is such that average power (an actual time average) = RMS voltage * RMS current.

It is true that the instantaneous product of V and I will equal the instantaneous power drawn from the circuit by the component. For energy storage components like capacitors and inductors, there is instantaneous power that is both positive and negative, indicating storage and release. The net time average power for an ideal inductor or capacitor is zero, and all power is reactive. If there is resistance involved as well, there will be a net release of power that is converted into heat. The time average of this net release per cycle, is the real power. The energy stored and released cyclically, is the reactive power.

 

[GeorgeB]

The difference between sqrt3 and 1.732 (sqrt3 to 3 significant digits)

[sarcasm mode]
My math teacher called that 4 significant digits.

sqrt3 ~ 2 (1 significant digit)
sqrt3 ~ 1.7 (2 significant digits)
sqrt3 ~ 1.73 (3 significant digits)
sqrt3 ~ 1.732 (4 significant digits)

[/sarcasm mode]

but this has nothing to do with your excellent point, just one of my peeves like dampen and dampening vs the correct damp and damping

 

[Besoeker]

[sarcasm mode]
My math teacher called that 4 significant digits.

sqrt3 ~ 2 (1 significant digit)
sqrt3 ~ 1.7 (2 significant digits)
sqrt3 ~ 1.73 (3 significant digits)
sqrt3 ~ 1.732 (4 significant digits)

[/sarcasm mode]

but this has nothing to do with your excellent point, just one of my peeves like dampen and dampening vs the correct damp and damping

And most calculators have a sqrt function anyway.

 

[Sahib]

Further the instantaneous product of voltage and current producing instantaneous power applies to any waveform and any load, linear or non-linear. This product may vary from + to 0 to - from one instant to another
.

.

That is not disputed. What is disputed is the unit for product of instant voltage and current. I say it is VA. KBMHZ says it is watt applicable for any circuit.

 

[Sahib]

, I would think that with a resistive load, Vinst x Iinst = Vrms x Irms, .

Sorry wrong.

 

[gar]

170709-1104 EDT

Sahib:

I am in agreement with K8MHZ. The unit of instantaneous power is watt. And this can be +, 0, or -.

The word power describes doing work, for example raising a weight, or generating heat, and the unit for this has been watt for a rather long time.

.

 

[Sahib]

Sorry wrong.

170709-1104 EDT

Sahib:

I am in agreement with K8MHZ. The unit of instantaneous power is watt. And this can be +, 0, or -.

The word power describes doing work, for example raising a weight, or generating heat, and the unit for this has been watt for a rather long time.

.

Well, in that sense my contention is the product of instantaneous voltage and current need not be always 'power'.

 

[mivey]

I was thinking more along the lines of visualizing the voltage and current waveforms on an oscilloscope. A phase shift would change the result of V x I at any instant and time, which I presumed would be equal to real power.

That is to say, I would think that with a resistive load, Vinst x Iinst = Vrms x Irms, wheras that would not be true with a reactive component.

Vinst x Inst with a resistive load creates a sinusoidal waveform that is above the zero axis (for a sinusoidal input) . It is a sub-cycle consideration.

Vrms x Irms is constant because it looks at an average across the complete cycle (for the resistive load we leave out the pf).

With a mixed load, the reactive components cause the instantaneous power trace to cross the zero axis. The portion below the axis indicates there is reactive power going back and forth over a complete cycle (that cycles equally above and below the zero axis).

Remove that reactive power portion and you are left with an instantaneous power wave that traces above the zero axis and the average over the cycle is what we call the real power we get from Vrms x Irms x power factor.

 

[GoldDigger]

Well, in that sense my contention is the product of instantaneous voltage and current need not be always 'power'.

In the sense of physics, the product of instantaneous current and voltage is a rate of change (flow) of energy. That, by definition, is power. It is not average power or reactive power. It is the current time value of power.
It is only when you integrate it over time that you get energy. Total energy over a time period divided by the time gives average power, which is inherently real power.
When you try to multiply some measure of the periodic current and the periodic voltage, you have thrown away information and may not have a valid measure of average power. In those cases we refer to that numeric product as VA.

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[Carultch]

The unit for this has been watt for a rather long time.

Interestingly enough, the man who is the namesake for the unit of Watt, never even used it. James Watt coined the horsepower unit, which measures the same concept. The SI unit of the Watt wasn't a concept until after his death. The same is true for Jimmy Joule, the namesake of the energy unit the Watt is based upon, as well.

 

[Sahib]

In the sense of physics, the product of instantaneous current and voltage is a rate of change (flow) of energy. That, by definition, is power. It is not average power or reactive power. It is the current time value of power.
It is only when you integrate it over time that you get energy. Total energy over a time period divided by the time gives average power, which is inherently real power.
When you try to multiply some measure of the periodic current and the periodic voltage, you have thrown away information and may not have a valid measure of average power. In those cases we refer to that numeric product as VA.

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Interestingly, in a circuit with inductance or capacitance or both and resistance in series, multiplication of power factor into the product of the circuit instantaneous voltage and current will yield the instantaneous power dissipated in the resistance.