Ohms law

[gar]

170708-1611 EDT

Sahib:

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.

Power factor of a two terminal load is defined as:
PF = Average Real Power consumed in said load over a time T/Vrms*Irms
where Vrms and Irms are individually averaged over the same time T, and
the load is a stationary process

Power factor by definition involves time average values, instantaneous power involves no time factors.

Instantaneous power, v(t)*i(t), is what it is, and it can not tell you how much average power is consumed in some resistive portion of a load over some averaging time. Integration of instantaneous power over some time T is required to obtain total energy to the load over that time T, and dividing by T gives you average power for that time.

I have never seen a definition of power factor on an instantaneous basis. Of what useful value would it have?

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[mivey]

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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Correct. VA is used when we consider the whole period and use RMS values. We then have apparent power defined with VA units (same base units as watts but we use VA for identification purposes). We can look at the complex power and break it into its real and reactive components.

For instantaneous values, the unit is watts. Here we do not have apparent power so no VA units.

 

[mivey]

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.

There is no power factor in an instant.

Add: I see gar covered this on the next page.

 

[Sahib]

gar and mivey:

I consider a revolving voltage phasor and current phasor at the same frequency. Though both are revolving, there is fixed phase angle between them depending on the values of resistance, capacitance and inductance in a circuit. The cosine of that phase angle is the power factor for the instantaneous case just as for the RMS case. So what is wrong in determining the instantaneous power from the multiplication of instantaneous voltage,instantaneous current and power factor?

 

[Besoeker]

if they don't give pf assume it to be 1

But state the assumption you have made.

Most us use Ohm's for the Amps, Volts and Ohms proportionality. But that's not quite how Geogie boy stated it. Not a lot of folks know that to paraphase Michael Caine,

 

[Sahib]

gar and mivey:

what is wrong in determining the instantaneous power from the multiplication of instantaneous voltage,instantaneous current and power factor?

Oh! I realize my error. Sorry.

 

[mivey]

Oh! I realize my error. Sorry.

Good. You are mixing up period and instantaneous numbers and will get the wrong answer.

As one example with a non-unity power factor: there are times the voltage is zero but the current is non-zero due to lead or lag. The v(t)*i(t) product is zero but the real power dissipation in the resistor is i(t)*i(t)*R, a non-zero value.

 

[Sahib]

As one example with a non-unity power factor: there are times the voltage is zero but the current is non-zero due to lead or lag. The v(t)*i(t) product is zero but the real power dissipation in the resistor is i(t)*i(t)*R, a non-zero value.

Are you saying ohm law does not hold for instantaneous voltage and current, which is really wrong? I think you are carried away by your imagination too far.

 

[mivey]

Are you saying ohm law does not hold for instantaneous voltage and current, which is really wrong? I think you are carried away by your imagination too far.

I may have given a bad example for you but you still can't mix instantaneous with RMS.

 

[Carultch]

Are you saying ohm law does not hold for instantaneous voltage and current, which is really wrong? I think you are carried away by your imagination too far.

Ohm's law only holds true for linear resistive components.


Capacitors and inductors have a different relationship between voltage and current. Inductors develop a voltage that depends upon the rate of change in current. Capacitors develop a voltage that depends upon the time accumulation of current, which ends up being the charge stored on their plates in equal and opposite amounts.

 

[Sahib]

Carultch: There exist linear inductance and capacitance also obeying ohm law.

 

[Ingenieur]

Ohm's law only holds true for linear resistive components.


Capacitors and inductors have a different relationship between voltage and current. Inductors develop a voltage that depends upon the rate of change in current. Capacitors develop a voltage that depends upon the time accumulation of current, which ends up being the charge stored on their plates in equal and opposite amounts.

in a steady state and constant freq they can be converted to Xc=1/(j 2 Pi f C) Ohms and Xl = j 2 Pi f L Ohms
Ohms law and KVL/KCL will hold true in the complex domain

 

[mivey]

in a steady state and constant freq they can be converted to Xc=1/(j 2 Pi f C) Ohms and Xl = j 2 Pi f L Ohms
Ohms law and KVL/KCL will hold true in the complex domain

But Sahib is trying to mix things in with instantaneous values. Seems like he forgot AC circuits class so I'm not sure what examples would be good for him.

At one point he said he found his error then he was back off the wagon right after that and confused again.

Not sure what he needs to at this point as it has been explained multiple times.

 

[GoldDigger]

in a steady state and constant freq they can be converted to Xc=1/(j 2 Pi f C) Ohms and Xl = j 2 Pi f L Ohms
Ohms law and KVL/KCL will hold true in the complex domain

The direct equations for current v. voltage on a capacitor or inductor involve time derivatives. It is only because the derivative and integral of a sine wave are still (phase shifted) sine waves of the same frequency that this "linear" formulation can be used.

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[mivey]

Are you saying ohm law does not hold for instantaneous voltage and current, which is really wrong? I think you are carried away by your imagination too far.

To find the instantaneous value for a circuit with something like an inductor you can't make a simple ohm's law calc.

Suppose you have a series RL circuit. You know the voltage at time t. Find the current at time t using the phase shift. Knowing the current you can calculate the voltage across R at time t. Then V_L(t) = V_source(t) - V_R(t).

You could also find a very small delta t, then divide the change in current by the change in time and multiply by L to get an approximate V_L. Similar to what you would do on a scope to measure L.

Don't know why you don't remember all this from school.

Add: You may not realize that the power dissipated in the resistor at the instant source voltage is zero all comes from the inductor in a series RL circuit. The voltage across the resistor is exactly opposite of the voltage across the inductor.

 

[Carultch]

Carultch: There exist linear inductance and capacitance also obeying ohm law.

But they don't really obey Ohm's law, in its original form. Instead, they obey the definitions of inductance and capacitance.

It is only when you use a modified Ohm's law in the special case of AC systems, with the generalized concept of impedance replacing resistance, that you can say they "obey Ohm's law".

 

[Sahib]

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

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.

 

[GoldDigger]

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.

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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[Sahib]

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.

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

 

[GoldDigger]

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

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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