THE PHYSICS OF... POWER

[GoldDigger]

See the last paragraph of that post.

Cheers, Wayne

There is only one paragraph! :?

 

[wwhitney]

so two different amps at the same time? how is this shown in post 753? the amps waveform is shown as V(t)/Z for amplitude shifted by phi.

Sorry, I thought you were referring to post 754.

P(t) = I(t)V(t). For a resistor, V(t) across the resistor is proportional to the current I(t), at that time. For an inductor, V(t) across the inductor is proportional to I(t) at a later time (1/4 cycle later). So then for the power across an inductor, you get that P(t) is proportional to I(t)I(t+pi/2ω). Same current waveform, just evaluated at two different times. Post 753 shows what you get when you multiple two sinewaves that are 1/4 cycle apart.

Cheers, Wayne

 

[FionaZuppa]

Sorry, I thought you were referring to post 754.

Cheers, Wayne

are you ok with this plot of a real inductor. scales not shown, lets say 60Hz, but you ok with this graph, voltage leads amps, R X_L Z are constants. at any given time t the voltage will have phase shift of 37.016, a constant. you are also ok with saying that for any time t that t can be any vertical line on this graph, yes?

 

[wwhitney]

are you ok with this plot of a real inductor.

Well, it appears that X_L is greater than R, which means the phase shift should be greater than 45 degrees. So I'm not OK with that part. Also, if it is a real inductor, would R be that high?

at any given time t the voltage will have phase shift of 37.016

Since phase shift has to do with the relationship of how two different quantities vary in time, it doesn't make sense to say "at any given time t". At any given time t, you just have a voltage and a current, there's no notion of phase shift at a single point in time.

you are also ok with saying that for any time t that t can be any vertical line on this graph, yes?

I don't follow the question.

Cheers, Wayne

[/QUOTE]

 

[FionaZuppa]

Well, it appears that X_L is greater than R, which means the phase shift should be greater than 45 degrees. So I'm not OK with that part. Also, if it is a real inductor, would R be that high?


Since phase shift has to do with the relationship of how two different quantities vary in time, it doesn't make sense to say "at any given time t". At any given time t, you just have a voltage and a current, there's no notion of phase shift at a single point in time.


I don't follow the question.

Cheers, Wayne

a plot can have many scales all at once. but for ease of viewing to your point, created new graph X_L < R, phase less than 45

there is a notion of V(t) and I(t), its the instantaneous measurement of Voltage and Amps at the same time, and the results are shown in this graph. i used real #'s, they should all jive within the math. X-scale #'s not in radians

blue voltage = 3sin(x)
red amps = 1.2sin(x-pi/4.65)
yellow X_L = 1.5
green R = 2
lt blue Z = 2.5
phase angle = 38.69

so at pi/2 the volts is 3sin(pi/2) = 3
at the same time (pi/2) the amps are 1.2sin(pi/2 - pi/4.65) = 1.2sin(90 - 38.69) = 0.9366 amps
notice i said same time, we dont shift the time of observance

 

[FionaZuppa]

so the instantaneous power at any given time (radians) is
[(1.2sin(Θ - pi/4.65))²*R] + [(1.2sin(Θ - pi/4.65))²*X_L]

simplify by T=[1.2sin(Θ - pi/4.65)]²


P(Θ)= T*(R+X_L)


at pi/2 VA = 3*0.9366 = 2.81
this does not agree with I²(R+X_L) = 3.07

@ pi/2 VA != T(R+X_L)

 

[Besoeker]

How long does this need to go on????
Power is poewer. That's all there is to it. The Peggy Lee song summs it up.
But some here want to make a meal of it. The why is fairly obvious.

 

[ggunn]

Since phase shift has to do with the relationship of how two different quantities vary in time, it doesn't make sense to say "at any given time t". At any given time t, you just have a voltage and a current, there's no notion of phase shift at a single point in time.

At any time t there are values for dI/dt and dV/dt. If, at the time you pick, one is positive and the other is negative that is at least a notion of phase shift.

 

[wwhitney]

At any time t there are values for dI/dt and dV/dt. If, at the time you pick, one is positive and the other is negative that is at least a notion of phase shift.

That's a good point. In fact, if you had outside knowledge that I and V are sinewaves, then each one is determined by just three parameters. So most any three values, e.g. I(t), I'(t), and I''(t) at a particular time t, will determine the sinewave.

Of course, knowledge of I'(t) and I''(t) requires knowledge of I(t) on an infinitesimal interval around t, so it is more than just knowing the value of I at time t.

Cheers, Wayne

 

[wwhitney]

blue voltage = 3sin(x)
red amps = 1.2sin(x-pi/4.65)
yellow X_L = 1.5
green R = 2
lt blue Z = 2.5
phase angle = 38.69

This one's close, but arctan(1.5/2) = 36.87 degrees, so for the given R and X_L, the phase shift should be 36.87 degrees = pi/4.88 radians.

Cheers, Wayne

 

[wwhitney]

so the instantaneous power at any given time (radians) is
[(1.2sin(Θ - pi/4.65))²*R] + [(1.2sin(Θ - pi/4.65))²*X_L]

Nope. As previously discussed, the power flow at time t across the inductor is not I(t)²X_L. It is I(t) * I(t + 1/4 cycle) * X_L. If you fix that, plus the phase angle issue, your calculations should agree.

Of course, the above power flow across the inductor is a formula that uses current data at two different times. This is just to make the analogy to the I²R formula for a resistor, and as a convenient way to find the voltage drop across the inductor at time t.

You can, of course, use the voltage drop across the inductor directly to say that P_inductor(t) = I(t) * V_inductor(t). Likewise, P_resistor(t) = I(t) * V_resistor(t).

Adding these together gives you P(t) = I(t) * V(t), because the current is the same through both elements, while the total voltage drop (or power flow) is just the sum of the voltage drops (or power flows) across the two elements.

Cheers, Wayne

 

[FionaZuppa]

Nope. As previously discussed, the power flow at time t across the inductor is not I(t)²X_L. It is I(t) * I(t + 1/4 cycle) * X_L. If you fix that, plus the phase angle issue, your calculations should agree.


This one's close, but arctan(1.5/2) = 36.87 degrees
Cheers, Wayne

i not following, the graph already shows the phase diff, dropping in a vertical line to represent time "t" gives us V(t) and I(t), the phase shift is already accounted for.

notice at any time t V(t)/Z(t) != I(t)

ah yeah, i read my calculator small lcd wrong. 36.87 it is.

 

[wwhitney]

i not following, the graph already shows the phase diff, dropping in a vertical line to represent time "t" gives us V(t) and I(t), the phase shift is already accounted for.

Great, I agree with the above. Once you've done that, P(t) = I(t) * V(t).

But in your post, you wanted to calculate the power drop across the resistor and across the inductor separately, add them up, and check it matches the total power. To do that, you need to use the voltage drop across the inductor, V_inductor. Unlike a resistor, the voltage drop across the inductor is NOT IX_L. So the formula for power flow across an inductor is NOT I²X_L.

Cheers, Wayne

 

[FionaZuppa]

Great, I agree with the above. Once you've done that, P(t) = I(t) * V(t).

But in your post, you wanted to calculate the power drop across the resistor and across the inductor separately, add them up, and check it matches the total power. To do that, you need to use the voltage drop across the inductor, V_inductor. Unlike a resistor, the voltage drop across the inductor is NOT IX_L. So the formula for power flow across an inductor is NOT I²X_L.

Cheers, Wayne

i have one discrete component, the R is built into the coils of the wire that make up the inductor. i have a clamp-on amps "meter", i have two o-scope probes, one is attached across the inductor, the other is attached to the clamp on "meter". my scope shows me the graph i posted. the voltage waveform is in-phase with the src waveform, the amps have same waveform shape as V but amps waveform is phase shifted by some amount.

the graph shows us exactly what the voltage and amps are across the coil at any time (t), no need to muck around with having to shift V to make any calculation, the amps waveform function i provided has phase shift in it, etc.

I(t + 1/4 cycle) ?? yes, for just a pure inductor w/o R, thus we would see amps waveform shifted over 90 degrees from V. my graph shows exactly your equations for amps, but not as a pure inductor, a real inductor that has R.

 

[wwhitney]

i have one discrete component, the R is built into the coils of the wire that make up the inductor.

OK, my answers have been based on a circuit consisting of a pure resistance in series with a pure inductance.

For a single component, you measure the current through the component, I(t), and you measure the voltage across the component, V(t). Then P(t) = I(t) * V(t).

So, is there a further question?

Cheers, Wayne

 

[Ingenieur]

Best thread ever!

 

[FionaZuppa]

OK, my answers have been based on a circuit consisting of a pure resistance in series with a pure inductance.

For a single component, you measure the current through the component, I(t), and you measure the voltage across the component, V(t). Then P(t) = I(t) * V(t).

So, is there a further question?

Cheers, Wayne

Then P(t) = I(t) * V(t)

but due to the phase shift, didnt i show that V(t)I(t) != I(t)²(R+X_L)

R & X_L are constants, I(t)=I(t)

 

[wwhitney]

OK, my answers have been based on a circuit consisting of a pure resistance in series with a pure inductance.

And you can still treat your single component as such, so the answers are applicable.

Cheers, Wayne

 

[wwhitney]

but due to the phase shift, didnt i show that V(t)I(t) != I(t)²(R+X_L)

That's right, we agree. The power across an inductance is NOT I²X_L.

Cheers, Wayne

 

[FionaZuppa]

And you can still treat your single component as such, so the answers are applicable.

That's right, we agree. The power across an inductance is NOT I²X_L
Cheers, Wayne

sure, ok, and for any time t the amps flowing through both is the same I(t), and the total voltage across both is I(t)R+I(t)X_L, knowing that I(t) is a sin function with phase shift already in it, as seen in the graph.