THE PHYSICS OF... POWER

[wwhitney]

So if you want to be able to combine the resistance R and the inductance X_L into a single impedance number Z, and get formulas that look like the DC formulas for power and resistance, you need to use complex numbers. E.g. Z = R + j X, where R and X are real numbers. You get nicer looking formulas at the cost of some more complicated mathematics. But it's entirely optional, you can treat everything using real sinewaves as we have been doing recently.

Cheers, Wayne

 

[wwhitney]

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

No, that's voltage formula is not correct.

I(t) R is the voltage drop across the resistor at any time t.

I(t) X_L is NOT the voltage drop across the inductor at time t. The voltage drop across the inductor is not in phase with the current, it is 90 degrees out of phase. There is no formula of the form V_inductor(t) = I(t) * constant, because I(t) and V_inductor(t) are not in phase.

Cheers, Wayne

Edit: in the above, I wrote resistor and inductor instead of resistance and inductance. I am modeling the "real inductor" as a series combination of a pure resistance and a pure inductance. So the V_inductor should be V_inductance, the voltage drop across just the pure inductance, not the total voltage drop across the "real inductor".

 

[FionaZuppa]

So if you want to be able to combine the resistance R and the inductance X_L into a single impedance number Z, and get formulas that look like the DC formulas for power and resistance, you need to use complex numbers. E.g. Z = R + j X, where R and X are real numbers. You get nicer looking formulas at the cost of some more complicated mathematics. But it's entirely optional, you can treat everything using real sinewaves as we have been doing recently.

Cheers, Wayne

notice i didnt use Z in the math. two wave functions representing V and I, with phase shift due to reactance

because I(t) and V_inductor(t) are not in phase

so, you use wrong syntax, I(t) and V(t) are measurements at time t, drop in a vertical line on this graph, read the values. if you want to say V an I are not in phase, sure, as seen in the graph and shown in the waveform equations.

 

[wwhitney]

In my last post (782), I am treating the real inductor as a pure resistance in series with a pure inductance. If you actually could separate the two and measured the voltage across the inductance, you would find it is 90 degrees out of phase with the current. When you add that 90 degree out of phase voltage to the in phase voltage across the resistance, you end up with the phase shift of 37 or so degrees.

Note that any phase shifted sine wave can be decomposed as the sum of an in-phase sine wave and a 90 degree shifted sine wave:

cos(t + φ) = cos(φ) cos(t) + sin(φ) cos(t + pi/2)

Cheers, Wayne

 

[GoldDigger]

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.

In which case you should not use the same I() symbol for both!

You could call it I₁(t) and I₂(t) if you do not want to explicitly show the phase angle. But you cannot use I(t) in two places in the same formula and casually say that it has two different values in the two places at a single time, t.

 

[wwhitney]

In which case you should not use the same I() symbol for both!

I think FionaZuppa was just referring to the phase shift between I(t) and V(t), not to using I(t) plus a phase shifted I(t).

Cheers, Wayne

 

[wwhitney]

so, you use wrong syntax, I(t) and V(t) are measurements at time t

Usually, the term "I(t)" means the function that gives you the current I for any time t, i.e. all times possible times t. Now if you say "at a point in time t, I(t)", you are referring to just that particular function value at the particular time t. It is a somewhat sloppy shortcut to use the same letter t in both cases, it would be better to say "at a point in time t₀, I(t₀)."

Cheers, Wayne

 

[GoldDigger]

I think FionaZuppa was just referring to the phase shift between I(t) and V(t), not to using I(t) plus a phase shifted I(t).

Cheers, Wayne

... I(t) is a sin function with phase shift already in it....

At that point, he calls the voltage "I(t)R + I(t)Z", which as you pointed out is only true if the two I() functions have different values at time t. He cannot have it both ways. :happysad:

 

[wwhitney]

if you want to say V an I are not in phase, sure, as seen in the graph and shown in the waveform equations.

In post 782, when I wrote V_inductor, I didn't mean the total voltage drop V, I meant the voltage drop across the inductance X_L. I added a postscript to post 782 to clarify that.

Cheers, Wayne

 

[FionaZuppa]

At that point, he calls the voltage "I(t)R + I(t)Z", which as you pointed out is only true if the two I() functions have different values at time t. He cannot have it both ways. :happysad:

i didnt use Z in the math

amps = 1.2sin(x-pi/4.65)

x being any radial time, you want seconds then drop in freq term

there is one and only one current at any time x

i later swapped x for theta (graphing apps be it on calculator or online like to use f(x), etc.)

 

[GoldDigger]

i didnt use Z in the math

amps = 1.2sin(x-pi/4.65)

x being any radial time, you want seconds then drop in freq term

there is one and only one current at any time x

i later swapped x for theta (graphing apps be it on calculator or online like to use f(x), etc.

My typo. You used X_L, but that does not change my argument in the least.
If you want to say that you are using a phase angle that is in between the pure R term and the pure X term to represent the timing of the total current, you can, but then your expression will still have the amplitude of that current wrong.

 

[wwhitney]

Hi FionaZuppa,

Here's a graph that may be helpful:

Blue: I(t) = cos(t)
Red: V_resistance(t) = 2 cos(t) (no phase shift relative to I)
Green: V_inductance(t) = 1.5 cos(t + pi/2) (90 degree shift relative to I)
Yellow: V_total = Red + Green (36.87 degree shift relative to I)

This is for R = 2 ohms and X_L = 1.5 ohms, per the functions V_resistance and V_inductance

Cheers, Wayne

 

[FionaZuppa]

do i need to have voltage to have power? isnt that the crux of it.
with the separation as you show it, V_XL=0 Amps=max at 0pi
at any time t X_L = some constant (this is a function of freq & henries)
so how do you escape Power = I²*ohms

at 0pi, your chart shows zero for V_XL(0pi)*I(0pi)
but I(0pi)² * X_L is not zero

just because voltage=0 that does not mean amps are not flowing, as shown by you graph.

 

[GoldDigger]

do i need to have voltage to have power? isnt that the crux of it.
with the separation as you show it, V_XL=0 Amps=max at 0pi
at any time t X_L = some constant (this is a function of freq & henries)
so how do you escape Power = I²*ohms

at 0pi, your chart shows zero for V_XL(0pi)*I(0pi)
but I(0pi)² * X_L is not zero

just because voltage=0 that does not mean amps are not flowing, as shown by you graph.

The average power is indeed just I²R, where I is now the RMS value of I(t). No problem there at all.
But that does not mean that at any moment of time the power is I(t)²R.
Even when you leave out that the amplitude, I, in I(t) = I*sin(something+something*t) is not the same as the RMS value of I we are used to working with.

 

[mbrooke]

do i need to have voltage to have power? isnt that the crux of it.
with the separation as you show it, V_XL=0 Amps=max at 0pi
at any time t X_L = some constant (this is a function of freq & henries)
so how do you escape Power = I²*ohms

at 0pi, your chart shows zero for V_XL(0pi)*I(0pi)
but I(0pi)² * X_L is not zero

just because voltage=0 that does not mean amps are not flowing, as shown by you graph.

You need voltage to hold a charge and you need voltage to move amps. Without voltage PF would not exist.

 

[wwhitney]

do i need to have voltage to have power? isnt that the crux of it.

Yep, you do. At a time when V = 0, then P = I*V = 0. Absent a superconductor, I don't think you can have current flow without a voltage drop over any finite period of time. But voltage can instantaneously be zero, at which point power is instantaneously zero as well.

so how do you escape Power = I²*ohms

That statement is true for a resistance only, both instantaneously, and on average (when you use the RMS average).

For an inductance, it is not true, either instantaneously or on average (on average it is 0).

but I(0pi)² * X_L is not zero

Indeed. Which just goes to shower than Power is NOT I² * X_L, as the power over the inductance is 0 at time t=0.

just because voltage=0 that does not mean amps are not flowing, as shown by you graph.

Sure, amps are flowing, but instantaneously, there is no voltage drop over the inductance, so there is no power input/output to the inductance.

Remember when I(t) is a maximum, its derivative dI/dt = 0. And for an inductance of L henrys, the voltage drop V = L dI/dt. So no dI/dt, no voltage drop.

So that's another way to draw the distinction between the power flow for a resistor and an inductor:

P_resistance = I * I * R
P_inductance = I * dI/dt * L

Cheers, Wayne

 

[FionaZuppa]

Yep, you do. At a time when V = 0, then P = I*V = 0. Absent a superconductor, I don't think you can have current flow without a voltage drop over any finite period of time. But voltage can instantaneously be zero, at which point power is instantaneously zero as well.

for just the reactance component your graph shows V=0 and amps !=0, so how it that possible if your graph already accounts for the relative voltage phase shifting?

 

[Besoeker]

Well, my good golly gosh, this can't still be going on............
Power is still power. My dog is still black and white.

 

[GoldDigger]

Well, my good golly gosh, this can't still be going on............
Power is still power. My dog is still black and white.

And this thread is read all over.

 

[wwhitney]

for just the reactance component your graph shows V=0 and amps !=0, so how it that possible if your graph already accounts for the relative voltage phase shifting?

I don't quite follow your question. But let me point out that in your graph, when V = 0, I is also non-zero. So having a point in time where V = 0, I is non-zero shouldn't be surprising. It happens any time V and I are out of phase.

Cheers, Wayne