This course explains how to analyze circuits that have alternating current (AC) voltage or current sources. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.
4.5 AC Power Transfer
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Del curso dictado por Georgia Institute of Technology
Linear Circuits 2: AC Analysis
53 calificaciones
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De la lección
Module 4: Power
The description goes here
Conoce a los instructores
Dr. Bonnie H. FerriProfessor
Electrical and Computer Engineering
Dr. Joyelle HarrisDirector of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Welcome back to our in linear circuits.
Today we'll be talking about maximum power transfer in cases of sinusoidal systems or
AC system.
We will contrast them with the behavior of the DC maximum power transfer that
we calculated earlier.
We will also show how to use extra reactive components to offset for
loads by doing power factor correction.
We'll talk about what that is and why you might want to do it.
In the previous lesson we showed how power factors or
power is calculated in AC circuits.
And we define all of the important values that are associated with
power triangles and how to use them.
In this module we will apply that knowledge to doing maximum power transfer.
The objectives for this lesson are to be able to find
the load impedance that gives the maximum power transfer and
then find the average power that is being consumed by that load.
You want to be able to find the optimal purely resistant impedance
that gives the largest possible power transfer when being constrained by using
purely resistors.
And then to be able to power factor correction
using additional reactive components.
Going back to our analysis techniques, we know that a linear system can be
represented by a Thévenin or a Norton equivalent.
And in AC systems we can do the same thing but now our sources become phasor sources,
and our resistances become impedances.
But we can still use the same techniques.
In this lesson we will be using Thevenin equivalent circuit for doing our analysis.
So here we have taken the Thevenin equivalent circuit and
we have attached a load impedance to it.
I want to find the maximum power transfer for the system.
First of all, we have a voltage expression which we can calculate by doing
a voltage divider.
So that's ZL divided by the sum of Z Thevenin plus ZL multiplied by V Thevenin.
You can also calculate the current phasor by taking the current and
voltage and dividing it by the some of those two impedances.
Going back to our calculations for complex power, we know that S is equal to one-half
times the phasor of the voltage times the complex conjugate of the current phasor.
So in this system to do our calculation, we see that S is gonnas be equal
to one-half and then we put in our voltage here.
ZL over ZTH plus ZL times V Thevenin
multiplied by one over ZTh plus ZL conjugate and
then we'll put up right the Thevenin conjugate.
Because I'm taking the conjugate of this,
it's the same thing as taking the top of the conjugate at the bottom.
Multiplying that out, we get a result of one-half times VTh squared.
Divided by the amplitude of ZTh plus ZL quantity
squared, all multiplied by ZL.
Now, this is no longer a phasor, this is the amplitude of the Thevenin of a phasor.
And then here this is the amplitude of the sum of the two impedances.
ZL is just ZL is found before, it turns out that to optimize this we are going to
want ZL to be equal to Z Thevenin conjugate.
And when we do that, this sum of Z Thevenin plus ZL
becomes equal to two times the real part of Z Thevenin.
And then this ZL becomes the real part of Z Thevenin
times the imaginary part of Z Thevenin.
So this gives us, well, I'll make it one over eight.
VTh squared divided by the real
part of Z Thevenin squared.
We're going to be multiplying that by the real part of
Z Thevenin plus j times the imaginary part of Z Thevenin.
When we do that, if we want to know the P of the system, we just find the real
part of this whole expression which would be this portion multiplied by this.
In that case the real part of the Z Thevenin squared,
one of those cancel that with one of these.
And then if you multiply this by the imaginary part we can find Q.
So P is going to be equal to that times that.
So I'll clear all of that off.
Again, that the optimal Thevenin or
the optimal load is equal to Thevenin, complex conjugate.
So P is then equal to V Thevenin over eight times the real part of Z Thevenin.
It's useful to contrast this with the result that we found with the DC system,
where P was equal to one-fourth times VI.
Or could do V squared divided by R.
And here we have an eight instead of a four but that's because the way that
sinusoids behaves with average power, so that should not be too much of a surprise.
But we can also calculate the reactive power this way and
we see that here the reactive power's not zero.
This because the reactive power in the load is used to offset some of
the reactive power in the Thevenin impedance.
So you might have expected the reactive power to be zero but for
maximum power transfer it's not.
Let's look at an example.
Supposed we have this system, and
we want to find the optimal load impedance to get maximum power transfer.
Well, we remember that ZL is equal to complex conjugate of Z Thevenin and
power can be calculated this way.
So we should let ZL be equal to
ZTh star or 4 minus j3 Omega.
And then the average power is then going to
be equal to 4 squared divided by 8 times 4,
which is equal to one half of the watt.
So it's pretty easy to apply to this technique.
Now suppose that we have a bit of a constraint.
Maybe we don't have any reactive components to stick in our load.
Maybe our load is periodic resistant.
So for this example this is maximum power transfer for
when we've constraint our load phase.
Which is to say that we are stuck with having a particular theta for
our load, in this case zero because it's a resistor, there's no reactive part to it.
The optimal power transfer for this system occurs when the amplitude
of our load impedance is equal to the amplitude of our Thevenin impedance.
So this is any time we have a fixed load angle, load phase angle theta.
So in this example, doing our calculation,
we find that R is equal to Z Thevenin, which is 5 ohms.
And using S is equal to one half
the I star, we are going again to use the same technique.
We do a voltage divider and same thing with the current, and
putting them all together we discover that this is four-ninths volt amps.
And since P is equal the real part of S.
That's going to be equal to four-ninths of a watt.
So it turns out that in this case, the power is purely real.
The complex power is purely real, and there's zero reactive power component.
And we also see that it has less average power than the previous example.
Another way of applying these ideas that we've been discussing is for
doing power factor corrections.
Power factor corrections where we apply an extra reactive component to a system to
try and control what the reactive power is in our system.
In this case we're going to want 100% power factor, or zero reactive components.
So how do we go about doing the calculation?
Well, one way that we are going to simplify this a little bit is,
identifying that if we want to calculate complex power, we can use the V equals IZ
in phasor forms to find a couple of equivalent expressions for complex power.
In this example we will be using this one since we'll be working with the voltage.
So if I want to find the complex power through our load
to the resister and reactor we're going to find that
S is equal to one half times V squared divided by Z star.
And putting that all together, we find that this is equal to V squared
divided by 2 times R squared plus omega squared L squared
multiplied by R plus j omega L.
Now everything here is real and so if I want to find the power
which is equal to the real part of S, I take this section and multiply it by R.
So maybe call this alpha.
If I want to find Q, which is what I'm more interested in in this
particular example, I take all of this and
multiply it by omega L and it gets us our Q.
Because this is a system with an inductive load.
To offset that we're going to want put in a capacitor but
we don't know what the capacitance and that's what we're looking to find.
Now the way that we're going to find this capacitors, we want to find a system where
this reactive power is being offset by reactive power on the capacitor.
So what we want to do then is calculate the complex power for this capacitor.
Which is going to be equal to again,
V squared divided by 2Z star so that gives us
minus J times V squared omega C divided by 2.
And this is purely imaginary, which means that
Q in this example is going to be equal to minus V squared omega C divided by 2.
This should be maybe Sc to
indicate that this is the complex power through our capacitor and Qc.
The desired quantity is to find value c
such that Q is equal to minus Qc.
And doing that calculation and solving for C, we discovered that C
should be equal to L divided by R squared plus omega squared L squared.
A couple of comments.
First of all, we did not need to use the line impedance anywhere in here,
because we're not trying to offset the line impedance, only the load.
We also never needed to use the phasor for our source or Thevenin source.
All we really cared about was the frequency.
So if I took the same system and connected it to a different amount of
voltage with a different load resistance or load impedance but it was operating at
the same frequency, this would still give this proper power factor correction.
There would be zero reactive power being used in the system.
A couple other comments.
Purely resistive devices do no use any real power.
They're just stored up temporarily.
So we haven't actually use anymore power by putting this capacitor into our system.
It might be possible to correct for line impedance.
Now if you're taking your power from power company
then you wouldn't necessarily want to do that.
But if you're doing it in an embedded system where you have control over
the sources and the load lines it could be valuable for you to
take that into consideration so you don't waste any extra power than necessary.
Adding the capacitor does change the voltage behavior.
So we'll see a different wave form for
our voltage, but it does not change the actual power that's being consumed
by the resistive part of the system, part that's actually doing the work.
And then using capacitors is actually a common practice for
heavy industrial applications where they don't want to pay for
the reactive power from the power company.
They'd prefer to have large capacitors that store up energy temporarily
to be used internally.
So instead of sending all of this extra power back and forth to and
from the power company, they keep it all internally and use it there themselves.
To summarize, we found impedance that gives maximum power transfer, and
found the impedance that gives the best power
transfer given that our load is purely resistive.
We also used a capacitor for power factor correction of an inductive load, and
found what capacitance was necessary to get a 100% power factor correction.
In the next lesson we will be looking at a new device, transformers.
They are quite useful devices for changing voltages in systems, and
they're very important in power transmission.
So we will look at those devices and
a couple of applications of how they're used.
Until then.
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