Welcome back to our linear cirucuits class.
And today we're going ot be covering the liner model for transformers.
We'll use the Linear Model of Transformers to analyze a transformer
circuit. In the previous lesson we introduced
tranformers and how they work. And then we identified some analysis
models that exist for transformers. Mainly the linear and ideal models.
And the objectives for this lesson are to identify the linear model of transformers
so to be able to express that linear model, to use circuit analysis to analyze
the behavior of a transformer system, and then apply this analysis to solving a
transformer circuit problem. The linear model of transformers treats
each of the two inductors as inductance's, and because we're operating
in AC, you can put this as J omega L1 and J omega L2.
There's also this mutual inductance, j omega M, that relates the one to the
other, and so the way that we do the analysis here is we.
Can make a current here, I1 and a current here, I2, to express the two currents
that are going through each of the portions of the system, and then we can
use Kirchoff's Laws to come up with systems of equations to find the
relationship between these. So here using
[INAUDIBLE]
is going to be equal to I1 MC7 plus I1 times jambic L1.
But the thing is, there's still this mutual inductance term that we need to
account for. Now, remember that these dots are
representative of the reference directions.
Here, my current is going in the dot and out the not dot.
Here I'm going in the not dot and out the dot, so they're referenced in opposite
directions. So, this will be minus I, 2, times, J,
Omega, L, 2. Because it's mutual.
It's based on the current that's moving through, this, conductor.
This should not be L, 2, it should be M. Those based on this [UNKNOWN] M the
mutual invectance based on this current. The other equation for this side, again
using Kurkot's voltage lot is zero is equal to I2 times J omega L2 plus ZL.
But again we have to account for the mutual inductions term minus i1 j omega
n. Again because of the opposite reference
directions of the dots. If this dot were down at the bottom
instead of at the top these then would become pluses, so using this and some
linear algebra we can actually find and solve for the values of i1 and i2.
So these are the type set equations that we were using.
And then here's an expression of i1. It's a fairly lengthy and convoluted
equation. But it turns out that it might sometimes
be a little bit easier to work with an equivalent impedence.
So if I want to know the impedence. That this source sees in all of this.
That is going to be equal to the [UNKNOWN] voltage divided by i 1.
Since the impedence is defined to be that ratio between the voltage and the, the
current. And putting all of that together, we can
find that z is zq is going to be equivalent to Zth7 plus j omega L1, plus
this term the omega squared M squared over ZL plus j omega L2.
We notice that this zth7 and j omega L1 correspond to this and this piece.
And then this final portion Is sometimes referred to as the reflected impedance.
This is the impedance that results by the impedance on this side being intracted
through the mutual inductance twice, and so this side does actually impact this
side even though it's not directly connected, and so that gives us an idea
of how this type of circuit is analyzed. Now let's actually use these calculations
to solve an example problem. So, consider this example.
If we want to calculate the values we are going to be calculating a few different
things. First of all, we're going to calculate
I1. We're going to calculate V1, calculate
V2, and then we'll calculate I2. Now, to make it a little bit easier,
we're not going to do the whole derivation.
We're just going to make use of the derived values of I1 and I2, so we can
just plug in some values. And so, they're here listed.
And you can do this just by the same analysis that you've already used for
other phasors. So, in doing this we just need to plug in
values for I1. So, on the top we have ZL.
Just 50 minus j100 plus j omega l2 plus j100 divided by z7 plus j omega l1 is 4
plus j2 plus j1. Multiply by zl plus j omega l2, which is
the same thing that's in the numerator here, so 50 minus j 100 plus j 100 is
simply plus 50. And then there's this plus omega square,
n square. We know that this is j omega m is equal
to j 10. So, that means that omega m is 10, so
that should be plus 100. Then all of that gets multiplied by
[UNKNOWN] which is 150, 150 arc zero or just 150.
Putting all of that together, we get a, an I 1 value of 20 minus j10 in amps,
which is equal to 22.4, with an angle of minus 26.6 degrees.
To find I two, we plug into this equation.
So on the top we have j10, that's divided by and it turns out the denominator here
is the same as the denominator for I1. 4 plus j2 plus j1 times 50 plus 100.
All times 150 is going to be 7, giving a value of 4 minus j2 amps or 4.5, an angle
of minus 26.6 degrees. If I want to find my voltage 1.
We have not only this inductance to consider, if I multiply my current times
my impedence, but we also have the mutual inductance to consider.
So it's actually going to be easier for us to calcualte V1 as V7 minus what I'll
call V zed. Which corresponds to the voltage across
the z
[INAUDIBLE].
So it's going to be equal to 150 minus 4 plus j2 times 20 minus j10.
And typically what you're going to do is convert these each into polar, multiply
them together, convert them back into rectangular to do the addition, but I'll
save the algebra there. What you get is 50 with an angle of zero
degrees false. The 2 can simply be calculated as i2
times zl. If I did i2 times this j omega l, I could
end up getting the wrong value. Because there's also the mutual
inductance to consider. But because these two are in parallel, we
can simply say, b2 is equal to i2 times set l.
This is going to be equal to 50 minus j100 times 4 minus j2, which is equal to
500 [INAUDIBLE] angle of minus 90 degrees.
And so now we know the voltage here v1. the voltage here v2, and then the two
currents as well. So it turns out that you can then
calculate the power by multiplying the the rms values of the, the various things
that use all of the same techniques that we've already used.
to find out how the system works. But one thing that's worth noting is that
v2 has a high voltage with a low current on this side.
Here we have a higher current with a lower voltage.
And that's something that will be common through transformers.
We don't get any more power by sending something through a transformer But it
converts more of the power from being in voltage to being in energy.
To summarize we presented the linear model, and derived the phenomenon of
reflected impedance and then used circuit analysis to analyze an example
transformer circuit. In the next lesson, we will look at the
ideal transformer model and contrast it with the linear transformer model until
then.