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.