Welcome back to our class on linear circuits.
Today we're going to be talking about the ideal transformer model.
So we're going to use the ideal transformer model to analyze basic
transformer circuits and then see the relationships between voltages and
currents. So in the previous lesson we looked at
the linear model. It's a fairly complicated model that
makes use of phasor analysis and mutual inductance to do these calculations.
In the ideal model, we're going to make a few assumptions that are never completely
true, but they allow us to get a decent ideal of how a transformer's going to
behave without having to do more complicated analysis.
So first of all, we need to identify the asusmptions that are used for the ideal
transformer model, and then use the ideal transformer model to do some basic
circuit analysis. Then we will describe the importance of
transformers in power transmission. First of all we need to define the
coefficient of coupling. So the coefficient of coupling is going
to be represented by a lower case k. And the coefficient of coupling is the
value that gives this relationship between the mutual inductance, M, and the
self inductance is L1 and L2 in the transformer.
So, it's always possible to find this k, but some k's are valid in a physical
sense and some k's are not. K's basically have to be similar between
0 and 1, 0 meaning that there's no mutual inductance between the two coils, and
they're completely independent of each other.
And one, meaning that they're very tightly coupled.
There's a limit to, a physical limitation to how tightly this coupling can be.
So to do an example problem, consider that L1 is 4mH, L2 is 9mH and M is 2mH.
Or, we can simply use this equation to see that k is equal to M divided by the
square root of L1 times L2. So that equals 2mH divided by 4mH times
9mH, so it's 36mH squared. Take the square root of that to get 6mH
under 2mH. So that means that k is going to be equal
to one third. So it is quite simple to calculate simple
to calculate the coefficient of coupling. it's going to be very important when we
start talking about ideal transformers, though, because we are going to make a
certain assumption about what this coefficient of coupling happens to be.
In the ideal transformer case the coupling coefficient k is equal to 1,
which means that these two coils are very tightly coupled.
L2 and L1 are assumed to go to infinity, which means that so is our mutual
inductance. Now, this is a limit, it's not that it's
equal to the value infinity, that their going to approach very, very large
numbers. And the reason that we make this
assumption, is that, it allows the analysis that leads to the transformer
equations. We're going to skip over the actual
analysis. It's available.
You can find it if you're interested. But we just need to, to make use of it.
But we clearly know that having this infinite in-, inductance is not possible.
Finally, we assume that losses from coil [UNKNOWN] resistances are negligible.
Sometimes when we're doing analysis of transformers, we're going to stick a
resistor here and a resistor here, to correspond to the resistance of the two
wires that are placed here and here. it gives a better representation, but the
ideal transformer case we're going to assume that both of those go to zero.
Now, the implications of this ideal transformer model are the following.
First of all v1 over N1 is equal to v2 over N2 where v1 and v2 are the voltages
across the primary and secondary coils respectively.
And then N1 and N2 are the number of rotations of the coil, or how many reps
of the coil in the primary and the secondary coils respectively.
We also have the relationship that N1 times i1 is equal to N2 times i2.
And these equations come from the uh,Faraday's law of induction so what we
have is we're making a circle like this. Then we have a changing B field through
that surface. But as I wrap coils around again, and
again, and again, that corresponds to additional surfaces.
And so every time we have another wrap it's just another one of these surfaces.
And so we take the number of these wraps and multiple it by the voltage that's
induced, by the changing magnetic flux, in each of these sections.
Which is why we get this, N [INAUDIBLE] behavior.
But this equation is very simple to work with.
They are quite easy to, to use for analysis.
And so that's the most important thing that we're going to need to take out of
this. So let's do an example.
Here we're going to start by writing down our two equations.
That v1 over N1 is equal to v2 over N2, and that N1 times i1 is equal to N2 times
i2. We have a phasor voltage here, and I'm
going to do all of my analysis for this problem in terms of RMS voltages and
currents, because it's entirely possible to do that without any kind of confusion,
but I want to point that out. So, we have 120 kilo Volts rms here,
leading to 120 Volts rms here. Now we don't know the number of coils
because we don't have a number for either of these, but what we do know is the
relationship between them. So frequently, with ideal transformers
will show something like this, a ratio of the number of coils from one to the
other. Because the actual number themselves
don't really matter as the relationship between them.
And in this case, we see that there's 100 coils to one over here.
Because, if I take 120 and multiply it by 100, we get 12 kilo-volts, over here.
And if I want to find the current, it's going through over here, we had 120 volt
rms, 240 ohms. So that means that i2 is going to be
equal to half an amp with phase angle of zero.
On this side i1, we're going to have the same relationship using this i2 and i1,
that means that i1 is going to be equal to 5mA, yeah, 5mA rms.
And so what we can then do is we can find the power that's being consumed in each
of these devices. So first of all, we'll look at the
resistor, i2 is one half, amp RMS and the voltage it crosses is equal to 120.
And since this is a completely real device, p is going to be equal to the
apparent power which is one half times 120 the rms current in the rms voltage
together, so 60 watts. So this is basically 60 watts.
If I calculate the power that's being consumed here we find that p is equal to
minus 60 watts based on the reference structure and the current and the
voltage. So this transformer is generating 60
watts of power and that's because this side if we take this voltage and this
current and multiply them together. We find that that p is also equal to 60
watts. So we'll call this p of the secondary.
And we'll call this p of the primary. And this is p of the resistor.
And that's because this source is going to be generating 60 watts as well.
So we see that that's why we need to have this relationship between voltage and
current because the power is being transferred from one side to the other.
And that's one of the nice things about ideal transformer is it's converting all
of the power that's going in over here, pushing it to the other side to be used
to do some type of work. So essentially what we're doing is we're
allowing ourselves to change the voltage to operate at a different voltage.
So consequently transformers are often used in computers where you have 120
volts coming out of the wall but you need 5 volts DC to run your components.
You need a transformer to change the voltage.
It's also used in power applications. In this case, this is what we're
illustrating here. This is like the power that being sent by
the power company, this transformer is a big power transformer that's owned by the
power company that changes the voltage they're transmitting to a smaller
voltage. The reason for this is that if you have
high currents flowing through wires, it leads to, basically, heaters.
And you lose a lot of power as heat lost in the wires.
Power companies don't want to lose all of that heat in their wires, so they operate
their transmission lines at very high voltages so that the currents can remain
small. And they use transformers to go from
those very high voltages to smaller voltages that are used in residential
homes. So what are the implications?
Well transformers allow change from one voltage to another voltage.
With a high-voltage and low-current power and transform it into using long-distance
power distribution through transformers into something that is a low-voltage and
high-current so you can still get the power to all of the homes.
And it turns out that this is a very important phenomenon.
Before this event it was required for power stations to be placed very close to
where the power was being consumed. By being able to use transformers the
power can now be sent over very long distances, so things like the Niagara
Falls power plant can now send power all across the United States and Canada
without having to have a whole bunch of power stations all over the place.
So in summary, we showed the ideal transformer model, and used this model to
solve an example system, and identified that transformers are useful for power
transmission because they make possible for sending power using very high
voltages and low currents to be able to avoid getting high heat through high
currents, through transmission links. In the next lesson we'll be looking at a
sensor that makes use of this concept of mutual inductance, linear-variable
differential transformer. And see places that they can find useful
applications. Until then.