Welcome back to our course on linear circuits.
And today, we're going to be talking about, a couple of more advanced
techniques. The y delta transform and the Wheatstone
Bridge. And so, it's kind of two separate things
that we're going to be looking at. But they kind of have some relation to
each other. And we'll see where that is.
From the previous lesson we talked about how we systematically obtain circuit
equations and then the maximum power transfer.
Finding the load resistance that gives the best power transfer.
So this is only mildly related to some of this previous material.
After we've completed this lesson there will be an application showing something
about using Wheatstone bridges and using sensors.
So there will be some more demos that we'll complete this module.The objectives
of this lesson are to transform resistor circuits between wye configurations and
delta configurations. To specify a test resistor which balances
a Wheatstone bridge. And identify whether the resistor under
test in a Wheatstone bridge is below or above the target resistance.
So, first of all, the Wye-Delta Transform.
If I have a circuit, a system where I have three resistors that kind of come
into point, I can take this whole set and replace it with three resistors in kind
of this triangle configuration. This is called a Wye-delta transformation
or sometimes a Y-D transformation since the y, this kind of looks like a y, this
kind of looks like Greek letter delta, which is just a triangle.
In this system, this a corresponds to this a, this b to this b, this c to this
c. And then we add an extra node d in the y
configuration. You can use these equations to take you
from a Wye into a delta and so the equations are all listed here.
You can actually also go the other direction.
If you have a delta and you want a wye you use this set of equations.
Now I'm going to skip over the derivation.
It takes a little while but if you're interested we can make those available as
resources to you. But we will make use of the actual, of
these actual equations and use it to solve and do a transformation.
So suppose we have this circuit and we want to know the equivalent resistance
from the top node to the bottom node. So we have several different things here
and we'll recognize that nothing here is actually in parallel or in series.
So we cant use the parallel and series combinations we were using before.
However one method to use would be to stick a known source, so in this case I
use a one amp current source and then measure the voltage that we see.
And then you can use other analysis techniques that have already been
presented. But since we now have the delta transform
we can take this top triangle, which is a delta and we can turn it into a y.
So we can call this node a, this node b, this node c.
So transforming it, going to get a Oops. The resistor there.
Now this is b and this is c. Now this 210 ohm resistor is part of the
delta. So you see we no, don't have it there
anymore. But these two resistors, the 10 and the
30. So that's 10, that's 30.
Now we need to figure out these three resistances.
So for the top one, this resistor using the equations that were presented before
that r is going to be equal to 168 times 63 divided by the sum of 168, 63 and 210.
So calculating the out the solution is equal to 24.
To find the resistor here you take 168 times 63 divided by the sum.
That gives us 24. To find b we're going to basically do the
same thing. Now it's 168 times 210 over the sum of
all of these. So doing that calculation, we see that b
is going to be 80 ohms. And with c, 63 times 210 over the sum
giving us 30. One we've gotten it in this configuration
we see that it becomes very simple. Now, this 80 ohm resistor and this 10 ohm
resistor are in series so that becomes an equivalent 90 ohm's.
The 30 and the 30 are in series becoming an equivalent 60 ohm's.
Combining these two together and this 24 ohm here, we discover that the final
resistance is 60 ohms. And so now instead of having to do one of
these other analysis techniques, we were able to just combine resistors and by
doing a transform and simplifying a system.
Now it turns out that this does have some application.
You might say, you might think this is a little bit a bit of work for what we're
doing with it. But it turns out that in certain
circumstances particularly with power, three-phase power, power generation,
power consumption that these types of configurations are very common.
And being able to convert from a delta to a y back and forth can make the analysis
a lot easier. The other thing we're going to talk about
is a Wheatstone Bridge. A Wheatstone Bridge is a device that is
used to measure a resistance that's under test.
So in this case, RX here is the resist, resistor we wish to test.
So we have R1, R2 and R3. These little arrows, going through R2 and
R3 indicate that these are resistors that we can change.
Sometimes referred to as potenciometers. Where we, we can change what the
resistances and we can measure what it is.
We know what it is. There's some way of identifying it.
The way that we do this type of measurement and test these Wheatstone
bridges by taking a measurement here demonstrated by this little graphic.
That measures the current that goes across what's called the bridge.
This wire right here. We're going to say that this bridge is
balanced when this current is equal to 0. No current is flowing across the gap.
So what do we need to do to I get that current to be 0.
Well firstly, what we need is for this voltage, which it will be 1.
And this voltage, it will be 2. With a voltage across R3 and the voltage
across RX have to be equal. Because when they're equal.
Using Kirchhoff's voltage law we see that there's zero voltage drop between these
two places. There's going to be no, no pressure for
this to try and drive this current across the bridge.
When I'm, discovering what type of scenario I want to get to see that
through, you can look at this like a voltage divider.
So ther's no vol, no currents going across this bridge at this point.
You end up having two independent voltage divider circuits.
Since I know that the voltage from this top node to this bottom node is specified
by the s. So, consequently V1 using a voltage
divider is equal to R3 divided by our 1 plus our 3 times the s and the 2 is going
to be equal to Rx over R2 plus Rx. And then I know that they have to be
equal. So by setting these two things equal and
then, we can, factorise the VS's and doing a little bit of algebra.
We get the result that R1 over R2 is equal to R3 over Rx.
So when this happens, then this bridge said to be balanced, no currents going to
flow across. And, because our x is an unknown value.
All we do is we take the know values, R1, R2, and R3 and then solve for Rx because
we've already seen in practice that this current is 0 we know what we're
expecting. If this resistance is smaller than the
balanced resistance, there's going to be some current flowing across this bridge,
down this way. And so with this reference direction
there'll be a positive current flowing. If it's larger, there's going to be some
current flowing through here across the bridge this way.
So with this reference direction, we would measure it in negative current
flowing across the bridge. So in summary we showed the delta y
transform or the y delta transform, to simplify circuits.
And we balanced a Wheatstone bridge and identified as we change the test resistor
if we see a positive current or negative current what that means for our resister
under test. In the next lesson, we'll be able to look
at an application of resisters in sensors as well do an experiment showing a
Wheatstone bridge. And see exactly how they're used in
practice which is a much better way of just looking at it from an analysis
perspective because you can actually see how Wheatstone bridge can be used in
practice. So after those two lab demos that will
conclude module two and so I will see you again on the forums and in module three.