0:02
The topic of this problem is mesh analysis.
And we're going to be working with circuits with independent voltage and
current sources.
The problem is to determine V sub 0 in the circuit shown below.
Circuit has three independent sources, two current sources and a voltage source.
We're measuring the output voltage, V sub 0, across the load resister,
a 4 kilo ohm load resister, level resistor in our problem.
It's on the right most side of our circuit.
0:46
So the first thing we do when we have a problem and
we want to solve it using mesh analysis is we assign loops or meshes to the circuit.
And we also assign loop or mesh currents as well.
So let’s do that first.
First loop I’m going to assign is the upper left hand loop and
I'm going to give it mesh current I1.
1:26
So now we've assigned our meshes and
we've assigned our mesh currents that's the first step in mesh analysis.
Then the second step is to go to each one of these meshes and to add up or
sum up the voltages around the loop that close path, which we have.
Starting at the, typically I start at the lower left hand corner and
going around the loop, back to that same corner.
So we'll do that for each one of our loops.
We'll go all the way around it from start to finish,
ending in the same point, summing up the voltages.
What we're going to find in this problem is that we have a couple of
issues with using this straightforward approach and doing the problem.
So we have to use some of our additional rules of mesh analysis and
Kirchhoff's voltage law.
So let's start with Loop 1 which is the upper left hand corner.
We need to sum up the voltages around this loop.
So if we start in the lower left hand corner,
we start up, we see that we run into a 3 milliamp source.
We don't know what the voltage drop is across this, so we'd have to assign
another variable D sub 3 milliamps for the voltage drop across it.
So we'd add another variable in addition to our Mesh current.
Our four mesh currents would have a fifth variable in that case.
And we could go on around and we can find the voltage drop
across the two kilo ohm with respect to the loop current I1 and I sub 2.
And the 4k we could do the same using I1 and
I4 and the current as, to find the current through the 4k.
But we have this problem with the 3 milliamp resistor or
3 milliamp source on the left hand side of the circuit.
3:11
Once we have those, we can find any other value that we want in the circuit,
any voltage drop, any power whatever else might be of interest to us.
So that's really what we're after.
If we look at this loop one and we inspect it closely we see that,
I sub 1 is equal to 3 milliamps.
So the problem itself gives us I1 and we don't need to,
3:49
Let's go to the second loop and see if we can come up with an equation for it.
So starting in the lower left hand corner,
we encounter first the voltage drop across the 2 kilo ohm resistor.
That voltage drop is going to be 2K.
4:01
And since we're going clockwise,
the current which is going the same direction as I2, it's going to be I2- I1,
which is also flowing through the 2 kilo ohm resistor, I2- I1.
Because it's flowing through in the opposite direction of I2.
4:21
And we continue on around the loop and we
encounter the four kilo ohm resistor which just has a current I2 flowing through it.
So the voltage drop is 4k(I2) and
continuing down through the 12 kilo ohm resistor at the bottom of
loop two the voltage drop is 12 kilo ohms times I2 minus I3.
4:49
And that's our last voltage drop before we get back to our starting point in our loop
that we had for loop two, back to the lower left-hand corner of it.
So let's go down and look at loop three.
If you look at loop three,
we encounter really the same type of problem that we did up with loop one.
But this time we see that when we hit this
current source that we don't know the voltage drop across.
We don't have this current source assigned to just one loop,
as shared between loop three and loop four.
So we have an equation that we can write for that and
it will be that 1 milliamp = I3, which is following
the same direction as the 1 milliamp source,- I4.
So we can at least write that.
5:40
So, that gives us another equation, but we need four equations to solve
this because we have four unknowns I1, I2, I3 and I4.
So, we need one more independent mesh,
6:05
In order to get around this 2 milliamp source, I'm sorry the 1 milliamp source,
in the center of the these two loops I'm going to assign a mesh which
we can sum the voltages around just like we could any other mesh.
But is independent of mesh one and
mesh two above, so we get our, another independent equation.
6:36
And we can again sum up our voltages,
starting in the lower left hand corner of this mesh.
Going around this mesh, we have the ability to write an equation for
all the voltage drops from start to finish.
And this is, again, the supermesh.
We use this concept when we have a current source in a mesh analysis
problem that is shared between two adjacent loops.
Whenever we have that the case we have to rely on mesh analysis and
supermesh analysis in order to solve the problem.
So let's write our last equation for the lower two loops.
So starting in the lower left hand corner we run into the negative polarity of
the 6v source first.
It's a -6v for that voltage drop.
And then we encounter the 4 kilo ohm resistor.
Since we're going in the clockwise fashion, I4 is also travelling
the same direction through that loop, it's going to be 4k I4- I1.
8:11
And that's our last voltage drop before we get back to the starting point.
So that's equal to zero, so now we have four equations and four unknowns.
What we're really after in this problem is we're after V sub out.
Which is the voltage drop across the four kilo ohm load resistor.
We know V out = 4K times I sub 2,
pass of sign convention tells us the positive I2 would flow
into the positive of the voltage drop across the 4K resister.
So it's 4K( I sub 2) for V out.
So really the only loop, or mesh, current that we're interested in is I sub 2.
So that's the one we're going to solve for.
If we solve for I sub 2 from this equations above,
we get a I2 = 31 15ths milliamps.
Dr. Bonnie H. FerriProfessor
Dr. Joyelle HarrisDirector of the Engineering for Social Innovation Center
Dr Mary Ann Weitnauer
