In this next set of videos,
we're going to introduce a circuit analysis method known as mesh analysis.
My name is Lee Brinton,
I'm an Electrical Engineering Instructor at Salt Lake Community College.
In this videos we'll make the distinction between a loop and
a mesh, we'll make distinctions between branch currents and mesh currents
We'll demonstrate the process of analyzing circuits using mesh analysis.
Look at how dependent sources impact our analysis approach.
And we'll also introduce concept of supermeshes.
We're gonna find that a supermesh is analagous to a supernode that we ran into
doing node analysis First of all the difference between a loop and a mesh.
A loop is any closed path.
In this circuit we have a closed path here, a closed path there,
and yet one more closed path.
So this circuit has three different loops.
Let us formally define a mesh now.
A mesh is a closed Path that contains no other closed paths within it.
So in this case, we have a mesh here,
a mesh here, but this outer loop,
because it contains two loops within it, does not qualify as a mesh.
Lets also make the distinction between branch currents and mesh currents.
As we know, a branch current is the current flowing in
any single branch of a circuit.
This circuit has three different branches, there is this branch here,
I sub a, this branch here that has the current to find I sub b, and
this branch here that has the current I sub c.
A mesh current is somewhat different.
A mesh current is something of a contrived or a tool that we had define to
assist in this analysis and to understand it think simply about or
imagine that there was a current That flowed around this loop.
That is a mesh current.
Similarly imagine that thereâ€™s a current flowing around this loop.
Those two mesh currents.
We're now gonna see that those two miss current are all that we need to represent
the minimum number of variables that we need to analyze the circuit.
So what the difference between i1 and i2?
Don't they both impact this branch?
And in fact they do.
Is defined as the current going down r3, it's a branch current.
In terms of our mesh currents, I sub b
would equal our mesh current I1 flowing down in the same direction as I sub b
minus the I2 mesh current that's flowing or opposing that.
So I sub b then would equal I1 minus I2.
On the other hand,
I1 is this current flowing through here.
And that current is in fact the branch current Ia.
So we can say, then, that Ia = I1.
Or I guess more appropriately, more intuitively,
we'd say that I1 is simply the branch current Ia.
And similarly over here, this mesh current I2 Is
the same as the branch current I sub c.
So we can then see that I sub 2 is equal to I sub c.
Thus, we can define any branch current in this circuit in terms of
the mesh currents I1 and I2.
Now let's see how we
would go about writing The equations to solve this,
or to analyze this kind of circuit.In node analysis we write KCL equations.
One KCL for each node.
In mesh analysis, we're going to write mesh equations or
KVL equations, summing the voltage drops around the closed loops.
One for each mesh.
So for example, let's start with this left hand mesh, and
write a KVL equation starting right here.
a we go across the voltage source, we're increasing V0 volt, and we're
gonna sum the voltage drops, therefore the voltage increase would be negative.
And we have then -Vo continuing on around here the full stop across that resistor.
In terms of I1 and I2, our mesh currency is going to be,
and because we're going in the direction of I1,
it'll be a plus R1I1 Continuing on down here,
the voltage drop across R3 in terms of mesh currency is going to
be R3 times the current flowing in that direction, which is I1 minus I2.
And the sum of those three terms must equal zero.
Now look at the right-hand mesh.
Starting right here, and let's go in the direction that I2 has referenced.
We have, then, the voltage drop going plus to minus,
going in the voltage, direction of current flow, we will have R3 times
the current going up that branch Which is (I2- I1).
Continuing on around this loop we have the voltage
drop across R2 is going to be + R2 times I2 and
the sum of those two terms then must equal 0
We have two equations with two unknowns, we've got everything we need to solve so
the next step with then simply be to combine like terms.
The first equation, lets factor or combine the I1 terms and factor I1 out.
In doing so we're left with an R1 + R3.
Similarly, for I2, there's only one I2 term.
It's got a -R3 times that.
So -R3 there.
Bringing this -V0 to the other side of the positive,
we have on the left hand side = V0.
The second equation will similarly factor out I1.
When we do, we got a negative R3
now working with I2 we have, what is that?
R2 plus R3.
And the sum of those terms has to equal 0.
So, we have two equations, two unknowns if we knew what R1, R2, and R3 and
V0 are we could plug in those values and we would be able t solve for I1.
And I2.
Given those two values, we could then go back and calculate any branch voltage or
branch currents that we were interested in in this circuit.