0:09

This lesson is on mesh analysis.

Â The objective is to introduce mesh analysis in order to solve

Â circuit problems.

Â Mesh analysis is one of my go to methods for solving circuited problems.

Â I can look at a circuit is fairly complicated,

Â I've got multiple sources in there and lots of resistors and

Â I want to come up with a systematic way of solving it.

Â Maybe I'm solving for a particular voltage or a particular current.

Â 0:35

The mesh analysis relies on the prior concept of the Kirchhoff's voltage law,

Â which is a sum of the voltages around any loop is zero.

Â And in fact the mesh analysis is a systematic application of

Â the Kirchhoff's voltage law.

Â There are three basic steps in applying mesh analysis.

Â First to define mesh currents, one for each non-inclusive loop.

Â By non-inclusive, I mean a loop doesn't include any other loops.

Â So I've got three non-inclusive loops here.

Â And then define these mesh currents to be I1, I2, and I3.

Â And I give it a direction of the current flow around that loop.

Â And then I do a KVL for around each of these loops.

Â And then I solve for the mesh currents.

Â In this particular case,

Â the only unknowns in my equation should be these mesh currents.

Â So even if I have another variable floating around in there,

Â I ignore that variable for the time being,

Â I try to write my equations only in terms of those mesh currents.

Â Now, let's take a look at coming up with the equations for this particular circuit.

Â Now in coming up with the equations for

Â this example, we're going to need to use the KVL.

Â And in using the KVL we're going to need to be able to do

Â find the voltage across a resistor like this.

Â Well, that means that we really want to know what the current

Â is through this resistor.

Â Now I'm going to put something off to the side here.

Â Suppose we have a resistor like this and

Â the voltage is equal to IR, Ohm's law.

Â But suppose our current is shown in terms of these branch currents, so we've got

Â one branch current going this way and then another branch current going this way.

Â The combination i is equal to I1 which is

Â going in the same direction as i minus I2 which is going in the opposite direction.

Â So then this voltage across that resistor is R(I1-I2).

Â 2:38

Like that.

Â So, we use that in being able to do the KVL around these loops.

Â So, let me start with the I1 loop, Applying the KVL around that.

Â Let me start at this point here and I'll do it in this direction And

Â I've got a -V1 + R1 times this mesh current,

Â + R2 times the current going in this direction.

Â With that current going in this direction, will be equal to the I1 which is in

Â the same direction, minus I3 which is in the opposite direction.

Â All right let's see, I've got three terms here and

Â I've got three terms around the loop so I'm done with that loop, is equal to 0.

Â Now, I2 loop.

Â 3:31

We do not do a KVL around the I2 loop because I already know what

Â this current is.

Â The whole purpose of the doing a KVL is to find the mesh current,

Â well I know the mesh current.

Â I2 is equal to I sub s, because it's going in the same direction, so i sub s.

Â So then I just go on to the I3 loop.

Â 3:54

And let me go ahead and start, say right at V2.

Â So I've got V2 going around here.

Â In this case I'm going to be looking at the voltage from here to

Â here and that means the current is going to be going in this direction.

Â If the current is going in this direction, then I'm going to be looking at

Â 4:20

I3 which is in the same direction minus I1 in the opposite direction.

Â That's the total voltage drop across that resistor from this direction.

Â And similarly, up here, if I'm going in this direction right here,

Â I want this voltage drop from here to here.

Â And I'm going to be looking at that branch current there.

Â 4:51

And then coming right here I've got one more term and the Ohm's law on that

Â term is R0 times I3 is equal to 0.

Â And let's see I've got one, two, three, four elements across here and

Â I've got four terms in my [INAUDIBLE] and that's it.

Â So right know I've got really two equations and two unknowns,

Â because I2 is known, I assume my source is known so I2 is known, wherever I

Â see an I2 I substitute in isoface and then I just end up with just two equations and

Â two unknowns, and I can solve that for I sub 3, and I sub 4.

Â And once I solve for I sub 3 I can go back and solve for V0.

Â 5:33

So let's look at this example right here.

Â I want to first to find my two mesh currents, I1 and I2.

Â And then, I want to do a KVL around, say, the I1 loop.

Â So, I start with -15 + 10,

Â going around in this direction,

Â I would have 10 times I1 plus this resistor,

Â let me give it 5 ohms, so if this is 5 ohms,

Â then I would want to solve for 5(I1-I2)=0.

Â If I simplify that a little bit I would

Â find that I've got 3I1- I2 = 3.

Â Then I do the I2 loop same sort of thing,

Â I'm going to start out with over here,

Â 5V + 5 ohms (I2- I1).

Â Going up here plus 4 times I2 = 0.

Â And if I solve these, and

Â I've got two equations and two unknowns,

Â I solve them I will get an I1 = 1A, I2 = 0A.

Â 7:03

Which is kind of interesting.

Â It means that there's zero current going through here and

Â all the current goes in that direction.

Â Now suppose the original question had been, solve for this current right here.

Â Well, if I'm going to use mesh analysis, I first solve for I1 and I2,

Â those are my unknowns.

Â After I solve for I1 and I2 then I go back and I solve for

Â whatever voltages or currents I need to solve for.

Â So everything has to be written in terms of my mesh currents and

Â those are my unknowns.

Â We're going to set up a quiz here for this particular circuit.

Â How many unknown mesh currents do you need to solve this?

Â 7:42

Okay, in this case the answer is one, because we've got three mesh loops here,

Â but two of these are known, we are assuming that the sources are known.

Â So this is the only mesh current that you're going to need to solve for.

Â Here's another quiz I want you to do.

Â I want you to solve for

Â the equations that result from using mesh analysis of this circuit.

Â Now here I've used a dependent source up here.

Â Don't let that scare you.

Â It's a current source.

Â Pretend that we know this value when setting this up.

Â In other words don't do a KVL around this loop.

Â 8:19

All right, the solution is, I do start with a KVL around this other loop.

Â I get -3 + R2 times, everything's

Â gotta be in terms of mesh currents.

Â So I've got I2- I1

Â + R4 I2 = 0.

Â Okay, and I need one more equation, because I've got two unknowns.

Â What my equation is that

Â I've got I1 = -2i1.

Â Because it's in the opposite direction.

Â 8:59

And I1 is the net current through this, which is equal to, let me put the -2.

Â I1 is equal to I2, which is in the same direction, minus I1,

Â which is in the opposite direction.

Â So now, I've got two equations, I1 = -2 (I2-I1).

Â 9:19

So it's one equation in terms of my mesh currents.

Â And then this is the other equation in terms of again my mesh currents.

Â The key concepts we've covered in this lesson,

Â first of all is that mesh analysis is a systematic application of the KVL.

Â 9:34

The number of mesh currents is equal to the number of equations meet it.

Â The variables are mesh current.

Â So, make sure that whenever you write your equations you ignore any other variables.

Â The only variables should be those mesh currents.

Â Avoid common mistakes that students make.

Â One is the polarity and

Â the current directions very easy to get mixed up on those.

Â Do not do a KVL for a loop with a current source.

Â You don't need to.

Â The current source already defines what that mesh current is.

Â And also, while you're writing the equations, and you're trying to solve for

Â those mesh currents, ignore any other variables than those mesh currents.

Â Solve for those afterwards once you know what your mesh currents are.

Â All right, thank you.

Â