0:10

The problem is to write the mesh equations for the circuit that we have shown below.

Â This is our first example of mesh analysis, so

Â weâ€™re going to go through a few points of interest and important

Â steps in performing mesh analysis and coming up with the mesh equations.

Â First thing we want to know is that mesh analysis is the same thing as loop

Â analysis.

Â The terms are used interchangeably.

Â 0:34

In this problem, we can identify several meshes.

Â We have a mesh on the left-hand side of the circuit, which is the left loop,

Â if you will, but also the mesh on the right-hand side, which is the right mesh,

Â or right loop of the circuit.

Â We also have mesh which is around the outside of the circuit, and

Â so we have three different meshes.

Â They're not independent of one another because

Â the outside mesh composes the left and the right meshes, so

Â we're going to write mesh equations for two different meshes.

Â Mesh 1 and Mesh 2 for this problem.

Â So, first of all, we're going to identify the meshes.

Â This is our Mesh 1 and this is the mesh current associated with Mesh 1,

Â and this is Mesh 2, the right-hand side mesh, and

Â we're going to assign currents for Mesh 2 to I sub 2.

Â So, we have two meshes, Mesh 1 and Mesh 2, and mesh currents I1 and I2.

Â So, the first step in writing the Mesh equation is to

Â use the passive sign convention to assign polarity is across the different elements.

Â Our voltage sources already have polarities assigned so

Â we don't have to do that.

Â But we have to identify polarities for the resistors and

Â the voltage drop across the resistors.

Â So using the passive sign convention, if we start with the left most loop and

Â we're looking at the current I1 It flow through R1.

Â And positive sign convention tells us that the positive current flows

Â into the positive polarity of the voltage drop across those resistors.

Â So this would be identified like this, as V R1.

Â So V R1. We also have when

Â we continue around that loop, past resistor one,

Â we have resistor three So we have a voltage drop across it, V R3.

Â And if we continue around that mesh, we have R2 at the bottom of that loop.

Â And the voltage drop, again using the passive sign convention is V R2

Â across the resistor at the bottom of the left hand loop.

Â 2:43

Okay.

Â So now we're going to turn to the right hand loop and

Â we're going to assign polarities for it.

Â So if we look at that, and starting at the lower left hand side of that loop, and

Â going around the loop.

Â We already have a polarity assigned for R3 so we don't need to do that.

Â We have our voltage drop at the top VS2 As a polarity already assigned.

Â And then we come around to the right-hand side of the circuit, we have resistor R4.

Â Using the pass of we know that the positive direction of the current

Â 3:14

meets the positive polarity of the voltage drop across R4 first.

Â So he assigned.

Â That voltage drop, VR4, as shown.

Â And then we go around that loop, continuing around the loop to R5.

Â Same technique, we have a voltage drop VR5 across resistor 5.

Â So that's what our polarities and

Â our voltage drop assignments Are based on the passive sign convention.

Â The next thing we're going to do is we're going to add up,

Â 3:55

as a way of solving These problems in mesh analysis.

Â And Kirchoffâ€™s voltage law tells us that, the sum of the voltage

Â around any closed loop is equal to zero at any instant in time.

Â So whether we take loop one,

Â loop two, or if we even choose the outside loop around the outside of the circuit.

Â If we add up those voltage drops,

Â the sum of them is going to be zero at any instant of time.

Â So let's do that, let's start mesh one and let's add up the voltage Around that loop.

Â 4:28

Starting with the lower left hand corner of that mesh,

Â travelling upward in a clockwise direction.

Â The first thing we come across is voltage source VS1 and

Â we, in fact hit the negative polarity of that voltage source first.

Â So we have minus VS1 For our first voltage rock.

Â We continue on that circuit and we encounter resister R 1 with

Â voltage drop V R 1 and counting the positive polarity of it first.

Â We have plus V R1 and we have

Â continuing around that circuit the voltage drop across R3 so we have a plus VR3.

Â And we continue around that loop and

Â we encounter resistor R2 and the drop across it.

Â And that's V R2.

Â We hit the positive polarity of each one of those resistor

Â 5:42

Again, doing the same thing with loop two, starting at the lower left and

Â corner of loop two, traveling clockwise around the loop.

Â The first thing we encounter as we start that journey is we encounter the voltage

Â up across resister R three.

Â We get the negative polarity first, so it's minus VR3.

Â 6:53

Okay.

Â So now we have these two equations and ultimately these will allow us to

Â solve for The loop currents I1 and I2.

Â We can make some exchanges here and they're easy in some cases.

Â We know that Ohm's Law V = IR can be used wherever

Â we have a resistor we can replace The voltage,

Â for instance, VR1, we can replace that by

Â 8:19

R3 has A current I1 flowing through it, and it also has a current

Â I2 flowing through it, but they're flowing in opposite directions.

Â So if you remember, using the passive sign convention, we assign the positive

Â polarity across R3, the voltage drop across R3 to be at the top of R3,

Â based on A positive current I one flowing in to the top of R three.

Â 9:10

And we chose I1 as the positive current because using the passive sign convention

Â That was our positive current flowing into that voltage drop across R3.

Â And the I2 is the opposite direction, so it has the negative sign in front of it.

Â So now we have R1, R2, R3, R4, and R5 identified, we see that

Â 9:32

if we plug those into equation one and equation two, That we have two unknowns.

Â Our only unknowns would be I1 and I2.

Â So we'd have two equations which are independent of each other, and

Â we'd have two unknowns, I1 and I2.

Â We could solve for I1 and I2 if we knew values for the resistors and for

Â the voltage sources.

Â 9:54

And if we solve for I1 and I2,

Â then we have everything else that we would need for the circuit.

Â We could solve for the power, absorb the power supply.

Â We could solve for the voltage drops across any element in the circuit.

Â So, that's all we need in order to able to solve for all the values of interest for

Â this circuit.

Â