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Alright, let's formalize our[br]approach to mesh analysis.

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When analyzing a circuit using mesh[br]analysis we first of all identify

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the meshes.

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Once again, a mesh is a closed loop[br]that contains no other loops within it.

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We'll next assign variable names for[br]each of the mesh currents.

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We'll then write KVL equations around[br]each mesh in terms of the mesh current.

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And then at that point, we have a system[br]of equations in terms of our mesh currents

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that simply need to be solved, and[br]we've got the problem taken care of.

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Let's take a look at this circuit here.

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First of all, identify the meshes.

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We've got a mesh here on the left,[br]we've got this upper mash, a lower mesh.

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And again, this outer loop is not a mesh[br]because in this case it contains three

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different meshes within it.

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So we identify then or we define three[br]different mesh currents, one for

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each of the meshes.

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We'll define the current[br]In this mesh to be i1 and

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referenced in that direction,[br]the mesh in the other,

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mesh current in this mesh we call i2 and[br]the mesh current in this mesh we call i3.

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Lets go ahead now then and write the three

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Mesh equations in terms of I 1 and[br]I 2 and I 3 starting right here.

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Going up this way we cross the voltage[br]source going from minus to plus

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therefore it's a voltage increase and[br]again that would be a negative V 0.

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Plus the votage drop across[br]R1 is just R1 times I1.

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Now coming down across R2,[br]we need to be carefully.

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The current through R2

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involves two different mis currents[br]because we are going down at this point

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The current through here is gonna be the[br]mesh current reference down which is I 1,

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minus I 2, which is referenced[br]in the opposite direction.

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So we'll have then plus[br]R 2 times I 1 minus I 2.

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Now continuing on down, across R5.

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The current flowing through[br]R5 is going to be I1- I3.

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And those terms must then add to 0.

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Let's take a look at the top mesh.

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Starting right here and

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going up we've got R2 times the current

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flowing up which is I2 minus I1.

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And again let us just point[br]out that the current going up

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I2 minus I1 In this equation[br]is the opposite current that

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was flowing down which was I1- I2 and[br]this equation.

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All right continuing on[br]around we have in plus

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R3 times the current[br]through R3 is simply I2.

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Coming back to the left through R4[br]we'll have plus R4 times the current

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flowing in R4, which

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is I2- I3.

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That brings us back to where we started so[br]the sum of those terms must equal zero.

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And finally we do a KVL around this bottom

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mesh starting here and[br]we'll have R five times

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the current flowing through R five in the[br]direction we're going which is I three.

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Minus I1

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plus coming across here, plus R4 times

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the current flowing left to right in R4.

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In terms of the mesh[br]currents is I3 minus i2 and

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than finally the current coming down[br]here through that r6 will be r6 times i3

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and than some of those[br]currents equal zero.

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Up here, lets just go ahead and

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complete this by factoring out the mesh[br]currents and combining like terms.

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We have, for the top one,[br]the top equation we've got I1 times R1.

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Let's see what we've got.

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R1 there, there and there.

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So, we've R1 plus R2 plus R5.

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+ I2 times, we have one I2 term there,

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it's got a negative.

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R2 + I3 We've got one I3 term here[br]with a negative sign in front of it.

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So that would be I + I3[br]times a negative R5.

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And the sum of those then equals,[br]we've got the negative V0 that we

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need to bring over to[br]the other side as a positive.

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V 0 combining like terms in the second[br]equation we factor out the I 2 I'm

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sorry the I 1 here and I've got one

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I 1 term, it's got a negative on it so[br]it'd be a negative

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R 2 Plus I2 times.

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You've got I2 here, here, and here.

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All positives will be R2 plus R3 plus R4.

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Plus R3 plus R4 and

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then our I3 turns There is only one.

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It's got a negative, so[br]it'll be a negative R4.

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And the sum of those[br]terms have to equal 0.

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And finally, the bottom equation here,[br]factoring out the I1s to begin with.

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I've got an I1 term here.

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Times what is that?

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That's R5 isn't it?

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So I1's got a negative R5.

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Plus I2.

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once again there's only one I2[br][INAUDIBLE] has a negative R4.

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Plus I3.

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And there are three I3 terms,

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I got R4 plus R5 plus R6,

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R4 + R5 + R6.

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And the sum of those has to equal 0.

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Plug it in, well, of course, with[br]the values of the resistors and the V0,

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plug it in to your matrix solver or[br]your solve button on your calculator,

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and you've got everything you need[br]to calculate those mesh currents.

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Once you know the mesh currents you[br]can calculate any branch voltage or

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branch current that you[br]might be interested in.