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In this video,[br]we introduce the concept of a super mesh.

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Here we have a circuit that[br]involves three meshes.

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The left-hand mesh, which has a mesh[br]current associated with it of I1,

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a mesh here with a mesh current defined of[br]I2, and a mesh current here defined of I3.

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The problem arises when[br]we have a branch that

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contains an independent current source.

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By definition, an independent current[br]source produces, in this case,

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I0 amps of current ,no matter[br]what the voltage is across it.

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In other words, there is no relationship[br]between a current source current and

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its voltage.

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Let's demonstrate the problem by[br]attempting to write the mesh equation

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around this loop right here starting[br]in the lower left-hand corner.

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We've got the voltage drop across R2,

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which would be R2 x the current through R2

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going in this direction, is I2- I1.

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Coming down the right-hand side, we have[br]the voltage drop across R3 is R3 X I2.

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And now we get to this independent source.

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We're attempting to[br]write the voltage drops

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around this mesh, but[br]we have no expression for

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the voltage drop across[br]I0 in terms of I2 and I3.

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How do we resolve this?

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Well, to resolve it, we take[br]advantage of the fact that the sum of

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the voltage drops around any[br]closed loop must equal zero.

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So instead of going around each of these[br]meshes separately, we're going to make

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a trip around the outside of[br]both of these loops combined.

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Let's show what we mean here[br]by now analyzing this circuit.

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Let's start by writing the loop[br]equation around this left-hand mesh.

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We'll have -V0 + R1I1 + now

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coming down this branch here,

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voltage drop across R2[br]will be R2 x I1- I2.

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Plus the voltage drop across R5,

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which will be R5 x I1- I3.

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The sum of those voltage drops[br]must equal 0, nothing new there.

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Now let's write the KDL equation[br]going around our combined super mesh.

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Starting down here in the lower left-hand[br]corner and going in a clockwise direction,

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we'll have voltage drop across R5 will[br]be R5 times, let's be careful here.

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We're going up across R5.

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The current going up is I3- I1.

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So R5 x I3- I1,

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plus the voltage drop across R2

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going in this direction will be

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I2- I1 + the voltage drop across

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R3 is just gonna be R3 x I2.

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And finally, we've got the voltage[br]drop across R6, which will be R6 x I3.

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And the sum of those[br]voltage drops will equal 0.

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There, we've got the two equations.

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Again, we're gonna get our[br]third equation by taking

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advantage of the fact that[br]the current in this branch is I0.

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And that that current I0,[br]in terms of the two mesh currents,

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is equal to the mesh current[br]flowing in this direction,

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minus the mesh current[br]flowing in that direction.

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Or, I0 will equal I2- I3.

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Thus, we have the three equations that we[br]need to cover the three variables that we

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defined.

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Again, our first equation came from going[br]around this mesh, as we have always done.

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The second equation came from[br]going around the super mesh,

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effectively avoiding the need[br]to go across the volt or

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to know the voltage across[br]that current source.

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And then the third equation came from[br]the fact that the branch current in this

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sources is in fact I0.

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And in terms of the mesh currents,[br]the current in that branch is I2- I3.

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Three equations, three unknowns,[br]be ready to solve it,

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using whatever method for solving[br]systems equations you choose to use.

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In the next video, we'll do[br]a numerical example of a super mesh.