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In this video, we're gonna[br]demonstrate the super mesh technique.

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As you look at this,

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you'll notice that there are three meshes,[br]this one here to the left.

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And we'll label the current[br]in that mesh current I1.

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We have a mesh up here.

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And we'll call this mesh current I2, and

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we have a mesh here with a mesh[br]current that we'll label I3.

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And further before we even get started[br]we'll notice that separating mesh one from

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mesh three is a branch that[br]contains a current source.

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And as we've mentioned in the past,[br]we don't have a.

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A numerical or mathematical[br]relationship between the voltage across

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a current source and the current itself.

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That will then require us to[br]do a super mesh which we will

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Be writing around there.

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Let's get started first of all[br]though by writing the mesh

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current equation around[br]this mesh number one.

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Starting down here in the lower left hand[br]corner, we have going minus to plus.

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It will be a negative 12 volts.

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And then across this resistor of a voltage[br]drop of five resistance five times current

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rate through that which is I1-12

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plus the volts struck across this[br]film resistor which is going to be 3

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times I 1- I3 and

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the sum of those voltage[br]drop must equal 0.

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Now, the Super Mesh starting here at this[br]point going up and around the 15 ohm, and

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when we're across the 15 ohm resistor now[br]the voltage drop there being 15 times I2.

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Now continuing on down here,

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not going around this branch here because[br]that's where the current source is.

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Continuing on down here across the 10[br]ohm resistor we have a voltage drop

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of +10 x (I3) Plus,

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now continuing around here,[br]across the 3 ohm resistor, +3 times.

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Now the current going in[br]this direction is I3-I1.

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I3-I1+ the voltage

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drop across that 5 ohm[br]resistor Is five times.

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And again,[br]because we're going from right to left,

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the current going in that[br]direction will be I2 minus I1.

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I2 minus I1,[br]the sum of those terms must equal 0.

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And finally the third equation we need for

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these three unknowns comes from[br]the relationship between I2, I3.

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And this current source.

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In this case, I2 is going in[br]the direction of the current source, so

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we're going to have I2 minus I3,

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which is the current going through that[br]branch in terms of the mesh currents.

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It's going to equal 2 Amps.

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So we have our three equations and[br]three unknowns.

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Let's go ahead and combine terms.

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Taking the first equation[br]we have I1 times,

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we got 5 there plus 3 is 8 plus

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I2 times We have a negative 5

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Plus I3, we have a negative 3.

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And then we have this negative 12 volts on[br]the left-hand side we bring to the other

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side as a positive 12 volts.

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And there is our first equation.

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Second equation combining like terms,[br]we have I1.

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And in this we have -3 and a -5.

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That makes -8 times I1 + I2 times,

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we got 15

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plus five is 20 times i2 plus i3, we've

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got 10i3 plus 3 is 13i3 equals, and

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there are no constants there So

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the sum of those terms equals 0.

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Finally, our third equation is I2,

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Times 1 + I3(-1) = 2.

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And we have our three equations and[br]three unknowns so

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get ready with your calculator or whatever[br]you choose to solve through your systems

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through your equations and three unknowns.

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And when you do we get the following.

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I1=2.03 amps.

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I two is equal to one point two eight amps

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and I three is equal to negative

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point seven two amps.

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LEts do a couple of consistency checks[br]here just to make sure we didn't make any

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mistakes not to prove that our work was[br]correct but at least to look at a couple

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of calculations that would be consistent[br]with what we know to be true.

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For example we know that i two[br]minus i three equals two amps.

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So I2, that is 1.28

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minus I3, well I3 is a -0.72 amps.

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The sum of those,[br]they're supposed to equal 2 amps,

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and sure enough that works there.

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And of course we could plug those values[br]into any one of those equations And

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proves that our solutions are correct.

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But now that we know what I1,[br]I2, and I3 are.

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Let's use them to say for

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example now determine what the voltages[br]drop is across this current source.

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We said that we couldn't numerically[br]calculate it based upon the current going

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through it, but now that we know[br]the currents around it, we can write a KVL

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once again around that loop, and[br]solve for the voltage across the current.

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Starting here and going clockwise,[br]across that 3 ohm resistor,

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we're going to have Three[br]times I three minus I one.

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Now, going across this current source with[br]the voltage as referenced plus to minus.

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That represents a voltage drop,[br]so it would be plus v.

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Now, coming down across here,[br]that would be plus 10.

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Oops, plus 10 times I3

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equals that's where we started from, 0.

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Now I plug in our values for I1, I2,

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and I3 and we have 3 times I [SOUND] So

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3 times I3, well, I3 is a -0.72 amps,

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-0.72,- I1, which is

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2.03 amps, + V + 10 times I3,

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which is a negative 0.72 amps.

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The sum of those things has to equal 0.

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Solving this for V gives us

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V = 15.45 volts.

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All righty, now looking at this source,[br]we see that the current is reference into

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the negative terminal of our voltage and[br]leaving the positive terminal.

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So it's acting as a source.

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And we can then say that the power being[br]generated by that source is equal to

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negaitive i times v Is equal to negative i

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which is two times v which is 15.45 and[br]the power

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then would be equal to[br]an negative what is that

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30.9 watts the fact that

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it's negative means that it's energy being[br]put into the circuit from the source.

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So once again, once we know all[br]three of the mass/f currents,

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we can determine any voltage or[br]current or for that matter power or

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any other quantity that we might[br]want to know within this circuit.