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In this video we're gonna take a look at[br]using node analysis techniques to analyze

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a circuit that's got a dependent source.

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You'll notice right off the bat that[br]we got a dependent current source

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here in this branch.

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This dependent current source,[br]the amount of current that this current is

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producing depends upon the current i[br]flowing through the 6 ohm resistor and

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this current source, then,[br]is two times that I.

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We're gonna take that into account as we[br]go through and define our node voltages

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and we'll come up with an expression for[br]I in terms of our node voltages.

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So let's go ahead and get started.

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We've got three critical nodes again.

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One here, one here, and[br]one along the bottom.

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Once again, we'll go ahead and[br]call this our reference node,

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where we've got v equals 0.

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And once again, we've got a voltage[br]source tied to the reference,

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so we know the voltage here at this point[br]of our circuit is equal to 5.3 volts.

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We'll define this node here to be V1,[br]this node here to be V2.

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And with those two defined[br]voltages now we can express i,

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the current i In terms of V1 and V2.

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In fact I is equal to V1[br]minus V2 divided by 6 ohms,

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it's just the voltage[br]here on the left minus

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the voltage on the right[br]divided by 6 ohms.

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When we're all done with this analysis,

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we're going to use that expression[br]to calculate I let's say

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to calculate the power being[br]produced by this dependent source.

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With those definitions, let's go ahead and

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write the node equations[br]at these two nodes.

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First one here on the left[br]the current leaving.

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The first node here going[br]to the left is going to be,

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V1-5.3 divided by 4 ohms[br]+ the current coming down

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here through the 3 ohm resistor[br]is going to be V1 divided by

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3 + the current leaving this[br]node going to the right.

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Well notice that that's that[br]current I has been defined for us.

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We don't want to use I we want to use or[br]we want an expression for

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that current in terms of[br]the node voltages V1 and V2, and

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we've already expressed that I over[br]here as V1 minus V2 divided by 6.

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So plus V1-V2 divided by 6.

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The sum of those three[br]currents has to equal 0.

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Now let's write the node[br]equation of the V2 node.

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Current going to the left is going to

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be V2-V1 divided by 6[br]+ V2-0 divided by 12.

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Minus, okay, now let's be careful.

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Minus two times I, it's minus because[br]it's referenced into the node.

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Two times I is this dependent source,[br]but I don't want to put I in here.

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We want to write everything in[br]terms of our node variables or

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node voltage variables V1 and V2.

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So I've got i is equal to two,[br]or v1 minus v2 divided by 6.

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This current here is two times I so[br]it's going

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to be -2 times v1 minus v2 divided by 6,

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the sum of those three currents equals 0.

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I suggest that you stop the video right[br]now and just take a look at that,

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there's a couple of things going on here.

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The fact that it's going into[br]the node gives us the minus sign.

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It's gonna be minus[br]the current flowing in.

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Well, the current flowing in is 2 times or

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has a value of 2 times whatever I is.

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We defined I as V1- V2 divided by 6.

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So we have, then,[br]-2 times V1- V2 divided by 6.

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Alrighty, now that you've taken a moment[br]to make sure you understand what we did

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there with the signs, let's go ahead and

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solve this system of two equations with[br]two unknowns by combining like terms.

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For the first equation we've got

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V1 times one-fourth + one-third

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+ one- sixth + V2 times.

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This is just one V2 term that's got[br]a negative one- sixth with it is

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equal to negative 5 divided by 4 take it[br]to the other side as I'm sorry negative.

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Let's see, it's -5.3 divided by 4,

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take it to the other side[br]is a +5.3 divided by 4

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will write it like that for[br]now 5.3 divided by 4.

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Alrighty, the second equation and

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here we're going to need to be careful[br]because we've got this stuff going on

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there with the last term but[br]once again factoring out the V1 terms.

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Here I've got a V1 term and I've got[br]a V1 term here, let's be careful there.

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The first one is negative one- sixth.

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But what do I actually have over here?

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I've got -2 times V1 divided by 6.

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So that's negative -2 over 6,[br]that's negative one third V1.

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Factor out the V1, and that leaves me

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with a negative one-third there.

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Alrighty, now for[br]the V2 terms, + V2 times,

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I've got a V2 over 6, factoring[br]out the V2 leaves me a one-sixth.

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I've got a V2 over 12, so + one-twelfth.

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Now once again, slow down here.

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You've got signs and[br]signs like red flags waving.

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That means slow down, you've got[br]an opportunity to make a sign error and

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we don't like sign errors.

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We've got -2 times a -V2 over 6.

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Negative times a negative[br]is a positive to V2 over 6,

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so that's gonna be positive[br]factor out the V2,

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I've got 6 that's one-third, and[br]the sum of those three currents equals 0.

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Once again, I'd encourage you to[br]stop the video at this point and

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just make sure that you understand[br]what we did with the signs.

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Alrighty, let's just Be careful here[br]that we keep our equations separate,

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and now let's go ahead and[br]combine the fractions.

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We've got here then for[br]the first equation we have v1 times

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one-fourth plus one-third plus[br]one-sixth is three-forth.

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Plus V2 time negative one-sixth

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= 5.3 divided by 4 is 1.325.

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And on the second equation[br]we have V1 times negative

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one-sixth minus one-third[br]is a negative one-half.

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Plus V2 times one-sixth + one-twelfths

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+ one-third is seven-twelfths and[br]it's positive equals 0.

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So we've got our two equations there.

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Once again,[br]plug that into your matrix solver,

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use the solve button on your calculator or[br]you can use MATLAB for that matter.

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And when you do, you'll get

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that V1 equals 2.1824,

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2.1824 and V2 is equal to 1.87.

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Now what can we do with V1 and[br]V2 while we can do anything we need to do.

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Just for example in this one, why do[br]we calculate the power of this being

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generated by this dependency source and

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let's be careful here because we've[br]got issues with reference direction.

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There is a dependence[br]source it has a value

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of 2 times I, reference going up.

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And what is the voltage across that power[br]supply, well is the reference plus to

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minus V2 that power supply that dependent[br]current source has V2 on the top and

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it's tied on the bottom to our reference[br]so the voltage across that is just V2.

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Now we know that power[br]is equal to I times V.

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And we also know that we got to be[br]careful about the sign in front of this.

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Our passive sign convention says that[br]if current is referenced going from

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the positive to the negative terminal,[br]it will be a positive power.

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That's not what we have here, in this[br]case, we have the current going up,

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referenced into the negative terminal and[br]out the positive terminal.

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So that means that we[br]need a minus sign there.

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Now, what is I?

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Well, we've got I up here,[br]I is V1- V2 divided by 6.

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So let's go ahead and calculate that.

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V1 was 2.1824 minus V2.

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Which was 1.87-1.87,

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and the whole thing divided by 6.

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That gives us a value for

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I of 0.052 amps, or

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52 milliamps, so 0.052 amps.

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Now this current source right[br]here is equal to if I sub x,

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the current that it generates[br]is equal to 2 times I,

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we now know I is 0.052.

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So that current source is producing

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2 times .052 or .104 amps.

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That's the current that we need[br]down here to calculate the power.

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This current source is[br]producing 0.104 amps going up.

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So the power down here is going to[br]be equal to negative I is 0.104.

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Times V, which V did we say it was?

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It was V2.

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So times V2, which is 1.87 volts,

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and when you multiple those two together,

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you get a negative 0.194 watts.

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Negative, what does the negative[br]sign mean on power?

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It means that it's producing energy,

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it's the source is putting[br]energy into the circuit.