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>> In this video, we're[br]going to demonstrate

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the node analysis method of analyzing[br]circuits in the Phasor Domain.

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You'll recall from[br]our previous discussions of

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node analysis that the first step[br]in analyzing circuits

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using the node voltage approach

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involves identifying[br]the critical or essential nodes.

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In this circuit we have got one,

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two, three, four essential nodes.

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Again, an essential node is a node where[br]three or more branches are connected.

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We choose one of these nodes[br]to then serve as

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our reference node where we by

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definition we're going to say[br]voltage is equal to zero.

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In this case, I've selected this[br]node here to be our reference node.

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So we'll set that voltage equal to zero.

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The other three nodes are

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the three essential nodes we will then[br]identify or assign variables to them.

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Variable names. This one[br]we're going to call V1.

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The voltage at this node will refer to as

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V2 and the voltage at this[br]node here will refer to as V3.

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Node voltage analysis then has this writing

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Kirchhoff's Current Law or node equations[br]at each of these essential nodes.

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Summing the currents leaving[br]each of those nodes in terms of

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the node voltage variables and[br]the circuit parameters in the circuit.

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So let's go ahead and do this.

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On this first node,

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let's write KCL at node one.

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The current leaving this node coming[br]down in this direction here will be

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V1 minus the voltage at this node[br]minus the voltage at this point.

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The voltage here by definition is zero,

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so V1 minus 0. We won't even bother.

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Well, start let's write[br]it V1 minus 0 divided by

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the impedance there which is 3 minus J4.

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Plus the current leaving this[br]node going in this direction,

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which is V1 minus V3 divided by

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the impedance here of two ohms plus

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the current leaving this node[br]going in this direction here.

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We need to be a little bit careful here.

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It's going to be the current[br]leaving this node going in

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that direction will be[br]the current through that resistor.

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The current through that resistor[br]is equal to the voltage drop

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across that resistor divided[br]by the resistance of two ohms.

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The voltage drop across that resistor[br]is the voltage on this side which is V1

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minus the voltage on

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this side of the resistor and[br]here's where we need to be careful.

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The voltage here in terms[br]of node voltages can be

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thought of as being starting here at V2.

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V2 we drop going from here to here[br]crossing this voltage source of 12 volts.

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We drop 12 volts..

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So the voltage at this point then[br]is going to be V2 minus 12 volts.

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So then the current leaving[br]this branch going in

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that direction will be the voltage of

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the left-hand side of that resistor which

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is V1 minus the voltage[br]of the right-hand side of

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the resistor which is V2 minus 12.

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We throw in the parentheses there[br]to remind ourselves that we're

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subtracting the voltage[br]there which is V2 minus 12.

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Finally, the current in that branch then[br]will be that voltage drop divided by

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two and that equals zero.

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Now, KCL at node two.

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The current leaving this node[br]going to the left, again,

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is going to be the current through[br]this resistor only

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this time going from right to

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left and that voltage[br]will be the voltage here,

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which we've already established[br]to be V2 minus 12 volts minus

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the voltage over here which is V1[br]minus V1 divided by the two ohms.

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The current coming down to the[br]node or leaving this node coming

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down is going to be then[br]added in and that's

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Simply V2 divided by the impedance[br]of two plus one J or simply J.

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Then the current leaving node to[br]going to the right will be equal

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to V2 minus V3 divided by two ohms.

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The sum of those three currents equal zero.

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Now, at the third node we've[br]got KCL at node three.

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The current leaving node three and[br]come into the left is going to be

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V3 minus V2 divided by two ohms.

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Plus the current leaving this node going[br]up and over to the left is going to

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be V3 minus V1 divided by two ohms.

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Finally, the current leaving node[br]three coming down this

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way is going to be the voltage drop[br]across that 2-ohm resistor,

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which will be V3 minus[br]the voltage right there.

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Now, the voltage at that point[br]is equal to starting here

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the reference node zero minus to plus

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we go up a value of negative J6 volts.

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So the voltage right here then is

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zero minus J6 and then[br]the current through here will be V3

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minus a minus J6

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divided by the two ohms and the sum[br]of those three currents equal zero.

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Those are the three note note equations.

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All that's left now is to[br]calculate the node voltages V1,

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V2 and V3 is a little bit of algebra.

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Let's start by combining like terms.

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Let's do that on up here.

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The V1 terms in the first equation.

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I've got V1 over.

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Let's see 3 minus J4.

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I've got V1 over two and I've[br]got another V1 over two.

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So plus one over two plus one[br]over two gives us the V1 terms.

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Plus the V2 terms here we've

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got only one right here and[br]that's a minus V2 over two.

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So if we pull the V2 out we'll have

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a minus one half there and[br]then our V3 terms there.

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Again, is only one of them.

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It's a minus V3 over two.

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So we have V3 times a negative one half.

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That then will equal.

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Let's see, we've got a 12 volts here.

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12 divided by two is six.

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A minus times a minus is a plus.

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I've got plus six on the left-hand side.

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We subtract six from

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both sides to bring our constant[br]over there on the right-hand side.

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Now, combining like terms[br]in the second equation,

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we have V1 times.

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This one we've only got one term that's[br]going to be a negative V1 over two.

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So we pull the V out or[br]the V1 out or left with

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the negative one half plus V2 times.

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I've got V2 over two,

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V2 over two plus J and V2 over two.

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So V2 or let's see V2 would be one half

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plus one over two plus J plus another[br]one half and then the V3 terms,

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again, there's only one and that[br]will be a negative one half.

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Now, here we've got a minus 12 over two.

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So I've got a negative six volts[br]on the left-hand side.

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We add six volts to plus[br]to our both sides and that

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gives us then equaling positive six volts[br]on the right-hand side.

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Now, equation three, we have V1[br]times a negative one half plus

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V2 times got a negative one half again

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plus V3 times I've got one half[br]plus one half plus one half.

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That's three halves, V3.

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Now, here we've got a constant[br]of negative times negative is

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a positive J6 over two.

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So on the left-hand side[br]we have a positive J3,

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which we subtract J3[br]from both sides gives us

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then a negative J3 over here on this side.

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All right. Moving ahead now[br]just cleaning things up

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so that we're ready to put[br]it into a matrix solver.

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Let's combine all those fractions[br]here and when we do

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that we've got V1 times 1.12

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plus 0.16J plus V2 times a negative

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0.5 plus V3 times a

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negative 0.5 equals negative six volts.

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The second equation we have V1[br]times a negative 0.5 plus V2 times

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1.4 minus 0.2J plus

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V3 times a negative 0.5[br]equals a positive six.

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Finally, the third equation,

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V1 times a negative 0.5[br]plus V2 times a negative

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0.5 plus V3 times

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1.5 is equal to a negative J3.

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When you plug this into[br]a matrix solver of some sort,

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you come up with that V1 is equal[br]to in rectangular coordinates,

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a negative 4.72 minus 0.88J,

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which in polar coordinates[br]is 4.8 angle negative 169.5.

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V2 is equal to 2.46 minus 0.87J,

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which in polar coordinates is 2.6[br]angle negative 19.84 and V3 we

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find to be equal to negative 0.76 minus

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J2.59 in rectangular coordinates in polar

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coordinates it's 2.7 angle negative 106.29.

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Of course, are all degrees.

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Now, that we have the voltages[br]at each of the nodes we can

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calculate any current or voltage[br]that we might want to calculate.

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For example, let's determine

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the current leaving node two[br]coming down this branch here.

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Well, it's called in this circuit[br]here which has been identified

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as I5 or I sub L. So, I5 then,

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is equal to V2 divided by two plus J,

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which is equal to V2 is 2.6

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angle negative 19.84 divided by two plus J.

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We find then if that is equal[br]to in polar coordinates 1.1.

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six angle negative 46.4 degrees.