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100:00:00 --> 00:00:01Good morning,OK.

200:00:01 --> 00:00:05Let's get started.We have one handout today.

300:00:05 --> 00:00:10That's your lecture notes.There's some copies still

400:00:10 --> 00:00:14outside for those who haven'tpicked one up.

500:00:14 --> 00:00:19In general, what I do is,in the lecture notes,

600:00:19 --> 00:00:22I leave out large amounts ofmaterial.

700:00:22 --> 00:00:28So, this will enable you tokeep your hands busy while I'm

800:00:28 --> 00:00:34

lecturing and take down somenotes and so on.

900:00:34 --> 00:00:41So, don't assume thateverything that I talk about is

1000:00:41 --> 00:00:44on here.Please follow along.

1100:00:44 --> 00:00:52OK, so as is my usual practice,let me start with a quick

1200:00:52 --> 00:00:58review of what we covered sofar.

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1300:00:58 --> 00:01:03So what we did primarily waslooked at this discipline that

1400:01:03 --> 00:01:07we call the lump matterdiscipline, which was very

1500:01:07 --> 00:01:13similar, very reminiscent of thepoint mass simplification in

1600:01:13 --> 00:01:15physics.And this discipline,

1700:01:15 --> 00:01:20this set of constraints weimposed on ourselves,

1800:01:20 --> 00:01:25allowed us to move fromMaxwell's equations to a very,

1900:01:25 --> 00:01:30very simple form of algebraicequations.

2000:01:30 --> 00:01:35

And specifically,the discipline took two forms.

2100:01:35 --> 00:01:42One is, we said that we willdeal with elements for whom the

2200:01:42 --> 00:01:50rate of change of magnetic fluxis zero outside of the elements,

2300:01:50 --> 00:01:57and for whom the rate of changeof charge I want to charge

2400:01:57 --> 00:02:03inside the element was zero.So, if I took any element,

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2500:02:03 --> 00:02:07any element that I called alump circuit element,

2600:02:07 --> 00:02:11like a resistor or a voltagesource, and I put a black box

2700:02:11 --> 00:02:15around it, then what I'm sayingis that the net charge inside

2800:02:15 --> 00:02:18that is going to be zero.And this is not true in

2900:02:18 --> 00:02:20general.We will see examples where,

3000:02:20 --> 00:02:24if you choose some piece of anelement for example,

3100:02:24 --> 00:02:27there might be charge buildup,but net inside the,

3200:02:27 --> 00:02:30

if I put a box around theentire element,

3300:02:30 --> 00:02:34I am going to assume that therate of change of charge is

3400:02:34 --> 00:02:39going to be zero.So, what this did was it

3500:02:39 --> 00:02:44enabled us to create the lumpcircuit abstraction,

3600:02:44 --> 00:02:49where I could take elements,some element of the sort,

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3700:02:49 --> 00:02:54this could be a resistor,a voltage source,

3800:02:54 --> 00:02:57or whatever,and I could now ascribe a

3900:02:57 --> 00:03:04voltage, some voltage across anelement, and also some current,

4000:03:04 --> 00:03:09"i," that was going into theelement.

4100:03:09 --> 00:03:12And as I go forward,when I label the voltages and

4200:03:12 --> 00:03:16currents across and throughelements, I'm going to be

4300:03:16 --> 00:03:21following a convention.OK, the convention is that I'm

4400:03:21 --> 00:03:24

going to label,if I label V in the following

4500:03:24 --> 00:03:28manner, then I'm going to label"i" for that element as a

4600:03:28 --> 00:03:33current flowing into thepositive terminal.

4700:03:33 --> 00:03:36It's just a convention.By doing this,

4800:03:36 --> 00:03:40it turns out that the powerconsumed by the element is "vi"

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4900:03:40 --> 00:03:44is positive.OK, so by choosing I going in

5000:03:44 --> 00:03:48this way into the positiveterminal, the power consumed by

5100:03:48 --> 00:03:51the element is going to bepositive.

5200:03:51 --> 00:03:55OK, so in general of evensimply following this

5300:03:55 --> 00:03:59convention, when I labelvoltages and currents,

5400:03:59 --> 00:04:03I'll be labeling the currentinto an element entering in

5500:04:03 --> 00:04:08through the plus terminal.Remember, of course,

5600:04:08 --> 00:04:12

if the current is going thisway, let's have one amp of

5700:04:12 --> 00:04:15current flowing this way,then when I compute the

5800:04:15 --> 00:04:18current, "i" will come out to benegative.

5900:04:18 --> 00:04:22OK, so by making theseassumptions, the assumptions of

6000:04:22 --> 00:04:26the lumped matter discipline,I said I was able to simplify

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6100:04:26 --> 00:04:31my life tremendously.And, in particular what it did

6200:04:31 --> 00:04:36was it allowed me to takeMaxwell's equations,

6300:04:36 --> 00:04:41OK, and simplify them into avery simple algebraic form,

6400:04:41 --> 00:04:46which has both a voltage lawand a current law that I call

6500:04:46 --> 00:04:51Kirchhoff's voltage law,and Kirchhoff's current law.

6600:04:51 --> 00:04:55KVL simply states that if Ihave some circuit,

6700:04:55 --> 00:05:01and if I measured the voltagesin any loop in the circuit,

6800:05:01 --> 00:05:07

so if I look at the voltages inany loop, then the voltages in

6900:05:07 --> 00:05:13the loop would sum to zero.OK, so I measure voltages in

7000:05:13 --> 00:05:16the loop, and they will sum tozero.

7100:05:16 --> 00:05:21Similarly, for the current,if I take a node of a circuit,

7200:05:21 --> 00:05:25if I build the circuit,a node is a point in the

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7300:05:25 --> 00:05:29circuit where multiple edgesconnect.

7400:05:29 --> 00:05:32If I take a node,then the current coming into

7500:05:32 --> 00:05:37that node, the net currentcoming into a node is going to

7600:05:37 --> 00:05:40be zero.OK, so if I take any node of

7700:05:40 --> 00:05:45the circuit and sum up all thecurrents going into that node,

7800:05:45 --> 00:05:51they will all net sum to zero.So, notice what I've done is by

7900:05:51 --> 00:05:55this discipline,by this constraint I imposed on

8000:05:55 --> 00:06:00

myself, I was able to make thisincredible leap from Maxwell's

8100:06:00 --> 00:06:04equations to these really,really simple algebraic

8200:06:04 --> 00:06:09equations, KVL and KCL.And I promise you,

8300:06:09 --> 00:06:13going forward to the rest of6.002, if this is all you know,

8400:06:13 --> 00:06:18you can pretty much solve anycircuit using these two very

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8500:06:18 --> 00:06:20simple relations.It's actually really,

8600:06:20 --> 00:06:24really simple.It's all very simple algebra,

8700:06:24 --> 00:06:26OK?So, just to show you an

8800:06:26 --> 00:06:29example, let me do a littledemonstration.

8900:06:29 --> 00:06:33Let me build let me build asmall circuit and measure some

9000:06:33 --> 00:06:37voltages for you,and show you that the voltages,

9100:06:37 --> 00:06:42indeed, add up to zero.So, here's my little circuit.

9200:06:42 --> 00:07:25

9300:07:25 --> 00:07:29So, I'm going to show you asimple circuit that looks like

9400:07:29 --> 00:07:33this, and let's go ahead andmeasure some voltages and

95

00:07:33 --> 00:07:35currents.In terms of terminology to

9600:07:35 --> 00:07:40remember, this is called a loop.So if I start from the point C

97

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00:07:40 --> 00:07:45and I travel through the voltagesource, come to the node A down

9800:07:45 --> 00:07:49through R1 and all the way downthrough R2 back to C,

9900:07:49 --> 00:07:52that's a loop.Similarly, this point A is a

10000:07:52 --> 00:07:55node where resistor R1 thevoltage source V0,

10100:07:55 --> 00:07:58and R4 are connected.OK, just make sure your

10200:07:58 --> 00:08:04terminology is correct.So, what I'll do is I'll make

10300:08:04 --> 00:08:10some quick measurements for you,and show you that these KVL and

10400:08:10 --> 00:08:14KCL are indeed true.

So, the circuits up there,

10500:08:14 --> 00:08:18could I have a volunteer?Any volunteer?

10600:08:18 --> 00:08:22All you have to do is writethings on the board.

107

00:08:22 --> 00:08:26Come on over.OK, so let me take some

10800:08:26 --> 00:08:30measurements,and why don't you write down

109

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00:08:30 --> 00:08:35what I measure on the board?What I'll do is,

11000:08:35 --> 00:08:39let me borrow another piece ofchalk here.

11100:08:39 --> 00:08:45What I'll do is focus on thisloop here, and focus on this

11200:08:45 --> 00:08:49node and make some measurements.

11300:08:49 --> 00:09:00

11400:09:00 --> 00:09:05

All right, so you see thecircuit up there.

11500:09:05 --> 00:09:11OK, so I get 3 volts for thevoltage from C to A.

11600:09:11 --> 00:09:16so why don't you write down 3volts?

11700:09:16 --> 00:09:30

11800:09:30 --> 00:09:44OK, so the next one is -1.6.And so that will be,

11900:09:44 --> 00:09:50I'm doing AB,V_AB.

12000:09:50 --> 00:10:01OK, and then let me do the lastone.

12100:10:01 --> 00:10:08It is -1.37.The measurements,

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12200:10:08 --> 00:10:14I guess, have been this way.So, what's written is V_AC.

12300:10:14 --> 00:10:18But it's OK for now.Don't worry about it.

12400:10:18 --> 00:10:23So, well, thank you.I appreciate your help here.

12500:10:23 --> 00:10:27OK, so within the bonds ofexperimental error,

12600:10:27 --> 00:10:32noticed that if I add up these

three voltages,

12700:10:32 --> 00:10:38they nicely sum up to zero.OK, next let me focus on this

12800:10:38 --> 00:10:41node here.And at this node,

129

00:10:41 --> 00:10:44let me go ahead and measuresome currents.

13000:10:44 --> 00:10:50What I'll do now is change toan AC voltage so that I can go

13100:10:50 --> 00:10:55ahead and measure the currentwithout breaking my circuit.

13200:10:55 --> 00:10:59OK, this time around,you'll get to see the

13300:10:59 --> 00:11:04measurements that I'm taking aswell.

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13400:11:04 --> 00:11:09So, what I have here,I guess you can see it this

13500:11:09 --> 00:11:14way.What I have here is three wires

13600:11:14 --> 00:11:20that I have pulled out from D.And this is the node D,

13700:11:20 --> 00:11:24OK?So, I have three wires coming

13800:11:24 --> 00:11:32into the node D just to make it

a little bit easier for me to

13900:11:32 --> 00:11:37measure stuff.OK, so everybody keep your

14000:11:37 --> 00:11:42fingers crossed so I don't looklike a fool here.

141

00:11:42 --> 00:11:46I hope this works out.So, you roughly get,

14200:11:46 --> 00:11:48what's that,10 mV.

14300:11:48 --> 00:11:53OK, so it's about 10 mV peak topeak out there,

14400:11:53 --> 00:11:59and let's say that if thewaveform raises on the left-hand

14500:11:59 --> 00:12:05side, it's positive.So, it's positive 10 mV.

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14600:12:05 --> 00:12:08And another positive 10 mV,so that's 20 mV.

14700:12:08 --> 00:12:11And this time,it's a negative,

14800:12:11 --> 00:12:13roughly 20, I guess,-20.

14900:12:13 --> 00:12:17So, I'm getting,in terms of currents,

15000:12:17 --> 00:12:19I have a -10,

-10, I'm sorry,

15100:12:19 --> 00:12:22positive 10,positive 10,

15200:12:22 --> 00:12:27and a -20 that adds up to zero.But more interestingly,

153

00:12:27 --> 00:12:31I can show you the same thingby holding this current

15400:12:31 --> 00:12:37measuring probe directly acrossthe node.

15500:12:37 --> 00:12:43And, notice that the netcurrent that is entering into

15600:12:43 --> 00:12:49this node here is zero.OK, so that should just show

15700:12:49 --> 00:12:56you that KCL does indeed hold inpractice, and it is not just a

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15800:12:56 --> 00:13:02figment of our imaginations.So, before I go on,

15900:13:02 --> 00:13:05I wanted to point one otherthing out.

16000:13:05 --> 00:13:09Notice that I've written downtwo assumptions of the lumped

16100:13:09 --> 00:13:11matter discipline,OK?

16200:13:11 --> 00:13:16There is a total assumption of

the lump matter discipline,

16300:13:16 --> 00:13:19and that assumption is,in spirit, at least,

16400:13:19 --> 00:13:23shared by the point masssimplification in physics as

165

00:13:23 --> 00:13:26well.Can someone tell me what that

16600:13:26 --> 00:13:28assumption is?A total assumption,

16700:13:28 --> 00:13:32which I did not mention,which you can read in your

16800:13:32 --> 00:13:36notes in section 8.2 in theappendix, what's a total

16900:13:36 --> 00:13:41assumption that is shared inspirit with the point mass

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17000:13:41 --> 00:13:44simplification?Anybody?

17100:13:44 --> 00:13:49A total assumption to be madehere is that in all the signals

17200:13:49 --> 00:13:53that we will study in thiscourse, we've made the

17300:13:53 --> 00:13:57assumption that the signalspeeds of interest,

17400:13:57 --> 00:14:01transition speeds,

and so on, are much slower than

17500:14:01 --> 00:14:05the speed of light.OK, that my signal transition

17600:14:05 --> 00:14:11speeds of interest are muchslower than the speed of light.

177

00:14:11 --> 00:14:14Remember, the laws of motion,the discrete laws of motion

17800:14:14 --> 00:14:18break down if your objects beginmoving at the speed of light.

17900:14:18 --> 00:14:22OK, the same token here,our lump circuit abstraction

18000:14:22 --> 00:14:24breaks down if we approach thespeed of light.

18100:14:24 --> 00:14:28And there are follow on coursesthat talk about waveguides and

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18200:14:28 --> 00:14:32other distributed analysistechniques that deal with

18300:14:32 --> 00:14:36signals that travel close tospeeds of light.

18400:14:36 --> 00:14:41OK, so with that,let me go on to talking about

18500:14:41 --> 00:14:48method one of circuit analysis.This is called the basic KVL

18600:14:48 --> 00:14:53KCL method.

So just based on those two

18700:14:53 --> 00:15:00simple algebraic relations,I can analyze very interesting

18800:15:00 --> 00:15:05and complicated circuits.The method goes as follows.

189

00:15:05 --> 00:15:09So, let's say our goal is,given a circuit like this,

19000:15:09 --> 00:15:12our goal is to solve it.OK, in this course,

19100:15:12 --> 00:15:15we will do two kinds of things:analysis and synthesis.

19200:15:15 --> 00:15:17Analysis says,given a circuit,

19300:15:17 --> 00:15:20OK, what can you tell me aboutthe circuit?

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19400:15:20 --> 00:15:24OK, so we'll solve existingcircuits for all the voltages

19500:15:24 --> 00:15:26and currents,voltages across elements,

19600:15:26 --> 00:15:30and currents through thoseelements.

19700:15:30 --> 00:15:32Synthesis says,given a function,

19800:15:32 --> 00:15:35I may ask you to go and build

circuits.

19900:15:35 --> 00:15:40OK, so for analysis here,we can apply this method that I

20000:15:40 --> 00:15:44want to show you.And the idea here is that,

201

00:15:44 --> 00:15:48given a circuit like this,let us figure out all the

20200:15:48 --> 00:15:53voltages and currents that are afunction of the way these

20300:15:53 --> 00:15:57elements are connected.So, the basic KVL and KCL

20400:15:57 --> 00:16:02method has the following steps.The first step is to write down

20500:16:02 --> 00:16:08the element VI relationships.OK, right down the element VI

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20600:16:08 --> 00:16:10relationships for all theelements.

20700:16:10 --> 00:16:14The second step is write KCLfor all the nodes,

20800:16:14 --> 00:16:19and the third step is to writeKVL for all the loops in the

20900:16:19 --> 00:16:20circuit.That's it.

21000:16:20 --> 00:16:23Just go ahead and write down

element rules,

21100:16:23 --> 00:16:27KVL, and KCL,and then go ahead and solve the

21200:16:27 --> 00:16:29circuit.So, what we'll do,

213

00:16:29 --> 00:16:33we'll do an example,of course.

21400:16:33 --> 00:16:39But, just as a refresher,we've looked at a bunch of

21500:16:39 --> 00:16:43elements so far,and for the resistor,

21600:16:43 --> 00:16:50the element relation says thatV is pi R, where R is the

21700:16:50 --> 00:16:56resistance of the element here.For a voltage source,

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21800:16:56 --> 00:17:01V is equal to V nought.That's the element

21900:17:01 --> 00:17:06relationship.And for a current source,

22000:17:06 --> 00:17:12the element is the relation is,"i" is simply the current

22100:17:12 --> 00:17:19flowing through the element.OK, so these are some of the

22200:17:19 --> 00:17:25simple element rules for the

devices that the current source,

22300:17:25 --> 00:17:30voltage source,and the resistor.

22400:17:30 --> 00:17:34So let's go ahead and solvethis simple circuit.

225

00:17:34 --> 00:17:40And what I'll do is go aheadand solve the circuit for you.

22600:17:40 --> 00:17:46OK, if you turn to page five ofyour notes, I'm going to go

22700:17:46 --> 00:17:53ahead and edit the circuit here.You can scribble the values on

22800:17:53 --> 00:17:58your notes on page five.OK, so as a first step of my

22900:17:58 --> 00:18:03KVL KCL method,I need to write down all my

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23000:18:03 --> 00:18:08element VI relationships.So, before I do that,

23100:18:08 --> 00:18:13let me go ahead and label allthe voltages and currents that

23200:18:13 --> 00:18:17are unknowns in the circuit.So, let me label the voltages

23300:18:17 --> 00:18:22and currents associated with thevoltage source as here.

23400:18:22 --> 00:18:26Notice, I continue to follow

this convention where whenever I

23500:18:26 --> 00:18:31label voltages and currents foran element, I will show the

23600:18:31 --> 00:18:35current going into the positiveterminal of the element

237

00:18:35 --> 00:18:40variable, OK,after element variable voltage.

23800:18:40 --> 00:18:42So here, I have V nought and Inought.

23900:18:42 --> 00:18:47Let me pause here for fiveseconds and show you a point of

24000:18:47 --> 00:18:49confusion that happenssometimes.

24100:18:49 --> 00:18:53Often times,people confuse between what is

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24200:18:53 --> 00:18:57called the variable that isassociated with the element

24300:18:57 --> 00:19:01versus the element value.OK, notice that here,

24400:19:01 --> 00:19:07capital V nought is the voltagethat this voltage source

24500:19:07 --> 00:19:12provides, while this name here,v nought, is simply a variable

24600:19:12 --> 00:19:17that we've used to label the

voltage across that element.

24700:19:17 --> 00:19:21So, similarly,I can label v1 as the voltage

24800:19:21 --> 00:19:26across the resistor,and i1 is the current flowing

249

00:19:26 --> 00:19:30through the resistor.So this method of labeling,

25000:19:30 --> 00:19:36where I follow the convention,that the current flows into the

25100:19:36 --> 00:19:44positive terminal is called theassociated variables discipline.

25200:19:44 --> 00:19:48I was trying to use the worddiscipline in situations where

25300:19:48 --> 00:19:50you have a choice,OK, and of a variety of

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25400:19:50 --> 00:19:54possible choices,you pick one as the convention.

25500:19:54 --> 00:19:56OK, so here,as a convention,

25600:19:56 --> 00:20:00we use the associated variablesdiscipline, and use that method

25700:20:00 --> 00:20:04to consistently label theunknown voltages and currents in

25800:20:04 --> 00:20:09our circuits.

OK, so let me continue the

25900:20:09 --> 00:20:13labeling here,v4, i4, i3, v3 here,

26000:20:13 --> 00:20:16and v2 and i2,v5, and i5.

261

00:20:16 --> 00:20:22I think that's it.So, I've gone ahead and labeled

26200:20:22 --> 00:20:27all my unknowns.So each of these voltages and

26300:20:27 --> 00:20:35currents are the voltages andcurrents associated with each of

26400:20:35 --> 00:20:40the elements.And my goal is to solve for

26500:20:40 --> 00:20:44these.OK, so in terms of our solution

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26600:20:44 --> 00:20:49here, let's follow the methodthat I outlined for you.

26700:20:49 --> 00:20:54So, as the first step I amsimply going to go ahead and

26800:20:54 --> 00:21:00write down all the element VIrelationships.

26900:21:00 --> 00:21:05OK, so as a first step,I'm going to go ahead and write

27000:21:05 --> 00:21:12down all the VI relationships.

So, can someone yell out for me

27100:21:12 --> 00:21:16the VI relationship for thevoltage source?

27200:21:16 --> 00:21:20OK, good.So, v0 is capital V nought,

273

00:21:20 --> 00:21:25that is that the variable Vnought is simply equal to the

27400:21:25 --> 00:21:29voltage, v0.Similarly, I can write the

27500:21:29 --> 00:21:32others.v1 is i1, R1.

27600:21:32 --> 00:21:36v2 is i2, R2,and so on.

27700:21:36 --> 00:21:39OK, and I have one,two, three, four,

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27800:21:39 --> 00:21:44five, six elements.So, I will get six such

27900:21:44 --> 00:21:48equations.Step two, I'm going to go ahead

28000:21:48 --> 00:21:52and write KCL for the nodes inmy system.

28100:21:52 --> 00:21:57So, let me start with node A.So, for node A,

28200:21:57 --> 00:22:05let me take as positive the

currents going out of the node.

28300:22:05 --> 00:22:10So, I get i nought flowing out,plus i1 flowing out,

28400:22:10 --> 00:22:16plus i4 flowing out,and they must sum to zero for

285

00:22:16 --> 00:22:20node A.Then, I can go ahead and do the

28600:22:20 --> 00:22:24other nodes, let's say,for example,

28700:22:24 --> 00:22:28I do node B.For node B, I have i2 going

28800:22:28 --> 00:22:31out.That's positive,

28900:22:31 --> 00:22:36i3, and i1 is coming in,so I get -i1 equals zero.

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29000:22:36 --> 00:22:39OK, so I have one,two, three, four,

29100:22:39 --> 00:22:44I have four nodes.OK, so I would get four

29200:22:44 --> 00:22:47equations.It turns out that the fourth

29300:22:47 --> 00:22:53equation is not independent.You can derive it from the

29400:22:53 --> 00:22:56others.

So, I get three independent

29500:22:56 --> 00:23:01equations out of this.I can then write KVL.

29600:23:01 --> 00:23:05And for KVL,I just go down my loops here.

297

00:23:05 --> 00:23:10And let me go through thisfirst loop here in this manner.

29800:23:10 --> 00:23:15OK, and a simple trick that Iuse, you have to be incredibly

29900:23:15 --> 00:23:20careful when you go through thisin keeping your minuses and

30000:23:20 --> 00:23:23pluses correct.Otherwise you can get

30100:23:23 --> 00:23:26hopelessly muddled.Once you label it,

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30200:23:26 --> 00:23:32you need to be sure that youget all your minuses and pluses

30300:23:32 --> 00:23:34correct.So, for KVL,

30400:23:34 --> 00:23:38what I'd like to do is,let's say I start at C,

30500:23:38 --> 00:23:40and from C I'm going to go toA.

30600:23:40 --> 00:23:44For A I go to B,

and from B I'm going to come

30700:23:44 --> 00:23:46back to C.OK, that's how I traverse my

30800:23:46 --> 00:23:49loop.And, the trick that I'm going

309

00:23:49 --> 00:23:52to follow is,as my finger walks through that

31000:23:52 --> 00:23:57loop, I'm going to label thevoltage as the first sign that I

31100:23:57 --> 00:24:02see for that voltage.OK, so I'm going to start with

31200:24:02 --> 00:24:05C, and I go up.I start by punching into the

31300:24:05 --> 00:24:08voltage source element,and then punch into it,

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31400:24:08 --> 00:24:10I hit the minus sign for the Vnought.

31500:24:10 --> 00:24:14OK, so I'm just going to writedown minus V nought,

31600:24:14 --> 00:24:18plus then I go through and as Icome up to A and go down to B,

31700:24:18 --> 00:24:21I punch to the plus sign of theV1.

31800:24:21 --> 00:24:24So, that's plus V1.

And then I punch into the plus

31900:24:24 --> 00:24:26sign of the V2,and so I get plus V2,

32000:24:26 --> 00:24:30and that is zero.OK, good.

321

00:24:30 --> 00:24:33So, that matches what you havein your notes as well.

32200:24:33 --> 00:24:36So, this is the first equation.Similarly, I can go through my

32300:24:36 --> 00:24:40other loops and write downequations for each of the loops.

32400:24:40 --> 00:24:43OK, and the convention that Ilike to follow is as I go

32500:24:43 --> 00:24:45through the loop,I write down as a sign for the

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32600:24:45 --> 00:24:49voltage the first sign that Icounter for that element.

32700:24:49 --> 00:24:51OK, you can do the exactopposite, if you want,

32800:24:51 --> 00:24:54just to be different.But, as long as you stay

32900:24:54 --> 00:24:57consistent, you'll be OK.All right, so in the same

33000:24:57 --> 00:25:00manner here, there are four

loops that I can have,

33100:25:00 --> 00:25:03so four equations.Again, one of them turns out to

33200:25:03 --> 00:25:08be dependent on the others.So I end up getting three

333

00:25:08 --> 00:25:12independent equations.So, I get a total of 12

33400:25:12 --> 00:25:15equations.I get 12 equations.

33500:25:15 --> 00:25:20There are six elements,OK, voltage source,

33600:25:20 --> 00:25:24and five resistors.So, there are six unknown

33700:25:24 --> 00:25:27voltages, and six unknowncurrents.

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33800:25:27 --> 00:25:33So, I have 12 equations,and 12 unknowns.

33900:25:33 --> 00:25:38OK, I can take all of theequations and put them through a

34000:25:38 --> 00:25:42big crank, and sit there andgrind.

34100:25:42 --> 00:25:47And if I was really cruel,I'd give this as a homework

34200:25:47 --> 00:25:53problem, and have you grind,

and grind, and grind until you

34300:25:53 --> 00:25:57get your six voltages and sixcurrents.

34400:25:57 --> 00:26:01OK, it works.OK, so you get 12 equations,

345

00:26:01 --> 00:26:07and this method just works.However, notice that this is

34600:26:07 --> 00:26:10quite a grubby method.It's quite grungy.

34700:26:10 --> 00:26:14I get 12 equations,and it's quite a pain even for

34800:26:14 --> 00:26:18a simple circuit like this.However, suffice it to say that

34900:26:18 --> 00:26:22this fundamental method is onestep away from Maxwell's

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35000:26:22 --> 00:26:25equations, simply works.OK?

35100:26:25 --> 00:26:28So what you'll do is the restof this lecture,

35200:26:28 --> 00:26:33I'll introduce you to a couplemore methods.

35300:26:33 --> 00:26:40One is an intuitive method,and another one called the node

35400:26:40 --> 00:26:46method is a little bit more

formal, but is much more,

35500:26:46 --> 00:26:51I guess, terse Than the KVL KCLmethod.

35600:26:51 --> 00:26:56Method 2.So the relevant section to read

357

00:26:56 --> 2.4.in the course notes is section

3582.4. --> 00:27:02

35900:27:02 --> 00:27:06One of the things that I willbe stressing this semester is

360

00:27:06 --> 00:27:08intuition.What you'll find is that as you

36100:27:08 --> 00:27:11become EECS majors,and so on, and go on,

362

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00:27:11 --> 00:27:15or if you talk to your TAs oryour professors and so on,

36300:27:15 --> 00:27:19you will find that very rarelydo they actually go ahead and

36400:27:19 --> 00:27:21apply the formal methods ofanalysis.

36500:27:21 --> 00:27:25OK, by and large,engineers are able to look at a

36600:27:25 --> 00:27:28circuit and simply byobservation write down an

36700:27:28 --> 00:27:31answer.And usually in the past,

36800:27:31 --> 00:27:35what we have tried to do iskind of ignore that process and

36900:27:35 --> 00:27:37told our students,

look, we teach you all the

37000:27:37 --> 00:27:40formal methods,and you will develop your own

37100:27:40 --> 00:27:44intuition and be able to do it.What we'll try to do this term

372

00:27:44 --> 00:27:47is try to stress the intuitivemethods, and try to show you how

37300:27:47 --> 00:27:51the intuitive process goes,so you can very quickly solve

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00:27:51 --> 00:27:54many of these circuits simply byinspection.

37500:27:54 --> 00:27:57OK, so this method that I'mgoing to show you here is one

37600:27:57 --> 00:28:02such an intuitive method.And I'll call it element

37700:28:02 --> 00:28:08combination tools.OK, for many simple circuits,

37800:28:08 --> 00:28:14you can solve them very quicklyby applying this method.

37900:28:14 --> 00:28:18The components of this methodare these.

38000:28:18 --> 00:28:24I learned about how to composea bunch of elements.

38100:28:24 --> 00:28:27So, let's say,

for example,

38200:28:27 --> 00:28:32I have a set of resistors,R1 through RN,

38300:28:32 --> 00:28:38in series.OK, you can use KVL and KCL to

384

00:28:38 --> 00:28:44show that this is equivalent toa single resistor whose value is

38500:28:44 --> 00:28:48given by the sum of theresistances.

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00:28:48 --> 00:28:54OK, so if I have resistors inseries, then effectively it's

38700:28:54 --> 00:28:59the same as if there was asingle resistor whose value is

38800:28:59 --> 00:29:07the sum of all the resistances.OK, you can look at the course

38900:29:07 --> 00:29:11notes for a proof for derivationof this fact.

39000:29:11 --> 00:29:16Similarly, if I haveresistances in parallel,

39100:29:16 --> 00:29:20so let me call themconductances.

39200:29:20 --> 00:29:24A conductance is the reciprocalof a resistance.

39300:29:24 --> 00:29:31If resistance is measured in

ohms, conductance is measured in

39400:29:31 --> 00:29:36mhos, M-H-O-S.OK, so that's the conductance

39500:29:36 --> 00:29:39is G1, G2, and G3.And effectively,

396

00:29:39 --> 00:29:44this is the same as having asingle conductance whose

39700:29:44 --> 00:29:49effective value is given by thesum of the conductances.

398

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00:29:49 --> 00:29:53OK, the conductances inparallel add,

39900:29:53 --> 00:30:00and resistances in series add.Similarly, for voltage sources,

40000:30:00 --> 00:30:06if I have voltage sources inseries, then they are tantamount

40100:30:06 --> 00:30:10to the sum of the voltages.And similarly,

40200:30:10 --> 00:30:14for currents,if I have currents in parallel,

40300:30:14 --> 00:30:19then they can be viewed as asingle current source,

40400:30:19 --> 00:30:24whose currents are the sum ofthe individual parallel

40500:30:24 --> 00:30:30currents.

So, let's do a quick example.

40600:30:30 --> 00:30:36So let's do this example.So, let's say I have a circuit

40700:30:36 --> 00:30:41that looks like this,and three resistances.

408

00:30:41 --> 00:30:47And let's say all I care aboutis the current,

40900:30:47 --> 00:30:50I, that flows through thiswire.

410

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00:30:50 --> 00:30:54All I care about is thatcurrent.

41100:30:54 --> 00:31:02Of course, you can go ahead andwrite KVL and KCL.

41200:31:02 --> 00:31:06You will get four equations,and there are four unknowns.

41300:31:06 --> 00:31:09And you can solve it.But, I can apply my element

41400:31:09 --> 00:31:12combination rules,and very quickly figure out

41500:31:12 --> 00:31:16what the current I is,using the following technique.

41600:31:16 --> 00:31:19So, what I can do is,I can, first of all,

41700:31:19 --> 00:31:23take this circuit.

And, I can compose these two

41800:31:23 --> 00:31:27resistances and show that thecircuit is equivalent as far as

41900:31:27 --> 00:31:30this current,I, is concerned to the

420

00:31:30 --> 00:31:33following circuit,R1.

42100:31:33 --> 00:31:39And I take the sum of the twoconductances,

422

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00:31:39 --> 00:31:46OK, and that comes out to beR1, R2, R3, R2 plus R3.

42300:31:46 --> 00:31:54And then, I can furthersimplify it, and I get a single

42400:31:54 --> 00:32:01resistance, whose value is givenby R1 plus R2,

42500:32:01 --> 00:32:06R3, R3.OK, I'm just simplifying the

42600:32:06 --> 00:32:08circuit.Now, from this circuit,

42700:32:08 --> 00:32:11I can get the answer that Ineed.

42800:32:11 --> 00:32:15I is simply the voltage,V, divided by R1 plus.

42900:32:15 --> 00:32:20OK, so in situations like this

where I'm looking for a single

43000:32:20 --> 00:32:25current, I can very quickly getto the answer by applying some

43100:32:25 --> 00:32:28of these element combinationrules.

432

00:32:28 --> 00:32:34And, I can get rid of having togo through formal steps.

43300:32:34 --> 00:32:36So, in general,whenever you encounter a

434

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00:32:36 --> 00:32:41circuit, by and large attempt touse intuitive methods to solve

43500:32:41 --> 00:32:43it.And go to a formal method only

43600:32:43 --> 00:32:47if some intuitive method fails.Even in your homework,

43700:32:47 --> 00:32:51by and large,the homeworks are not meant to

43800:32:51 --> 00:32:54be grungy.OK, if you find a lot of grunge

43900:32:54 --> 00:32:57in your homework,just remember you're probably

44000:32:57 --> 00:33:04not using some intuitive method.OK, so just be cautious.

44100:33:04 --> 00:33:11All right, so let me go on to

the third method of circuit

44200:33:11 --> 00:33:19analysis, and the third methodis called the node method.

44300:33:19 --> 00:33:28So, the node method is simply aspecific application of the KVL

444

00:33:28 --> 00:33:36KCL method and results in amuch, much more compact form of

44500:33:36 --> 00:33:42the final equations.If there's one method that you

446

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00:33:42 --> 00:33:46have to remember for life,then I would say just remember

44700:33:46 --> 00:33:48this method.OK, the node method is a

44800:33:48 --> 00:33:50workhorse of the easiestindustry.

44900:33:50 --> 00:33:55OK, if there's one method thatyou want to consistently apply,

45000:33:55 --> 00:33:58then this is the one toremember.

45100:33:58 --> 00:34:01So, let me quickly outline foryou to method,

45200:34:01 --> 00:34:04and then work out an examplefor you.

45300:34:04 --> 00:34:09The first step of the node

method will be to select a

45400:34:09 --> 00:34:14reference or a ground node.This is the symbol for a ground

45500:34:14 --> 00:34:16node.The ground node simply says

456

00:34:16 --> 00:34:21that I'm going to denotevoltages at that point to be

45700:34:21 --> 00:34:26zero, and measure all my othervoltages with reference to that

458

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00:34:26 --> 00:34:30point.So, I'm going to select a

45900:34:30 --> 00:34:36ground node in my circuit.Second, I want to label the

46000:34:36 --> 00:34:41remaining voltages with respectto the ground node.

46100:34:41 --> 00:34:48So, label voltages for all theother nodes with respect to the

46200:34:48 --> 00:34:52ground node.Next, write KCL for each of the

46300:34:52 --> 00:34:57nodes write KCL.OK, but don't write KCL for the

46400:34:57 --> 00:35:02ground node.Remember, if you have N nodes,

46500:35:02 --> 00:35:07the node equations will give

you N-1 independent equations.

46600:35:07 --> 00:35:12So, write KCL for the nodes,but don't do so for the ground

46700:35:12 --> 00:35:15node.Then, solve for the node

468

00:35:15 --> 00:35:18voltages.So, let's say when we label

46900:35:18 --> 00:35:21voltages.I want to be labeling them as E

470

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00:35:21 --> 00:35:26something or the other.So, solve for the unknown node

47100:35:26 --> 00:35:31voltages.And then, once I know all the

47200:35:31 --> 00:35:37voltages associated with thenodes, I can then back solve for

47300:35:37 --> 00:35:41all the branch voltages andcurrents.

47400:35:41 --> 00:35:47OK, once I know all the nodevoltages, I can then go ahead

47500:35:47 --> 00:35:52and figure out all the branchvoltages and the branch

47600:35:52 --> 00:35:56currents.So, let's go ahead and apply

47700:35:56 --> 00:36:02this method, and work out an

example.

47800:36:02 --> 00:36:05Again, remember,if there's one method that you

47900:36:05 --> 00:36:08should remember,it's the node method.

480

00:36:08 --> 00:36:11OK, and when in doubt,consistently apply the node

48100:36:11 --> 00:36:15method and it will work whetheryour circuit is linear or

482

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00:36:15 --> 00:36:20nonlinear, if the resistors arebuilt in the US or the USSR it

48300:36:20 --> 00:36:23doesn't matter.OK, the node method will simply

48400:36:23 --> 00:36:26work, linear or nonlinear,OK?

48500:36:26 --> 00:36:30So, what I'm going to do is I'mgoing to build a circuit that's

48600:36:30 --> 00:36:36my old faithful.It's our old faithful,

48700:36:36 --> 00:36:45plus I'll make it a little bitmore complicated by adding in

48800:36:45 --> 00:36:52the current source.So, let's go have some fun.

48900:36:52 --> 00:36:59Let's do this.

So here's my voltage source,

49000:36:59 --> 00:37:04as before.OK, what I'll do is for fun,

49100:37:04 --> 00:37:13add a current source out there.And, you can convince

492

00:37:13 --> 00:37:20yourselves that if you go aheadand apply the KVL KCL method,

49300:37:20 --> 00:37:25it'll really be a mess ofequations.

494

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00:37:25 --> 00:37:28OK, so R1, R3,R4, R2, R5.

49500:37:28 --> 00:37:36OK, so let's follow our methodand just plug and chug here.

49600:37:36 --> 00:37:39So let's apply the first step.I select a ground node.

49700:37:39 --> 00:37:42It's a reference node fromwhich I'll measure all my other

49800:37:42 --> 00:37:44voltages.OK, now without knowing

49900:37:44 --> 00:37:48anything about the node method,try to use intuition as to

50000:37:48 --> 00:37:51which node you should choose asa ground node.

50100:37:51 --> 00:37:55Remember, you want to label the

ground node with the voltage

50200:37:55 --> 00:37:58zero, and measure all the othervoltages with respect to that

50300:37:58 --> 00:38:02node.OK, a usual trick is to pick a

504

00:38:02 --> 00:38:07node which has the largestnumber of elements connected to

50500:38:07 --> 00:38:11it as the ground node.OK, and in particular,

506

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00:38:11 --> 00:38:16you will find out later it'suseful to pick a node in which

50700:38:16 --> 00:38:20all your voltage sources,the maximum number of your

50800:38:20 --> 00:38:23voltage sources are alsoconnected.

50900:38:23 --> 00:38:27OK, so in this instance,I'm going to choose this as my

51000:38:27 --> 00:38:32ground node.OK, that's my first step.

51100:38:32 --> 00:38:38I chose that as my ground node.And I'm going to label that as

51200:38:38 --> 00:38:41having a voltage zero.Second step,

51300:38:41 --> 00:38:47I'll label voltages of the

other branches with respect to

51400:38:47 --> 00:38:52the ground node.OK, so what I'll do is add this

51500:38:52 --> 00:38:55node here.So I'm going to label that

516

00:38:55 --> 00:39:00voltage E1.These are my unknowns.

51700:39:00 --> 00:39:05Remember, node method,because my node voltages are my

518

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00:39:05 --> 00:39:09unknowns.So, I label this as E1.

51900:39:09 --> 00:39:14I label this one as my unknownvoltage, E2.

52000:39:14 --> 00:39:19What about this one here?Is that voltage unknown?

52100:39:19 --> 00:39:23No.I know what the voltage is

52200:39:23 --> 00:39:28because I know that this node isat a voltage,

52300:39:28 --> 00:39:33V0, higher than the groundnode.

52400:39:33 --> 00:39:38OK, notice that to go from hereto here, I directly go through a

52500:39:38 --> 00:39:42voltage source.

And so, this node has voltage

52600:39:42 --> 00:39:45V0.And I'll simply write down V0.

52700:39:45 --> 00:39:50OK, try to simplify the numberof steps that you have to go

528

00:39:50 --> 00:39:55through, so directly go aheadand write down the voltage,

52900:39:55 --> 00:39:58V0, for that node.What I will also do,

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00:39:58 --> 00:40:02is for convenience,I'm going to write down,

53100:40:02 --> 00:40:08I'm going to use conductances.So I'm going to use GI in the

53200:40:08 --> 00:40:12place of one by RI,and write down a bunch of node

53300:40:12 --> 00:40:14equations.OK, so step one,

53400:40:14 --> 00:40:19I've chosen my ground node.Step two, I've labeled my node

53500:40:19 --> 00:40:23voltages, E, OK?I've done that with two of my

53600:40:23 --> 00:40:27steps.Now, let me go ahead and --

53700:40:27 --> 00:40:41

53800:40:41 --> 00:40:44OK, so let me go ahead andapply step three.

53900:40:44 --> 00:40:50And, step three says go aheadand apply KCL for each of the

54000:40:50 --> 00:40:54

nodes at which you have anunknown node voltage.

54100:40:54 --> 00:40:59And then that will give youyour equations.

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So let me start by applying KCLat E1.

54300:41:02 --> 00:41:06So, let me write KCL at E1.I do one more thing.

54400:41:06 --> 00:41:09Notice, I don't have anycurrents there.

54500:41:09 --> 00:41:14OK, so how do I write KCL?KCL simply says the sum of

54600:41:14 --> 00:41:18currents into a node is zeroagain, remember,

547

00:41:18 --> 00:41:23by the lump matter discipline.So, if I don't have currents in

54800:41:23 --> 00:41:28there, so the trick that I adoptis that to write KCL,

54900:41:28 --> 00:41:34I use the node voltages,and implicitly substitute for

55000:41:34 --> 00:41:37the node voltages,divide by the elemental the

55100:41:37 --> 00:41:41resistance, for instance,so I take the node voltages,

55200:41:41 --> 00:41:44

and divide by the resistance,get the current.

55300:41:44 --> 00:41:48OK, so I implicitly applyelement relationships to get the

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node currents.So, the example that make it

55500:41:51 --> 00:41:55clear, so I take node E1 and,again, for currents going out

55600:41:55 --> 00:42:00I'm going to assume to have,to be positive.

55700:42:00 --> 00:42:05So, the current going up is E1minus V nought,

55800:42:05 --> 00:42:09divide by R1,so I multiplied by the G1.

559

00:42:09 --> 00:42:16That's the current going up.Plus, the current going down is

56000:42:16 --> 00:42:22E1 minus zero where the groundnode potential is zero,

56100:42:22 --> 00:42:28G2, OK, plus the current thatis going to resistor R3,

56200:42:28 --> 00:42:35which is simply E1 minus E2,divide by R3.

56300:42:35 --> 00:42:37So, E1 minus E2,divide by R3,

56400:42:37 --> 00:42:40

or multiplied by G3 is equal tozero.

56500:42:40 --> 00:42:44OK, see how I got this?This is simply KCL,

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but to get my currents,I simply take the differences

56700:42:49 --> 00:42:54of voltages across elements,and divide by the element of

56800:42:54 --> 00:42:58resistance, and I get thecurrents.

56900:42:58 --> 00:43:01OK, so I can similarly writeKCL at E2.

57000:43:01 --> 00:43:06So, at KCL at E2,again, let me go outwards.

571

00:43:06 --> 00:43:14So, the current going up is E2minus V nought multiplied by G4.

57200:43:14 --> 00:43:22The current going left is E2minus E1 divided by R3 or

57300:43:22 --> 00:43:29multiplied by G3.The current going down is E2

57400:43:29 --> 00:43:37minus zero multiplied by G5.And, the current going down is

57500:43:37 --> 00:43:40-I1.OK, you've got to be careful

57600:43:40 --> 00:43:46

with your polarities here.So all the currents going out

57700:43:46 --> 00:43:50sum to zero.And here are the currents that

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are going out at this point.So what I do next is I can move

57900:43:56 --> 00:44:01the constant terms to theleft-hand side and collect my

58000:44:01 --> 00:44:07unknowns.So, let me write them out here.

58100:44:07 --> 00:44:14So, let's say I get E1 here,OK, and from this equation,

58200:44:14 --> 00:44:20I have a V nought,G1, which comes out here.

583

00:44:20 --> 00:44:26So, minus V nought G1 comesover to the other side.

58400:44:26 --> 00:44:34And, let me collect all thevalues that multiply E1.

58500:44:34 --> 00:44:39So I get, G1 is one example.I have G2, and I have G3.

58600:44:39 --> 00:44:43And then, for E2,I have minus G3.

58700:44:43 --> 00:44:49OK, so I'll simply express thisas the element voltages

58800:44:49 --> 00:44:55

multiplied by some terms inparentheses, and I put my

58900:44:55 --> 00:44:59external sources on the righthand side.

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Similarly, I go ahead and dothe same thing here.

59100:45:06 --> 00:45:10In this instance,let me move my sources to the

59200:45:10 --> 00:45:13right.So, I get I1 coming out there,

59300:45:13 --> 00:45:16and I get V nought G4 comingout there.

59400:45:16 --> 00:45:21By the way, I just want tomention to you that if you're

595

00:45:21 --> 00:45:25looking to fall asleep,this is a good time to do so

59600:45:25 --> 00:45:30because as soon as I write downthese two equations,

59700:45:30 --> 00:45:36OK, from now on it's nap time.There's nothing new that you're

59800:45:36 --> 00:45:40going to learn from here on.It's just Anant Agarwal having

59900:45:40 --> 00:45:43fun at the blackboard,pushing symbols around.

60000:45:43 --> 00:45:46

So, once you write down thesetwo node equations,

60100:45:46 --> 00:45:48the rest of it is just grubbymath.

60200:45:48 --> 00:45:52

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So, let me just have some fun.So let me just go ahead and do

60300:45:52 --> 00:45:54that.So, I moved my voltages and

60400:45:54 --> 00:45:58currents to the other side.And let me collect all the

60500:45:58 --> 00:46:01coefficients for E1 here.So, E1 minus G3,

60600:46:01 --> 00:46:04and that's it,I guess.

607

00:46:04 --> 00:46:08OK, and then I'll do the samefor E2.

60800:46:08 --> 00:46:12So, I get G4,and I get G3,

60900:46:12 --> 00:46:18and I get G5.OK, so notice here that I have

61000:46:18 --> 00:46:22two equations,and two unknowns.

61100:46:22 --> 00:46:29OK, the two equations are onthe right hand side,

61200:46:29 --> 00:46:35

I have some voltages andcurrents which are my dry

61300:46:35 --> 00:46:43voltages and dry currents.OK, so actually this is getting

61400:46:43 --> 00:46:46

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quite boring.I'm going to pause here,

61500:46:46 --> 00:46:52and talk about something else.So, you can take this and you

61600:46:52 --> 00:46:56can put it in matrix form,so I've done that for you on

61700:46:56 --> 00:47:00page ten.It's all matrix form.

61800:47:00 --> 00:47:03Yeah, I know that.You can use any technique to

619

00:47:03 --> 00:47:07solve it, use algebraictechniques, use linear algebraic

62000:47:07 --> 00:47:09methods to solve it,use a computer,

62100:47:09 --> 00:47:11whatever you want.And, computers,

62200:47:11 --> 00:47:15when computers analyzecircuits, they write down these

62300:47:15 --> 00:47:18equations, and deal with solvingmatrices.

62400:47:18 --> 00:47:21

So, when you take the linearalgebra across,

62500:47:21 --> 00:47:25how many people here have takena linear algebra class?

62600:47:25 --> 00:47:30

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How many people here have heardof Gaussian elimination?

62700:47:30 --> 00:47:34How can more people have heardof Gaussian elimination than

62800:47:34 --> 00:47:37took a linear algebra class?Well anyway,

62900:47:37 --> 00:47:42so now you know why you tookthose linear algebra classes.

63000:47:42 --> 00:47:47And so, if I just collectedthese into matrix form --

631

00:47:47 --> 00:48:06

63200:48:06 --> 00:48:09OK, so I just simply expressedthose two equations in linear

63300:48:09 --> 00:48:12algebraic form,and here's my column vector of

63400:48:12 --> 00:48:15unknowns, and you can apply anyof the techniques you've learned

63500:48:15 --> 00:48:17in linear algebra to solve forthis.

63600:48:17 --> 00:48:20Gaussian elimination works.

OK, and in computer,

63700:48:20 --> 00:48:23people doing research incomputer techniques,

63800:48:23 --> 00:48:26or solving such equations

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simply deals with huge equations

63900:48:26 --> 00:48:28like this, building computerprograms that,

64000:48:28 --> 00:48:33given equations like this,can go ahead and solve them.

64100:48:33 --> 00:48:36OK, so let me stop here andreemphasize that what you've

64200:48:36 --> 00:48:39done is made a huge leap fromMaxwell's equations to using the

64300:48:39 --> 00:48:43

lump matter discipline to KVLand KCL, which ended up giving a

64400:48:43 --> 00:48:46simple algebraic equation tosolve, and not having to worry

64500:48:46 --> 00:48:49about partial differentialequations that were the form of

64600:48:49 --> 48:52Maxwell's equations.