# Kirchhoff’s Loop Rule Is For The Birds

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I am Walter Lewin and I’m going to teach you

a little bit of physics at the request of your own physics teacher,

Jason Hafner. We really don’t have

the right equipment for this, so it’s highly improvised. But the goal is

that I want to cover with you something that is the most intuitive

of all of electricity and magnetism. So non-intuitive that almost all

college physics books have it wrong. Why do they have it wrong? Because the physics teachers

do not understand Faraday’s law. Yeah, you may be embarrassed

to hear that, but it’s true. Almost all physics books have it wrong. I will advise you to watch my lecture 16,

I will– Some of that I will cover now, but you definitely have to read

my lecture notes on lecture 16. I have here a simple circuit

with two resistors. One resistor, which I call R2

and the other resistor I call R1. All the wires here have no resistance,

super-conducting wire. And we have that at MIT. You may not have that at Rice,

but we have that at MIT. Perpendicular to the blackboard,

we have an electromagnet and we can turn that

electromagnet on. That’s all I am going to do because the physics when turning it off

is very similar, so I will only discuss with you

when we turn it on. So the electric– that magnetic field

goes up very rapidly and then stays constant. And let’s assume that that magnetic field

is pointing out of the blackboard. I could have chosen into the blackboard,

but I’ll take out of the blackboard. As long as the magnetic flux– and the magnetic flux is defined

as the integral of the magnetic field B dot ds

over an open surface. It’s very important, an open surface. And that open surface I will attach

to this closed loop. And you may choose that surface

anywhere you want to, but I want it to be attached

to this closed loop. I call this point here A1,

this point here A2, this one D2

and this one D1. I want you to understand and appreciate

that A1 and A2, from an electrical point of view,

are identical. They are the same. They are connected to a none–

to a wire that has no resistance. So you can think of A1 and A2,

if you want to, simply as A. They are one and the same. You’ll see there’s a reason

why I split them. And the same here–

D1 and D2 are also the same. So you can think of this as one point A

and one point D if you want that. When the magnetic flux,

the way I defined it, is changing in time

according to Faraday’s law, the closed loop integral

around this circuit is not 0. If the closed loop integral

of E dot dl is not 0, that means that Kirchhoff’s loop rule

does not work, cannot be used. Using it is absurd. Kirchhoff’s loop rule would say

that the closed loop integral of E dot dl– the closed loop integral

of E dot dl will be 0 according to Kirchhoff. I start here

and I go all the way around, and I come back here

and that should be 0. But that’s not the case

if there is a changing magnetic flux the way I defined it. The answer according to Faraday’s law

is that this is minus d phi– the B makes use to [?]

the magnetic field– dt. If the magnetic field is increasing, because I turned it on all of a sudden,

the magnet, then clearly the magnetic field is not constant,

magnetic flux is not constant. So I’m going to get an integral E dot dl

that is no longer 0. We call that also often the EMF,

Electromotive Force. Therefore, there’s a current

going to run and you can figure it out for yourself

because I’m sure you’ll cover it in class that, in this case,

where the magnetic field is increasing and therefore,

the flux is increasing, that the current is in this direction,

the current I. And so this EMF, which is this integral,

equals I times R1 plus R2. So if we just look

at our first eight minutes and we notice that the red color

that I’m using, which worked so well at MIT in 26-100,

resists camera, didn’t come through so well. And so whatever is red here, which was I,

the current was this arrow. I will put them in white. I wish I didn’t have to,

but I will do that. Otherwise, you may not see. So the current is going around

like this, if that B field pointing out of the blackboard,

is increasing. Now, what now is the potential difference

between A2 and D2, between here and here? Or I could say, what is the potential difference

between A and D? Because we agreed

we can call this point A and we can poll–

call this point D. So the potential difference

V A2 minus V D2 is the same as VA minus VD

because A2 is A, and D2 is D. And that must be I times R2. Very simple. What now is the potential difference

between A1 and D1, which, of course, is also the potential

difference between A and D? So the potential difference

between A1 and D1, which is also VA minus VD– now be careful. The current is in this direction. So this point has a lower potential

than that point. So it is not I times R1,

but it is minus I times R1 because I go from here to there. I will oppose the current. Whereas here,

I went with the current. So it is minus I R1. Now, it’s staring you in the face now

that Kirchhoff’s loop rule is for the birds, because if I measure here VA minus VD,

then I get I times R2. if I measure it here,

which are at the same point, I get minus I times R1. They even have different polarities. In other words, if I would put here

a voltmeter V1, that voltmeter would measure

this value. And if I put one on the other side,

for which I have little room anymore, that would measure

this value. And the voltmeter on this side

I will call V and I will call the voltmeter

on this side V1. And so what you see now is that V2

divided by V1 is minus R2 divided by R1. Think about this for a minute. Both voltmeters, this one and this imaginary

one that I haven’t drawn, they both measure the potential difference

between A and D. Both voltmeters I will have plus up and minus

down. I will do the same here. If the ratio of this resistance is 100, then one voltmeter will read a value

which is 100 times larger than the other voltmeter. But not only that,

the polarity is reversed. So this one might read

plus 10 volts. This one might read minus 1 volt. So it’s now staring you in the face

how non-intuitive this is. I can give you an example

by making certain assumptions. Suppose I have R1– I think I chose 10 ohms

and I have for R2 100 ohms. And let us assume

that at a particular moment in time, this induced current– it’s induced

by this changing magnetic flux, that that induced current at that moment

happens to be 10 milliamperes. Keep in mind it will change with time. Why will it change with time? Because the EMF

will change with time. Why will the EMF change with time? Because d phi dt

will change with time. I cannot keep the change of the magnetic flux

constant all the time. So let us assume

that at a particular moment, this induced I,

which changes with time, is 10 milliamperes. Then you see that this voltmeter

on the left side, V1, will read 10 milliamperes

times 10 ohms. That is minus 0.1 volts. Minus! Keep that in mind. It’s minus. This voltmeter, the plus is here,

the minus is here, will read a negative value. A has a lower potential than D. But V2 will read 100 times 10,

1 volt. And I do this experiment

in my lectures. You will see it if you watch lecture 16

and you won’t believe your eyes. And I will show you, we’ll have a small break

now here for our eight minutes, I will show you what actually you will see

on an oscilloscope when you put an oscilloscope

here for V1 and at the same time,

you demonstrate here the oscilloscope, which shows you V2. So we’re going to put that on the board

very shortly. I want to repeat once more that the larger

the value of d phi dt is, the larger the EMF. If the magnetic field

is increasing, then the current will be

in a different direction, the induced current, than if the magnetic field

is decreasing. But those are details that Jason

will discuss with you. That’s not really, at this moment,

so important. So let’s now assume that the EMF

is in this direction, that the current is going around

like this and I’m going to make you now

a sketch of what we will see. Look what I did in the meantime. I cleaned the blackboard

and so now I have a nice opportunity to add that voltmeter V2. Here is the plus side of V2 and here is the minus side

of that voltmeter. So they are both hooked up

in exactly the same way. They both measure the potential difference

between A and D and they both get totally

different values. Let me make a drawing,

a sketch of the magnetic field, which is not the magnetic flux, but simply in magnetic field

that we produce in the lecture hall as a function of time when we turn on the electromagnet. Um, I’ll make some rough guess. It starts with 0 and then this goes

something like this. And then if we keep

the magnet on, then, of course, the magnetic field

will reach a maximum, and we’ll flatten off. Notice that here,

dB dt is 0, so there’s no

longer any flux change. But here, it’s almost constant. There is a flux change. Let us first do V2

and this means plus. So I would say between these two points,

between here and here in time, um, the flux derivative

is roughly constant because this–

this is almost a straight line. And so it means that V2

will have a certain value. In our case, with the example

that we chose, V2 was– what was it?

1 volt? Yes, I think so,

plus 1 volt. That’s why I put that plus. And when you are here,

then d phi dt is 0, so you must end up here again. And I will just somehow sketch it

that it gets down to 0. So that’s what you will see

on the oscilloscope. Then we show students,

at exactly the same time, we show them

V1 as a function of time. And since V1 is negative, I can use the same

[INAUDIBLE], the same vertical. Now, in our special example,

V1 was 10 times smaller and negative. Well, 10 times smaller

is something like this. So you’ll get something like this. And so the moment that we turn on

the electromagnet, we show the students,

as a function of time, millisecond resolution,

V2 and V1 and these are the curves

that you will see. You see, in front of your eyes,

the fact that one voltmeter, in this case, reads a 10 times larger value

than the other, but also that

the polarities are reversed. When I did this experiment, in my audience,

was my good old friend, very bright physics professor,

Bob Ledoux. This was a long time ago. It was 1992, I think. And Bob came to me after the lecture

and he says, Walter, this cannot be. There must be something wrong. I said, Bob, I don’t think so. Give it some time. I went to my office

and the phone rang. Who was on the phone?

Bob Ledoux. He says, Walter,

I can’t believe it. It’s true. You see that even for physicists,

this is not so obvious, that the potential difference

between two points, measured one way

and measured another way, between exactly the same points, totally different in value

and different in polarity. And so I really want you to watch this

on lecture 16 and I want you to read my lecture

notes on number 16. If Kirchhoff’s loop rule were to hold,

which it doesn’t, then V1 and V2 should always measure

exactly the same value. I hope you realize that. So you see, to use Kirchhoff’s loop rule here,

it’s not even wrong, but it’s criminal. You get nonsense. But Kirchhoff will predict that the potential

difference between here and here and here and here, is the same

so that the closed loop integral of E dot dl, that becomes 0. All right. You have something now to think about. And now I will, again,

stick with Faraday’s law but change the topic a little bit. So maybe this is a good moment

for Saif to have a break and I will clean the blackboards. I will show you a very simple circuit

in which I put a solenoid, a self-inductor as we call it. It is an ideal self-inductor. And what that means is that the self-inductor

is made of super-conducting wire. So the self-inductor has no resistance and that means that there can never be

an electric field inside that self-inductor

because the resistance is 0. So keep that in mind,

ideal self-inductor, that the wires have no resistance. There can never be an electric field

in a self-inductor. I have here a simple battery. I chose just the simple battery model,

that of an alternating current because I’m sure that Jason

will cover with you these circuits with alternating current. I just want to stick

to the very basics. So here is a battery

and here is a resistance R. Here is a capacitor C. Here is a self-inductance L. And then we close the loop. And we have it here

and in here, I’m going to put a switch. So a self-inductor,

which has no internal resistance. The wire is a super-conductor,

capacitor, resistance. And let us assume for simplicity,

although it’s not important, that the battery

has no internal resistance itself, or negligently

small. And let’s assume that the value

of the battery is V0. So it remains constant. I throw the switch. Obviously,

what’s going to happen, a current is going to flow

in this direction because this is

the positive side. So I wish I could do it with red,

but that doesn’t work so well on the camera, so I will do it then with white. So a current is going to run

in this direction, through a resistor

in this direction, in this direction through the self-inductance, in this direction, in this directions

and in this direction. Now, what does Faraday say? Faraday says that the closed loop integral

of E dot dl, it was minus d phi dt. The magnetic flux is through an open surface,

as I discussed earlier and that open surface

must be attached to your closed loop. So the closed loop is this

and so we must attach a surface. Think of it as a soap film. It doesn’t have to be flat. It could be curved. It could be very strange in shape. But it has to be an open surface

that is attached to the closed loop. And if you want to see

how that looks inside the self-inductor, it’s very hard to see

becomes almost like a staircase that the soap will roam around because it must be

one open surface. All right. Now, we’re going to write down

the integral of E dot dl. I really want to try this

in a different color and if it doesn’t work,

then Saif will tell me. Do we agree that the E field here

is in this direction? Can you see it,

or not so well? The E field is in this direction

in the resistor. This side of the capacitor

will become plus because of the current flow and so the E field here

is also in this direction. What is the E field in the wire

of the self-inductor? What do I hear? What are you saying? It’s 0. Exactly, because there is no resistance. So E here is 0. So the integral E dot dl

through this wire is 0. The E field here

goes from plus to minus, so the E field here

is in that direction. So now we are ready to write down

the differential equation for this circuit. And I will start at this point,

I will go around and I will come back at this point. So the first value when I go in this direction

is minus V0. Right, I– It is the integral of E dot dl,

the angle between the two, is 180 degrees, so I get a minus sign. In the resistor,

it is plus I times R. I changes in time, but at a particular moment

in time it happens to be I. Remember, the capacitor C

equals Q divided by V. I call this V of C. And so we get plus Q

divided by C. That is E dot dl

over the capacitor. Yah, C is Q divided by V

and so the potential difference, which is, after all,

the integral of E dl, this VC is C divided by–

Q divided by C. Now I come to the self-inductor

and I’m crawling inside that wire. And I go through that wire

and I never see that E field. I never see that E field. In other words, the integral E dot dl

between this point and this point is 0. So I am back here now at this point

where I started. So according to Faraday’s law,

that is minus d phi dt, for which Jason will tell you, that

that is the same as minus L dI dt. I will not explain

why this takes this form. I can’t take that

away from him. He’s dying to explain that to you. So look at this equation now. This is the closed loop integral

of E dot dl. That, according to Kirchhoff,

would have to be 0, but it isn’t. It is minus L dI dt. Now I’ll tell you that almost all college

physics books cheat. They do it wrong, but they get the right

answer. Why would they get the right answer? Because they know the answer. They write the following. They write in their book– boy, [INAUDIBLE] this is–

they write minus V0 plus IR plus Q divided by C

plus L dI dt is 0. So this is the honest way it should be written,

according to Faraday’s law and this part is the closed loop integral

of E dot dl. The college physics books– most of them are written

by physics teachers who do not understand physics–

they write it this way. And then they say, do you know

why this is 0? Because of Kirchhoff. They say, look, this is

the closed loop integral of E dot dl and that has to be 0

according to Kirchhoff. This is not only wrong, but as I said earlier,

it is criminal. Because this is not the closed loop integral

of E dot dl. This is the closed loop integral

of E dot dl and that is not 0. So the simple fact

that they put this term here doesn’t make the equation wrong, but the physics stinks. And this is very, very serious

because if you start teaching students a concept that is completely wrong,

well, I have to calm down now. But this is emotionally

very important for me. I would advise you to also read,

definitely read, my lecture notes number 20. So I imagine by now

you are awfully confused. But at least you know, you’ve seen once,

how to do things the correct way, correct in terms of physics. The fact that these two equations

give the same answer, well, of course, because the authors

of the book know the answer. I’m now going to cause you

some sleepless nights. And I’m going to attach here

a voltmeter– and now I’m going to ask you, “What

will this voltmeter show?” And you will probably say, “Well,

didn’t you say that the integral E dot dl through the wires here is 0?” So you would think

that V here is 0. But that’s not true. We know what that V is going to be. It’s going to be

plus L dI dt. In other words,

this equation is now correct. And you will say, no, what happened now

with your d phi dt? Well, what I have done

in a clever way, by going around the circuit I have now no longer

any changing magnetic fluxes going through my open surface, which is now attached

to my new closed loop. And my new closed loop, folllow me,

is like this. And look when I come here. I bypass that self-inductance. I go through the voltmeter

and I’ll come here and I’ll come here. And so now Kirchhoff’s loop rule

should hold, and so the sum should be 0. Therefore, it shouldn’t surprise you

that this plus L dI dt is exactly what this voltmeter

then will show. I’m going to confuse you

even more. I’m going to remove this voltmeter– and you hold in your hand an instrument to measure the electric field E. And you start here, you go around

and you get this term, you get this term,

you get this term. And now when you arrive here,

you leave the circuit and you go this way. You can choose your path. I don’t care. And you come back

to the circuit here. But while you are going

this dotted line, which could be just through

the air like this, you measure the integral of E dot dl

over that path that you have chosen. And what will that integral show? Plus L dI dt. Again, you have now chosen a–

a loop whereby there is no longer any changing magnetic flux

going through that surface, and so Kirchhoff’s loop woo–

loop rule works. I would like Jason

to discuss with you why it is that that voltmeter

would measure plus L dI dt because clearly the L dI dt

has to do with the self-inductance. So I’ll leave that up to Jason. Got to leave something for him. So before I go off the air, I would like to say a few words

to the many people who are now taking this course,

this edX course, which is taught by Jason. I would like, to many of you who may also

have taken my 8.01x course, almost everyone who took 8.01x seriously passed the course, more than 85%,

for those who took it seriously. So my advice to you is, to all of you,

maybe tens of thousands all over the world who have been following this course,

this edX course, this Electricity and Magnetism course, if you take it seriously, that means

doing your homework seriously, work hard on the exams, you too are probably very likely

going to pass. I wish you good luck.

HL2199Post authorHello sir! I interpret Kirchoff's laws as a statement of the conservation of energy. Since it does not hold true in this case, where has the difference in potential energy gone/arisen from?

Sujoy BhattacharyaPost authorWhen electron passes through an inductor, does it face any potential difference?

Rakibul IslamPost authorI can not explain in words how much I love your lectures Mr. Lewin!!! I just love it!! THANK YOU for all the intuitive ideas and all! You will always be my favourite physics teacher! Love & Respect! 🙂

Teh Yong LipPost authori noticed a lot of college books try to resurrect kirchhoff loop rule by modifying the rule, avoiding the differential equation… ,i have never seen any books/websites that say we have to use Faraday's Law to solve the problem, and yet they are using electricity everyday which is not possible without faraday's law, so embarassing

chinamattPost authorWhat if the wires are not superconducting? E would not be zero across the self-inductor.

Kelvin TanPost authorwhat if i connect voltmeter across the point between A1 and A2 to the point between D1 and D2. what is the reading

Kanad Krishanu SenguptaPost authorSir I am from India.Will you please suggest me a good book for AC circuit theory.

LReBe7Post authorProfessor Lewin, ik snap één aspect van de demo met de spoel in het midden van de lus niet. Hoe "weet" de voltmeter welke van de twee potentiaalverschillen hij moet meten? Hoe weet de linkse dat hij positief en een factor 10 moet uitslaan en de rechtse dat hij negatief en een factor 1 moet uitslaan? Wordt dat bepaald door :

a) de polariteit waarmee de Voltmeter wordt aangesloten

of

b) de plaats waar de meter is aangesloten?

Ik heb er al enorm mijn hoofd op gebroken en ik heb helaas geen goede spoel om dit experiment zelf uit te voeren.

Thomas JimenezPost authorThank you sir. Writing Kirchhoff's rule was making me sick to my stomach. I just watched this completely through today.

Gustavo MerchanPost authorHelp please: How do the voltmeters "decide" which integral path to show?!!!

t chungPost authorDr Lewin: Thank you for your illuminating talk. I wonder if you would be kind enough to clear up the following point.

A1 being the same as A2 since the resistance in the wire is negligible, D1 being the same as D2, allow me to call them just A and D. If I (1) connect A and D to a voltmeter, put the voltmeter on the right side of the circuit, turn the magnet on and measure the voltage, and (2) turn the magnet off, put the voltmeter on the left side of the circuit without breaking the points of contact with the circuit (by alligator clips, say) and turn the magnet on again, it appears from your lecture that I would then measure a different voltage. The internal resistance of the voltmeter is assumed to be very large, and one can assume that no current flows through the voltmeter. The electrical contact points being the same for the two positions of the voltmeter, how does the circuit "know" that the voltmeter is on its right or on its left? What if I put the voltmeter out of the plane of the circuit, say, between R1 and R2, what would be the voltage?

I did read your supplementary note on lecture 3-15 on this demonstration, but still could not figure this out. How could the measured voltage be dependent on the position of the voltmeter? Thank you.

MANIKANTT MPost authorthere are 5 levels of appreciation above average, good ,very good, excellent ,professor walter lewin seriously the best physics professor of all time who concentrates more on getting right concepts rather than getting the answer like most would do

nicolas ariasPost authorProf Lewin, thank you for everything, from Argentina I sincerely admire your work and dedication.

Nevertheless i think Kirchoff s rule is applied in theory for electrostatic fields and not for electromagnetic ones. Actually, the "conservative" property of the electric field is only valid for electrostatic ones.

GonsPost author17:49 what kind of sorcery is that!

stefmazz97Post authorSir, why is the current clockwise if the magnetic field is coming out of the black board? Woudn´t it be counter clockwise to follow the right hand rule?

Pan PanPost authorI really have learned a lot from your lectures. I have a question about the derivation of E in Jason's lecture: https://www.youtube.com/watch?v=Qc4g6-Ncn-w&feature=youtu.be

The formula Jason has is E=(r/2)* (dB/dt). My question: if B is everywhere, and dB/dt is everywhere the same. E should be everywhere the same, E should not depend on r. Is my understanding correct? Could you please help with my question? Thanks so much!

Arunaava GhoshPost authorYour lectures not only made me love Physics .They made me love Physics passionately.

mac1082003Post authorI am big fan of your lectures sir. sir I am not that smart in physics but I love it and always try to learn it. sir I have a question it may sound a little silly may be but am confused about how potential difference and emf are different things. please help

RADHE SHYAMPost authorYou are an inspiration for us. I pray to God that good teachers like you must always stay on earth. They must never die. I do not know the progress of this kind of technology that can make you young again. We love your teaching.

Phil LiberatorePost authorThank you, Professor. You are always interesting and challenging.

I see it, I understand the math (I think). But it will take me a while to believe it.

Anakin StinsonPost authorMy physics teacher says that kirchoffs law is the father of all the laws and is applicable everywhere. how do i explain this to him?

kshitiz rimalPost authorI was first introduced to kirchhoff's theorem in my discrete math class.

Manav SivaramPost authorDoes the absence of electric field due to 0 resistance follow from ohm's law and V=E dl

pioneer80Post authorI have been an electrical engineer for many years, and I get the general idea about what kirchoff's Loop Rule says, but after watching this interesting video, it seems to me that Kirchoff's Loop Rule is another way of restating what we already know about conservative electric fields, which is that the line integral around a closed loop of a conservative electric field is always zero. So, in my opinion, it should not be used when we have non-conservative electric fields along the circuit. Because, if it's used (like it is taught today) when we have a non-conservative electric field like that inside a battery, why not using it when we have another non conservative field like that inside the wire of a coil? What I mean is that, if we can't use kirchoff's Loop Rule when we have a non-conservative caused by a changing magnetic flux, then we could not use it either with a non-conservative field inside a battery. Or we can, but one (with a different name than "Kirchoff's Loop Rule") which says Sum(E)=Sum(R*i). Superb video. Thank you Professor

Chitti Satish RaoPost authorsir which type of book should I have to use sir for numericals i

ASHISH JHAPost authorat 36:32 you asked a question why volmeter will read plus l(di/dt) how can i get the ans ??

ASHISH JHAPost authordoes that kind of confusion do occur in krichoof's current law

maths sosoPost authorhi professor, a current flowing in a wire means that there is a force acting on the charges, and it means that there is an E field, so a potential difference in that same wire,?

ayeshee sarkarPost authorSir did you make videoes on optics?

Niranjan BiswalPost authorThis cleared my head

Kalash BaisoyaPost authorawesome lectures given by you sir I am an Indian and I think that I will be the most lucky person if one day I will be your student in your college but the only thing is my financial condition is not that much good and my father can't afford it….

Kalash BaisoyaPost authoryou are just like God for me….u are the bestest teacher I had ever seen… I love you a lottt sir…and really want to be your student one day….I will love to see your lectures till the end of my life…

Rick TjepkemaPost authorThe EMF between wire A1-A2 and wire D1-D2 in a changing B field is not 0.

If you account for the induced voltages between A1-A2 and D1-D2, Kirchoff's law is valid again.

Also you can't reduce the wire A1-A2 into a single point A and D1-D2 into a single point D,

since this would result in no open surface loop at all.

Resistor R1 would just lie on top of R2, and there would be no loop surface through which the B field runs.

Munish PatialPost authorGod is supreme as he has the complete knowledge of everything. I wish, I could ever meet this man to see God on our planet…………….

Naman KarnPost authorSir,

Where can I get the Lecture notes mentioned in the video?

ElectroBOOMPost authorI don't understand. Why in this video are we ignoring the voltage across inductances not a part of the Kirchhoff's loop? Doesn't the law say the sum of voltages in a loop is zero? So for example in the first example with two resistors, he is not showing the existing inductance of the loop in his model, which can be modeled between the two series resistor (series with them) and is what causing the voltage difference between the resistors. Considering the voltage across inductance in a loop, KVL still stands.

[Edit] Never mind, at the end he somewhat says the law holds, and he is intentionally making us confused to thing…

MissDocPost authorSo can we say that work done against electric field is actually path dependent ?

Gokul KrishnaPost authorFantastic lecture professor!!!!

One doubt….

Does this have anything to do with the fact that the induced electric field is non conservative?

Siti MahsahirunPost authorI love this professor! he is amazing…I wish I go to mit and become his student in physics. But I'm dumb enough to get into Such a highly prestigious college and I don't even interested to go there to waste money that my family doesn't own. But now I'm very lucky, I have reached on the lectures from Youtube/

I'm an electrical engineer, the truth is, I do face a lot of confusing problems when dealing with circuit analysis. We don't really go into the deep physics on KVL. We just apply it on the surface. According to our circuit textbook (we use Alexander Sadiku book) the total algebraic sum of voltages in a closed loop circuit is equal to zero. It usually always works with simple planar (2D) circuit. But when the circuit consist of supernode and multiple branches, the confusing equations will kicks in. We will have to look into the equivalent circuit derivation. Circuit analysis on such circuit will only worth if the circuit is correct.

The assumptions on KVL is not really well defined by our lecturers during my undergrads and I still have a fuzzy idea on sign convention and definition of supernode in KVL circuit analysis, for 6 years graduated as electrical engineer. Usually we just use SPICE to do circuit analysis. SPICE is computerized program to do circuit analysis based on Kirchhoff's laws. But the circuit must be correct first. If not, I will lost in the loop forever. I have a lot of problem when dealing with nonlinear electrical device such as transistor and motor. Because there is no simple circuit elements configuration to represent it. My basic is not strong. I hope someday, someone will change our teaching module; improvise with proper understanding on the physics on equivalent circuit derivation for nonlinear devices

李愚Post authorI really wanna become a professor of MIT, the great professor like Walter Lewin, Eric Lander, Donald Sadoway Gil Strang, Eric Grimson, John Guttag (so many i don't know yet)…MIT's course compared to Harvard (more diverse)looks simpler but more insightful, impressive, logical,memorable, and unparalleled…study from the best, must makes me a better teacher! and prof Walter Leuwin devoted to teaching and his lifelong goal and hobbies! that's so originally fascinating ,for a student never like physics from primary school to high school ,wow

always lives with love, passion and curiosity!

ProtoexPost authorProfessor, I get quite confused on when we put a voltmeter in parallel with the self-inductor. I think my best guess is the following:

When I look only to the loop coil-voltmeter, the complicated surface I would get near the coil (if I sunk it in the soap) would be the same as in the V-RCL loop but with flipped sides. A nice imagination trick to view why its flipped is if I take the surface in the V-RCL loop and paint say the "facing me" side with red, and the other with green. than I release all but the coil from this surface and fold it to the right and maybe cut it a bit and glue to the voltmeter & wires in such a way that the green surface is facing me.

I tried to guess why in loop V-RC-voltmeter there is no changing flux. I was not successful, and just now as I write it that I realized that the red-green surface explain it.

there is any simpler solution?

witch good books about electromagnetism and about Theory of relativity you recommend?

thanks.

Suman ChowdhuryPost authorSir, from where can I get the lecture notes of your lecture number 16???

habib ahmedPost authorhello prof. I am pretty well at physics 2 than physics 1 where i lac in knowledge and in concept topics like dynamics , dynamics of rigid body relative motion . i will be very thankful to you sir if you can provide me some source to prepare all that stuff

marc rubinPost authorDr Lewin. If I use the same value of resistors should I not be able to feed a set of buck converters and make a bipolar power supply? I shall try it. If it works Can I call it the the Lewin bipolar buck converter? My biggest question is where to assign ground!

Nedaa AlzkimiPost authordo not understand those people, who argued with about this matter…. it is one of the most obvious things that Kirchhoff's Loop does not suit at all with the induction and that is so obvious in RL circuits when we get by measurement many differential values… of course I am not sure if it's the reason but I think so…. anyway Faraday law is always right…..

Michael LawPost authorLove your lectures and I love Physics, always have….going back to my first word: "why?"

Karthik G SPost authorHello sir once again a great lectures from you thank you

TheBlueRavenPost authorCan anybody please explain me at 7:00 how is EMF=I(R1+R2).Because as professor Walter Lewin said A1 and A2 can be referred to as same point and D1 and D2 can be referred to as same point. So that means these two resistance R1 and R2 should be in parallel connection. Then if that's the case then EMF should not be equal to I(R1+R2) right?

Lectures by Walter Lewin. They will make you ♥ Physics.Post authorThere are people who object religiously to the correctness of the mind-boggling demo that I do at the end of lecture 16 of 8.02. They all conveniently overlook that this demo showed PRECISELY what was predicted by my derivations. Two identical voltmeters are attached to the exact same 2 points in my circuit. The Curl(E)=-dB/dt caused an induced time variable emf in the circuit and consequently at any moment in time (my time resolution was 0.1 msec) voltmeter #2 showed values 9 times larger than voltmeter 1 and the two voltages had reversed polarities. END OF STORY!Therefore, all comments and objections are meaningless and without merit; they are wrong at best. It's well known to all people with a decent background in Physics that in the case of induced emfs potential differences between 2 points in a circuit are not uniquely determined; they depend on the path. This is also the case in the closed loops of the secondary windings of ALL TRANSFORMERS! My demo can therefore easily be repeated using an AC power supply and a transformer (see website below). The ratio of the resistors in the transformer demo was a factor of 20 (in my case a factor of 9). Notice that the difference in readings of the voltmeter in the transformer demo was a factor of 20 (depending on the location of the voltmeter) as dictated by Maxwell's eqs. Unfortunately the reversal in polarity cannot be demonstrated with the transformer demo as it uses AC current. That's the reason why I decided to use an electromagnet (DC current); it shows the polarity reversal beautifully. I too, in my early days at MIT, used only 1 voltmeter which I moved from one side to the other as is done in the transformer demo. But two voltmeters simultaneously used in my lecture #16 with a time resolution of 0.1 msec is more impressive as students can then see the results (including the polarity reversal) of both voltmeters simultaneously.Transformer demo

https://www.youtube.com/watch?v=b7i2uMx7gHo&list=PLyQSN7X0ro22zanLOcvkaSY-IZqheFYM5&index=8

Lect 16, 8.02

https://www.youtube.com/watch?v=nGQbA2jwkWI&index=17&list=PLyQSN7X0ro2314mKyUiOILaOC2hk6Pc3j

Believing and Science are very Different

https://www.youtube.com/watch?v=wz_GqO-Urk4&index=10&list=PLyQSN7X0ro22zanLOcvkaSY-IZqheFYM5

Matthew Ardito-ProulxPost authorProfessor… another sleepless night :/

mim zimPost authorif we remove the solenoid, the current is changing as the capacitor charges. So is there not a change of the magnetic flux through the surface? Should there not be an additional term on the right for this RLC circuit that would be present even in a RC circuit?

Sriram .V.SPost authorShould the direction of dl be in the direction of current?

Or else the right hand side will become +d(phi)/dt right?

Sriram .V.SPost authorIs there any qualitative explanation for why the voltage leads in case of an inductor and lags in case of a capacitor when they are connected to an AC source?

INNOVATION & INITIATIVE DiyPost author7:20 impressive catch !!

INNOVATION & INITIATIVE DiyPost authorcan i do this experiment with circuit simulation in comp ????

Nasir HajiniPost authorSir won't there be any induced electric field in and around the inductor. Induced E field because of changing magnetic flux .

drago slyPost authorSir please make a video on Kirchoff's law … Problem solving

don carlosPost authorHello Dr. Lewin, I came acrss this popular EE site please see the question and the discussion at this link they are writing about your video: https://electronics.stackexchange.com/questions/309286/circuit-yields-a-peculiar-contradiction-between-kcl-kvl-and-faradays-law Its a site for electrical engineers and some do not agree with you. Why is this topic not agreed upon by all? I mean there must be a declaration from physicists to close this case.

Abdullah AlsakkaPost authorSir were can I your lecture notes

Ahmed ThoufeekPost authorSir how is kirchoff's laws applicable when i have 2 cells say E1 & E2 where +terminal of E1 is connected to -terminal of E2 and + terminal of E2 to – terminal of E1,connected together by a superconductor wire. By the way sir may God further increase your knowledge and thus make you of those close to him

voora venkata mohana vamsiPost authormay be kirrchoff's law can be modified and said a 'the closed loop electrostatic field is zero'.

Shreyash MeshramPost authorWhy do we need to have any surface attached to it?????

sir me bubul bohot badhiya h sir GhoshPost authorcan u please make a lecture of chapter heat and thermodinamics

PLASMIC TECHNOLOGYPost authorwhat a superb chance to watch your lectures on youtube sir thank you so much

cychoiPost authorthank you sir this video is awesome

Emmanuel IrizarryPost authorTeach me how to make your perfect lines before such technique gets lost! Im fact, make a tutorial and explanation of the physics behind of it!!!

tapash debPost authora huge doubt has been cleared

riyan15979Post authorIf einstein was alive he would probably talk like this!!!

Tarang RajvanshiPost authorSir, at 27:45 if we had moved around the other loop the other way, then would the integral E. dl would still be Zero and equal to – L di/dt? Because in this case the sign of the potential differences would be reversed but the sign of – L di/dt would still be the same.

Tarang RajvanshiPost authorYou are truly an inspiration, Sir.

balkrishna aroraPost authorSir, I am unable to digest that electric field inside a superconductor is 0. Because it is the electric field that is causing the current flow in it(in the inductor) and that is opposing the change in megnetic field.

Will you please explain. It….?

lucas LarsPost authorMy book states that the closed loop integral E *dL = the change of flux holds ONLY if the path around which we integrate is stationary. Is this true?

pusat berkPost authorBoom

Violent ZaniPost authorMr Lewin, regarding the topic i think you have to look at this video https://www.youtube.com/watch?v=0TTEFF0D8SA. Wish you all the best

Oolon ColluphidPost authorSo is this equivalent to saying : "The voltage/potential difference (E dot dl) from one end to the other end of an ideal-inductor is always zero" ? Mainly because this ideal inductor has no resistance and E is 0. ???

AlchemyOfDeathBandPost authorAt 27:03 he states that the electric field in the inductor is 0. According to Faraday's law, the electric field of the inductor is nonzero (unless steady-state dc). Google "electric field inside a solenoid". The E field inside the wire of the self-inductor is (u0 * n/2) * (di/dt) * R where R is the radius of the inductor.

Mateo Ceballos QuerolPost authorAmazing video. Shouldn't the current be going the other way around the loop? using the right hand rule if the current is going out of the board the current goes anti-clockwise

Momir ZecevicPost authorSo you still have no math proof that Kirchhoff's is not valid. And your experimental values are simply WRONG. Do it properly and you will see where you are making mistake.

I am still not sure is that wrong experiment lead you to wrong conclusion and now out of shame you can not go back, or this is a intentionally made wrong for publicity purpose.

Flávio CoziPost authorHello mister Lewin. It is nice see you saying "the curl of electric field is equal to variation of density of magnetic flux inside of this curl and if the variation of density of magnetic field is different of zero then electric field is not equal to zero too ( nabla x E = – dB/dt). This is the another way to see the third Maxwell equation or Faraday's law.

Thank you for showed it!

Best regards

simplic1000Post authorSuppose you take the voltmeter V1 and, keeping its contacts at A & D, move it to the other side of the circuit (quickly enough that dPhi/dt has not changed appreciably). Presumably the EMF it registers will change continuously from IR1 to IR2… registering zero somewhere in the middle of the loop? So it seems like the exact details of where the voltmeter cables are, are relevant too.

Flávio CoziPost authorHello mister Lewin. With all respect to your knowledge and history but I gotta tell you that you have forgotten the indutance parameter on you circuit of resistors because you have only induced voltage when you change the density of magnetic field when you have indutance and in this case this indutance is distribuited along the circuit but you can represent it by only one inductor.

I have maked too the practical experience using a little transformer, wires, a magnet and a voltimeter shorting the terminals of this transformer and watching the voltage when I was changing the density of magnetic flux by the magnet and the result is zero for voltage and not for electric current.

simplic1000Post authorHow does the voltmeter know which side of the circuit it's on?

Piotr ZazakownyPost authorIR1=-IR2 since A1 is the same as A2. The same applies to D. Moreover R>0 and I>0 so this formula is true only for I*Ri=0. If my algebra isn't wrong where is the issue?

JohnnyPost authorLove your videos professor. I hope you're still making them. You need a better microphone, other than that, keep them coming!

Devam JangraPost authorSir you are the best

Benjamin LloydPost authorLove that mind-numbing background whirr.

ThomasHaberkornPost authorso, electric fields are electric (E dot dl) and magnetic fields are magnetic (-dPhi/dt). No suprise there. KVL describes only the electric part.

Chris BagnallPost authorCan someone explain to me why he measures the voltage across R1 in the opposite direction to R2? When he measures R2 he does it with the direction of current, but for R1 it's against the current. I'm sure there's a good reason, I just don't see it.

Vaidhyanathan APost authorTHE BEST AND THE WISEST PHYSICS TEACHER,THANKS A LOT AND LOT SIR,YOU ARE JUST GIVING GOOSEBUMPS

MR TECPost authorThanks sir your method of teaching is impressive

Juraj DobrikPost authorPeople just denied electromagnetic current generation just to prove you wrong, and they are still engineers, that is kind of sad.

Lectures by Walter Lewin. They will make you ♥ Physics.Post authorI demonstrated at the end of my lecture #16 of my 8.02 E&M course at MIT that two identical voltmeters attached to the same 2 points in a circuit can show very different values.https://www.youtube.com/watch?v=nGQbA2jwkWI&list=PLyQSN7X0ro2314mKyUiOILaOC2hk6Pc3j&index=17The reason is that in the case of an induced EMF (Faraday's Law) potential differences are no longer determined; they depend on the path. This also applies to the secondary windings of transformers as the EMF in the closed loop of secondary windings is induced.https://www.youtube.com/watch?v=b7i2uMx7gHo&list=PLyQSN7X0ro22zanLOcvkaSY-IZqheFYM5&index=8Of course, in cases where Kirchhoff's loop rule (KVL) applies, 2 voltmeters attached to the same 2 points in a circuit will always show the same value.My demo was first suggested and published by Romer in December 1982 in the American Journal of Physics.This demo has now become a classic; it's done all over the world at many colleges and universities.Kirchhoff's original text can be found in the following link, pages 497-514:https://books.google.de/books?id=Ig8t8yIz20UC.Clearly he was fully aware of the prerequisite for his "loop rule". KVL is a special case of Faraday's Law. That's why Faraday's Law is one of Maxwell's equations and KVL is not.By teaching students that KVL always works without telling them when it does not work, makes many believe that the closed loop integral of E dot dL is always zero. ElectroBOOM and Dirk Van Meirvenne therefore believe that 2 voltmeters attached to the same 2 points in a circuit must always show the same value which is not true as demonstrated in my lectures.They each posted a video on their channel to prove what cannot be proven as their ideas violate Maxwell's equations.Apparently they do not know, or do not understand, that in the case of an induced EMF potential differences are no longer determined; they depend on the path.MIT students who took my 8.02 course (Electricity and Magnetism) would not make this mistake!I therefore believe that to introduce a "modern version" of KVL and then teach students that KVL always holds is the wrong thing to do as you set them up for making the same embarrassing mistake that both Dirk Van Meirvenne and ElectroBOOM made.Carlos SotoPost authorThe issue in short is that due to changing magnetic field virtual inductors not physical are created in the loop due to Lenz Law.This is what is called self inductance.Piece of cake.

gnan tejaPost authorSir could you please explain why is the electric field in a perfect inductor zero? Doesn't the changing magnetic flux induce an electric field in the super conducting material which makes the inductor?

rajas walavalkarPost authorHi sir,

Have a question, because of the induced EMFs we have an induced current flowing through the circuit. Now the polarity of the voltage is different on both the resistors, but the current is flowing in the same loop. Thus for one of the resisitor the current is flowing from low voltage to high voltage end, is that possible. Please let me know your thoughts on this Sir

RK GAMING ytPost authorHe made us understnd why do phy teachers even phy teacher may not khlnow this

.. 😂 naming single line with 2 diff. Words

Jayraj KareliaPost authorJason on the next day

Where are the rest of the students

Child:aaaaa they went for the eightox.?..

chandan sinhaPost author🙂

Jaideep SharmaPost authorYou are a great legend sir. You have really made me love Physics!

Gilbert PPost authorIsn't it the convention that a pure resistance modelized by "R" shall be passive hence not sensitive to magnetic effect ? So if you want to modelize a system which is sensitive to magnetic flux variation, don't you need to add an inductance "L" attached to each "R" ?