This website contains all my 94 course lectures (8.01, 8.02 and 8.03) with improved resolution. They also include all my homework problem sets, my exams and the solutions. Also included are lecture notes and 143 short videos in which I discuss basic problems. ENJOY!
That RLC demonstration, especially the part where you insert the ferromagnetic material into the solenoid, was wonderful! Even though I am quite used to RLC circuits being a senior EE student, that was still insightful and enjoyable to watch :D
the bulb used for the resonance demo (40:59) appears to have fringes!! can it be some kind of interference i'm not aware of? or maybe it has something to do with the camera...
I have a question about relationship between I max and Z = (sqrt(R^2 + (wL - (1/wc))^2). Throughout your experiment during the lecture, I understood that I max’s value could only be interfered with the R, not Z. However, according to 8:18, I max is determined also with Z value, which lead me to confusion that different i max values could consequence different L or C value. I tried to figure out the question by using external resources, but I couldn’t end up with a solution. If possible, please answer my question so that I could stay on the right track! Thank you always
why will there be no current even in a high quality system when slighty off resonance? 20:30 I get that the width is decreased but there is still current going up according to the resonance curve. Isnt it only when above resonance its start to die down due to the dominance of the inductor fighting back as you mentioned?
Dear Walter Lewin, i have seen all Lectures in this Playlist(8.02x) and read the Lecture Notes and I have been thinking about one Problem for 3 hours now. If in a L-R circuit there cannot be an Eletric-Field in the Solanoid, where is the Force coming from, thats "slowing the electons down" because of the Solanoid when you switch the circuit on. Doesnt the changing Magnetic field, that is induced in the Solanoid by the initial Current, induce again an Electric FIeld in the Solanoid wich opposes the moving electrons, therefore slowing them down? Thanks for your Lectures they help me a lot! Maxi
E field in an ideal solenoid is ALWAYS zero as an ideal solenoid is made of superconductive material. Maxwell's equations tell you what will happen. Newton's equations tell you how gravity works. Newton did not know why his equations work. The same can be said of Maxwell's eqs. If I correctly used Maxwell's eqs, then you should be happy. NO changing current in a solenoid can produce an E field in the solenoid (MAXWELL!!).
Thanks for the fast answer! I by no means doubt your correct use of Maxwells Equations. But doesnt Maxwell himself say, that a change in Magneticfield, which certainly is the case here, induces an Electricfield. As far as i know the Efield will be perpendicular to the Magneticfield. In a Solanoid this means the induced Efield will oppose the Current. However as you said there cannot be an Electricfield in an ideal Solanoid. So my conclusion so far is, that there somehow has to be another Electricfield that cancels out the induced Electricfield. Is this correct? And could you describe, where this second electricfield is coming from? I hope i described my Problem good enough. Kind regards Maxi
The closed loop integral of E dot dl in the RL circuit must be --dphi/dt at any moment in time. Any E field induced or not induced must be in the resistance not in an ideal self inductor.
If I have a superconductive wire vertically up, any E-field perpendicular to the wire (pointing, e.g., from left to right) inside the wire would cause a charge polarization across the write: access electrons on the left, access protons on the right. That would cause an opposing E field and the polarizing current across the wire will immediately stop on time scales less than pico sec and the net E-field across the wire would be zero This is very different from an E-field in the direction of the wire. That is NOT allowed as the conductivity is infinitely high.
Again thanks for your fast reply. I came up with an explanation to my Problem. I guess that the changing magneticfield in the solanoid will induce an efield1 that is tangential to the current in the solanoid (not outside of the solanoid) and opposes it. However this efield1 will immediately by cancled out by another efield2, created by an higher density of electrons in the wire outside the solanoid. These two Efields are dependent on each other and will always in total be zero. Therefore Faradays Law is fulfilled. Do you think this is a reasonable conception? Regards Maxi
1:50 Sir , How do the electrons move inside a current carrying conductor if there is no electric field inside conductors ? I agree that if resistance is zero , potential difference is zero ..which means electric field has to be zero . From what I understand ,in regular current carrying wire ,there is small resistance ==> small potential difference ==> small electric field .so electron will move in opposite direction of field and current will be established (please , correct me if I'm wrong) . But if it were super conductor R = 0 ==> V= 0 ==> E =0 , still currents flows through it . How does it do so , if there is no electric field ?? ^.^
U said that by putting that rod u lower the resonance frequency by increasing inductance but if we say that when I took out that rod L decreases and impedance decreases hence current increases can't we say like this please tell
Dear Lewin This question is regarding the RLC resonance circuit. Say I have a voltage input (AC) at 50HZ. and My RLC circuit has a resonant frequency at 75hz. As i am varying L and C such that the system natural frequency can match the input frequency. So by changing L and C for five different sets I attained natural frequency of RLC system as ( 75, 70 , 65, 60, 55 and finally 50HZ) during this process, my capacitor is getting charged and storing some energy. As it reaches 50HZ my system attains resonance mode, and the reactance of inductor and capacitor gets nullified among themselves.So my question is whats happening to the energy stored in the capacitor and inductor when it reached the resonance mode.
Will the capacitor starts discharging? after attaining resonance? I mean, what happens to the energy stored in both capacitor and inductor after the resonance mode is attained?
I(t)=(V_o/R)cos(wt). w is 2pi*f=314 rad/sec. The power(t) dissipated in the resistor is R*[I(t)^2], I(t) is dQ(t)/dt, the voltage(t) over the capacitor is Q(t)/C. Thus the energy(t) stored in the capacitor is 0.5*C*V(t)^2. The energy(t) stored in the magnetic field of the self inductor is 0.5*L*[I(t|^2.
The charge Q on the capacitor changes with a function as A*sin(wt). A can be calculated as I(t) = dQ(t)/dt and I gave you the solution of I(t). Thus every 1/50 sec, the charge on the capacitor will go through zero. In other words twice in a cycle will the capacitor have no charge and in one cycle the plates will twice have a maximum possible charge. However, the polarity changes once per cycle.
Sir, does the electric field in the closed loop integral of (E.dl) mean the superposition of conservative and nonconservative electricfield ? Please sir , specify the kind of electric field we integrate in Faraday's law..
integral of (E.dl) means exactly what it says. IR is the integral of E dot dL through a resistor. The closed loop integral of E dot dL = -dphi/dt or -LdI/dt, -dphi/dt or -LdI/dt is the induced EMF! This is my last msg on this topic.
Hi Professor, thank you for the great lecture. I was wondering, if we want to find the magnetic field generated by the inductor in the circuit (as a formula), must we set up another set of equations because the changing magnetic field has an effect on the current through the circuit? or can we simply take B=mu0*I*N/Length like it is a solenoid (and we calculate I = Imax*cos(wt-phi) as that effect of the changing magnetic field is already accounted for in the equation I=Imax*cos(wt-phi)? Thank you for your help.
Very interesting. Sir, what about the crack detector? How it work? Crack detector is mostly used to detect if any small crack (big crack moreover) in a metal, probably outside look fine but inside is not.
Dear Professor, I ve made an experiment when I had an RLC circuit connected in series and I subjected it to different frequencies(1-50kHz). I run it with Vrms =1 sin wave on signal generator. Ive also added a voltmeter parallel to the whole circuit(the way that it measures the Vr+Vl+Vc, voltage of whole circuit). Ive noticed that when I changed frequency, the voltage on voltmeter also changed (by really small amount something like 5/1000 V with every 1kHz), even if I kept Vrms=1V on singal generator. So at the end of the experiment, to make the circuit run on Vrms=1V i had to set the signal generator to Vrms=1.05, so the voltmeter shows voltage = 1V. Do you know why this happens? The overall circuit voltage from what I know should not be dependent on frequency.
>>>the way that it measures the Vr+Vl+Vc, voltage of whole circuit>>> *was your voltmeter attached to the poles of your AC power source?* >>>to make the circuit run on Vrms=1V i had to set the signal generator to Vrms=1.05 so the voltmeter shows voltage = 1V>>> *unclear what you did and what this means.*
1. I have attached it to the wires, which were connected to the signal generator. 2. I wanted to keep constant voltage of the circuit throughout the experiment V=1V. To check if that is true, I attached a voltmeter parallel to the circuit. During the experiment I noticed that actual voltage (the one I measured on voltmeter) is different from the one i set it on signal generator and it changes with change of frequency. For, lets say f=1kHz, the voltmeter value was equal to set value on signal generator, but with every increase of frequency, I always had to increase the voltage on generator for the curcuit to run on exactly at 1V (not 0.99V), cus the actual voltage (measured on voltmeter) was decreasing every time I increased the frequency by 1kHz. So at 50kHz I had to set my Vrms to 1.05V to keep the circuit work on voltage of 1V
>>> I have attached it to the wires, which were connected to the signal generator.>>> I really do not know what this means. Is this your AC power supply?
Beste professor, in de afleiding van de differentiaalvergelijking bij het RLC-circuit zegt u dat er in de inductiespoel en in de geleiders geen elektrisch veld is en deze dus niet bijdragen aan de integraal rond de kring. Nu moet er toch wel een elektrisch veld zijn daar, anders kan er toch geen stroom lopen aangezien j = o * E?
Sir, this question I am about to ask may not have anything related to the content of this lecture. I always wonder where are the transmitter and receiver of signals in a cellphone. Handheld radio has an antenna type thing but in a smartphone I dont see any such things.
Regarding TVs and radios: If an LRC circuit can only have one resonance frequency, how can we view many different channels and stations. Does it have to do with the amount of satellites that transmit signal?
as you turn the knob that changes channels you are changing the value of the capacitor and that changes the resonance frequency. I suggest you google "variable capacitors" and click on "images".
Will the capacitor in the LRC circuit blow? I thought they can only be charged one way and if we expose it to an AC power supply it will get charged the other way sometimes? Thank you in advance.
Lectures by Walter Lewin. They will make you ♥ Physics. In the LRC Circuit, the current goes oscillates to and fro... but i thought that if you charge a capacitor in the reverse direction it might explode.. In this case, won't putting a capacitor in an AC Circuit make it blow? Thank you.
Sir, if we convert cos to sin in formula, then tangent of phi becomes R/X from X/R. and that means phase angle has changed.But it shouldn't change because inductance and capacitance doesn't change. How Come it changes? thank you
Hi Professor. I love your lectures!! I had a doubt regarding the Electric Field inside an inductor. If the E field is 0, doesn't that mean the electrons (when inside the inductor) no longer have a particular direction to flow, as opposed to a resistor?
YOU misquoted me. The E-field inside an ideal self inductor is ALWAYS zero as it has no resistance. Thus it is also zero when current is going through it. But conductors that have resistance, the E-field is ONLY zero when there is no current going through.
Sir, you mentioned that we cannot have electric field inside an inductor but what about that induced electric field due to the changing current through the inductor ?
@Mahmoud Desokey The current in a circuit which has an induced EMF is I=EMF/R. R is the total resistance of the circuit. *An ideal inductor has zero resistance. There can be no E-field inside a superconductor but, of course, any current can go through a superconductor. That's the basic idea in the LHC! use google for more information.*
Please sir ,Does the impedance of RLC circuit change with time ? I think that it changes since in the beginning the opposition of the inductor is maximum and then the opposition becomes zero when di\dt is zero .for the capacitor the the opposition increases as it charges and then decreases .
Sorry, but I cannot change Maxwell's equations. I can therefore not anything to the clarity of this lecture. This is very basic 1st year college physics.
Dear Professor i follow book electricity and magnetism by "edward mills purcell" along with your lectures and he was a noble prize winner. However in rlc circuit he uses kirchoff loop equation. Does some kind of modified kirchoff exists or its simply wrong?
Purcell could not possibly have made a mistake. In RLC circuit he must have used Faraday's Law. If you then bring the -dphi/dt turn from the right to the left in the eq. You have 3 terms that add up to ZERO. But that is NOT KVL; it is still Faraday's Law. 5 + 3 = 8 that is Faraday's law. 5 + 3 -8 = 0 is still Faraday's law. Send me the page of Purcell. I may have a copy at home.
I am looking at page 298 of Purcell. his LdI/dt term is Faraday. *He nowhere uses the word Kirchhoff.* His eq.2 is Faraday's Law. I treat the RLC circuit slightly diffrently. The closed loop integral of E dot dL = --dphi/dt = --LdI/dt . *IR + Q/C = - LdI/dt* Look at Purcell's eq (1) and (2). That's what Purell also has. He brought the -LdI/dt term to the left. That's Faraday's Law.
The book i am refering to is the 3rd edition where morin has co authored. Page 403 reads Our strategy will be the following. We will write kirchhoff loop equation as we did above but instead of solving it directly,we will solve a slightly modified equation in which .... (In the next paragraph) The kirchoff loop equation for the series RLC circuit in Fig.8.10 is LdI/dt+RI(t)+Q(t)/C=E0coswt.(eq 8.46)
I told you that Purcell could not have done that wrong and I showed you how Purcell deals with RLC circuits and he did it right. I can not make any judgements about books that I do not have.
sir in resonance condition the LCR circuit behave a resistive circuit , then why we need to put L and C in the circuit , just with a resitor only we can get a resistive circuit?
+Hrishikesh... Part of the reason is to explain and demonstrate how L and C impedance components behave in an AC circuit... but the answer to your question is to ask you a question myself: how do you think basic filters (for audio/radio/microwave signals) do what they do?
LRC circuit is also commonly known as a filter, designed to allow a narrow band of frequencies through the filter and block those above and below the band. A resistor alone would have no filtering effect. An (imperfect) mechanical analogy is a two screen filter through which gravel of many sizes including sand is poured. The capacitor is the first screen with medium openings that blocks the largest rocks. Medium rocks and sand pass through. The inductor is a fine screen that allows only sand to pass through. In between the screens one gets only medium rocks, the desired size. I say imperfect analogy because there also must be a mechanism to continually clear away the large rocks blocked by the capacitor screen, and the medium rocks must be continually removed after a short time to more rocks and sand can come in and the sand then filtered out by the inductor screen. You could also use long trough shaped screens set at an angle, say 30 deg, so that the large rocks run all the way down the capacitor trough into a buckt while medium and sand fall through to the second inductor screen. The inductor screen must also be set at an angle so that all the sand falls through while the medium rocks flow all the way down the trough into a bucket. The resistor slows the rate at which the mixed gravel enters the two screen assembly. Then you have to use your imagination to handle the reverse of voltage and current....flip the positons of first and second screens, and .then at resonance the maximum amount of medium gravel will flow into it's bucket
Sir i think kirchoff law is valid because E bar is a conservative field so its cyclic integral must be equal to zero... also -dibydt is a non conservative so if possible please explain
you are mistaken. I am sorry but I cannot help you. I realize that Faraday's Law is difficult for most people but I cannot change Maxwell's eqs. The Emf generated by Faraday's Law creates a non-conservative E-field. Therefore the closed loop integral of E dot dL is not zero; it's -L*di/dt. If you need more help I suggest you use google.
@@lecturesbywalterlewin.they9259 sir yes i agree with u on that but in the video u said kirchoff is not valid which i think kirchoff is E conservative so it must be zero..
I suggest you brush up on your Physics knowledge. I realize that Faraday's Law is not easy for many students. I am sorry but I cannot help you. Use google. Kirchhoff is only valid if the closed loop integral of E dot dL is zero.
maybe this is of some use to you. These are my lecture notes of Lect #20 of 8.02. Professor Belcher's notes of MIT: Most Physics College Books have it WRONG!: freepdfhosting.com/0813df09f5.pdf
yes they arte active, I use them every day. Many students do not know how to get them. I therefore made 3 playlists: "8.01 Homework, Exams, Solutions & Notes" same for 8.02 and 8.03
Sir since we know that in pure inductive circuit current lags behind by voltage what if we have two inductor then what will be the phase difference between input and output voltage in AC
Sir , shouldnot we consider the induced current in the resistor when setting up the differential equation ? And if we consider that induced current the integral of (E.dl) through the resistor will be (iR_induced current .R), am i right ?
@@lecturesbywalterlewin.they9259But shouldnot we take into account the induced current through the resistor while doing the integral of E.dl or do you mean that the induced emf represent this current ? Thanks
Yes! I've seen your lectures. They are really really great. I'm an physics student and i also teach to other students. The most significant thing I've learn from your lectures it's the correct interpretation of Faraday's Law. I had to re-learn a lot of things. And I can't yet understand what I asked you before. If the electric field is not conservative, how can i speak of a Vc potential? Thank you very much!
Potential difference is no longer a meaningful concept if the E-field is not conservative. The integral of E dot dL between point A and B can be very different depending on the path. Thus, it's meaningless (even misleading) to call that integral the potential difference because you will have 2 different values. I suggest you read my Lecture Notes of Lect #16. I posted them below the thumbnails of the videos.
RLC circtui. If you place a Voltmeter over the capacitor, you measure V_c(t)=Q(t)/C. I discuss the LRC circuit in great detail in my lectures. Emf*cos(wt)+ IR + V_c + 0 = -LdI/dt. The integral of E dot dL through the wire of an ideal self inductor is ZERO!
Sir when I set differential equation of LC circuit I get -Vc=-Ldi/dt that is equal to Vc-Ldi/dt=0 I am wrong somewhere kindly guide because actually right ans is Vc+ Ldi/dt=0
if the inductor is an "ideal" inductor, than it's conductivity is infinitely high (superconductive materials have zero ohmic resistance). A current can flow through it, but there cannot be any E-field inside as that would mean that there is a potential difference over the inductor which would lead to an infinite current. I mention this in several of my 8.02 lectures. Also use google or ask Quora.
Thank you, Professor! Your lectures are so detailed that I rewind the video and play it again, even if I miss a second of it! Thank you for giving us the opportunity to access them! Could you please give me your email ID so that I can ask you some questions? Thanks, once again!! :)
Professor, I have a few questions. 1) What is the difference between an ideal conductor and a superconductor? 2) Why do we assume that IDEAL INDUCTORS are made of superconducting material and not using ideal conducting wire ( which also has R = 0 )? 3) Sir, you mentioned in one of your earlier lectures that in an IDEAL INDUCTOR, the Electric Field = 0, ALWAYS. If it is so, how do the electrons know in which direction they are supposed to move?
It's a matter of semantics. Yes in an superconducting inductor the E-field is always zero. But a current can go through it depending on the circuit because there is an EMF somewhere else and thus a current. Thus electrons know which way to go.
Lectures by Walter Lewin. They will make you ♥ Physics.Professor,why is resistance of a self inductor always zero?ANd can the AC VOLTAGE be taken as Vzero(sine omega*t)?
It's ONLY zero if the self inductor is made of super-conductive material. We call that an "IDEAL" self inductor. The E-field inside super-conductive material is always ZERO. If you have an AC power supply (like your electricity at home) then it can be described as Vzero(sine omega*t) or (cos omega*t).
8.01 Physics Hans C. Ohanian Physics Volume 1 2nd edition W.W. Norton & Company ISBN 0-393-95748-9 8.02 Physics for Scientists & Engineers by Douglas C. Giancoli. Prentice Hall Third Edition ISBN 0-13-021517-18 8.03 Vibrations and Waves by Anthony French CRC Press ISBN 9780748744473 8.03 Electromagnetic Vibrations, Waves and Radiation by Bekefi and Barrett. The MIT Press ISBN 0-262-52047-8
This website contains all my 94 course lectures (8.01, 8.02 and 8.03) with improved resolution. They also include all my homework problem sets, my exams and the solutions. Also included are lecture notes and 143 short videos in which I discuss basic problems.
ENJOY!
I just cannot express how much i appreciate your ability to teach!!
+Jamil F Thank you!
Tremendous preparation before delivering lecture. Simplicity shows painstaking work in knitting thing together.
👍👍👍👍
indeed - 60-80 hr of preparation for me for each lecture
Yes sir, completely agree
World can not be more grateful for this series. Thank you prof. Lewin.
That was quite a large part of subject "Circuit Analysis" covered in 50 minutes! Brilliant!
:)
That RLC demonstration, especially the part where you insert the ferromagnetic material into the solenoid, was wonderful! Even though I am quite used to RLC circuits being a senior EE student, that was still insightful and enjoyable to watch :D
3db AND RMS explained....MIND BLOWN!!! @19:45 -THANKS!
Beautiful lecture! Very conceptual with a fantastic demonstration. I am loving Physics with Professor Walter Lewin teaching it. Great!!
the bulb used for the resonance demo (40:59) appears to have fringes!!
can it be some kind of interference i'm not aware of?
or maybe it has something to do with the camera...
I can't see any fringes - sorry
@@lecturesbywalterlewin.they9259
i misused the word fringe, my bad!
Prof Lewin is the greatest teacher of Physics. Thanks prof Lewin.
I have a question about relationship between I max and Z = (sqrt(R^2 + (wL - (1/wc))^2).
Throughout your experiment during the lecture, I understood that I max’s value could only be interfered with the R, not Z.
However, according to 8:18, I max is determined also with Z value, which lead me to confusion that different i max values could consequence different L or C value.
I tried to figure out the question by using external resources, but I couldn’t end up with a solution.
If possible, please answer my question so that I could stay on the right track!
Thank you always
why will there be no current even in a high quality system when slighty off resonance? 20:30
I get that the width is decreased but there is still current going up according to the resonance curve. Isnt it only when above resonance its start to die down due to the dominance of the inductor fighting back as you mentioned?
Excellent way of teaching Physics. Thanks and Regards 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
Dear Walter Lewin, i have seen all Lectures in this Playlist(8.02x) and read the Lecture Notes and I have been thinking about one Problem for 3 hours now. If in a L-R circuit there cannot be an Eletric-Field in the Solanoid, where is the Force coming from, thats "slowing the electons down" because of the Solanoid when you switch the circuit on. Doesnt the changing Magnetic field, that is induced in the Solanoid by the initial Current, induce again an Electric FIeld in the Solanoid wich opposes the moving electrons, therefore slowing them down?
Thanks for your Lectures they help me a lot! Maxi
E field in an ideal solenoid is ALWAYS zero as an ideal solenoid is made of superconductive material. Maxwell's equations tell you what will happen. Newton's equations tell you how gravity works. Newton did not know why his equations work. The same can be said of Maxwell's eqs. If I correctly used Maxwell's eqs, then you should be happy. NO changing current in a solenoid can produce an E field in the solenoid (MAXWELL!!).
Thanks for the fast answer! I by no means doubt your correct use of Maxwells Equations. But doesnt Maxwell himself say, that a change in Magneticfield, which certainly is the case here, induces an Electricfield. As far as i know the Efield will be perpendicular to the Magneticfield. In a Solanoid this means the induced Efield will oppose the Current. However as you said there cannot be an Electricfield in an ideal Solanoid. So my conclusion so far is, that there somehow has to be another Electricfield that cancels out the induced Electricfield. Is this correct? And could you describe, where this second electricfield is coming from? I hope i described my Problem good enough. Kind regards Maxi
The closed loop integral of E dot dl in the RL circuit must be --dphi/dt at any moment in time. Any E field induced or not induced must be in the resistance not in an ideal self inductor.
If I have a superconductive wire vertically up, any E-field perpendicular to the wire (pointing, e.g., from left to right) inside the wire would cause a charge polarization across the write: access electrons on the left, access protons on the right. That would cause an opposing E field and the polarizing current across the wire will immediately stop on time scales less than pico sec and the net E-field across the wire would be zero This is very different from an E-field in the direction of the wire. That is NOT allowed as the conductivity is infinitely high.
Again thanks for your fast reply. I came up with an explanation to my Problem. I guess that the changing magneticfield in the solanoid will induce an efield1 that is tangential to the current in the solanoid (not outside of the solanoid) and opposes it. However this efield1 will immediately by cancled out by another efield2, created by an higher density of electrons in the wire outside the solanoid. These two Efields are dependent on each other and will always in total be zero. Therefore Faradays Law is fulfilled. Do you think this is a reasonable conception? Regards Maxi
Dr. Lewin, do you know the physical reason why the resonance bandwidth is independent of capacitor C in a series LRC circuit. Thank you.
What a huge subject these LRC circuits are! Thank you for presenting it in such an interesting manner!
Glad you like them!
32:30 Is it not just the rms value of current?
1:50 Sir , How do the electrons move inside a current carrying conductor if there is no electric field inside conductors ? I agree that if resistance is zero , potential difference is zero ..which means electric field has to be zero . From what I understand ,in regular current carrying wire ,there is small resistance ==> small potential difference ==> small electric field .so electron will move in opposite direction of field and current will be established (please , correct me if I'm wrong) .
But if it were super conductor R = 0 ==> V= 0 ==> E =0 , still currents flows through it . How does it do so , if there is no electric field ??
^.^
I cover this in detail in one of my 8.02 lectures. Please watch it.
U said that by putting that rod u lower the resonance frequency by increasing inductance but if we say that when I took out that rod L decreases and impedance decreases hence current increases can't we say like this please tell
I know that talent is nothing. Your hard work proves that you are not talented but your hard work made you so called "Talented sir "
Love from INDIA.
Dear Lewin
This question is regarding the RLC resonance circuit.
Say I have a voltage input (AC) at 50HZ. and My RLC circuit has a resonant frequency at 75hz. As i am varying L and C such that the system natural frequency can match the input frequency. So by changing L and C for five different sets I attained natural frequency of RLC system as ( 75, 70 , 65, 60, 55 and finally 50HZ) during this process, my capacitor is getting charged and storing some energy. As it reaches 50HZ my system attains resonance mode, and the reactance of inductor and capacitor gets nullified among themselves.So my question is whats happening to the energy stored in the capacitor and inductor when it reached the resonance mode.
Will the capacitor starts discharging? after attaining resonance?
I mean, what happens to the energy stored in both capacitor and inductor after the resonance mode is attained?
the charge on the capacitor is Q(t)
I(t) is dQ(t)/dt
You keep asking questions that I have already answered in my first msg. The solution I sent you is only valid for resonance as I used wL=1/wC.
I(t)=(V_o/R)cos(wt). w is 2pi*f=314 rad/sec. The power(t) dissipated in the resistor is R*[I(t)^2],
I(t) is dQ(t)/dt, the voltage(t) over the capacitor is Q(t)/C. Thus the energy(t) stored in the capacitor is 0.5*C*V(t)^2.
The energy(t) stored in the magnetic field of the self inductor is 0.5*L*[I(t|^2.
The charge Q on the capacitor changes with a function as A*sin(wt). A can be calculated as I(t) = dQ(t)/dt
and I gave you the solution of I(t). Thus every 1/50 sec, the charge on the capacitor will go through zero. In other words twice in a cycle will the capacitor have no charge and in one cycle the plates will twice have a maximum possible charge. However, the polarity changes once per cycle.
Sir, in the demonstration why the ferromagnetic material wasn't attracted and went in as it did in a previous lecture?
Sir, does the electric field in the closed loop integral of (E.dl) mean the superposition of conservative and nonconservative electricfield ? Please sir , specify the kind of electric field we integrate in Faraday's law..
integral of (E.dl) means exactly what it says. IR is the integral of E dot dL through a resistor. The closed loop integral of E dot dL = -dphi/dt or -LdI/dt, -dphi/dt or -LdI/dt is the induced EMF! This is my last msg on this topic.
Hi Professor, thank you for the great lecture.
I was wondering, if we want to find the magnetic field generated by the inductor in the circuit (as a formula), must we set up another set of equations because the changing magnetic field has an effect on the current through the circuit? or can we simply take B=mu0*I*N/Length like it is a solenoid (and we calculate I = Imax*cos(wt-phi) as that effect of the changing magnetic field is already accounted for in the equation I=Imax*cos(wt-phi)? Thank you for your help.
at any given current the field is a dipole field. The stronger the current the stringer the field
Lectures by Walter Lewin. They will make you ♥ Physics. Thank you Professor!
Thank you for the good work.
Very interesting. Sir, what about the crack detector? How it work? Crack detector is mostly used to detect if any small crack (big crack moreover) in a metal, probably outside look fine but inside is not.
Learn to teach from the best.
World's best physicist and professor
:)
Dear Professor,
I ve made an experiment when I had an RLC circuit connected in series and I subjected it to different frequencies(1-50kHz). I run it with Vrms =1 sin wave on signal generator. Ive also added a voltmeter parallel to the whole circuit(the way that it measures the Vr+Vl+Vc, voltage of whole circuit). Ive noticed that when I changed frequency, the voltage on voltmeter also changed (by really small amount something like 5/1000 V with every 1kHz), even if I kept Vrms=1V on singal generator. So at the end of the experiment, to make the circuit run on Vrms=1V i had to set the signal generator to Vrms=1.05, so the voltmeter shows voltage = 1V. Do you know why this happens? The overall circuit voltage from what I know should not be dependent on frequency.
>>>the way that it measures the Vr+Vl+Vc, voltage of whole circuit>>>
*was your voltmeter attached to the poles of your AC power source?*
>>>to make the circuit run on Vrms=1V i had to set the signal generator to Vrms=1.05 so the voltmeter shows voltage = 1V>>>
*unclear what you did and what this means.*
1. I have attached it to the wires, which were connected to the signal generator.
2. I wanted to keep constant voltage of the circuit throughout the experiment V=1V. To check if that is true, I attached a voltmeter parallel to the circuit. During the experiment I noticed that actual voltage (the one I measured on voltmeter) is different from the one i set it on signal generator and it changes with change of frequency. For, lets say f=1kHz, the voltmeter value was equal to set value on signal generator, but with every increase of frequency, I always had to increase the voltage on generator for the curcuit to run on exactly at 1V (not 0.99V), cus the actual voltage (measured on voltmeter) was decreasing every time I increased the frequency by 1kHz. So at 50kHz I had to set my Vrms to 1.05V to keep the circuit work on voltage of 1V
>>> I have attached it to the wires, which were connected to the signal generator.>>>
I really do not know what this means. Is this your AC power supply?
Yes, signal generator is my AC power supply.
I will have to see a diagram of your complete set up
21:35 When I teach to my students, I give them the same example 🤣🤣🤣
If there is no potential difference across a solenoid, then what does the term "Irms*XL" even mean? (In a alternating voltage applied to a inductor)
The integral of E dot dL through the wires of an ideal conductor is ZERO at any moment in time. The induced emf in the circuit is -dphi/dt= -L*dI/dt
Lectures by Walter Lewin. They will make you ♥ Physics.
But what actually is V sub L?
V = Irms* XL = Irms * omega* L
watch my lecture and learn about notations. All what counts is the solution to the diff eq. That's the physics
Beste professor, in de afleiding van de differentiaalvergelijking bij het RLC-circuit zegt u dat er in de inductiespoel en in de geleiders geen elektrisch veld is en deze dus niet bijdragen aan de integraal rond de kring. Nu moet er toch wel een elektrisch veld zijn daar, anders kan er toch geen stroom lopen aangezien j = o * E?
there is an E field in the R and in the C but not in an ideal conductor.
Sir, this question I am about to ask may not have anything related to the content of this lecture.
I always wonder where are the transmitter and receiver of signals in a cellphone. Handheld radio has an antenna type thing but in a smartphone I dont see any such things.
use google all your answers are there
@@lecturesbywalterlewin.they9259 thank you
Regarding TVs and radios: If an LRC circuit can only have one resonance frequency, how can we view many different channels and stations. Does it have to do with the amount of satellites that transmit signal?
as you turn the knob that changes channels you are changing the value of the capacitor and that changes the resonance frequency. I suggest you google "variable capacitors" and click on "images".
Understood, thank you :)
Thanks for the speedy replies by the way! Very convenient for the student!
Will the capacitor in the LRC circuit blow? I thought they can only be charged one way and if we expose it to an AC power supply it will get charged the other way sometimes? Thank you in advance.
how many minutes into the lecture?
2:03
Thanks. :)
question not clear.
Lectures by Walter Lewin. They will make you ♥ Physics. In the LRC Circuit, the current goes oscillates to and fro... but i thought that if you charge a capacitor in the reverse direction it might explode.. In this case, won't putting a capacitor in an AC Circuit make it blow? Thank you.
Watch my lectures about RC RLC and CL DC and AC circuits.
Sir, if we convert cos to sin in formula, then tangent of phi becomes R/X from X/R. and that means phase angle has changed.But it shouldn't change because inductance and capacitance doesn't change. How Come it changes? thank you
Hi Professor. I love your lectures!! I had a doubt regarding the Electric Field inside an inductor. If the E field is 0, doesn't that mean the electrons (when inside the inductor) no longer have a particular direction to flow, as opposed to a resistor?
E-field is ONLY zero in a conductor if there is NO current going through it.
At 1:52 you mention that E is always 0 in an Inductor, doesn't that mean that a current never flows in an inductor?
E is ONLY zero in a conductor is no current is going through it.
YOU misquoted me. The E-field inside an ideal self inductor is ALWAYS zero as it has no resistance. Thus it is also zero when current is going through it. But conductors that have resistance, the E-field is ONLY zero when there is no current going through.
Lectures by Walter Lewin. They will make you ♥ Physics. Thanks a lot, sir.
why the ferromagnetic material did not get attracted by the magnetic field produced inside an inductor? can someone please explain?
Sir, you mentioned that we cannot have electric field inside an inductor but what about that induced electric field due to the changing current through the inductor ?
an ideal inductor has no resistance. Thus there can be no E-field inside the wires of the inductor.
@@lecturesbywalterlewin.they9259 Then , what about that induced current in the inductor due to nonconservative efield
@Mahmoud Desokey The current in a circuit which has an induced EMF is I=EMF/R. R is the total resistance of the circuit. *An ideal inductor has zero resistance. There can be no E-field inside a superconductor but, of course, any current can go through a superconductor. That's the basic idea in the LHC! use google for more information.*
Is that not a parallel resonance curve rather than a series one?
It's a plot of current, not impedance vs frequency therefore correct. My bad
how can we prove that delta omega=R/L?
but why not inr=troduce complex impedances? It makes understanding LRC circuits very easy
Please sir ,Does the impedance of RLC circuit change with time ? I think that it changes since in the beginning the opposition of the inductor is maximum and then the opposition becomes zero when di\dt is zero .for the capacitor the the opposition increases as it charges and then decreases .
Sorry, but I cannot change Maxwell's equations. I can therefore not anything to the clarity of this lecture. This is very basic 1st year college physics.
@@lecturesbywalterlewin.they9259 I know that it depends on frequency but it was just an intuition .thanks
what's the relationship between the inductor having no resistance and zero electric field inside?
correct if the inductor is made of superconducting wires
use google
Dear Professor i follow book electricity and magnetism by "edward mills purcell" along with your lectures and he was a noble prize winner. However in rlc circuit he uses kirchoff loop equation. Does some kind of modified kirchoff exists or its simply wrong?
Purcell could not possibly have made a mistake.
In RLC circuit he must have used Faraday's Law. If you then bring the -dphi/dt turn from the right to the left in the eq. You have 3 terms that add up to ZERO. But that is NOT KVL; it is still Faraday's Law. 5 + 3 = 8 that is Faraday's law. 5 + 3 -8 = 0 is still Faraday's law. Send me the page of Purcell. I may have a copy at home.
I am looking at page 298 of Purcell. his LdI/dt term is Faraday. *He nowhere uses the word Kirchhoff.* His eq.2 is Faraday's Law. I treat the RLC circuit slightly diffrently. The closed loop integral of E dot dL = --dphi/dt = --LdI/dt . *IR + Q/C = - LdI/dt* Look at Purcell's eq (1) and (2). That's what Purell also has. He brought the -LdI/dt term to the left. That's Faraday's Law.
@@lecturesbywalterlewin.they9259 professor,book's page number 403, eq 8.46. On top of that he says Kirchoff loop equation ... thank you for replying
The book i am refering to is the 3rd edition where morin has co authored. Page 403 reads
Our strategy will be the following. We will write kirchhoff loop equation as we did above but instead of solving it directly,we will solve a slightly modified equation in which ....
(In the next paragraph)
The kirchoff loop equation for the series RLC circuit in Fig.8.10 is LdI/dt+RI(t)+Q(t)/C=E0coswt.(eq 8.46)
I told you that Purcell could not have done that wrong and I showed you how Purcell deals with RLC circuits and he did it right. I can not make any judgements about books that I do not have.
sir in resonance condition the LCR circuit behave a resistive circuit , then why we need to put L and C in the circuit , just with a resitor only we can get a resistive circuit?
use google
Lectures by Walter Lewin. They will make you ♥ Physics. sir i didnot get this answer anywhere pls help me
+Hrishikesh...
Part of the reason is to explain and demonstrate how L and C impedance components behave in an AC circuit... but the answer to your question is to ask you a question myself: how do you think basic filters (for audio/radio/microwave signals) do what they do?
LRC circuit is also commonly known as a filter, designed to allow a narrow band of frequencies through the filter and block those above and below the band. A resistor alone would have no filtering effect.
An (imperfect) mechanical analogy is a two screen filter through which gravel of many sizes including sand is poured. The capacitor is the first screen with medium openings that blocks the largest rocks. Medium rocks and sand pass through. The inductor is a fine screen that allows only sand to pass through. In between the screens one gets only medium rocks, the desired size.
I say imperfect analogy because there also must be a mechanism to continually clear away the large rocks blocked by the capacitor screen, and the medium rocks must be continually removed after a short time to more rocks and sand can come in and the sand then filtered out by the inductor screen.
You could also use long trough shaped screens set at an angle, say 30 deg, so that the large rocks run all the way down the capacitor trough into a buckt while medium and sand fall through to the second inductor screen. The inductor screen must also be set at an angle so that all the sand falls through while the medium rocks flow all the way down the trough into a bucket.
The resistor slows the rate at which the mixed gravel enters the two screen assembly.
Then you have to use your imagination to handle the reverse of voltage and current....flip the positons of first and second screens, and .then at resonance the maximum amount of medium gravel will flow into it's bucket
Sir i think kirchoff law is valid because E bar is a conservative field so its cyclic integral must be equal to zero... also -dibydt is a non conservative so if possible please explain
you are mistaken. I am sorry but I cannot help you. I realize that Faraday's Law is difficult for most people but I cannot change Maxwell's eqs. The Emf generated by Faraday's Law creates a non-conservative E-field. Therefore the closed loop integral of E dot dL is not zero; it's -L*di/dt. If you need more help I suggest you use google.
@@lecturesbywalterlewin.they9259 sir yes i agree with u on that but in the video u said kirchoff is not valid which i think kirchoff is E conservative so it must be zero..
I suggest you brush up on your Physics knowledge. I realize that Faraday's Law is not easy for many students. I am sorry but I cannot help you. Use google. Kirchhoff is only valid if the closed loop integral of E dot dL is zero.
maybe this is of some use to you. These are my lecture notes of Lect #20 of 8.02. Professor Belcher's notes of MIT: Most Physics College Books have it WRONG!: freepdfhosting.com/0813df09f5.pdf
@@lecturesbywalterlewin.they9259 thanks a lot sir i will try my level best to understand it sir!!
Sir assignment and notes link are not active , i can't able to download .
yes they arte active, I use them every day. Many students do not know how to get them. I therefore made 3 playlists: "8.01 Homework, Exams, Solutions & Notes"
same for 8.02 and 8.03
Sir since we know that in pure inductive circuit current lags behind by voltage what if we have two inductor then what will be the phase difference between input and output voltage in AC
2 inductors (L1 and L2) in series will add up to L=L1+L2.
Pure inductive circuit: Inductor current lags inductor voltage by 90o.
Lectures by Walter Lewin. They will make you ♥ Physics. But what happen to phase difference in case of two inductor is it become zero
For L=0 watch my lectures
why current is proportional with capacitance in RC circuits but inverse proportional in RLC circuits?
solve thw diff eq
@@lecturesbywalterlewin.they9259 what is thw?
@@ugursoydan8187 solve the diff eqs
@@lecturesbywalterlewin.they9259 I have solved them but I only ask to understand it intuitively. thank you
how can we find the derivation of this formula?
Respect to Sir Walter Lewin from Pakistan 🎉
Sir , shouldnot we consider the induced current in the resistor when setting up the differential equation ? And if we consider that induced current the integral of (E.dl) through the resistor will be (iR_induced current .R), am i right ?
I cannot add to the clarity of this lecture. In the case of an induced emf in a closed loop of a circuit, the EMF is part of the diff eq.
@@lecturesbywalterlewin.they9259But shouldnot we take into account the induced current through the resistor while doing the integral of E.dl or do you mean that the induced emf represent this current ? Thanks
how many minutes into my lecture?
2.19
diff eq is 100% correct. It's an application of Maxwell's eqs. I cannot help you any more. Use google. Faraday's Law in action!
Wonderful lecture and thanks.
If the electric field is not conservative. How can you uniquely define a Vc potential? Thanks! :D
watch my lectures
Yes! I've seen your lectures. They are really really great. I'm an physics student and i also teach to other students. The most significant thing I've learn from your lectures it's the correct interpretation of Faraday's Law. I had to re-learn a lot of things. And I can't yet understand what I asked you before. If the electric field is not conservative, how can i speak of a Vc potential? Thank you very much!
Potential difference is no longer a meaningful concept if the E-field is not conservative. The integral of E dot dL between point A and B can be very different depending on the path. Thus, it's meaningless (even misleading) to call that integral the potential difference because you will have 2 different values. I suggest you read my Lecture Notes of Lect #16. I posted them below the thumbnails of the videos.
Thank you for your answer! Aha. Ok. I understand. So, when you conect a Voltimeter to a Capacitor in a LRC cirtcuit, What are you measuring?
RLC circtui. If you place a Voltmeter over the capacitor, you measure V_c(t)=Q(t)/C. I discuss the LRC circuit in great detail in my lectures. Emf*cos(wt)+ IR + V_c + 0 = -LdI/dt. The integral of E dot dL through the wire of an ideal self inductor is ZERO!
Sir when I set differential equation of LC circuit I get -Vc=-Ldi/dt that is equal to Vc-Ldi/dt=0 I am wrong somewhere kindly guide because actually right ans is Vc+ Ldi/dt=0
Watch my lecture about LRC make R=0
+Lectures by Walter Lewin. They will make you ♥ Physics. sir I have confusion about LC circuit without power supply with capacitor fully charged
en.wikipedia.org/wiki/LC_circuit
why is the field inside an inductor zero?
if the inductor is an "ideal" inductor, than it's conductivity is infinitely high (superconductive materials have zero ohmic resistance). A current can flow through it, but there cannot be any E-field inside as that would mean that there is a potential difference over the inductor which would lead to an infinite current. I mention this in several of my 8.02 lectures. Also use google or ask Quora.
All this we have learnt in class 11th while preparing for JEE ADVANCED
why not discuss constant voltage case first?
I always wished I could study under this guy,....
I don't think I'll be going to MIT soon
Best lecture...best teacher...thank you sir very much😍Sir please make a detailed video on LC circuit😓😓
it's in my 8.02 lectures
Thank you, Professor! Your lectures are so detailed that I rewind the video and play it again, even if I miss a second of it! Thank you for giving us the opportunity to access them! Could you please give me your email ID so that I can ask you some questions?
Thanks, once again!! :)
:)
Professor, I have a few questions.
1) What is the difference between an ideal conductor and a superconductor?
2) Why do we assume that IDEAL INDUCTORS are made of superconducting material and not using ideal conducting wire ( which also has R = 0 )?
3) Sir, you mentioned in one of your earlier lectures that in an IDEAL INDUCTOR, the Electric Field = 0, ALWAYS. If it is so, how do the electrons know in which direction they are supposed to move?
It's a matter of semantics.
Yes in an superconducting inductor the E-field is always zero.
But a current can go through it depending on the circuit because there is an EMF somewhere else and thus a current. Thus electrons know which way to go.
Thanks a lot, Sir. Could you please elaborate a little more on the answer to the first two questions?
1.it's a matter of semantics
2. If someone write "ideal conductor" it means zero ohmic resistance - thus superconducting.
Lectures by Walter Lewin. They will make you ♥ Physics.Professor,why is resistance of a self inductor always zero?ANd can the AC VOLTAGE be taken as Vzero(sine omega*t)?
It's ONLY zero if the self inductor is made of super-conductive material. We call that an "IDEAL" self inductor. The E-field inside super-conductive material is always ZERO. If you have an AC power supply (like your electricity at home) then it can be described as Vzero(sine omega*t) or (cos omega*t).
thank you
Lectures by Walter Lewin. They will make you ♥ Physics.
May I know the reference books sir..? Please
8.01
Physics
Hans C. Ohanian
Physics
Volume 1
2nd edition
W.W. Norton & Company
ISBN 0-393-95748-9
8.02
Physics for Scientists & Engineers by Douglas C. Giancoli.
Prentice Hall
Third Edition
ISBN 0-13-021517-18
8.03
Vibrations and Waves by
Anthony French
CRC Press
ISBN 9780748744473
8.03
Electromagnetic Vibrations, Waves and Radiation
by Bekefi and Barrett.
The MIT Press
ISBN 0-262-52047-8
EXCELLENT
21:35 words of wisdom
I love you sir!
41:35 so amazing
" Physics WORK and you CAN'T see ! "
You Rock man
4:56
Electrons flow from negative to positive
Thanks professor
You are welcome
How how so good
I wish i could derive solutions for the second order differential equations, i cant find any help from google...
www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf
www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt1.pdf
Thank you very much professor
Later he will reach 1 million
Who is here back
i NEED SOMEBODY... HELP... NOT JUST ANYBODY.. HELP! WALTER LEWINNNNN
th-cam.com/video/6O0aPrSlqV8/w-d-xo.html
MIT must have really watered down their courses since I went there.
when did you go to MIT?
Thanks professor
You are welcome