Greetings from Poland. I really enjoy watching your experiences. Your lectures open minds thanks to their examples. I wish you a lot of health and perseverance.
Good to see you sir.....I will be in class 9th sir...day after tomorrow..till now in 8th...can you please tell me how to understand Electrochemistry......
Greetings sir. I have Neet examination going to held on 5th May this year. It has been around 13 months when i started preparing and it was going smooth until now. But whenever i tried hard for something i always end up messing things up. And the reason is that i lost confidence at last moments and I'm feeling the same this time. From the completion of my 12th exams last year , i gave my everything for this examination whether it time, health etc. Sir please help me out. I am loosing my hold on concepts which i used to have mastered .
Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Impressive. I should have done the math. The aproximate sign needs to be there as the integration is not perfect. The last page of his solution does not account for the sawtooth signal going below zero as it will in time but that is exactly what the aproximate sign is about. I'll study the math, i'll learn a lot front that ...
Thanks. One thing that I try hard to do is keep videos short otherwise I would ramble on and on. I now use a stopwatch while recording them. Guess who I copied that idea from!
Hello Walter Sir , Since you have crossed your 80's you had lots of experience and incidents that has taught you something. Please Walter sir , make a video explaining all those lessons and what our generation can do to make better life . Please walter sir , this will be life changing video for us .
One thing I'd like to point out is that I was able to post two links to desmos where I set up interactive charts for both Vin and Vout. Various parameters can be set in the charts, like Vmax, period and so on... I cannot repost the links, so you'd have to check my solution in his previous video showing the problem. Another thing is that I posted the actual equation that you can use the generate Vout. Here it is again: The output of the square wave is a triangular wave (isosceles), with an equation of the form: (2a/π)arcsin(sin(2πt/τ-π/2))+1, for some a. There was a lot in my solution and maybe it was too complicated, so it probably confused someone, but here some conclusions from it: Current out: i'ₖ = (1/L)·ₒ∫⁼ͭ vₖ dt Thus the output voltage is Vout = v'ₖ= i'ₖR = (R/L) ₒ∫⁼ͭvₖ dt For the time t< 1st half- period T₁: Vout =(R/L) ₒ∫⁼ͭvₖ dt =(R/L)vₖ ₒ∫⁼ͭ dt , as vₖ is constant over ½period =(R/L)v₁ t₁, with 0
Assumed Vout = R/L*integral(Vin)dt with Vout = R * i. Then derivation on each side: R * di /dt = R/L * Vin or Vin = L * di / dt. If i look at your first page with the DGL Vin = i * R + L * di / dt, then Vout would be zero ?
I have nearly a week left for my JEE exam and i feel like i am losing the foothold I've gained all 2 years of my 11th and 12th for mastering few topics of physics... More the progressing days, more i feel like i have forgotten everything.... Please recommend me some ways to atleast revise the concept in this week
Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
But there is a voltage drop across an inductor, which is proportional to the rate of change of the current passing through it. So, why does Keith take it zero??
there is no E-field in an ideal inductor. *Thus the integral of E dot dL through the inductor is zero.* However, if you attach a voltmeter over the inductor you read a value minus*dphi/dt bcoz you have now created a new loop which includes the voltmeter and the inductor.
Good question. There is an apparent voltage drop across the inductor as you will measure with your voltmeter, but it it is due to magnetic induction, rather than electrostatic interactions of charges. The field is not conservative, and does not add up to zero around a closed loop. One way to approach the inductor, is to treat its apparent voltage drop, as if it were a real voltage drop, ∆V = L*dꞮ/dt. Or a variant of this formula, ∆V=Ɪ*Z_L, where Z_L is the impedance of the inductor, Z_L = j*ω*L for the Fourier domain and Z_L = s*L in the Laplace domain. You then use it with Kirchhoff's laws, exactly as you would if it were any other kind of circuit. This approach is mathematically correct, but not physically correct. Because this apparent voltage drop across the inductor, isn't really a drop in electric potential. Instead, it is a consequence of a non-conservative electric field, which is more accurately modeled by Faraday's law of induction, rather than Kirchhoff's voltage loop law. A similar comparison of methods (Newton vs D'Alembert) happens for forces in an accelerating environment. One approach is to add up forces and equate them to m*a, which is the more physically accurate way to do it. Another approach is D'Alembert's principle, treating the apparent inertial force (-m*a) as if it were a force, and summing it with the other real forces, to add to zero. Both get you the same result mathematically, one just reflects the physics a lot more accurately.
@@lecturesbywalterlewin.they9259 Thank you sir for the clarification! Is it that we have a square wave as an input and for that part of the square wave where we have a constant voltage like for the time T/2 I think, the di/dt for the inductor is zero. Love from India!!
@@carultch Thank you sir for the clarification! Is it that we have a square wave as an input and for that part of the square wave where we have a constant voltage like for the time T/2 I think, the di/dt for the inductor is zero.
@@amitkumarchejara6042 The cutoff frequency is R/L, in rad/sec. Also called corner frequency, since this frequency appears as an approximate corner on a Bode plot. At frequencies much lower than R/L, this circuit is a low pass filter, very closely letting the original input voltage through directly. Instead of abruptly alternating like the input, it follows a quick exponential approach, and then settles to match the input. Then, the input abruptly changes again, and it exponentially approaches to the opposite phase of the square wave. At frequencies much higher than R/L, the circuit is an integrator, turning the square wave into a triangle wave. The exponential approach is very close to a diagonal line, and the opposite sloped diagonal line, after the input alternates. The output ultimately is a pattern of alternating exponential approaches, just cut short to look like a triangle wave before curvature can be noticeable. Close in frequency to R/L, and the output clearly appears as exponential approaches of alternating curvature. I call it as a sharkfin wave.
Sir I am preparing for JEE 2024. But I can't understand the Topic- *Entropy* and *Gibbs free Energy* from Thermodynamics . I am confused in these topics. Can you tell me in which lecture, have you covered those Topics? It will really help me sir😊❤
Hello sir I am your big fan I know you have devoted entire life to physics . But I am student of class 8 and how can I get faster in physics tell you something sir
Hi sir.. I am from Pakistan and big fan of you. Sir i study physics but i cannot solve problems myself Can you please guide me how can i good be in solving problems please 🥺
@@muhammadyousuf4248 Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
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Greetings from Poland. I really enjoy watching your experiences. Your lectures open minds thanks to their examples. I wish you a lot of health and perseverance.
Dr. Lewin, I visited MIT a few days ago and found 26-100: the best classroom ever!
Great channel, Dr. Lewin!
Good to see you sir.....I will be in class 9th sir...day after tomorrow..till now in 8th...can you please tell me how to understand Electrochemistry......
Dude specialised in physic
Electrochemistry is the study of electricity and how it relates to chemical reactions. hope this helps🐺
Always kind of love for u sir.. ❤
Greetings sir. I have Neet examination going to held on 5th May this year. It has been around 13 months when i started preparing and it was going smooth until now. But whenever i tried hard for something i always end up messing things up. And the reason is that i lost confidence at last moments and I'm feeling the same this time. From the completion of my 12th exams last year , i gave my everything for this examination whether it time, health etc. Sir please help me out. I am loosing my hold on concepts which i used to have mastered .
Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Hey, I love ur vids
Very good. Thanks 🙏
Thank you sir
First hello sir love your lectures
Impressive. I should have done the math. The aproximate sign needs to be there as the integration is not perfect. The last page of his solution does not account for the sawtooth signal going below zero as it will in time but that is exactly what the aproximate sign is about. I'll study the math, i'll learn a lot front that ...
Thanks. One thing that I try hard to do is keep videos short otherwise I would ramble on and on. I now use a stopwatch while recording them. Guess who I copied that idea from!
Hello Walter Sir ,
Since you have crossed your 80's you had lots of experience and incidents that has taught you something.
Please Walter sir , make a video explaining all those lessons and what our generation can do to make better life .
Please walter sir , this will be life changing video for us .
Thank you. I hope you are blessed through Jesus Name. Amen.
who is Jesus?
One thing I'd like to point out is that I was able to post two links to desmos where I set up interactive charts for both Vin and Vout.
Various parameters can be set in the charts, like Vmax, period and so on...
I cannot repost the links, so you'd have to check my solution in his previous video showing the problem.
Another thing is that I posted the actual equation that you can use the generate Vout.
Here it is again:
The output of the square wave is a triangular wave (isosceles), with an equation of the form:
(2a/π)arcsin(sin(2πt/τ-π/2))+1, for some a.
There was a lot in my solution and maybe it was too complicated, so it probably confused someone, but here some conclusions from it:
Current out:
i'ₖ = (1/L)·ₒ∫⁼ͭ vₖ dt
Thus the output voltage is
Vout = v'ₖ= i'ₖR = (R/L) ₒ∫⁼ͭvₖ dt
For the time t< 1st half- period T₁:
Vout
=(R/L) ₒ∫⁼ͭvₖ dt
=(R/L)vₖ ₒ∫⁼ͭ dt , as vₖ is constant over ½period
=(R/L)v₁ t₁, with 0
thank you
You're welcome
Matched
As a physics enthusiast could you tell me the career options in this field
Walter sir!! I want you to start teaching syllabus of JEE Advanced
Assumed Vout = R/L*integral(Vin)dt with Vout = R * i.
Then derivation on each side: R * di /dt = R/L * Vin or Vin = L * di / dt.
If i look at your first page with the DGL Vin = i * R + L * di / dt, then Vout would be zero ?
sorry for my ignorance sir.For me ,this seems like a disaster.I wish you well
I have nearly a week left for my JEE exam and i feel like i am losing the foothold I've gained all 2 years of my 11th and 12th for mastering few topics of physics... More the progressing days, more i feel like i have forgotten everything.... Please recommend me some ways to atleast revise the concept in this week
Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
But there is a voltage drop across an inductor, which is proportional to the rate of change of the current passing through it. So, why does Keith take it zero??
there is no E-field in an ideal inductor. *Thus the integral of E dot dL through the inductor is zero.* However, if you attach a voltmeter over the inductor you read a value minus*dphi/dt bcoz you have now created a new loop which includes the voltmeter and the inductor.
Good question. There is an apparent voltage drop across the inductor as you will measure with your voltmeter, but it it is due to magnetic induction, rather than electrostatic interactions of charges. The field is not conservative, and does not add up to zero around a closed loop.
One way to approach the inductor, is to treat its apparent voltage drop, as if it were a real voltage drop, ∆V = L*dꞮ/dt. Or a variant of this formula, ∆V=Ɪ*Z_L, where Z_L is the impedance of the inductor, Z_L = j*ω*L for the Fourier domain and Z_L = s*L in the Laplace domain. You then use it with Kirchhoff's laws, exactly as you would if it were any other kind of circuit.
This approach is mathematically correct, but not physically correct. Because this apparent voltage drop across the inductor, isn't really a drop in electric potential. Instead, it is a consequence of a non-conservative electric field, which is more accurately modeled by Faraday's law of induction, rather than Kirchhoff's voltage loop law.
A similar comparison of methods (Newton vs D'Alembert) happens for forces in an accelerating environment. One approach is to add up forces and equate them to m*a, which is the more physically accurate way to do it. Another approach is D'Alembert's principle, treating the apparent inertial force (-m*a) as if it were a force, and summing it with the other real forces, to add to zero. Both get you the same result mathematically, one just reflects the physics a lot more accurately.
@@lecturesbywalterlewin.they9259 Thank you sir for the clarification!
Is it that we have a square wave as an input and for that part of the square wave where we have a constant voltage like for the time T/2 I think, the di/dt for the inductor is zero.
Love from India!!
@@carultch Thank you sir for the clarification!
Is it that we have a square wave as an input and for that part of the square wave where we have a constant voltage like for the time T/2 I think, the di/dt for the inductor is zero.
@@amitkumarchejara6042 The cutoff frequency is R/L, in rad/sec. Also called corner frequency, since this frequency appears as an approximate corner on a Bode plot.
At frequencies much lower than R/L, this circuit is a low pass filter, very closely letting the original input voltage through directly. Instead of abruptly alternating like the input, it follows a quick exponential approach, and then settles to match the input. Then, the input abruptly changes again, and it exponentially approaches to the opposite phase of the square wave.
At frequencies much higher than R/L, the circuit is an integrator, turning the square wave into a triangle wave. The exponential approach is very close to a diagonal line, and the opposite sloped diagonal line, after the input alternates. The output ultimately is a pattern of alternating exponential approaches, just cut short to look like a triangle wave before curvature can be noticeable.
Close in frequency to R/L, and the output clearly appears as exponential approaches of alternating curvature. I call it as a sharkfin wave.
I'm in class 8th and I'm having some problems in understanding the Black- Body radiation...can you please explain it to me......
use google
@@lecturesbywalterlewin.they9259 😆😅😁 Greatest Reply from The greatest Mam😊❤
Hello sir ❤
Sir I am preparing for JEE 2024. But I can't understand the Topic- *Entropy* and *Gibbs free Energy* from Thermodynamics . I am confused in these topics. Can you tell me in which lecture, have you covered those Topics? It will really help me sir😊❤
use google
save earth save ladakh #climatefast
Where can we send solutions like this
Post a video? He doesn't give out his email to everyone, so that's not an option...
Links sometimes work.
They did for the problem,but don't here...
❤
Sir take some rest
Hello sir I am your big fan
I know you have devoted entire life to physics . But I am student of class 8 and how can I get faster in physics tell you something sir
eat yogurt every day but *never on Fridays* that also worked well for Einstein and for me
@@lecturesbywalterlewin.they9259legend. Replying to comments on a 13 year old video haha
hello professor
Welcome
How are you?
Hello👋
Hi sir..
I am from Pakistan and big fan of you.
Sir i study physics but i cannot solve problems myself
Can you please guide me how can i good be in solving problems please 🥺
eat yogurt every day but *never on Fridays* That also worked well for Einstein and for me.
@@lecturesbywalterlewin.they9259 sir i eat daily but it made no difference please give proper solution
I will be very thankful
@@muhammadyousuf4248 Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Great channel, Dr. Lewin!