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 .
Hello Sir, I'm a BIG fan of you. I also want a teacher like you in my school, so that physics become a piece of cake for me, but you are only one in this planet. Please give me some advice to perform well in JEE exam. Please reply.
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.
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
I think both solutions are not correct. V is not a constant, it is V(t) and so V(t) / L * exp(R/L*t) can not be integrated in this way. In the beginning, we can assume a switch on problem with a solution i(t) = V/R * (1 - exp(-R/L*t), but half a period later the Voltage drops from V0 to -V0 and there I(T/2) 0.
You forget that the time is small, t«L/R, so the integration is valid for the 1st half-period, and then for all half periods that have this same V(in)max, the form of the solution is the same if you integrate in those regions. If you go to the V(in)min= -V(in)max regions, then you just multiply the slope by -1.
If you take a closer look, I specifically mentioned the ODE is setup for the time period when Vin is constant and equal to V. For what the problem asks I would say is enough as one can easily draw conclusions for one full cycle. Note the input signal is not a continuous function as the problem set it up (at each half a cycle Vin has two values V and -V) so it is an unnecessary hassle to analyze it for a full cycle.
Professor while solving a physics question if we face in difficulty in making the answer to question should we leave the physics or give more time to it to solve the question What do we do in this type of situation leave or work on the question more time
2:10 I did pretty much the same thing, yet I was "way off"...❔😕 The integration constant isn't required if you just use the fundamental theorem of calculus and the notation ₒ∫⁼ͭf'(t)dt=f(t)-f(0)=∆f. The problem set the time t«L/R, so I used this and said exp(Rt/L)→exp(0)=1, then integrated. Much easier and it works!
we were fully aware of the fact that you copied the solution from the solution book of Giancoli. We had problems which that solution. On a separate issue: *There is no E-field in an ideal inductor.* *Thus the integral of E dot DL through the wire of the inductor from one side of the inductor to the other side is always zero.* Yet, if you atach a voltmeter over the inductor you have created a closed loop with the inductor and the voltmeter. The integral of E dot DL through the inductor is zero, but according to Fraday's Law the closed loop integral of E dot DL in the closed loop that you have created is -dphi/dt = -Ldi/dt and that's what the voltmeter will show.
@@lecturesbywalterlewin.they9259 Well, I did the solution first, then I looked up the solution which happened to be the same (it was obvious to me that I could ignore the exponential) but I found it didn't explain anything which is why I elaborated on everything and it was so long and perhaps confusing. I found the actual equations for the square wave and triangular wave that could be used to plot them over many cycles.Giancoli didn't do that. I was also trying to make it fit my simulation, but I'm not sure if the voltage that the proto app measures the voltage on the inductor correctly as it just came up with 2V, same as the power source, but I did notice that on one side of the wire, if I put a voltage probe there the sawtooth/triangle wave shows up. I played around with the app and until I had a perfect square wave on the emf voltage at about 20μs timestep for that timescale and I did get the isosceles triangular wave, but on a different timescale. I attempted to get di/dt from the current but it did not meet expectations. Kept getting around 9.5V for Ldi/dt for some reason which I think is too high, or is it? Here is some values: I had L=100H, R=1mΩ, Vin=±2V, f(in)=250Hz iR+Vin= -Ldi/dt 15μA(1mΩ)±2= ±100H(30μA/t) => Should have ~±2V= -Ldi/dt L=100H=> For 320μs should have: |di|~6.4μA, but had ~30 1/t=(15/1000)±2)/3000 →1000×((15÷1000)±2)÷3 ~671.67 Hz , 661.67 Hz ~ 1/1500μ=1000/1.5~666.67 Hz t=±3000μ/((15μ/1000) ±2)~1500μs 100(30/t)= 29.8/320=0.093125 di= 31.4μA, dt=330μs di/dt=31.4/330=0.095151515151515151 A/s Ldi/dt=9.5151 V ( 😕????) di= 39.8 di= ±(39.8μA - 200nA)=±39.6μA di/dt=39.6μA/408μs=0.0976 A/s , 1st ½cyc 39.8/412=0.096601941747572815 di/dt= -0.0976 A/s, 2nd ½cyc Ldi/dt=100*0.0976 A/s=9.76 V
@@lecturesbywalterlewin.they9259I accept the Faraday's law explanation but my app is just giving me the input voltage on the inductor. I think it's because the change in current is so small in comparison. Also, I try to get slope di/dt of the current over one ½ cycle and it seems to be off because I end up ~300 to 400μs for dt (with di~ 36 to 38.2μA), and it should be ~1500μs for the di I was looking at.
Well, I did the solution first, then I looked up the solution which happened to be the same (it was obvious to me that I could ignore the exponential) but I found it didn't explain anything which is why I elaborated on everything and it was so long and perhaps confusing. I found the actual equations for the square wave and triangular wave that could be used to plot them over many cycles.Giancoli didn't do that.
I was also trying to make it fit my simulation, but I'm not sure if the voltage that the proto app measures the voltage on the inductor correctly as it just came up with 2V, same as the power source, but I did notice that on one side of the wire, if I put a voltage probe there the sawtooth/triangle wave shows up. I played around with the app and until I had a perfect square wave on the emf voltage at about 20μs timestep for that timescale and I did get the isosceles triangular wave, but on a different timescale. I attempted to get di/dt from the current but it did not meet expectations. Kept getting around 9.5V for Ldi/dt for some reason which I think is too high, or is it? Here is some values: I had L=100H, R=1mΩ, Vin=±2V, f(in)=250Hz iR+Vin= -Ldi/dt Take iR as average i value on ½ cycle 1: 15μA(1mΩ)±2= ±100H(30μA/t) => Should have ~±2V= -Ldi/dt L=100H=> For 320μs should have: di~6.4μA, but had ~30 di=30μA i(1mΩ)±2V=±(100H)(30μA/300μs) => ±(10-2)V=±8V=i(1mΩ) => i=±8(1000)A !!! i(1mΩ)±2V= (100H)(-30μA/300μs) => -(10±2V)/(1mΩ)={-12000A, -8000A} (30μA)(1mΩ)±2V=-(100H)(di/300μs) => -(90·μ²·mA)+6μA=di => di~ 6μA 1/t=(15/1000)±2)/3000 →1000×((15÷1000)±2)÷3 ~671.67 Hz , 661.67 Hz ~ 1/1500μ=1000/1.5~666.67 Hz t=±3000μ/((15μ/1000) ±2)~1500μs 100(30/t)= 29.8/320=0.093125 di= 31.4μA, dt=330μs di/dt=31.4/330=0.095151515151515151 A/s Ldi/dt=9.5151 V ( 😕????) imax= 39.8 di= ±(39.8μA - 200nA)=±39.6μA max & min di/dt=39.6μA/408μs=0.0976 A/s , 1st ½cyc 39.8/412=0.096601941747572815 di/dt= -0.0976 A/s, 2nd ½cyc Ldi/dt=100*0.0976 A/s=9.76 V
Try taking the square foundational sum of a number, adding one to it, then add it to the original number and repeat step one again. See a pattern? A square foundational sum is the sum of a number's two closest factors which are both 1.) closest in value to one another, and 2.) sufficient to produce the inspected number.
