Problem 210

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Lectures by Walter Lewin. They will make you ♥ Physics.

Күн бұрын

Пікірлер: 83

@mauriziobianconi9480 2 ай бұрын

Under the given conditions, it’s zero in both cases. One can compute the current through each branch as: I_left=V_source/(Zc+Z1) I_right=V_source/(Z2+ZL) So that: Va=V_source Z1/(Zc+Z1) Vb=V_source ZL/(ZL+Z2) V_diff=V_a-V_b This gets a factor at the numerator equal to Z1Z2-ZLZc=R1R2-L/C=0

@anglaislangletaire2098 3 ай бұрын

Thank you and thanks to all teachers of the world.

@Sonu-hv4ef 3 ай бұрын

Potential difference is coming out to be zero irrespective of applied voltage. I solved using phasor diagram.

@oldtvnut 2 ай бұрын

Yes, phasor diagram makes this easy. Not sure why the professor expects it to be difficult.

@yuganshigupta2299 3 ай бұрын

Sir i love your videos… I am prep for IIT JEE EXAM and I am in class 11th and i have been watching your real life projectile ,circular motion demonstration…And i think no one can be as better as youu.If you are reading this comment please let yourself guide me for visualising the physics like you.❤(Eat yogurt everyday except on fridays-Do not say)

@AbdullohDoniyorov 3 ай бұрын

I envy you, Mr. Lewin, because if you make a discovery, you have the opportunity to make it public and perfect it. And I have... But now I create these opportunities for myself. (I used Google translate)

@hanslepoeter5167 2 ай бұрын

I gave it some more thought : Time constant τa = R1*C1. τb=L/R2. R1*R2=L/C is just a formula in disguise that says the time constants are the same. This alone proves that voltage a will always be the same as voltage b. You might think differently but imho that is a perfect and way simple solution. Of course, if the component values would be randomly chosen and the condition was not given this get's you nowhere and my previous calculation would have provided the answer.

@RohanPerera-d4w 2 ай бұрын

Please take good care sir ❤️ I was done with physics and you made me love physics. Truely grateful ❤️

@ulfhaller6818 3 ай бұрын

Given: V(t) = V₀ sin ωt and R₁R₂ = L/C => R₁C = L/R₂ … (1) We see two highpass filters in parallell with time constants R₁C and L/R₂ respectively. Using the jω-method to calculate V_a: V_a = V·R₁/[R₁ + 1/(jωC)] = V· jωR₁C/(1 + jωR₁C) From eqn (1) we get: V_a = V· jω(L/R₂)/[1 + jω(L/R₂)] => V_a = V· jωL/(R₂ + jωL) … (2) Using the jω-method to calculate V_b: V_b = V· jωL/(R₂ + jωL) … (3) Since eqn (2) and eqn (3) gives the same result: V_a = V_b the potential difference between point a and b is = 0 regardless of the frequency of the voltage supply. In question (a): V(t) = V₀ sin (15t) => ω₁ = 15 rad/s In question (b): V(t) = V₀ cos (-9t) = V₀ cos (+9t) = sin (90° - 9t) This means just a phase shift. => ω₂ = 9 rad/s Answer: In both question (a) and (b) the potential difference between point a and point b becomes zero.

@ulfhaller6818 3 ай бұрын

A more simple explanation: Two HP-filters with the same time constant. In the frequency domain they do the same thing. Thus Va(ω) is always the same as Vb(ω).

@Z-Xee 3 ай бұрын

I will be Learning the AC CURRENT in next month and will try this problem after completing it... till now I've covered till electromagnetic induction in my class 12...lots of Love from india ♥️

@killersergenyt5379 3 ай бұрын

Always a special thing teach by you sir ❤

@trevorkearney3088 2 ай бұрын

Yes the answer is zero irrespective of source frequency. Consider a step DC input at t=0, rather than a single frequency sinusoid. The transient response at nodes a and b (relative to the source common node) would be identical given (as noted earlier), the parallel path time constants are identical, if R1*R2=L/C. At t=0 in the step DC case, the voltage at both nodes a and b (relative to the source common node) would be equal to the input DC step voltage level. A single DC step would have a notionally 'infinite' spectrum. Nice question.

@darkomalic5950 3 ай бұрын

I get the answer that Vb - Va (or for that, Va - Vb) is 0 and is independent of omega. I guess you gave us a hint when you said that it somehow has a smell of a Wheatstone bridge. Using the supstitution R1C=L/R2 everything cancelled out for me so that Vb - Va = 0 at all times and is independent of omega.

@quinn-bg3ok 3 ай бұрын

Exactly! My first thought was was to check the condition for a balanced whetstone bridge and it matched! There’s the answer! I’m not sure why this problem is “difficult”.

@quinn-bg3ok 3 ай бұрын

It’s more similar to a Hay’s bridge actually. But still the same condition!

@hannaedwards5197 3 ай бұрын

Same happened to me more than one way so wonder what i do wrong

@thinker_123s 3 ай бұрын

Awesome videos continue this series

@Aafaqmalik-u3t 3 ай бұрын

Dear Professor,Lots of love and Blessings From KASHMIR 🌹🌹

@radhe_radhe-am 3 ай бұрын

huh?🤔

@carultch 2 ай бұрын

Answer for both A & B: Voltage difference is zero, regardless of the input waveform. Supporting calculations: Both circuit branches are first order high pass filters. Use the transfer functions for each type of filter. A transfer function in general is defined as: H(s) = £{V_out(t)}/£{V_in(t)} Where £{f(t)} is the Laplace transform of function f(t), and s is the Laplace domain variable. This means output voltages are given as: £{Va(t)} = £{V(t)}*H_rc(s) £{Vb(t)} = £{V(t)}*H_lr(s) Transfer function for an RC high pass filter: H_rc(s) = R1*C*s/(R1*C*s + 1) Transfer function for an RL high pass filter: H_lr(s) = L*s/(L*s + R2) Both transfer functions have the same form, with the only difference being the coefficients. So if all coefficients are equal, £{Va(t)} will equal £{Vb(t)}, implying that Va(t) = Vb(t), and the voltage difference will be zero. Indeed this is the case for the condition given. Rearrange the given condition, R1*R2 = L/C, to get L = R1*R2*C, and substitute H_lr(s) = R1*R2*C*s/(R1*R2*C*s + R2) = H_lr(s) = R1*C*s/(R1*C*s + 1) H_lr(s) = H_rc(s) Thus Va(t) = Vb(t) We can see that both transfer functions are identical, and thus there is no voltage difference across the two marked points.

@surendrakverma555 2 ай бұрын

Sir, My answer is ZERO in both cases based on phasor network analysis and given condition. Thanks and Regards

@ericpilboue3276 3 ай бұрын

C'est un pont de wheaston équilibré puisque r1*r2=l omega/comega donc ca doit se comporter comme un pont équilibré. Pour aller plus loin, faudrait que je quite le bar😂😂😂😂 Je veux une vraie reponse chiffrée 🙏

@KeithandBridget 3 ай бұрын

Go further. Your answer is...

@apurvkumar8223 3 ай бұрын

@@KeithandBridget 0 his way is beautiful

@apurvkumar8223 3 ай бұрын

shalom prof. walter lewin. I visited your channel probably after 2 years. Previously when i visited I was preparing for JEE and now I am studying in NIT second year now( its actually a class lower than IIT). 1. :- Sir I am failing terribly in college. I am actually not able to cope up with the daily routine and the classes are boring. It may be that the problem is in me as I am not able to secure good marks and my grades are degrading. I did try to understand what is being taught in the classroom sometimes I even grab the concepts but most of the times I fail. And then the environment -there are studious students but many study 'day before' exam, the problem is that if you sit in front and then try to focus on what is being taught and if you fail, they will start making fun. They say 'I studied 2 days before exam and still I got 8 CG but you focussed and you are still behind me. shalom prof Apurv

@lecturesbywalterlewin.they9259 3 ай бұрын

sorry to hear that - talk to your teachers about it

@The_Green_Man_OAP 3 ай бұрын

You could try taking vitamins, but check for signs of deficiencies first. Get advice from a nutritionist.

@dogscats8141 2 ай бұрын

I am 51 years old. I completed my first year in the specialty "Nuclear Physics" with A's in physics and chemistry and B's in mathematics.It was difficult for me to study, as I had completely forgotten the school curriculum, and I often watched KZbin videos on physics, both for schoolchildren and for first-year students.

@yisahak 3 ай бұрын

We love you Prof WL

@Aryan_IIT 3 ай бұрын

Been watching your lectures. Obviously love those ❤ Writing, just to ask how are you sir ?🙏🏻 And can i know your age right now 😅🙏🏻

@ISLAMICWORLD-tp9vx 3 ай бұрын

I hope to join your live sessions...

@exactempire7854 3 ай бұрын

sir could you give me the pdf of the version of the book you use in 8.02

@exactempire7854 3 ай бұрын

I already bought the book but another version and I can't keep track of the problems

@lecturesbywalterlewin.they9259 3 ай бұрын

I don't think a pdf of the book exists - The publisher of the book wants to sell the book

@AsitSarkar-lj2yg 3 ай бұрын

You are the best physicist I have ever seen❤.Is there any way to contact you?? Please tell me

@Pritam_Jana_u 2 ай бұрын

Hi sir, How are you..... Hope you are well.

@yetimoglu_ 3 ай бұрын

thanks walter

@lightsout4339 3 ай бұрын

Sir, Question did you draw the picture on the thumbnail ?

@saiarvind8513 3 ай бұрын

Sir so what I've done is that i first calculated the potential difference across R1 which is Va -0 = VR1/[R1² + (1/wc)²]. Now the potential difference across R2 is V - Vb = VR2/[R2² + (WL)²]. Adding these 2 equations we should get the answer because the resistors are always in phase with each other and with the voltage

@hanslepoeter5167 3 ай бұрын

Worked on it a bit ... head is spinning and i got no result .... Now that seems to be the correct answer but how to prove it .....

@prof.sonandrwer 3 ай бұрын

Potential difference between a and be is zero?

@hanslepoeter5167 3 ай бұрын

Hard, even for me .... that's a joke .. maybe ...

@quinn-bg3ok 3 ай бұрын

Where to submit solutions?

@hanslepoeter5167 3 ай бұрын

Can't believe it myself but i got it worked out : Va = V(Zr1/(Zr1+Zc) = V(R1/(R1-j(1/ωC))) Vb = V(Zl/(Zl+Zr2) = V(jωL/(R2+jωL)) Va-Vb = V((R1/(R1-j(1/ωC)))-jωL/(R2+jωL)) For both fractions to be subtracted the denominator must be made equal as usual. That means multiplying numerator and denominator with the denominator of the other fraction. I start with the numerators on both sides and already subtract these, which is the numerator of the result of the subtraction ( which then can be multiplied by V to get the numerator of the answer ) : So, for numerator of the result of fractions subtracted only : R1(R2+ jωL)- jωL(R1-j(1/ωC)) Substitute L with R1.R2.C as given produces : R1(R2+ jω R1 R2 C )- jω R1 R2 C (R1-j(1/ωC)) Working that out : R1 R2 + jω R1² R2 C - jω R1² R2 C + j² ωC R1 R2/ωC R1 R2 + jω R1² R2 C - jω R1² R2 C - R1 R2 Everything cancels out, the numerator is 0. I don’t have to bother working out the denominator cause whatever it is, the result will be 0 for this subtraction. 0 multiplied by V will still be 0, so 0 is the final answer for both questions. ω cancels out completely. It is funny to notice that the time constant for both a and b filters are the same. This means, whatever the waveform of the input voltage the answer will always be 0.

@Amos-fn7ie 3 ай бұрын

thanks prof W.

@ATHARVA_TIWARI 3 ай бұрын

Hello sir I'm 10 th grader student please give me any tips for study❤❤❤ I see your videos since class 7th

@imempire69 3 ай бұрын

Sir what would you tell to those students who dont grasp the beautiful concepts of physics rather memorize some equations and solve maths,? Can solve some maths but cant even explain Newton's law perfectly,

@KeithandBridget 3 ай бұрын

Understanding is way better than just knowing an answer. Understanding lets you solve problems you didn't think you could solve.

@kshitigupta5126 3 ай бұрын

Hello professor...i have a doubt...can you please help me solve it...Can't we say that since E due to dipole at any distance is inversely proportional to r^3 ,gauss law is not applicable for dipole?

@ArturoHernandez-y3k 2 ай бұрын

Vab = - jV0/(wC(sqrt(R1^2+(1/wC)^2))) Vab = (V0

@physicsmania310 3 ай бұрын

Va - Vb = 0 in both cases (a) and (b)

@yug9243 3 ай бұрын

Sir my mechanics part is very weak in physics please tell me how i make it strong 😢

@Sonu-hv4ef 3 ай бұрын

Go and study

@TheTanveerGaming 3 ай бұрын

Study.

@AbdulMoiz-l7e 3 ай бұрын

Hello Sir , i am new in physics can you tell me which book is good and easy please ❤

@lecturesbywalterlewin.they9259 3 ай бұрын

ask your teacher who knows you

@no-one406 3 ай бұрын

Hc Verma

@Nextyy 3 ай бұрын

love from India

@Ranjeetajaipal 3 ай бұрын

Sirrrrrrr❤❤

@ISLAMICWORLD-tp9vx 3 ай бұрын

Sir can i study in online from you........

@anasmansoori4969 3 ай бұрын

Dear professor I am in class 11 and I am preparing for NEET. But I am not able to solve tough questions in physics. Please give me some suggestions. And I am not able to remember anything like formula and derivation.

@lecturesbywalterlewin.they9259 3 ай бұрын

eat yogurt every day but *never on Fridays* That also worked for Einstwein and for me

@adawong2693 3 ай бұрын

@@lecturesbywalterlewin.they9259 Dear Sir, Do you think this will really work....... Do not ignore this question, answer please!!!!

@paullalau4286 3 ай бұрын

Chek Magnesium tauraat if works for you.

@gauri-s07 3 ай бұрын

Thanks sir!! sir tell me a solution i don't remember anything I studied and readed? What should I do?

@lecturesbywalterlewin.they9259 3 ай бұрын

eat yogurt every day but *never on Fridays* that also worked for Einstein and for me

@gauri-s07 3 ай бұрын

@@lecturesbywalterlewin.they9259 okay Thanks sir 🙏

@IsHuBhAm_19 3 ай бұрын

Thankyou sir You are great 👍

@paullalau4286 3 ай бұрын

Magnesium tauraat its better than yogurt my friend. Go to bibliotheek if you like to start study from physics books. You'll remember again. KZbin channels maybe can help also. Good luck.

@FutureStarSHORTS 3 ай бұрын

@@lecturesbywalterlewin.they9259why not Friday.