วีดีโอ

Wow❤

Thanks sir 🙏🏻

Amazing teaching. Helping me in 2024. Thanks

Solve Y = (A'+ B'C')' A'

Thank you! Very Helpful

Mr T, why didn't you group 1 3 9 11 in 1 group?

This was a great explanation.

The lord sent me here , thank you sir . Could you please share what textbook you use , im currently taking fundamentals of electrical engineering

great explanation : ))

thanks from france lol

This is 5star explanation🔥

Helpful

Thank you sir , you projected the theorem very well , All my confusions are gone

in you I found a solution. thanks a lot for the great work please!!!

this is awesome

The way i understand it at 27:19 is that Professor Tougaw is taking the magnitude of E to be Etheta times the complex conjugate of Etheta and then taking the squre root: |E| = [(E)*(E*)]ˆ1/2. This is the definition of the magnitude of a vector and this vector E has only one component along the theta direction. When you take the complex conjugate you change i wherever it appears to -i (or j -> -j for electrical engineers).

Thank you so much. Greetings from north Africa (Algerian desert)

Loved the tour. Valparaiso is a dream of a college! 😁

Thanks for this class, really helping me ❤

Again, thanks for this series of videos! :)

Thanks so much for posting these videos 🙏

At example 17.1 at 12:20 it take the charge to dissipate to 10% of its initial value 3.4 * 10 ^ - 19 sec And before that, you calculate that it take the charge to dissipate to 0% of its initial value 0.47 * 10 ^ - 19 sec! This means that it takes more time to dissipate to 10% than to dissipate to 0%! Isn't this a contradiction? It should takes less time to dissipate to 10% than to dissipate to ≈ 0%. Thank you for your efforts in creating this content This is my calculations For ≈ 0% , t= 7.375 × 10^-19 sec t= 5τ For 1% , t= 6.8 × 10^-19 sec For 10% , t= 3.4 × 10^-19 sec For 36.8% , t= τ = 0.47 * 10 ^-19 sec

What happened to C? HELP I DON'T UNDERSTAND

@ 13:52 The field lines of a charged particle moving at constant velocity (incl. v = 0) do not propagate or radiate. There is only radiation during the acceleration of a charged particle. Therefore, during the period t (>>> the period of acceleration dt) there can't be radiation all the time. Read this paragraph: @ 15:10. I don't see why the period t can't be equal to dt? The outcome is the same. But if I'm wrong in this, please correct me.

@ 6:38 In fact, dS = r-hat. dA, with r-hat being the dimensionless unit-vector pointing in the radial direction of the Poynting-vector. The dot-product (in eq. 29.1) is between the Poynting-vector S and this unit-vector, resulting in | S | . | r-hat | . cos(a) = | E | . | H | . cos(a), with 'a' being the angle between the E- and the H-vector; and remember that the length of r-hat = 1. The E- and H-vector are orthogonal and thus, sin(a) = 1. The result of a dot-product is a scalar, that we can take out of the integral. Then, in eq. 29.2 the surface-area A = integral of dA.

Thanks for everything

ESE MAQUINA

Thanks for this series. I have been watching them all. I am just wondering why in the final example squares 7 and 15 are not combined with square 3 to make a line of squares 3,7 and 15. I thought doubling up was right a 'proper' thing to do.

32:02 great animation, sir tougaw 47:49 another sweet animation of the Yagi-Uda transmission antenna

3:42 correction: the voltages are the same magnitude, just opposite polarities, so it's not quite accurate to describe the dipole ends as "high or low".

𝐆𝐫𝐞𝐚𝐭

Shouldn't A+(B•C) be (A+B)•(A+C)?

So far this playlist seems great for what I need. I took Physics 2 in 2019, but then the pandemic hit and I have 3 kids so my school progress was halted. Now in 2024 im taking Principles of Electrical Engineering as a mechanical engineering student who needs it to prereq into Vibration Analysis in the fall. Im way behind the ball now, and i had a harsh reality check in the first week. Thank you. A lot of the other videos dont seem to be what I need, or have either way too much, or not enough.

shouldn't I2 equal to Ix in the second example since they are in the same loop ?

I was following aong until you got rid of the C in the second equation.

Yes

thanks

thank you so so much

There's a rule that says voltage is same in parallel circuits🙂

What About AD

Thanks

Nice video

This was a useless video

You made two mistakes when solving Ex. 7.3. First of all when you say Vab and you are setting the integral limits from a to b (1 to 3) it means that you are moving in the +ar direction not negative. Second, the integral of 1/r^2 is minus 1/r, not plus.

Hey can someone please tell me where and how I can access the notes?

I love this

Thanks!

Sir let me explain something kindly, since we are following the clockwise loop in the circuit, the voltage source on the left is positive since it's following the clockwise from negative to positive, then the 2ohms resistor is a drop voltage from positive to negative not the opposite, becauseit'sa voltage drop from the battery clockwise. The 2A current source is going counter clockwise so it's negative too. So it's +V1-V2-V3=0. I checked it out in my electrical circuits book

i struggled so much with topic and i learned more from this 5 minute video than two years of studying computer science

fantastic