Fundamentals Of Electric Circuits Practice Problem 8.8

[@loyalleo467]

good explanation, you sound like a good person, and I understand it takes time to make these videos... however, the camera movement makes it harder to watch. It would be better if you kept the camera in one place even if it has bad quality. Nonetheless, thank you very much!

[@KhetzTutorials]

Thank you for your feedback. This is by far the best way that someone has highlighted this to me. These videos were made over two years ago and there are reasons behind the setup. I have since migrated to a new way of recording these and many others. You can search for the word "remake" on my channel to get a general idea of the new approach. Stay tuned for more in the coming years!

[@loyalleo467]

@@KhetzTutorials Thats Awesome Thank you! I was thinking of solving some of these myself too and post them on youtube but who knows, Currently an EE student trying to make it. Best of luck!

[@gamingnoobkurangpro8635]

Thank you very much for the explanation, I like the explanation, easy to understand, but I hope you add Indonesian subtitles so that my friends who are both from Indonesia can understand more easily, once again, thank you very much for your help sorry if there are words that are wrong keep up the spirit, create content like this

[@anshikashukla6315]

4:42 how is lL(0-)=0 ?

[@KhetzTutorials]

The current source is multiplied by the unit step, u(t)), which evaluates to 0 for t < 0 and 1 for t > 0 so it's behaviour affects the current source. The current is, therefore, only active after time 0

[@spooky_gamers4643]

thank you!!! great explanation sir !!!!! 👏👏👏👏

[@schneiderheathers7140]

When you wrote the KCL equation to find the initial condition for dV(0)/dt you took the value for Is as 12 but at t=0 isn't the value of Is zero due to u(t)?

[@KhetzTutorials]

Hello. That equation is at t=0+ which is just after t=0 and that would mean Is is "on".

[@madlife0231]

sir how do you eveb find to learn all of this im frustrated i cant find motivation any tips

[@KhetzTutorials]

Well, we all have different paths to follow, and there isn't one answer that fits everyone. My advice would be to aim for understanding instead of going through the textbook as a means to prepare for questions. Secondly, ask a lot of questions. There are no "stupid" questions. If there is any piece of the puzzle you don't get, ask.

[@sureshsalvi8393]

Thank you

[@내이름을볼수있는-n2i]

sir can i ask you a question? as i know, alpha=1/2RC in parallel RLC circuit. but in this case R=0. so i thought alpha ->infinite how can i know alpha equals 0?

[@KhetzTutorials]

There is no resistance here, so this is a lossless circuit. A lossless circuit has no attenuation, as I mentioned. For lossless circuits, the alpha will always be 0 and we do not have to substitute into that formula

[@내이름을볼수있는-n2i]

​@@KhetzTutorialsthank you so much!!!!!

[@kingofdice66]

Actually, here R=♾ => alpha = 1/(2RC) = 1/♾ = 0

[@KhetzTutorials]

That might make sense mathematically, but this is a lossless circuit. From the name, lossless, there is perfect conductance, which is in line with theory. The theory states that R=0 for lossless lines. An infinite resistance would lead to less conductance - making the circuit lossy

[@kingofdice66]

@@KhetzTutorials If R=0 => 1/(2RC) = 1/0 = ♾ Technically, it's not 1/0, it's 1 over something approaching 0. So no. Your math does not stand. The theory is based on math, so either the math is correct or not.

[@KhetzTutorials]

@kingofdice66 I was referring to your math. You can consult the textbook or do a quick Google search to get the correct value of R if you do not believe me

[@hanaqasim3132]

Thank you verey match 🌹🌹🌹👍👍👍

[@GoodEnough030]

this is very helpful but i dont like you keep the camera moving :))

[@KhetzTutorials]

Thank you for the feedback. I recently changed my method. You can look at the videos that say "REMAKE" in the title. If I get time later, I'll remake as many as possible

[@GoodEnough030]

yeah that is so nice the new video is very readable haha

[@bayzidhossain7422]

You should stabilise your camera