thank you thank you thank you

Thanks man, 2024 and still helped me :D

Very useful!

In summary, in a circuit with only dependent sources, Vth will be 0?

My professors is one of the authors of this book. Bro literally wrote the book and doesn't teach us how to do ts💀

Thank you. This is coming in handy for Circuits II class!

What is X supposed to be in this video?

4.19 am I blind or sth or did he just say that a is not = to c then divide everything and write -1,1,1/2 pretending like a = c?

Huh

Genius

Hello sir can i ask? On how can you determine the polarity of a conductor if not given

i was stuck on one problem for 8 hours today, i didint even need to get halfway through your vid to completely understand how to do it now. After 10 years you are still helping students thanks

thanks

THANK YOU KEEP UP THE GOOD WORK

White board you use for teaching

how com Voc double prime is equal to 4? plz answer this.

Legend

this is literally the same 2 questions i have for my first assessment. thank you

aren't you missing a particular solution, since we have the 10v voltage source which produces a forced response?

You just saved my ass on the last day before my endsems

Thank you for saving my grades

How Voc= -12V I mean shouldnt it be 16V since Voc= 2Va and Va= 8V ?

man its a 9 year old video but imma ask anyway.. So I applied Laplace transform to this example circuit, solved it and got the same expression of i(t)= 3e^(-5t). Same answer for less work it seems like. So what are the cases where the Laplace approach cannot be used?

u need to specify the direction of the current idiot. Doesn't matter the dots are.

You are the best!

Thanks, man, I really appreciate what you do <3

I've been struggling so much in this class & finding your video gives me hope thank you!!!

What if the switch is being closed instead of opened at t=0? how does this change the way you solve the problem?

10 years of helping people solve Material Science problems. Thank you!

Thank you so much for this video

Not to quibble, but if you have the Norton circuit you don't have to back solve to find the voltage drop across the load. V_Load = I_N÷(G_N + G_Load). In this case, - 3mA ÷(1/4k + 1/4k) = - 6 volts.

there is a miscalulation error for the total impedance the tolal will be 31+40j and the final voltage will be 1.897<127.78 deg V

At 13:55 i don't get why we were changing the sign of the capacitors voltage. Why does it not remain a minus? I mean the current is entering though the negative terminal so by my understanding it should remain that way. Thanks to anyone replying.

Hey any other way to find Rth with both dependent and independent souce...more like a short cut..?

The best explanation so far. Thanks a lot. Could you also explain how it relates to step by step switching? and how to find the exact capacitor cells/banks for same example. Thanks.

"this is the cheatsheet that I've created, you can pause the video and look at it but we'll come back to it throughout the video" every single student: 📸📸📸📸

What to do if there is a current source instead of 6v source.can we not skip its node?

I stop watching at 16:48. I’m upset that while you were deriving your characteristic equations that you didn’t show all the steps. Now I’m left wondering how you got the expression you found. You should have brought the equation you were working with over. If you are teaching people show all the steps. Everything is important to people trying to learn.

Awesome video! Thank you!

Great video can you help me to solve this problem

thanks and C is 0,001101F. Its equal 1,1001 mF(milifarad)

thanks 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

thanks this helped me a lot

tarihe not : ismail bunu sordu hacettepelysen çalış knk buna çıkar bak

what about thevenin voltage?

You can also apply KCL to solve this after finding out the input current.

The problem he does if #9.47 in the 11th edition, that's if anyone has the latest book.

Shouldn't the second kvl be -10 + 5 + 10 iL + L di/dt =0 ??

If in doubt with this physical mathematical model calculations, get some precision resistors and a calibrated multimeter and breadboard the circuit and measure all the parameters to VERIFY.

That's good thanks