RL & RC Circuits

RL & RC Circuits

[@stevenzmanx]

I like how savage this dude is. "I don't like to memorize anything so I derive it". This guy is the realest in the game.

[@nibzlegend99]

Makes sure everyone got a cheatsheet too

[@collinseib]

Really good instructor, very nice work. I thought I should point out a mistake at 9:10 mark he adds the two inductors, it should be 11/3H, not 8/3H, as the left hand inductor is 3H not 2H.

[@lyricalbrain2743]

Exactly

[@ebarbie5016]

This is is a great example on how NOT to solve simple RL/RC circuits. When you have a dependent source parallel to a resistor that carries the dependent current, the two are equivalent to a simple resistor... I took me 5 seconds to get the solution just by typing a key sequence into a calculator w/o having to write anything down...

[@joejoe2221hzi]

Love how this guy doesn't rush through explanations and shows every step !

[@haikal1680]

2021 and im still watching his video

[@bossmanbossman-iw7qc]

I’ve learned more here in 30+ minutes than I have in a month of lectures… amazing. Thank you so much! Just something about this lecture video that makes everything feel easy… unlike my professor, who messes up 3x per lecture and loses my attention.

[@SpoodrMin]

curious, what school? I have a similar situation

[@josuegialis8146]

You are amazing! I wish I was in your class. Thank you for saving me a couple days before my final. Your videos continue to help people to this day!

[@averagegamer1462]

on 9:30 it should be 11/3 not 8/3, I believe, no biggie.

[@gaurisethi5808]

yes it is 11/3 = 3.66

[@ronricogarcia3216]

you're right. He got confused...

[@marwansalah4910]

Mr.Zahi your videos have been saving me through out the semester thank you very much

[@ahrar9919]

awesome video clear conceptualization

[@hammamziadeh4093]

Your teaching skills make it easy for students to understand difficult topics. I admire your teaching skills sir, wish I had you as a teacher.

[@suhailasami2844]

at 9:30...shouldnt it be 11/3?

[@samynashabeh]

Yes, it should ..

[@sarba85528]

Yes you're right. it should be 11/3. I see that too

[@Daniel-rm6hj]

Amazing, thank you so much!

[@adilbekzhumabekov8303]

Sir, YOU ARE THE BEST!

[@huntar44ntokzin]

where could one acquire that cheat sheet?

[@fagbemitunji995]

the vc(o) could have been found easily by souce transforming the 10v source and the 6ohms resistor to give a current source in parallel to the 6ohms resistor. then we can apply nodal analysis since all the element will all be in parallel and will have a common voltage i.e in dis case vc(o) will be dere common voltage

[@sajjadulhaq4136]

wrong solution ..there should be 11/3

[@realaxis2562]

sajjad ul haq u can chk kzbin.info/www/bejne/nprTnJynd6eNbs0

[@richie1603]

Soldier attitude , dig it !

[@meghb.5003]

Could you make a video for the case when we have two switches?

[@aaronrasmussen16]

Very helpful thank you!

[@gaurisethi5808]

finally concept got solved , thnku sir :)

[@Janakiram_999]

Hi

[@alaazedan3794]

amazing thank you

[@kesavareddyshiva9206]

at 9:31 it should be 11/3 not 8/3 bcz 3+(2/3)=11/3.i think so sir

[@muhammadsabih4896]

Thank u so much :) made me clear :)

[@user-ry4ee4ev5u]

감사요..덕분에 좀 이해할수있겠네요

[@univerlearn]

Thank you, you made the question easy

[@g-xills7010]

which pen you have ?

[@awesomestuff8819]

Thank you! You're awesome!!

[@angelcuellar6613]

You saved my ass, thank you!

[@julesmendoza3096]

sir in the last part, i got confused because when I tried to calculate 1M *(2.5/8000) what I get is 3125 and not 312.5

[@nikai_ryosuke_2199]

What are the pens are you using?

[@ohmakademi]

thank you so much...

[@rezashah91]

thank you. it was great...

[@ridwanhusainishraq]

Product over sum which formula does it hold? Didn't get that

[@shibadhamija4305]

superb sir ..thnks

[@jadhajali2804]

"just the equations are uglier ! " lol hahahhaha

[@yosefkashkoul3472]

so nice pop hehehehehehe thanks dr zahi haddad it's seems like your an arab so شكرا كتير اهم اشي البوب هيهيهيهيه

[@tebogo4716]

Helpful indeed!

[@eric-jd1yl]

sir i think I1 at 0 different than IL at 0 so the equation for I1 (t) you need to find I1 at 0 then put it in the equation?.. im a student.. just asking so mind me if im wronf

[@wilmiecostiniano594]

i also agree

[@muhammadahmad96]

Sir, you are the Best!! May Allah(SWT) bless you!

[@vtgaming9204]

+Muhammad Ahmad ALLAAAAAAHHUUU AKBAAAAAR

[@muhammadahmad96]

Bro, do you know what Allah u Akbar means? It means "The Lord is Great" Don't you think that our Creater, our Sustainer, is the greatest of all?

[@muhammadahmad96]

+VT Gaming Ohh.. come on. Be a little respectful. He is not just my God. He is your Lord too. We all have the same God. And kindly think open mindedly without the media's propaganda influencing you. Read Quran... you ll surely dind numerous scientific facts mentioned 1400 years ago.. facts that have been found only very recently by scientists. some of them are: 1.The expansion of universe. 2.The fact that Sun is not stationary. 3.Moon has no light of its own. 4.Embryology 5.The sending down of Iron from Space. 6.Mountains as pegs. 7.The Ozone layer. 8.Space and time theory. 9.How honey ia made (The scientific method) 10.The Big bang ..... and numerous more. thank you.. i hope you ll enjoy reading Quran.. :)

[@parkerflop]

May Allah forgive us and guide us and save us from the fire.

[@vtgaming9204]

Muhammad Ahmad Lol, i'm not going to sit here and debate fiction regardless of predictions it has made, i'm stating that it's cruel and one evil religion to follow. Nostradamus, if i've spelled that correctly made many predictions about the future, he was not religious to my knowledge and we don't see us calling him god due to the fact that he predicted world war 1, 2, and the build up to world war 3 with accuracy down to the president's years before any of this happened. Is he a god now?

[@harikrishnakande4292]

excellent

[@loghmanhashemi2480]

at first thanks a lot for good and nice training but i have a question why you ignored from 8 ohm resistor when t>0 ???

[@lwallace758]

When the switch is open, there is no current flowing through the 8 ohm resistor. There for the 5A source and the 8 ohm resistor can be ignored from the circuit. Hope that helps

[@loghmanhashemi2480]

thanks very much bro

[@hananabd2224]

sir, i hope you make vedios for electronics :( you lectures so helpful :D

[@LazyMan1453]

What I should do to find i sub 1 if i can't find i sub 1 in terms of i sub L ?

[@amrnabil4508]

very good

[@ahrar9919]

it should be 11/3H not 8/3H

[@Peter_1986]

"Cheat-sheet". Haha, clever.

[@tylerwilson2280]

ty

[@bencam6720]

thank you

[@mohammedferoz4178]

thank you so much

[@BelayBirhane]

But i am confused cause dor t greater than zero inductance will be open ckt

[@cook1emnstr176]

@4:42 ish...Maybe I'm wrong but when you combined the 2nd set of resistors you forgot to multiply the (24/11) by 2 on the top. The resistance would be ((24/11)*2)/((24/11)+2) = 4.3636/4.1818 = 1.0434 ohm...for real though...please let me know if I'm wrong here...

[@OmErgeVU]

You aren't trying to find the voltage there, you are trying to find the current.

[@DaLeR080]

Окси форева!

[@RohanNair32]

The fornula for RC is Ke^-t/ RC. Using the formula for RL i was able to get the first example correct without derivation but i get like e^125t when doing the RC example with the formula. Is there a reason for this?

[@richie1603]

Im confused on the same thing , using the formula too , tricky !!

[@ergikaso9132]

Maybe you found the R wrong there because there is a dependent current source that cannot be killed

[@peruboyjose]

30:30 shouldn't that be (2.5/8k)/1u?, which would make alpha 3.125*10^-10?

[@loaikhaseeb1251]

Can I get the sheet as a pdf ???

[@andrewsaba2158]

You sum inductors in series ... 8:54

[@anugrahsingh1365]

L is 11/3

[@preme4371]

I dont get t, if yhe general form of RC circuit is v(t)= Vo e^-t/RC. Then you never get the final answer as yours.

[@cansnyollar5098]

The formula you are talking about assumes there are only a capacitor and a resistor but in his question you also need to account for the current source's contribution. The formula you are talking about also comes from simple 1st order diff. eq. I'd suggest you to stick with this proffessor's method

[@davidemelu1142]

where can get the cheat sheet?

[@SaadSheikh-uw6ew]

helpful indeed but you thought that was 2H inductor but that wasn't 2H LOL

[@oguzozum887]

sir you sound like mark hamill (joker)

[@micaelascanlon8761]

forever wishing my lectures were just as rich as his youtube videos my professors are so useless.

[@farawayskies]

Can I get that cheat sheet :D

[@tahaadel1718]

?what this guy name

[@shashankpandey1966]

its for dc or ac ?

[@abdallahmelhem9358]

If i see a switch then it is a DC?

[@loghmanhashemi2480]

what the students doing in class men????????

[@albiaclifordjanernib.5072]

3+⅔=11/3 for Inductor total equivalent

[@mynoobmw3]

where can i print this cheat sheet off? :(

[@mattdouglas8900]

You don't need that exact one... Google it to find others, or better yet: learn some diff eq. I would suggest Paul's Online Math Notes as well as PatrickJMT here on KZbin.

[@mynoobmw3]

Matt Douglas I alrdy took ODE. I just want to find one for reference just in case I need to look up something sometimes

[@nibzlegend99]

@9:35 should be 11/3 not 8/3

[@Janakiram_999]

Yes

[@bellalalakshmidevi9714]

you can tell in telugu zs21q`

[@raihanjami2936]

Tmi dhon er pora porao

[@sonlightstickbukucu]

am i stupid?

[@sridharchitta7321]

RC circuit time domain “The voltage across a capacitor cannot change instantaneously” is a statement one finds often in textbooks on circuit theory which discuss the application of a step voltage to an RC circuit. Most students memorise and apply this in circuits without understanding the physical processes involved. It will be useful to learn the operation of charging a capacitor using a unified approach to electrostatics and circuits. Electrostatics and circuits belong to one science not two. To learn the operation of circuits it is instructive to understand Current, the conduction process, resistors and Voltage at the fundamental level as in the following two videos: i. kzbin.info/www/bejne/ioXXpWVul5aXj9E and ii. kzbin.info/www/bejne/bnO0fpKurJeFnNE It is not possible in this post to discuss the charging of an uncharged capacitor. During the first few nanoseconds after switch ON, while the surface charges arrange themselves, there is no electric field E_cap and fringe field because there is no initial charge on its plates; it is as though the capacitor was not there - as though there were a continuous wire with no break in it. The last frame References in video #1 lists two textbooks 3 and 4 which discuss in detail with a series of diagrams (sequentially arranged) the physical processes in charging a capacitor. RC circuit frequency domain The current (sinusoidal steady-state) in a capacitor is due to the resultant electric field E_net (resultant of the applied field and an opposing electric field, the fringe field). If the capacitance of the capacitor C is made large, then the fringe field does not build as fast as it would have if C were to be smaller. With a large C, the charge sprays on the plates do not result in developing a large voltage in a given interval of time as evident from the capacitor voltage-charge relation Q = CV. The fringe field is smaller and the net field consequently is greater. Therefore, at a fixed frequency, the current increases as the size of the capacitor is increased. The current also increases as the frequency is increased. So, we say it passes higher frequencies of applied voltage. If the frequency is made smaller, the fringe field builds very rapidly and in the limit when it is dc, it blocks the applied voltage. It is not possible in this post to discuss in more detail current in capacitor circuits and capacitive reactance. The last frame References in video #1 lists textbook 4 which discusses these topics in more detail using a unified approach and provides an intuitive understanding of reactance. It also describes with sequential diagrams how the current leads the voltage across the capacitor by 90 degrees. RL circuit time domain When a magnetic field inside a solenoid varies with time, a curly non-coulomb electric field is observed both inside and outside the solenoid. A coulomb electric field results and we note that an attempt to change the current in the coil induces an emf in the same coil and makes the coil sluggish to respond to current changes. It is not possible in this post to discuss the production of induced emfs in inductors in detail. The last frame References in video #1 lists two textbooks 3 and 4 which discuss in detail with a series of sequential diagrams the physical processes to explain the operation of inductors and RL circuits. RL circuit frequency domain The existence of a sinusoidal current resulting from the application of a sinusoidal voltage to an inductor is a characteristic of the non-Coulomb electric field that is proportional to the rate of change in current causing a changing magnetic field. The current is a result of an opposing Coulomb electric field, which is a result of polarization by the non-Coulomb electric field associated with the changing magnetic field, and the current is a consequence of the resultant field of the applied field and the Coulomb electric field in the inductance coil. Inductors find applications as filters in power supplies and in resonant circuits in tuned amplifiers. If we increase the “frequency” of the input voltage to an inductor, the “rate of change” of the input voltage and the applied field is “greater than” the rates obtained with applied voltages at lower frequencies. At low frequencies, this causes a smaller induced opposing electric field and emf, therefore, large currents will flow within small intervals of time in the coils of the inductor. In the limit, if the input is a dc voltage, the current will become so large that the inductor will burn out. In an inductor, the opposition to the applied voltage which is changing the current is instantaneous and so, the current can only follow after the applied field has overcome the opposing emf. In an inductor for sinusoidal currents, the current lags the voltage by 90 degrees if the inductor is pure, and less if a resistance is in series with it; the inductor fights before current flows. If an inductor weren’t to fight, you will get energy for free!

[@diy-projects]

thank you so much