RC Circuits and Time Dependence | Doc Physics
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- Опубликовано: 12 сен 2024
- We derive the awesomely asymptotic time dependence of a capacitor's voltage as it is charged by opening or closing a switch. Hint - all real batteries have internal resistance, and this might be an easy way to measure it.
Better than a thousand days of diligent study ,is one day with a great teacher. Says so a Japanese proverb. ☆
Nora Ibrahimi You honor me with your kind words!
If more teachers in high school and college were like this guy America really would be great again.
ok Trevor....take it somewhere else dude
...so remember this feeling every time you know something and want to explain it to someone. Make sure you're explaining, and not showing off!
Thanks for watching and for writing the note. I appreciate your support!
You succeed where all have failed before you; you made physics fun to learn!
Doc Schuster / patrickJMT
for
President / VP 2016
Lol, look who got elected.
@@curiousbit9228 xD
I lovem both I cant choose
"Ohhhhhh Dang! I love this stuff!" - Best Quote Ever
It's like Barney from How I met Your Mother is teaching me, amazing.
(Just pointing out that voice and lighthearted way of speaking reminded me of him, also that chuckle 2:28 :'D )
I think you are the only person who is so AWESOME and a Physics teacher at the same time...!!🤣
you explain it in a waaay simpler way than my professor, thank you so much !!
Thank god for this. I've been staring at my notes for so long.
thank you again! you give joy to science and my life
Thanks, man. I obviously wish you were my student, too. But the nature of education is changing, and maybe you ARE. So, learn, student!
give that person a medal for his great work.nice
His talk is kinda entertaining, helpful vid. Thanks
Your videos are super entertaining and very helpful... I wish you taught physics at my school
So Deadpool was busy making these videos before he got ready for a movie. You sound a lot like him. lol. Great video (:
the integral doesn't come out the same if you initially started with dQ/dt-V/R.. and that is what gives you the exponential decrease in the end, no? So... Considering that's how we "should" have started it off, what gives?
2:53. Could you please explain why you took ln of dv/v? I don't understand why that's the integral.
Why is there a plus sign in I(cap)+I(res)=0? wouldn't there be a minus sign?
Uh oh. When's that?
Doc Schuster at 1:27. You said you could make that a minus but then said nevermind, thanks!
+Doc Schuster hey themosthelpful, i have a question, at 1:27 you say you could make that plus a minus but then the equation would change. it would transform to e^t/rc but in this expression the power of e is not negative which would effect things a great deal wouldnt it? plz plz plz reply quick i have an exam soon
+Doc Schuster Firstly thank you for the videos, they've always helped me clear up so many problems with understanding of physics! I think I'm a little confused (and others in the comments as well) about the 1:27 part, regarding the junction rule, just wondering if you could elaborate on that?
awesome! hahaha you make physics sounds actually fun :D thanks for the great classes
At about 3:50 I can't make out mathematically where the V_0 came from, since, when we exponentiate both sides (I'm including the +C since you didn't explicitly set it at 0), the equation renders V=exp(-t/RC)+exp(c). However, functionally it makes complete sense, since the only way to get V_0 equal to the initial potential at t=0 (since e^0=1) is to multiply it to e as you did. The fact that it functionally makes sense leads me to think that there was no mistake in the mathematics, but rather in my following. But if that's the case, then what happened to the +C? And furthermore what was the algebraic step to render the relationship V_0 * e(-t/RC)?
Technically it doesn't follow from the indefinite integral, however if you set up the definite integral of dV/V with parameters V and V_0 and -dt/RC with parameters t and t_0 (assuming t_0 = 0) then the V_0 appears where it's specified in the video. Hope that helps.
DOC SCHUSTER> without the LOAD
How much current in amps can a 4700uf hold store with a power supply DC voltage at +450vdc? the 4700uf filter cap will charge up to +450vdc but how much current is the 4700uf holding storing?
What formula can I use to calculate how much current the filter cap is storing in amps?
Ur really doing a great job!!
Your videos are too good to be true :')
Could you please do a video on Thevenan's and Norton's Theorem and, a video on Superposition theorem
Doc, Your tutes are just Jazz mate, love them heaps. Question though, why's Ic + Ir = 0 and not Ic = Ir and therefore Ic-Ir=0 via a version of the Kirchhoff Loop Rule? (Round 1:30). Thanks in advance Boss!
I really needed this video, thank you
That is beautiful!!
I'm getting a feeling that ryan renolds is teaching me physics
how cool would that be !!!!!!!
Dude, this is legen......wait for it.......daryyyy! :)
Thanks a lot!
Thank u D S. Amazing presentation, lot funny too.
Dr Schuster why does I cap + I res = 0 instead of I cap - I res = 0?
thanks that help me a lot
rc circuits are so cool....
this should have more views :)
Thank you so much sir!
great content subscribed.
tnx a lot mate
Can someone explain why the I(cap) + I(res) = 0??? He just skipped over it but I don’t get why it’s not I(cap) = I(res)
love you doc!!!! :D
Thank you so much!
You bet, sir.
you are weird but this video explained it perfectly
you sound like Cicero from Skyrim !
I wish i understood this.
foreverrrrrr
make it boldface? use different variable? naaaaa ill make the C fuzzy lol
3:35 batman
can i hit 115 now
you sound like deadpool
5:56 joker
Lost me at d. :(
I really wish you were my PHYS prof. :(