Wow, great video. Stumbled across it when finding examples for a friend. I like how concise it was, you managed to break down a fairly complicated problem, solve it without spending 10 minutes on the problem whilst keeping it extremely clear to what you were actually doing. Awesome job dude!
Well explained! Thankyou sir, it helped me a lot! That "coloring wires" are so effective to easily determine wherher the resistors are is series or in parallel.
Awesome! It's one of those things that seems almost childish, but it really works to help break down the circuit. The same idea can help when setting up circuits in labs, too. I should make a video of that some time... :-)
Yes it's really really work and so effective! I'm using that technique since I watched this video and it really helped me a lot with my electric circuit realate subjects!
Ha ha... thanks! Do you have any particular problems you'd like me to work through? I'm always looking for specific ideas for my videos, and I haven't chosen my topic for next week's (Oct 18, 2016) webinar yet. You can submit problems to info@redmondphysicstutoring.com. In the meantime, good luck and thanks for your comment!
Okay, I'll dig up an example for next week's webinar. If you'd like to attend live, you can register at www.redmondphysicstutoring.com, or you can also just wait for the recording to be posted on RUclips.
Hey Keston, I've prepared an example for tomorrow's webinar, but in the meantime you might like the following URLs: Rough overview of the process: hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html hyperphysics.phy-astr.gsu.edu/hbase/electric/norton.html Some nice worked examples: people.clarkson.edu/~jsvoboda/eta/dcWorkout/thevenin.pdf www.calvin.edu/~svleest/circuitExamples/TheveninNorton/ These aren't in any of my introductory physics textbooks, and since I haven't taught it I'm not an expert. Sometimes that makes my explanations better, though, since I can see what it's like from a typical student's perspective.
Hi Keston, FYI I recorded an example showing how to calculate the Thevenin equivalent circuit. You can find it on RUclips here: ruclips.net/video/lPdNl2J0Tyg/видео.html I hope that helps! If you have any more specific requests, let me know and I'll try to fit them in.
1:02, can you explain the importance of placing the NEWLY formed resistor? Because Why did it go on the side as the new combined resistor instead of choosing one of the 2 sides they were on as they get added onto each other?
Glad I could help! I'm on vacation for another couple of weeks so I suggest you check out this forum: www.physicsforums.com/forums/homework-and-coursework-questions.152/ They have a framework for submitting your question which seems tedious but actually helps you learn it faster & better than if someone simply answers your question. Good luck!
Thanks a lot buddy! It really helped and it's all due to your videos that now i can easily calculate the equivalent resistance .. Thank you again... Keep working like this dude..
It's possible to build each of these circuits in real life, and the voltage and current measurements will match what we can predict via theory. So it depicts how the current actually flows.
@@RedmondPhysicsTutoringVideo ie that current will flow from b to R2 R1 a on the other side from b to R5 R7 , at R7 it will go parallel split to R346 and the current from R1 will will split towards R4 and R3 ( parallel) and R6 will serve as parallel to R3 , so R346 facing current flow from both sides and concentrates on point a before entering the + end, but wont bidirectional current nullify the circuit ?
You have the right idea, that electrons will flow from b to a on both 'sides' of the circuit. The flow into the junction at point a adds up before going to the emf. You can think of it like water flowing, where small streams can add to form a bigger river. The emf acts like a charge escalator, pushing electrons towards the negative end. These electrons find their way through the circuit to the positive end of the emf. It's just like water flowing downhill. Sometimes islands form inside of rivers, and the current flows on both sides of the island and then rejoins.
This video literally saved me..wondering why I didn't think of colouring the wires before.. Thanks a lott sir..👍👍 I'm subscribing to your channel for sure!
Hello, what if there is a voltage source on one side and current source on another side with inductance in series with voltage source and resistance in parallel with current source?
If the resistors are all grouped together, you can replace them with an equivalent resistance to help simplify the circuit. If the voltage source, current source, and inductor are mixed up with the resistors, then you'll have to analyze the whole circuit. Good luck!
Yes it does, because each colour represents a wire (or connection of wires) that are all at the same potential. That said, there are some combinations of circuit elements that are neither in series nor in parallel and these combinations require different methods to simplify. That's beyond the scope of introductory courses though and at that point your prof or TA should be able to give strategies and specific advice. Good luck!
Hmm... Maybe watch it again. The definitions of parallel and series are _very_ specific so you'll need to go through the (sometimes frustrating) work of rewiring your neurons so that you can get these questions correct on exams and/or discuss these terms with other scientists. Best of luck to you!
R1 and R2 are in series because any electron that passes through R2 must also pass through R1. There are no other routes for the electricity to flow. Does that help?
I'm not sure I understand. Do you mean a short-circuit? If there's a short, you can treat the short as a continuation of the wire - so the wires connected by the short become a single wire. I could do an example with a switch, but maybe that's not what you mean...
Okay, I looked at an example of short circuits in this week's video: ruclips.net/video/0-OjO1YVCkw/видео.html I suspect that won't really answer your question though. If you can be more specific I can give it another try. The best would be to email me a picture of a problem that you're struggling with. :-)
+varun sai I'm not sure what to say... it works for me and a lot of other people, at least at the introductory physics level. I only recently encountered a case in which this method doesn't work: resistors set up along the edges of a 3D cube. There could very well be even more cases in which it doesn't work. Regardless, good luck!
i am talking about a case where there is triangle inside another triangle thats becoming more complicated nothing are like parallel or series DO you have any easy method for such situations?
+varun sai I'm sorry, but I don't have any easy methods for such situations... I would imagine you need to consider symmetry. These examples never came up at the level I was teaching.
Yes, that would make sense. Thank you for pointing that out! We never got to that in the college classes I taught, and I can see it being very helpful when solving complex circuits.
Yeah, I was trying hard to make these videos play quickly. If it helps, you can play it back at 0.5 or 0.75 speed (click on Settings, then Speed). Best of luck in your physics course!
Wow, great video. Stumbled across it when finding examples for a friend. I like how concise it was, you managed to break down a fairly complicated problem, solve it without spending 10 minutes on the problem whilst keeping it extremely clear to what you were actually doing. Awesome job dude!
+sharpenednoodles
Thank you so much for the feedback!!
Well explained! Thankyou sir, it helped me a lot! That "coloring wires" are so effective to easily determine wherher the resistors are is series or in parallel.
whether*
coloring of wires*
Awesome! It's one of those things that seems almost childish, but it really works to help break down the circuit. The same idea can help when setting up circuits in labs, too. I should make a video of that some time... :-)
Yes it's really really work and so effective! I'm using that technique since I watched this video and it really helped me a lot with my electric circuit realate subjects!
That's so great to hear! You rock!!
Many Many thanks, straight to the point yet detailed. Super useful.
+Francisco Borges
Thank you so much! I really appreciate it!
Really, I can't thank you enough
This was exactly what I needed ❤
Happy new year btw 🎉
You are the best teacher ever, please make more videos on Thevenin and Norton Equivalent Circuits
Ha ha... thanks! Do you have any particular problems you'd like me to work through? I'm always looking for specific ideas for my videos, and I haven't chosen my topic for next week's (Oct 18, 2016) webinar yet. You can submit problems to info@redmondphysicstutoring.com. In the meantime, good luck and thanks for your comment!
Thevenin Equivalent Circuits
Okay, I'll dig up an example for next week's webinar. If you'd like to attend live, you can register at www.redmondphysicstutoring.com, or you can also just wait for the recording to be posted on RUclips.
Hey Keston, I've prepared an example for tomorrow's webinar, but in the meantime you might like the following URLs:
Rough overview of the process:
hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html
hyperphysics.phy-astr.gsu.edu/hbase/electric/norton.html
Some nice worked examples:
people.clarkson.edu/~jsvoboda/eta/dcWorkout/thevenin.pdf
www.calvin.edu/~svleest/circuitExamples/TheveninNorton/
These aren't in any of my introductory physics textbooks, and since I haven't taught it I'm not an expert. Sometimes that makes my explanations better, though, since I can see what it's like from a typical student's perspective.
Hi Keston, FYI I recorded an example showing how to calculate the Thevenin equivalent circuit. You can find it on RUclips here:
ruclips.net/video/lPdNl2J0Tyg/видео.html
I hope that helps! If you have any more specific requests, let me know and I'll try to fit them in.
1:02, can you explain the importance of placing the NEWLY formed resistor? Because Why did it go on the side as the new combined resistor instead of choosing one of the 2 sides they were on as they get added onto each other?
it was so simple...which I made so complicated earlier...thanks a lot sir....
you made it simple sir thank!
I'm glad to hear it. Best of luck to you!
Thanks, I have a problem if you can help me !!! (how can send you the example ??)
Glad I could help! I'm on vacation for another couple of weeks so I suggest you check out this forum:
www.physicsforums.com/forums/homework-and-coursework-questions.152/
They have a framework for submitting your question which seems tedious but actually helps you learn it faster & better than if someone simply answers your question. Good luck!
Thank you very much
Great video, really helps demystify complex circuits.
Unique way to solve. That's pretty good!!!
Thanks a lot buddy! It really helped and it's all due to your videos that now i can easily calculate the equivalent resistance .. Thank you again... Keep working like this dude..
Fantastic! I'm glad to help. Good luck with your course!
Oh my God !
Thank you sooooo much dude ♥️
You helped me understand such a fundamental concept this fast 😃.
Once again, thanks a lot. :)
THANKS A LOT! Helped me with my test a ton!
Why didnt you change the colour of wire from r2 and r5,whereas you did in r7?
so clear and very useful app to explain examples
+Abd Barakat
I'm glad I could help. Thank you for the feedback!
Britishers visiting India - 0:11
😂 HAHAHA
Are these curcuts are a depiction of how actually current flows or an assumption of the probable path?
It's possible to build each of these circuits in real life, and the voltage and current measurements will match what we can predict via theory. So it depicts how the current actually flows.
@@RedmondPhysicsTutoringVideo ie that current will flow from b to R2 R1 a on the other side from b to R5 R7 , at R7 it will go parallel split to R346 and the current from R1 will will split towards R4 and R3 ( parallel) and R6 will serve as parallel to R3 , so R346 facing current flow from both sides and concentrates on point a before entering the + end, but wont bidirectional current nullify the circuit ?
You have the right idea, that electrons will flow from b to a on both 'sides' of the circuit. The flow into the junction at point a adds up before going to the emf. You can think of it like water flowing, where small streams can add to form a bigger river.
The emf acts like a charge escalator, pushing electrons towards the negative end. These electrons find their way through the circuit to the positive end of the emf.
It's just like water flowing downhill. Sometimes islands form inside of rivers, and the current flows on both sides of the island and then rejoins.
This video literally saved me..wondering why I didn't think of colouring the wires before.. Thanks a lott sir..👍👍 I'm subscribing to your channel for sure!
Thanks sir colored wire method really help me
Thank you very much for this, it removed my doubts in most question.
+greyosw22
Fantastic! I'm glad to be able to help, and I really appreciate you taking the time to let me know!
Why didn't I think of that coloring tech. so much easier to see it when starting out
Very helpful sir
Thank uu
Hello, what if there is a voltage source on one side and current source on another side with inductance in series with voltage source and resistance in parallel with current source?
If the resistors are all grouped together, you can replace them with an equivalent resistance to help simplify the circuit. If the voltage source, current source, and inductor are mixed up with the resistors, then you'll have to analyze the whole circuit. Good luck!
Thanks dude, really helpful and clear ! Please keep going ;)
+Samuel Drai
Thank _you_ for the feedback, and good luck with physics!
This helped me massively. Thanks
I just did a playback and decreased the video speed to 0.75x
And was amazing video!!!!!!!!!
Well done..subscribed!
please explain how to find current and voltage in each resistor
Thank u very much sir .
+Vimal raj
You're welcome!! Thank you for the feedback!
Does this method work for most circuits because it seems waay too good to be true?
I get it now thx for your help
Yes it does, because each colour represents a wire (or connection of wires) that are all at the same potential.
That said, there are some combinations of circuit elements that are neither in series nor in parallel and these combinations require different methods to simplify. That's beyond the scope of introductory courses though and at that point your prof or TA should be able to give strategies and specific advice.
Good luck!
Thanks a lot. This really helped
Good video!
Well Explained.
Spot on!
Help a lot, tq
im too fast im to furious, but gr8 explanation bro.
I still dont understand how to handle diagonal resistors.
GUYS why did he consider R3 R4 AND R6 as parallel combination?
thank you
Nice . Go ahead...
What if, one of the resistors is unknown?
Then the algebra gets tricky because you'll need to keep it as an unknown R in your equations. Even that gets easier with practice. Good luck!!
To my sense r 3,4,6 are parallel, and R 1,2,5,7 are in series
Hmm... Maybe watch it again. The definitions of parallel and series are _very_ specific so you'll need to go through the (sometimes frustrating) work of rewiring your neurons so that you can get these questions correct on exams and/or discuss these terms with other scientists. Best of luck to you!
Tq fr ur help dude
Why is R12 is Series ?
R1 and R2 are in series because any electron that passes through R2 must also pass through R1. There are no other routes for the electricity to flow. Does that help?
So you meant that the right wire of R1 & R2 doesnt let it connect each other and then they have only one way ?
thank you so much
calculators clear xplanation is missing
nice video
Thankyou.
thanks
well illustrated but you need to be little bit slow so that slow minded people get it all well. thanks
solve a circuit where have short. i have problems with that kind of circuit. :)
I'm not sure I understand. Do you mean a short-circuit? If there's a short, you can treat the short as a continuation of the wire - so the wires connected by the short become a single wire. I could do an example with a switch, but maybe that's not what you mean...
+Redmond Physics Tutoring yeah,i will be greatfull if you give me an complex example so that i can understand short-circuit easily
Okay, I looked at an example of short circuits in this week's video: ruclips.net/video/0-OjO1YVCkw/видео.html
I suspect that won't really answer your question though. If you can be more specific I can give it another try. The best would be to email me a picture of a problem that you're struggling with. :-)
i don't have your e-mail so that i can e-mail the photo to you :(
No problem: info@redmondphysicstutoring.com
Best
great
thank you :)
Nice
Tq sir
nice
nice give more exqmples
wow. it's 2024 but wow.
Bruh..... I can solve this problem within a minute
I got 57/21
This method doesnt work in most of cases
+varun sai
I'm not sure what to say... it works for me and a lot of other people, at least at the introductory physics level. I only recently encountered a case in which this method doesn't work: resistors set up along the edges of a 3D cube. There could very well be even more cases in which it doesn't work.
Regardless, good luck!
i am talking about a case where there is triangle inside another triangle
thats becoming more complicated nothing are like parallel or series
DO you have any easy method for such situations?
+varun sai
I'm sorry, but I don't have any easy methods for such situations... I would imagine you need to consider symmetry. These examples never came up at the level I was teaching.
Maybe he's talking about Kenelly Theorem
Yes, that would make sense. Thank you for pointing that out! We never got to that in the college classes I taught, and I can see it being very helpful when solving complex circuits.
not so tricky
You are speaking fully English and very fast I not understand anything
You speak fast like lightning but thanks
Yeah, I was trying hard to make these videos play quickly. If it helps, you can play it back at 0.5 or 0.75 speed (click on Settings, then Speed). Best of luck in your physics course!
@@RedmondPhysicsTutoringVideo thank you again this video help a lot.
Why call it tricky?