Capacitors - A Level Physics
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- Опубліковано 11 бер 2012
- Continuing the A Level revision series looking at Capacitors. Includes capacitance, how a capacitor works, the energy stored in a capacitor and the time for a capacitor to charge and discharge.
- Наука та технологія
The best physics teacher I have ever seen.
[Timestamps]
0:06 - Capacitance (Q=CV)
0:36 - Flow of electrons as a capacitor charges
2:23 - Q/V graph (measuring capacitance experimentally)
6:18 - Capacitors in series + parallel
9:50 - Energy stored in a capacitor
13:42 - Factors affecting magnitude of capacitance
15:14 - Charging + discharging curves (and equations)
19:31 - The time constant (𝜏=RC)
Thanks very much
This guy’s explanation is about as good as is humanly possible. I immediately subscribed. A great big thanks!
Thanks for kind words, on both counts. I guess I should say cells but used the rather more general ref to a battery. Hope the revision and the exams go well.
if we have a high resistance... last line please.. nice class sir
does it take more time to both charge and discharge?
C is fixed for a particular capacitor. It is Q which reduces as the charge flows (as current) thro the resistance. Since C (fixed) = Q/V then if Q falls, so does V.
Yes. I cover this at the end of the video at 16:29. The R in the circuit reduces the current, so smaller charge moved per second. I = Q/t. Takes longer for capacitor to charge. RC controls the timing factor.
Yes that is right. Well spotted. I have in fact added an annotation to the video to make this correction. I hope it is still there. But I understand that annotations to UA-cam videos do not appear on all viewing platforms which is a pity as I make a number of corrections and clarifications this way.
Congratulations. That sounds like a very good result. I hope it got you what you needed.
Explained beautifully! Thanks Dr P
U are Awesome Dr.Physics......................!!! Hats Off!!
This was very helpful! You explained it way better than my physics teacher.. Thanks :)!
Yes it is. Well spotted. I had put an annotation (I hope its still there) to correct the error.
I regret I did not found you earlier you are the best teacher of my life❤❤❤Thank you
Got a physics test tomorrow and haven't revised. You're a lifesaver.
Love your videos. They explain it a lot better than my own physics teacher(s).
Thank you! :)
A-level physics is usually done by those aged 17 to 18 and is the exam the results of which determine whether or not people go on to university. First-year college physics is likely to be one notch above this although there is often quite a lot of overlap.
Superb!!!! Everthing about capacitors done in 22minutes only....thank u so much!!!!.....
Hey! You are very amazing!! I just had a request.... Maybe you could just solve a few problems so that we students can understand what kind of approach should we possess while solving these kind of problems...... Thank you
Meanwhile the jet preparing to take off in the background at the start of the video... Great content btw
There is some basic info on cyclotron and linacs in my video "Nuclear Structure Physics". I dont think I've covered bubble chambers.
The battery is delivering a current which is charge per unit time (coulombs per second). The capacitor holds a certain charge so it takes time for the charge to build up on the capacitor. If you increase the resistance in the circuit the current will reduce and it will take longer for the same amount of charge to accumulate on the capacitor.
you're like the savior for a-level physics students....:D
These are really helpful videos thank you!
Good question. I'll try. Take 2 capacitors C1 and C2. C1 has +ve and -ve terminals A and B. C2 has +ve and -ve terminals C and D. Charge both to voltage V. Then C1 will have a charge Q1=C1*V and C2 will have a charge Q2=C2*V. Now disconnect the voltage and connect A to C and B to D. Now they are connected in parallel. Let's assume C1 is larger than C2. So Q1 is larger than Q2. You might expect charge to flow to even out the charge of each device.
if only you were my physics teacher :(
Yes altho in the graph at 12:20 I am not calculating C. I am using it to find energy stored. But you are right. On the graph as I have drawn it the gradient will be 1/C.
Excellent tutorial,really enjoyed watching it as it was so easy to to understand.A BIG Thank you.
I really love the sound of the wash machine behind, it transport me to the future of the video.
I use a wide range of books and my own knowledge. But for most of the A Level material any A Level revision guide will cover the ground that I cover. I understand you are from the US so I would expect you have a similar guide for the material taught to physics students aged 17-18.
Very well explained. Thank you Dr!
Not sure what you mean by errors while charging and discharging. Some large capacitors can be pretty dangerous because they carry a lot of charge. The key issue is to make sure you wire them correctly in the circuit and take care about charging and discharging.
Excellent delivery well done Dr Physics
Thank you, perfect compliment to my physics revision!
Its just an illustration, the kind of thing that might come up in an A level question. Since the exponent is -t/RC, then if t=tau=RC the exponent = -1. So Q/Qo = 1/e which is about 37%.
Thank you. It was v helpful you managed to cover the main points in such little time
I want to start off with a huge thank you for all your revision videos! They thoroughly cover everything in are simple and easy to grasp! I have a question regarding a discharging capacitor. Why does the current decrease during the discharge?
The current is the charge flowing per unit time. As the charge decreases on the capacitor so does the current. Alternatively you could say that the voltage decreases across the capacitor as it discharges and consequently the current similarly will decrease.
DrPhysicsA Thank you very much!
Because it discharges much more quickly according to the time constant and thus deposits a large amount of energy through the bulb in a short time.
Thank you so much I didn't really understand this before but now I do, keep it up
your explanations are spot on!
please keep up the good work sir :D
The formal title is Dr Robert Eagle CBE BSc PhD (London). Hope the assessment goes well.
Can you give me the time on the video where this occurs? The graph at 18:12 is a curve.
But that would mean that the voltages across the 2 capacitors would be different so charge would flow to even them out. So in effect I think no charge would flow. The 2 capacitors would just stay fully charged. But do others agree?
Thanks for this , I have an exam this week and I have missed quite a few lessons leading up to it through illness and didn't know about capacitance but I have just passed my mock exam and got all the questions in it correct so fingers crossed.
Thanks!
Thanks. I hope the final exam goes just as well. All good wishes.
Thank you so much! Your videos have been really helpful :D
I wonder how is he now? He truly helped me survive the A levels! Thanks so much Sir!
He's a DJ
DrPhysicsA aka Bob Eagle, CBE is now a DJ on local radio stations. He's on Twitter & he likes rain.
@@alwaysdisputin9930 glad to hear it. Living life well
You are definitely making me pass my exams...thank you so much. You're like...GOD!! I can't thank you enough :)
Good luck with the exams.
Thank you so much for clearing my doubts! :)
Amazing vid bro, many thanks!
Good idea. I'll try to introduce that. Or at least keep the annotation on the screen as long as the error is visible.
I'm not familiar with that particular book but it sounds as tho its a good introduction and will cover the right ground.
How right you are. Well spotted. My mistake, but interestingly, the book I was using to prepare the video has the same error. I've added an annotation.
Thx Dr .. Really thx ..
Am from jordan ... :)
I think you meant 17:50. You are right. I should have written V=Q/C. V is still proportional to Q. I have added an annotation. Thanks for pointing it out.
At 5:00 you have the axis labels reversed. If you are plotting voltage over time, which is what you want to do, then the horizontal axis, the abscissa, is time and the vertical axis, the ordinate, is voltage.
This is a very minor point, btw, this video is actually excellent.
sanjursan Thanks kind comments. I did it that way because I wanted to end up with a graph showing charge against voltage in order to obtain a gradient which was equal to the capacitance.
When you charge a capacitor initially the current will flow to build up a charge of electrons on one plate of the capacitor. But as charge builds up it has the tendency to repel the other electrons which are en route. Eventually the amount of charge on the plate suppresses any further build up of charge and no current flows. On discharging thro a resistance (R) the charge will gradually fall. Since C=Q/V, V will also fall. And since V=IR I will fall.
How come when the capacitor discharges through the bulb, it makes a flash of light stronger than when the bulb is connected to the battery?
Very good explanation 👍🏼
Need this physics for aerospace thank you so much for the tutorial
Thank you for this video 🙏
which value of voltage should we use in energy stored in capacitor the one across capacitor or between battery and capacitor
Thanks, I use "Fundamentals of Physics" by Resnick/Halliday as a reference for introductory Physics. Is this a good choice or is there any other specific book that you'd recommend?
Would you mind elaborating how the dielectric plays a role in retaining energy in a capacitor? That is, how is the energy stored in a cap? Is it the accumulation of electrons one plate, or is the dielectric affected somehow? What would happen if you remove the original plate from the dielectric (after charging) and replace it with a new set of plates? Can you still recover the charge? Sorry for the barrage of questions, I find this quite intriguing.
this video is extremely helpful and concise:)
I've got a question if you keep the voltage constant but increase the frequency what will happen to the out put power ???
Thank you very much. I have a crap physics teacher and this really helps
this is great, thank you so much for these!! :)
Thank you this is very helpful!
Thanks for your reply, I'm enjoying your videos! Regards.
This is really helpful!!! Made it seem so simple
He really is the best wish he would make more videos
Well spotted. Indeed it is.
I appreciate fully the essence of the outcome of the physics explained in the video but in practice how is the current maintained to be constant? The current must be changing continuously , so the rheostat must be adjusted continuously. Quite a challenge. As always a great video from DrPhysicsA ! :)
sorry, I meant that at 17:50 you have mentioned V=QC. Shouldn't it be V=Q/C?
VERY WELL-DONE !!
Hi, may I know whether you have videos on Particle accelerators such as cyclotron and linear accelerators?What about bubble chamber as well>?
these videos are great! I doing and online course so im pretty much self taught.
Is there a magnet being generated in the capacitor?
Brilliant!
You r a freakin life saver.....THANK YOU!!!!
Thanks. I've just uploaded a video with examples of A Level questions on capacitance. Hope that helps too.
Thank you so much Mr
Sir, one thing though....that if we are having a gap between the capactitor plates, then the circuit is not complete
.....so how is the current flowing in the circuit then?
YES! Nobody has explained this better to me. Thank you
help me pls!!why [Q] increase and [i] decrease when the capacitor is charging??thanks
Is V meant to be equal to V0(1 - e^(-t/RC)) which it equal to V0 - V0e^(-t/RC)
If on increasing time, the voltage is increasing like you said so shouldn't time be on x-axis?
Professor, can you start to give us the option to play the videos at a faster rate in the future? Thank you for your amazing videos.
Love the videos, please keep them rolling.
PS is that your washing machine.... =p
These are so good, thank you so much! Can you make revision any easier?!
I finally understand charging and discharging, the text books were not helpful. Thank you so much
What a KING!
Actually u should keep time on X axis and Voltage on y axis bcz Ur measuring voltage which is the dependent variable
The best explanation
You are amazing sir
ive derived the formula from scratch and i kept getting the charging V=Vo (1-e^t/RC). am i missing something. i can show the whole derivation
Hi sir i would like to ask that what are the possible errors while charging and discharging the capacitors and how to over come it..?
When you replace the proportional sign with = , shouldn't it be= k, constant
thank you so much amazing video amazing guiding :)
really helpful thank you so much!
What book are you using as a reference for these videos? Thanks
thanks .. this is very helpful
Hi, at 18:34 the formula for the charging phase should be V=Vo - Vo e^(-t/RC) instead of the one you wrote, V=1 - Vo e^(-t/RC), right? Thanks
Yes ur right. its been 10 years wonder how you are now compared to 10 years ago