Capacitors - A Level Physics

[@zakharenkoff]

The best physics teacher I have ever seen.

[@racheldoran9758]

[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)

[@freemanfrank7171]

Thanks very much

[@MegaSquiff]

This guy’s explanation is about as good as is humanly possible. I immediately subscribed. A great big thanks!

[@lbrandon1227]

I regret I did not found you earlier you are the best teacher of my life❤❤❤Thank you

[@DrPhysicsA]

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.

[@nivaabraham8535]

if we have a high resistance... last line please.. nice class sir

[@nivaabraham8535]

does it take more time to both charge and discharge?

[@DrPhysicsA]

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.

[@tamtube9758]

Got a physics test tomorrow and haven't revised. You're a lifesaver.

[@DrPhysicsA]

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.

[@DrPhysicsA]

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.

[@DrPhysicsA]

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.

[@DrPhysicsA]

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.

[@DrPhysicsA]

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.

[@urjabhosekar8326]

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

[@nonoview1856]

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?

[@DrPhysicsA]

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.

[@nonoview1856]

DrPhysicsA Thank you very much!

[@DrPhysicsA]

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 KZbin videos do not appear on all viewing platforms which is a pity as I make a number of corrections and clarifications this way.

[@hemalsharma9364]

I wonder how is he now? He truly helped me survive the A levels! Thanks so much Sir!

[@alwaysdisputin9930]

He's a DJ DrPhysicsA aka Bob Eagle, CBE is now a DJ on local radio stations. He's on Twitter & he likes rain.

[@kingrobertbaratheonofhouse6190]

@@alwaysdisputin9930 glad to hear it. Living life well

[@gertwallen]

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

[@adnankamal3697]

Yes ur right. its been 10 years wonder how you are now compared to 10 years ago

[@sKhan-cu9fe]

Superb!!!! Everthing about capacitors done in 22minutes only....thank u so much!!!!.....

[@DrPhysicsA]

Congratulations. That sounds like a very good result. I hope it got you what you needed.

[@DrPhysicsA]

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.

[@adnanjan101995]

you're like the savior for a-level physics students....:D

[@snowellrod]

I really love the sound of the wash machine behind, it transport me to the future of the video.

[@sanjursan]

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.

[@sanjursan]

This is a very minor point, btw, this video is actually excellent.

[@DrPhysicsA]

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.

[@DrPhysicsA]

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%.

[@DrPhysicsA]

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.

[@DrPhysicsA]

Yes it is. Well spotted. I had put an annotation (I hope its still there) to correct the error.

[@DrPhysicsA]

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?

[@mirmuneerhussainiamachange2862]

U are Awesome Dr.Physics......................!!! Hats Off!!

[@DrPhysicsA]

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.

[@DrPhysicsA]

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.

[@entententententente]

This was very helpful! You explained it way better than my physics teacher.. Thanks :)!

[@jcutey]

sorry, I meant that at 17:50 you have mentioned V=QC. Shouldn't it be V=Q/C?

[@DrPhysicsA]

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.

[@muhammadshayanusman5540]

Meanwhile the jet preparing to take off in the background at the start of the video... Great content btw

[@mattbfc2009]

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!

[@DrPhysicsA]

Thanks. I hope the final exam goes just as well. All good wishes.

[@MrKillbot216]

if only you were my physics teacher :(

[@DrPhysicsA]

There is some basic info on cyclotron and linacs in my video "Nuclear Structure Physics". I dont think I've covered bubble chambers.

[@mangotangochick]

Love your videos. They explain it a lot better than my own physics teacher(s). Thank you! :)

[@vitoscarletta9267]

You are definitely making me pass my exams...thank you so much. You're like...GOD!! I can't thank you enough :)

[@DrPhysicsA]

Good luck with the exams.

[@muhammadrafiq5768]

Very well explained about capacitor and its various parameters. However, I would suggest to write numerics clearly as written 2 seems to if it was Z, similarly V has to be clearly written. Don’t be offended from some sense comments as I do understand some people like presentation in power point but your way of demonstrations is very simple and less time consuming. Keep to serve the physics lovers.

[@chinamatt]

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?

[@adamhilmi761]

Very well explained. Thank you Dr!

[@RbtV92]

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.

[@ahmedmahomed6776]

Excellent tutorial,really enjoyed watching it as it was so easy to to understand.A BIG Thank you.

[@AntiProtonBoy]

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.

[@DrPhysicsA]

The formal title is Dr Robert Eagle CBE BSc PhD (London). Hope the assessment goes well.

[@DrPhysicsA]

Good idea. I'll try to introduce that. Or at least keep the annotation on the screen as long as the error is visible.

[@DarkLevis]

It deppends how you connect those capacitors when charging. If you connect 'em parallely when charging, no current will not flow after disconnecting the power supply because they already have same voltage. If you charge 'em so, that capacitors are in series with power supply and the capacity varies between both capacitors, then you'll get a current because the voltage is different. Can be fairly easily proven with the formula Q=CU.

[@SPV66]

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 ! :)

[@pumbaface]

Explained beautifully! Thanks Dr P

[@songyojung]

at 18:45 you said the charging of voltage is V=1-V.e^-t/RC. But isn't it V=Vmax(1-e^-t/RC) ?

[@DrPhysicsA]

I'm not familiar with that particular book but it sounds as tho its a good introduction and will cover the right ground.

[@DrPhysicsA]

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.

[@Coops99]

YES! Nobody has explained this better to me. Thank you

[@AI-oy8zm]

Excellent delivery well done Dr Physics

[@jvpineda6860]

Thank you very much. I have a crap physics teacher and this really helps

[@ham8utube]

Dr. Physics, Can I please confirm that the formula at 18:40, Vo = 1-Voe^-t/RC actually is Vo= Vo(1-e^-t/RC) ?

[@DrPhysicsA]

That is correct. I thought I had put an annotation to make the correction but I am aware that not everyone can see the allocations that I add subsequently.

[@ham8utube]

DrPhysicsA Oh yes, I just noticed that annotation now, couldn't attend to it earlier. Thanks anyway.

[@hereitszara1497]

This is really helpful!!! Made it seem so simple

[@cboy-ou2hr]

He really is the best wish he would make more videos

[@muhammadalimusani7393]

At 11:13 why is it that the electrons move anticlockwise? Or were you just generally indicating that they move towards the lamp? Thank you!

[@DrPhysicsA]

+Muhammad Ali Musani It was just a general movement. They would of course move clockwise.

[@anuragsuresh5867]

Very good explanation 👍🏼

[@sachureedas]

You r a freakin life saver.....THANK YOU!!!!

[@gayathrikumarir5670]

Actually u should keep time on X axis and Voltage on y axis bcz Ur measuring voltage which is the dependent variable

[@ham8utube]

Also can you please explain "Why the earth connection is used in electrical installations?" and "Do we need a neutral wire apart from the three phase wires for intercity transmission?"

[@DrPhysicsA]

The earth wire is usually connected to any metal frame of anything which works on electricity. That way, if the metal frame were to become live because of some fault in the equipment the current would quickly be conducted away to earth.

[@ham8utube]

DrPhysicsA Thanks.

[@edward11131]

Is V meant to be equal to V0(1 - e^(-t/RC)) which it equal to V0 - V0e^(-t/RC)

[@ashleysobers9626]

I finally understand charging and discharging, the text books were not helpful. Thank you so much

[@DrPhysicsA]

Well spotted. Indeed it is.

[@DrPhysicsA]

Thanks. I've just uploaded a video with examples of A Level questions on capacitance. Hope that helps too.

[@wownow6780]

Thank you so much Mr

[@321silversnake]

Thank you so much I didn't really understand this before but now I do, keep it up

[@DrPhysicsA]

Can you give me the time on the video where this occurs? The graph at 18:12 is a curve.

[@gertwallen]

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?

[@JackMitchinson]

At the end of the video why make tau=RC to gain 1/e? It's clear that by altering RC the time it takes for the capacitor to charge or discharge will be altered regardless of whether tau=RC; also what is the significance of finding t when charge is at 63% and 37%, it seems completely arbitrary?

[@TheAmmaraahmad]

If on increasing time, the voltage is increasing like you said so shouldn't time be on x-axis?

[@preeam108]

Considering that Time is the independent variable here and Voltage is the dependent variable, 5 Minutes in the video, should TIME not have been plotted along the X axis with VOLTAGE along the Y axis ?

[@DrPhysicsA]

preeam ghosh Yes. I was just showing that voltage increases with time.

[@sakifabdallah8765]

your explanations are spot on! please keep up the good work sir :D

[@jcutey]

At 17:48 you have mentioned Q=VC. Shouldn't it be Q=V/C?

[@darsanhirani8099]

yes i think so too

[@SuperKarmachameleon]

Thank you, perfect compliment to my physics revision!

[@crocetonify]

Why is the graph of t against V a straight line?My A level book displays it as a curve that levels off.

[@aminajameel5939]

Thank you. It was v helpful you managed to cover the main points in such little time

[@anjalidutta8471]

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?

[@DrPhysicsA]

Yep.

[@rifatchowdhury8043]

hello sir, in time 9:38 u wrote Q/C.eff = Q/C1+Q/C2. my question is how Q is same for all the capacitors? if C1 is larger than C2 than Q should be also different. shouldn't it be Q1/C1+Q2/C2 ?

[@Tristan2709]

Hey Dr Physics A, Great vid! But got a question about capacitors; So capacitors can have a voltage (given by Q = CV), and they Do Not have a current across them, but do they have a Resistance? Thanks

[@DrPhysicsA]

They have a reactance and you can find out more about that in my vid kzbin.info/www/bejne/fHaog6iAodx1rqc

[@hev2250]

great video however i think the charging equation is wrong it has v = vnought( 1-e^-t/RC)

[@FrozenArchLich]

Excuse me, I want to know why we can't use an equation Qeffctive = Q1+Q2 in a series circuit (8:29) to do the proof.

[@davegaming8674]

Since right side of C2 is connected to cathode of Battery, we can see that it'd have minus charge. So then the minus charge will make electrons in left side of C2 repel, which would make left side of C2 have positive charge. Then, the positive charge in left side of C2 will make right side of C1 negative, and so left side of C1 will have positive charge. So the effective charge is equal to charge stored in C1 and charge stored in C2. Therefore, only differences are capicitance of C1 and C2. so, since V = V1 + V2 and that means Q/C = Q/C1 + Q/C2, divide by Q gives you 1/C = 1/C1 + 1/C2 Therefore, C = C1C2/C1 + C2 Hope this was helpful to you.

[@FrozenArchLich]

do you mean that the Qeffective = Qc1 = Qc2 ?

[@davegaming8674]

Yes.

[@mahmoudm451]

think of charge as electrons flowing through a wire with 2 capacitors in series for example. Series mean 1 road only, the electrons have nowhere to go, other than to flow through this road only (compulsory way), and so the electrons flowing through the first capacitor will have to flow through the second capacitor and so the charge is equal in both ( which is equal to the charge from the battery). But in parallel, they have more than one way to go, there is a split of road, so now they're happy!, they can roam around, and so the electrons split through the upper capacitor and the lower capacitor, and thus the charge from the battery is now split between those 2 resistors. Hope that helps

[@FrozenArchLich]

Oh! YES! That's very intuitive! Thank you very much. ^_^

[@NurAlam-si5sq]

which value of voltage should we use in energy stored in capacitor the one across capacitor or between battery and capacitor

[@curtiscleary790]

Hello Bob, I absolutely love your Physics videos, me and my friend have found them very useful In our revision. In this video when you say a "couple of batteries", don't you mean to say a "couple of cells" and a single "battery". By the way we love your singing1

[@Miru_Man]

These are really helpful videos thank you!

[@RandomPerson-sh9tu]

Why are there NO capacitance videos explaining what you do if you have capacitors in BOTH series AND parallel?

[@jawwadadel3522]

At 18:32, shouldn't it be V=Vo(1 -- e^-t/RC) ??

[@DrPhysicsA]

+Jawwad Adel Yes. Well spotted. I had put an annotation correcting this mistake.

[@aeon450]

+Jawwad Adel Fam! Didnt expect to see your comment here xD

[@kurniyantobuang6012]

The best explanation

[@crocetonify]

The graph at 4.48.My book shows it as a curve whilst the capacitor is being charged.

[@TealeAndTheOtherOne]

These are so good, thank you so much! Can you make revision any easier?!

[@AmineChM21]

I didn't understand the whole time against voltage graph , could you elaborate on that ,please.

[@AmineChM21]

+Gunbnelch Does he actually mean v(t)???

[@ad2894]

+Gunbnelch Maui Yes. Time is always an independent variable. Unless dealing with general relativity maybe.

[@gertwallen]

What book are you using as a reference for these videos? Thanks

[@iamvideoguy]

Love the videos, please keep them rolling. PS is that your washing machine.... =p

[@robertoartachovila5502]

Amazing vid bro, many thanks!

[@shahriarakbar3102]

There's a slight mistake in the equation of charging capacitor, it should be V = V_0 - V_O e^(-t/RC). rather than V = 1 - V_o e^(-t/RC)

[@DrPhysicsA]

Yes thanks. Well spotted. I think I added an annotation to that effect.

[@supercoolmy]

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..?

[@fjficm]

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

[@TheLittleCoffeeFarmer]

I've got a question if you keep the voltage constant but increase the frequency what will happen to the out put power ???