Capacitors - A Level Physics

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.

[@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

[@zakharenkoff]

The best physics teacher I have ever seen.

[@MegaSquiff]

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

[@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.

[@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]

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.

[@lbrandon1227]

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

[@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]

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.

[@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

[@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.

[@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.

[@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.

[@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

[@DrPhysicsA]

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

[@MrKillbot216]

if only you were my physics teacher :(

[@tamtube9758]

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

[@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]

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?

[@cR4Sh6]

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

[@DrPhysicsA]

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

[@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]

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%.

[@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

[@entententententente]

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

[@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.

[@muhammadshayanusman5540]

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

[@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?

[@adnanjan101995]

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

[@mirmuneerhussainiamachange2862]

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

[@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]

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.

[@jcutey]

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

[@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]

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

[@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.

[@adamhilmi761]

Very well explained. Thank you Dr!

[@sKhan-cu9fe]

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

[@edward11131]

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

[@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]

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

[@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.

[@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.

[@DrPhysicsA]

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

[@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

[@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?

[@TheAmmaraahmad]

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

[@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 ?

[@jvpineda6860]

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

[@RandomPerson-sh9tu]

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

[@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?

[@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.

[@mangotangochick]

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

[@sachureedas]

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

[@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.

[@ahmedal-ebrashy3691]

How come are under graph gives you energy of capacitor? If that is true then Capacitance will be equal to Q.V not Q/V

[@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 ! :)

[@calieburger3113]

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

[@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.

[@gertwallen]

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

[@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.

[@crocetonify]

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

[@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.

[@huyvu5379]

help me pls!!why [Q] increase and [i] decrease when the capacitor is charging??thanks

[@NurAlam-si5sq]

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

[@jcutey]

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

[@darsanhirani8099]

yes i think so too

[@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.

[@gayathrikumarir5670]

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

[@jordanandino5332]

Can you explain why V=QC and Q=VC, I tried to rearrange them and they don't equate unless I'm making a stupid mistake or confused some stuff .

[@DrPhysicsA]

Jordan Andino Did I say that. If so it’s an error. As I say at the outset C=Q/V. So Q=VC. But V=Q/C.

[@DrPhysicsA]

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

[@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..?

[@alphie10]

i didnt quite understand the last part of the video why would you get a longer time for it to charge up? :S

[@luizinxz98]

Is there a magnet being generated in the capacitor?

[@Coops99]

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

[@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?

[@ahmedmahomed6776]

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

[@djsonu07]

Hello sir.. I would like to confirm the equation at 19:50 which says V=QC.. Is it not Q=CV-??

[@ashleysobers9626]

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

[@hev2250]

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

[@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

[@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.

[@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.

[@anuragsuresh5867]

Very good explanation 👍🏼

[@Ronnieht]

If a capacitor is fully charged, does no current flow through the circuit anymore?

[@DrPhysicsA]

No. Remember that no current flows across the capacitor at any time. But as the capacitor charges up the flow of electrons to the capacitor plate is reduced as the electrons on the plate effectively repel any further boarders.

[@leehauyuan]

Hi, may I know whether you have videos on Particle accelerators such as cyclotron and linear accelerators?What about bubble chamber as well>?

[@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.

[@natureandchange]

When you replace the proportional sign with = , shouldn't it be= k, constant

[@hereitszara1497]

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

[@cboy-ou2hr]

He really is the best wish he would make more videos

[@aminajameel5939]

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

[@danwilloughby728]

Why can't the method used to obtain effective capacitance in series be used when calculating effective capacitance in parallel?...I get that v varies in series so can't be done using the first method, but why can't the parallel capacitor be figured out by subbing in q?....I am obviously missing something?

[@danwilloughby728]

Nvm it's because v doesn't vary so the same method would find 2v not v, thanks great video.

[@AI-oy8zm]

Excellent delivery well done Dr Physics

[@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]

Well spotted. Indeed it is.

[@Miru_Man]

These are really helpful videos thank you!

[@ugi991]

Is the dielectric included in the edexcel 2016 syllabus?

[@hagoiTV]

I think its included for AQA

[@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.

[@supercoolmy]

Hello sir, I would like to ask why cant we use Energy stored in capacitor = (V)squared x C ^^^^ i derived that formula from; we know Q = VC and Q = E/V hence VC = E/V shift the V to other side then we get E the energy stored = (V)squared x C ----------------------------------------------------------------------------- Instead of Energy stored in capacitor = 1/2 x C x (V)squared

[@DrPhysicsA]

because the voltage across the capacitor is not constant throughout the charging. It builds up exponentially.

[@supercoolmy]

DrPhysicsA could you explain that a bit more? thanks

[@mohammadbasil5409]

Sohaib Hassan u know the graph of charging and discharging w/ time¿¿ it has an exponential equation I=I•*exponential^lambda*t