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.
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.
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 RU-vid videos do not appear on all viewing platforms which is a pity as I make a number of corrections and clarifications this way.
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.
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
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.
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.
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.
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.
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.
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!
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.
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%.
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.
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.
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 ! :)
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.
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.
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, 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?
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 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.
