 In this video, we're going to attach an alternating voltage generator to a capacitor and find out the relationship between the current and the voltage, and then eventually draw a graph for the current with respect to the voltage. So let's begin. One thing to clarify is we're imagining that this circuit only has capacitance, no inductance or no resistance. And although that's not really ideal, it's a nice way to learn how capacitors behave when you put an alternating voltage across them, and it'll help us to learn more realistic circuits using these insights. All right. So where do we begin? I want an expression for current, right? So where do we begin? Well, let's assume there's a current. At some moment in time, there is some current flowing this way. And let's say that the generator has, at that more point in time, has a positive voltage here and negative voltage here. It's continuously oscillating. So at some moment in time, let's say it's positive here and negative here. All right. Let's say I build an equation. Well, whenever we are dealing with such circuits, I think the way I like to think about it is in terms of voltage, I know that because there are no circuit elements in between, the potential at this point is the same as the potential at this point. And similarly, the potential at this point should be the same as the potential on this point. And therefore, I know that at any moment in time, the voltage across the capacitor should equal the generator voltage. And that's where I can start. So let's write that down. We can say, at any moment in time, the voltage across the capacitor should equal the generator voltage. So the source voltage, all right? Now comes the question, how do we figure out what's the voltage across the capacitor? Well, we've seen before from capacitor equation, voltage across capacitor is just the charge on the capacitor. Let me use pink for charge. Charge on the capacitor divided by the capacitance. This is the definition of the capacitance, right? So it's a charge by capacitance. This is basically saying that to generate a voltage, the capacitor must get charged. So there must be some charge right now. You can call it charge q. And that charge, because of that charge, there is a potential difference. And that voltage is equal to the generator voltage. So this should equal the generator voltage. The source voltage, which is v0 sine omega t. And so from this, I get an equation for charge. So I know charge should equal c times v0 sine omega t. So I found the expression for charge on the capacitor. And it's telling me that the charge on the capacitor is not a constant. It's continuously oscillating just like the voltage, which is not so surprising. I would expect the capacitor to charge and discharge and charge and discharge to the charge will continuously keep changing. So I found the expression for the charge. But I want the expression for the current, not the charge. How do I go from here to there? I want you to pause the video and think a little bit about how do you get current from this expression? Okay, let's see. Here's my question. Can I just say current equals charge divided by time? So if I divide this thing by time, I'll get the current. Can I just say that? Can you pause the video and think, is this right or wrong? If it's, yeah, with the reasons. Okay, I can't say this. This is not right. The reason I can't say this is because this would only work if the current was a constant. If the amount of charge flowing per second is a constant, only then I can just say it's charge divided by time. But clearly in our case, the current won't be a constant. It's continuously going to change its value. It's going to change its direction. So for this, we have to differentiate. So over here, the current is going to be dq over dt. So you have to consider very tiny amount of charge flowing through very tiny amount of time. And that would be a current at that moment in time. And just to clarify, one thing you might say is that, hey, this is the charge on the capacitor. So when you're differentiating, you're calculating how quickly the charge on the capacitor is changing. Is that the current? Yes, because the rate at which the capacitor charge changes is the same as the rate at which the charges are flowing here. If there are 10 coulombs flowing for a second, then 10 coulombs are getting deposited on the capacitor plate. Okay, so the rate at which the charge on the plate is changing is the same as the current. And so this makes sense. So again, if you couldn't do it before, now would be a great time to pause and see if you can differentiate and see what expression you get for current. Okay. So this will be c and v0 are constants. So you can pull them out. And differentiation of sine would be cos omega t. But that's not it. Remember, we're differentiating with respect to time. So we have to use a chain rule, and so then omega pops out. And so we're going to into omega, I'll write that omega over here. And ta-da, we've found the expression for current. But now we want to compare it with the voltage and draw a graph, right? So for that, let's try to bring this in the same format as the voltage equation is. So the first thing I see is that this portion over here, this part over here, this now represents our maximum current. Just like how this represents the maximum voltage. And immediately, this is telling us that even though there is no resistance in our circuit, our current is limited. There is a maximum value and it depends upon all these numbers. And we'll talk more about why or how all of that happens in future videos. Now let's focus on this part. This is the part that I'm really interested in. To compare, you know, what's happening with our current, it'll be great if you can have that same function over here. So here we have sine, here we have cos. It'll be great if you can convert this into sine function as well and then compare the phase angle and see what the current is doing with respect to the voltage. So again, it was a great time to see if you can pause the video and use some trigonometry and convert this into a sine function and eventually tell what the current is doing with respect to the voltage and maybe try to even figure out what the graph is going to look like. All right, okay. So we know how to convert cos to sine. We can say cos theta is converted as sine of 90 minus theta. So I can say this is sine pi by 2 minus omega t. Now the problem with me, I mean, sorry, the problem with this, not with me, the problem I have with this is that it's hard for me to compare this function with this function because there is a positive omega t over here and there is a negative omega t. I really don't know what to do with that. I can't tell just by looking at this, what's my current oscillation doing compared to the voltage oscillation? It's really hard for me. It would have been great if I could convert it into a sine function with a positive omega t. Then it would be really, really easy for me to tell what's this oscillation doing compared to this one. Then I can easily compare. So can I do that? The answer is yes. Because remember, sine of pi by 2 plus omega t is also cos omega t because in the second quadrant, sine is positive. Therefore, instead of doing this, I will write this as sine of pi by 2 plus omega t or I'll write as omega t plus pi by 2. And one thing to remember, it doesn't really matter whether you keep it this way or whether you change it, the graph is not going to change. It's just for our understanding, this is a more convenient way to put it. And you'll see now why this is convenient. Now when I look at this, I immediately understand, ah! So the difference is the current is having a plus pi by 2 here compared to this phase. That means the current is oscillating ahead with a phase angle of 90 degrees and that means it's oscillating a quarter cycle ahead of the voltage. And that's why we say in capacitor, current leads the voltage. So they're not oscillating in sync with each other. And in a second, we'll see the animation. But current leads the voltage by a phase angle of pi by 2 radians. And so if you were to look at the graph, this would have been the current graph if the current and the voltage were in sync with each other. But now that we know that the current is leading by pi by 2, I want you to again, last time I want you to pause and think about how would the current graph be shifted? Do you think it'll be shifted somewhere like this or do you think it'll be shifted somewhere like this? Can you pause the video and think a little bit about it? All right. So we want our current graph to be ahead of the voltage. And at first it might seem like, OK, ahead means, you know, go to the right because that's the time direction. But remember, this is the future. So if you shift it to the right, that means it's delayed. It's more in the future. So we need to shift it to the left so that we say that current comes before the voltage. You get what I mean? So that means our current will be shifted to the left. And how much? One half of a one quarter cycle. So this part will be here. So this will be somewhat like this. There we go. This will be how the current graph is going to look like. So this means current first reaches the maximum, then the voltage reaches the maximum. Current first reaches zero, then voltage reaches zero. Current first reaches negative maximum. So these are our positive and negative maximas. So this is minus I naught. This is plus I naught. You get the point. The current leads the voltage. And now let me show you how to visualize this. So here's our visualization. The way to visualize this, just like we've done in previous videos, is I'll make the graph go back. And then we'll concentrate over here. And we'll be able to visualize the oscillations. So I'll dim everything. And you can now clearly see that the voltage is chasing the pink current. Look at that. Look at that. And we can use arrow marks. The current first goes to maximum. And then the voltage goes to maximum. Can you see that? And therefore, we say that current is leading the voltage. OK. So the moral of the story is for a pure capacitive circuit, how can you find the expression for the current? Well, we can use the capacitor equation. And then once you get the equation for the charge, you can differentiate it to get the current. And what we find is that the current leads the voltage when it comes to oscillations by a phase angle of pi by 2. And I'm sure you'll be very curious to understand, why does it do that? Why is the current leading the voltage? What's going on? How can we understand it logically? We're going to explore all of those things in a future video.