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Current time:0:00Total duration:10:57

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 will 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 moment point in time has a positive voltage here and negative voltage here it's completely it's continuously oscillating so at some moment in time let's say it's positive here and negative here all right so how do 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 volt the 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 ping for charge charge on the capacitor divided by the capacitance this is the definition of the capacitance right so it's a charge by capacitor and this is basically saying that to gen to generate a voltage the capacitor must get charged so there must be some charge right now we 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 v naught sine omega t and so from this i get an equation for charge so i know charge should equal c times v naught 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 so the 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 there here to that 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 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 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 are differentiating you are 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 v naught are constants you can pull them out and differentiation of sine would be cos omega t but that's not it remember we differentiate with respect to time so we have to use a chain rule and so then omega pops out and so you know into omega i'll write that omega over here and ta-da we 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 but 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 we can have that same function over here so here we have sine here we have cos it'd be great if we 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 can written 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 but okay 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 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 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 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 is 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 this is the last last time i want to pause and think about how would this 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 okay ahead means you know go to the right because that's the time direction but remember this is the future so if 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 our 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 ah 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 which is zero then voltage which is zero current first which is negative maxima so these are our positive and negative maximas so this is minus i naught this is plus i naught and you get the point the current leads the voltage and now let me show you uh 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 changing the ping current look at that look at that and we can use error 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 okay so the model 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 two 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