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# Capacitor i-v equations

## Video transcript

we're going to talk about the equations that describe how a capacitor works and then I'll give you an example of how these equations work so the basic equation of a capacitor says the charge Q on a capacitor is equal to the capacitance value times the voltage across the capacitor okay so here's our capacitor over here let's say we have a voltage on it plus or minus V and it has a we say it has a capacitance value of C and that's a property of this device here and C is equal to just looking at the equation over there C is equal to the ratio of the charge stored in the capacitor divided by the voltage on the capacitor so what we mean by stored charge is if a current flows into this capacitor it can leave some excess charge on the top I'll just mark that with plus signs and there'll be a corresponding set of minus charges on the other other plate of the capacitor this collection of excess charge will will be Q plus and this down here will be Q minus and they're going to be the same value and what we say here is when the capacitors in this state we say it's storing this much charge and we'll just name one of these numbers here and they're going to be the same with opposite signs so that's what it means for a capacitor to store charge so what I want to do now is develop some sort of expression that relates the current through a capacitor to the voltage so we wanted to develop we want to develop an IV characteristic so this will correspond to sort of like Ohm's law for a capacitor what relates the current to the voltage so the way I'm going to do that is to exercise this equation by causing some changes and in particular will change the voltage on this capacitor and we'll see what happens over here so when we say we're going to change a voltage that means we're going to create something a condition of of DV DT change in voltage / change in time and so I can do that by taking the derivative of both sides of this equation here I've already done it for this side and over here what I'll have is DQ DT and I took the derivative of both sides just to be sure I treated both sides of the equation the same so let's look at this little expression right here this is kind of interesting this is change of charge with change of time and that's equal to that's what we mean by current that is current and the symbol for current is I so DQ DT is current essentially by definition we give it the symbol I and that's going to be equal to C DV DT and this is an important equation that's basically the the IV relationship between current and voltage in a capacitor and what it tells us that the current is actually proportional to and the proportionality constant is C the current is proportional to the rate of change of voltage not to the voltage itself but to the rate of change of voltage all right now what I want to do is find an expression that expresses V in terms of I so here here we have I in terms of DV DT let's figure out if we can express V in terms of some expression containing I the way I do that is I need to eliminate this derivative here and what I'm going to do that by taking the integral of this side of the equation and at the same time I'll take the integral of the other side of the equation to keep everything equal so what that looks like is the integral of AI with respect to time is equal to the integral of c dv/dt with respect to time dt and on this side i have basically a I do something like this and I have the integral of DV so this is this looks like an antiderivative this is an integral acting like an antiderivative and what is it what function has a derivative of DV and that would be just plain V so I can rewrite this side of the equation constant C comes out of the expression and we end up with V on this side just plain V and that equals the integral of I DT so we're partway through we're developing and what's going to be called an integral form of the capacitor IV equation what I need to look at next is what are the bounds on this integral so the bounds on this integral are basically minus time equals minus infinity to time equals some time T which is sort of like the time now and that equals capacitance times voltage and let me take the C over on the other side and actually I'm going to move V over here on to the left and then I can write this one over C this is the normal looking version of this equation I DT and minus infinity to time T T time Big T is time right now and what this says it says that the voltage on a capacitor has something to do with the summation or the integral of the current over its entire life all the way back to T equals minus infinity and this is not so convenient what we're going to do instead is we're going to pick a time let we'll call it we'll pick a time called T equals zero and we'll say that the voltage on the capacitor was equal to let's say V naught with some value and that what we'll do is we're going to change the limit on our integral here from minus infinity to time T equals zero and then then we'll use the integral from instead zero to the time we're interested in so that equation looks like this we're just going to change the limits on the on the integral and we have the integral now but we have to actually account for all the time before T equals zero and what we do there is we just basically add V not whatever V not is that's the starting point at time equals zero and then the integral takes us from time zero until time now this is the integral form of the capacitor equation and I want to actually make one more little change this is the current at V of as a function of big T what we would really want to write here is we want to write V of a little T just this is just stylistically this is what we like this equation to look like and so I want my the limits on my integral to be 0 to little T and now I need to sort of make up a new replacement for this T that's inside here I can call it something else I can call it I of will call it I'll call it tau this is basically just a little fake variable d tau plus V naught and this is now we finally have it this is the integral form of the capacitor equation we have the other form of the equation that goes with this which was I equals big c dv/dt so there's the two forms of the capacitor equation now I want to do an example with this one here just to see how how this it works when we have a capacitor circuit