[MUSIC] Welcome back to Linear Circuits, this is Dr. Weitnauer. This lesson is Introduction to Capacitors Part 2. Our objective is to understand the relationships between charge, current and voltage for capacitors. This builds on the passive convention and the Q V and the I V relationships for capacitors, which are q equals Cv and i equals C dv dt. Alternatively we can get voltage from current. We have this relationship, i = C dv/dt, but if we integrate those sides and then solve for the voltage we get this expression. Now, there's a few things that are important to note in this expression. First of all, don't forget to divide by the C in front of the integral. The other main point is, look at the lower limit. If the lower limit is naught minus infinity, here it's shown as being t naught. Then that t naught must be the argument of the voltage in the additive term. You can think of this v of t naught as the initial condition on the interval t naught to t. All right, here's an example, suppose the current through a 2 ferret capacitor is given as shown. Find the voltage for t greater than or equal to 0, given that v at 0 is -2 volts. And we will see this integral equation here, where you can see how I have used instead of t naught I have substituted in 0. The reason is because we were given the voltage at time 0, so want to set it up like that so I can use that given information. I'll begin by just getting the integral of the current, i(T)dT. Starting at 0. The important points in time are two and three. We observe that the current is constant over the interval between zero and two. That means the interval is going to be a straight line constant slope, and it will have a maximum value equal to the area of this box, which is eight. Then in the next interval between two and three we again are constant, but with the same magnitude but different polarity, which means that we're going to have another straight line and it will drop to half of its peak value. So it's going to drop to four, then after that the integral accumulates no more area, so it'll be straight after that. Now that's just the integral. You look at the formula, we have to now divide by the capacitance, which is 2. So that's going to scale. So we'll go ahead and do that scaling. So it's going to look exactly like the previous drawing, except the vertical axis will be relabeled. To be half of what it was, 4 and 2. Then finally what we're going to be doing is adding the initial voltage which is minus 2 volts. So that's going to have the effect of taking this curve and shifting it downwards by 2. So that'll give us, finally, the voltage In volts. As a function of time in seconds. And, instead of starting at zero as it does here, it's going to start at minus 2 volts. At 2, which is this point here, it will reach its maximum value of 2. So we're going to start here, go up to there. Then it's going to drop by two, which means it's going to drop back down to the horizontal axis and be constant after that. And so that would be the answer to this example. To sum up, we have the Q V characteristic, q = Cv, and in the passive convention, the I V characteristic, which is i = C dv dt. And finally, when using the integral relationship, don't forget to divide by C And then add the initial voltage. Thank you. [SOUND] [MUSIC]