[MUSIC] Welcome back to linear circuits this is Dr. Weitnauer. This lesson is Continuity In Reactive Elements. Our objectives are to introduce the unit step function and to know when the voltage or current cannot change discontinuously. It builds on the integral I V relationships for capacitors and inductors. We begin with the unit step function, also known as the Heaviside function. It's used in describing a source, a voltage source or a current source. It's defined symbolically, as I'm showing here, a function that takes the value zero for negative time and one for positive time. And here's the plot of it. And notice that it is discontinuous at the origin, and we will not be concerned, what is the exact value at the origin, only that it takes the value of zero for negative time, and one for positive time. Next I want to define the, what I call the running integral. In this function of time, y of t, we see t in two places. It's the argument of y and then on the right hand side, it appears only in one place which is the top of the upper limit of the integral. As t increases, more area is captured by the integral. Let's consider then, the integral of the unit step function. Here's the step function, as you saw before. And here is its integral. Notice that the interval is zero for negative time, that's because no area is captured for negative time. And then for positive time, as t grows, we're going to progressively and proportionally capture more and more area producing this ramp characteristic with slope one. Now the integral is continuous at t equals zero and we say that y of zero minus equals y of zero plus. Now, zero minus is that time that's infinitesimally less than t equals zero and zero plus is infinitesimally greater than t equals zero. We use this notation later when we talk about initial conditions for second order circuits. Now the voltage in a capacitor has this running integral form that we were just talking about. Notice that it's divided by c and therefore it's going, the voltage and the capacitor's going to be a continuous function of time. Let's consider this example, that c is two Fahrens the initial voltage is t equals zero is minus two vaults so, that means that for us t nought is going to be zero. And then, our current is given a shown. Now, the voltage that goes as with this, is shown here. We start off at minus two because that's the given initial condition at the origin. And then we see, because the current is constant from zero to two, our voltage has this ramp characteristic. So the only challenge for us is to determine what is the maximum value of the ramp. We can figure that out because at t equals two the integral will have captured all of this area which is two times four or eight. But don't forget your going to be dividing by c which is two. So the voltage over this interval of time has to increment by four, eight divided by two. And that's what it does. It goes from minus two to plus two. And then, since the current is zero after that point, the integral doesn't capture anymore area, and the voltage stays constant. So we notice that this is the voltage is continuous, even though, there is a discontinuity in the current, but because it's a capacitor, the voltage is continuous at that point. Alright, the same sort of thing holds true for inductors, but for current, so current is continuous in inductors And I'd like for you to try the following quiz. The mathematics is really the same as we just went through for the capacitor. So, let's suppose that we have the initial current at t equals zero is one amp. The inductance is one third Henrys and the voltage is as shown. Then I'd like for you to please choose one of the these three curves for current. In considering the choices here for the current, we can rule out the case in b because of it's discontinuity because we know that the current and doctor can not be discontinuous. So that rules that one out. All right so the other two are continuous and they both begin at one which is, what they ought to be starting at because of the initial condition. They both have a downward ramp which is correct for voltage being negative in that period from one to two. They differ in how much they drop. So let's consider how much the ramp should drop. The area if you consider what the current would be at time equals two, it's going to be the value of this area divided by the inductance. Since you're dividing by one-third, you're actually multiplying by three. So you take the area, which is minus one times three, gives you negative three. So our ramp should change by a net of minus three. This one only changes by minus one, so this one cannot be true. And that leaves a as your correct one. In summary, because of the integral in the V I relationship for capacitors and inductors, the following cannot change discontinuously, voltage in a capacitor, current in an inductor. Thank you. [SOUND]