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Â Welcome back to linear circuits this is Dr. Weitnauer.

Â This lesson is Continuity In Reactive Elements.

Â Our objectives are to introduce the unit step function and

Â to know when the voltage or current cannot change discontinuously.

Â It builds on the integral I V relationships for

Â capacitors and inductors.

Â 0:29

We begin with the unit step function, also known as the Heaviside function.

Â It's used in describing a source, a voltage source or a current source.

Â It's defined symbolically, as I'm showing here,

Â a function that takes the value zero for negative time and one for positive time.

Â And here's the plot of it.

Â And notice that it is discontinuous at the origin, and

Â we will not be concerned, what is the exact value at the origin,

Â only that it takes the value of zero for negative time, and one for positive time.

Â 1:06

Next I want to define the, what I call the running integral.

Â In this function of time, y of t, we see t in two places.

Â It's the argument of y and then on the right hand side,

Â it appears only in one place which is the top of the upper limit of the integral.

Â 1:43

Notice that the interval is zero for

Â negative time, that's because no area is captured for negative time.

Â And then for positive time, as t grows, we're going to progressively and

Â proportionally capture more and

Â more area producing this ramp characteristic with slope one.

Â 2:04

Now the integral is continuous at t equals zero and

Â we say that y of zero minus equals y of zero plus.

Â Now, zero minus is that time that's infinitesimally less than t equals

Â zero and zero plus is infinitesimally greater than t equals zero.

Â We use this notation later when we talk about initial conditions for

Â second order circuits.

Â 2:44

Notice that it's divided by c and therefore it's going,

Â the voltage and the capacitor's going to be a continuous function of time.

Â Let's consider this example, that c is two Fahrens the initial

Â voltage is t equals zero is minus two vaults so,

Â that means that for us t nought is going to be zero.

Â And then, our current is given a shown.

Â Now, the voltage that goes as with this, is shown here.

Â We start off at minus two because that's the given initial condition at the origin.

Â And then we see, because the current is constant from zero to two,

Â our voltage has this ramp characteristic.

Â So the only challenge for

Â us is to determine what is the maximum value of the ramp.

Â We can figure that out because at t equals two the integral will have captured all

Â of this area which is two times four or eight.

Â But don't forget your going to be dividing by c which is two.

Â So the voltage over this interval of time has to increment by four,

Â eight divided by two.

Â And that's what it does.

Â It goes from minus two to plus two.

Â And then, since the current is zero after that point,

Â the integral doesn't capture anymore area, and the voltage stays constant.

Â So we notice that this is the voltage is continuous,

Â even though, there is a discontinuity in the current, but

Â because it's a capacitor, the voltage is continuous at that point.

Â 4:24

Alright, the same sort of thing holds true for inductors, but for current, so

Â current is continuous in inductors And I'd like for you to try the following quiz.

Â The mathematics is really the same as we just went through for the capacitor.

Â 5:07

In considering the choices here for the current, we can rule out the case

Â in b because of it's discontinuity because we know that the current and

Â doctor can not be discontinuous.

Â So that rules that one out.

Â All right so the other two are continuous and they both begin at one which is,

Â what they ought to be starting at because of the initial condition.

Â They both have a downward ramp which is correct for

Â voltage being negative in that period from one to two.

Â They differ in how much they drop.

Â So let's consider how much the ramp should drop.

Â The area if you consider what the current would be at time equals two,

Â it's going to be the value of this area divided by the inductance.

Â Since you're dividing by one-third, you're actually multiplying by three.

Â So you take the area, which is minus one times three, gives you negative three.

Â So our ramp should change by a net of minus three.

Â This one only changes by minus one, so this one cannot be true.

Â And that leaves a as your correct one.

Â 6:22

In summary, because of the integral in the V I relationship for

Â capacitors and inductors, the following cannot change discontinuously,

Â voltage in a capacitor, current in an inductor.

Â Thank you.

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Â