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Current time:0:00Total duration:9:21

- [Voiceover] So now I have
my two capacitor equations, the two forms of this equation. One is I, in terms of V, and
the other is V, in terms of I. Now, we're gonna basically
look at this equation here, and do a little exercise
with it, to see how it works. I'm gonna draw a little circuit here. It's gonna have a current
source, and a capacitor. The value of the capacitor
is one microfarad. The value of our current
source, we'll call it I. It's actually gonna look
like a pulse, like that. It will go from zero to three milliamps, and then back to zero, over here. The amount of time it
takes, that's gonna be, this time here is gonna
be, three milliseconds. The question I wanna
answer is, what is V of T? Right here. We're gonna use this integral
equation to figure that out. Over here is where we're
gonna put our answer. This is gonna be T, and this will be I. Here's I. Our plot, right down here,
will be T, and here's V of T. We'll plot I this way, on here, in time. It's zero. Then, it goes up, to some
value, and then it goes over. So it's a pulse of current. We said that that was zero,
this is three milliamps... And this is three milliseconds. Now, what we want to do,
is we wanna find V of T. One of the things we have to do, we have to make an
assumption about V, not here. We're gonna assume, for our problem here, that V naught, equals zero volts. What that means is, there's no charge. There's zero charge
stored on this capacitor, when we start the experiment. Now then, let's look at
three different time periods. Let's look at the period before the pulse, during the pulse, and after the pulse. We'll break the problem into three parts. Part one is before, and we'll do that, by just looking up here. We decided that V naught, was
zero, in the before state. Put a little zero there. We decided that I, I is zero, so that means that the term
inside the integral, is zero. What that means, zero plus
zero, is equal to V of T. Before the pulse, before the pulse, the capacitor equation tells
us, the voltage is zero. Okay, now let's go during the pulse. Let's do the second of period of time. Let's do during the pulse. Now, we have to be a little more careful. We have V equals one
over C, integral from, now time equals zero, to
time equals sub-time T. What's I during the pulse? Well, it's sitting right here. I is a constant, so we
write in three milliamps. D tau... Plus, don't forget the starting voltage. What's our starting voltage? Well, we look right here, and
the starting voltage is zero. Plus zero. Now, we can solve this. V equals one over C, times
three milliamps comes out... Times the integral, from
zero to T, of D tau. That equals, let's plug in C this time, three milliamps, divided
by one microfarad, times, what does this evaluate to? The integral from zero to
T of D tau, is, just T. Let's do a little bit of
arithmetic here, to reduce this. Milliamps is 10 to the minus three, and microfarads, is 10 to the minus six. Let me move this up. We'll keep our plots on there. What we end up with is V equals three, 10 to the minus three,
10 to the minus six, so that's three times 10 to
the third, or 3,000 times time. What's that? That's the equation of a line,
and it has a slope of 3,000, what's the units here? This is 3,000 volts per second. I'll go over here, and
we'll sketch this in. This is gonna be a line,
that's a straight ramp, with a constant slope, like that. It's gonna be happening
all during this pulse. We can ask, what's this value right here? What's that voltage right there? Okay, let's work that out. T is three milliseconds. So let's plug in three
milliseconds, right here. V at three milliseconds,
equals 3,000 times three milliseconds, and that equals
three times three is nine. 3,000 times, this is an
exponent of minus three. It's gonna be nine volts. This value, right here,
right there, is nine volts. That says the voltage on our capacitor, during the pulse, during
the current pulse, rises in a straight
line, up to nine volts. We got that from this
integral that we did. The capacitor's integrating the current, adding up the current. It's integrating this pulse,
to get an ever-rising voltage. Okay, so now we've solved
the capacitor equation, during the pulse. Let's go back now, to what
happens after the pulse. We'll do that over in
the corner, over here. Now, let's solve what happens,
after the current pulse. What happens to this
voltage, from here on? Does it go down? Does it go up? Does it go straight sideways? Let's find out, and let's use our capacitor equation to do this. Now, what we're doing,
is we're gonna define, we're gonna do a new integral. We're gonna start at time
equals three milliseconds. What is our voltage at three milliseconds? What's V naught? V naught, in this case, equals nine volts. That's for this period of
time, after three milliseconds. We'll use our equation again. V is one over C, times
the integral of I of tau, D tau, plus V naught. It goes from time... It goes from time, three
milliseconds, to time T. Time T, now, is out here
somewhere on the time scale. We know the voltage right
here, we filled it in. There's the voltage right
there, we're gonna fill that in. We're gonna work out what's the voltage out here, after three milliseconds? To solve this integral, we look and see, "Well, what's I of tau? "What's I of tau, during this time?" Let's look at our chart. The pulse is at zero. Oh, look. So this whole term right here, this entire term right here is zero. What is V naught? V naught, we decided, was right there. V naught is nine volts. That's where we started from. The answer here, after the pulse goes away, is V equals nine volts. I can sketch that in
up here, and basically, it just goes straight
sideways, at the new voltage. That's how we solve a capacitor problem, a really simple one. It happens that we were
integrating a current pulse, and what we got was a voltage ramp.