Main content

## Electrical engineering

### Course: Electrical engineering > Unit 2

Lesson 4: Natural and forced response- Capacitor i-v equations
- A capacitor integrates current
- Capacitor i-v equation in action
- Inductor equations
- Inductor kickback (1 of 2)
- Inductor kickback (2 of 2)
- Inductor i-v equation in action
- RC natural response - intuition
- RC natural response - derivation
- RC natural response - example
- RC natural response
- RC step response - intuition
- RC step response setup (1 of 3)
- RC step response solve (2 of 3)
- RC step response example (3 of 3)
- RC step response
- RL natural response
- Sketching exponentials
- Sketching exponentials - examples
- LC natural response intuition 1
- LC natural response intuition 2
- LC natural response derivation 1
- LC natural response derivation 2
- LC natural response derivation 3
- LC natural response derivation 4
- LC natural response example
- LC natural response
- LC natural response - derivation
- RLC natural response - intuition
- RLC natural response - derivation
- RLC natural response - variations

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# A capacitor integrates current

A current flowing into a capacitor causes charge to accumulate. The voltage rises according to q = Cv and we say the capacitor integrates current. Created by Willy McAllister.

## Want to join the conversation?

- what would be the equation for voltage across a capacitor which is first charged some voltage and then the battery is removed and the capacitor is reconnected with a wire which has 0 resistance. I would also like to get some intuitive answer as to what happens with the electrons moving and all if possible in this Thank You in Advance.(3 votes)
- Hello Usman,

If you could find a zero resistance wire and if you could make a capacitor with zero parasitic resistance then an infinite current would flow instantaneously discharging the capacitor.

No such components are available in the real world. You will ALWAYS find the RC discharge curves for real world components.

Ref: http://www.allaboutcircuits.com/textbook/direct-current/chpt-16/voltage-current-calculations/

Regards,

APD(5 votes)

- 4:38Can someone please tell me how the integral is just t?(3 votes)
- It might help to write an explicit coefficient of 1 in front of dtau inside the integral.

The integral asks the question, "What is the integral of 1 (between 0 and t)?"

That's the same as asking, "What is the anti-derivative of 1 (between 0 and t)?"

To answer that, flip the anti-derivative question around, "Can I think of a function whose derivative is equal to 1?"

If you have studied derivatives you hopefully recall that the derivative of t = 1.

So the anti-derivative (and integral) of 1 is t.

https://www.wolframalpha.com/input/?i=integrate+dt+from+0+to+T(2 votes)

- At2:50, I'm not sure there is a need to consider the integral. The bounds of the integral go from 0 to t, so it doesn't seem valid to evaluate this integral for the current value given for times less than zero.(3 votes)
- You are correct. I used the integral with lower limit of t=0 to evaluate what happens before t=0. For the time before 0 I should have used the general version of the integral expression where the lower limit is t= -infinity. There is a more about this in the previous video "Capacitor i-v equation" at5:00.(1 vote)

- why we calculatin capacitor voltsge from -infinity to present

what is its significance

why cant we do from zero to present

what is it necesary to calculate response before applyin pulse(2 votes)- Correct me if I am wrong, but the point seems to be accounting for the starting condition. Thus, when V0 is introduced, the -infinity portion goes away because t0 is the starting time and V0 accounts for any charging done before t0 (aka from t-infinity to t0).(2 votes)

- 5:18sorry, but how is that an equation of a line? y=mx+b, there is not +b? How can a beginner identify that?(2 votes)
- In the video, the equation of the line is v = 3000 * t. As you said, the equation of the line is y=mx+b, so replacing the variables we have: y=v; m=3000; x=t and b=0. that's why v=3000t is the equation of a line, with a slope of 3000.(2 votes)

- does anyone use this app anymore? I can’t find anyone…i’m i…alone? Also, please do not respond if you’re a robot.(2 votes)
- please explain the math behind: "vt =VS −VS -e−t/RC or vt = VS (1 - e-t/RC )" where did the one come from? or how did the vs become one?(1 vote)
- This is the equation for an RC step response. You have an extra minus sign in front of the "e" term. That's not supposed to be there.

vt = VS - VS e^(-t/RC)

vt = VS ( 1 - e^(-t/RC))(2 votes)

- So that means if each current pulse lasts 3 ms and is at 3 mA, that a 1 microfarad capacitor will increase by 9 volts during each pulse until maximum voltage is reached and current no longer affects the voltage.

That makes sense since anything that stores charge, whether that be a battery, a capacitor, or some other element has a change in voltage in 1 form or another due to current. In the case of a battery, the voltage decreases as it gets closer to 0 electrical power.

So if I wanted to know the voltage drop from a battery with a certain starting voltage(say 3V if it is 2 AA batteries acting as 1), and I knew how many amps there were in the circuit and that it was a square wave pulse and not constant, would I basically take the negative of the capacitor formula?(1 vote)- Also with batteries they stop producing current well before zero volts so you need to look at its characteristic curve to see how many pulses it can accomplish. If it is rechargeable and you go below the recommended voltage you can damage the battery and it will lose its ability to recharge.(2 votes)

- Don't you need a resistor in series with the capacitor??(1 vote)
- In most practical circuits there is a resistor connected to the capacitor (either in series or parallel). The point of this video is to explore the i-v equation for the capacitor in isolation to see what it predicts. As you work your way into the RC Natural Response videos you will learn about the RC variation.(1 vote)

- Why did we take limit from "3ms to T" but not " 0 to T" when current again goes to zero?(1 vote)
- When solving an integral equation like this one you can put in i(t) as the pulse and integrate from 0 to T. But since I am a lazy EE I want to do something simpler than integrating a pulse. Instead I chop the problem up into little pieces of time. I pick time intervals such that i(t) is a constant during that time. A constant is easier to integrate than a pulse.

Between t = 0 and t = 3ms the current is constant 3mA. That's the first integral equation. It gives us the voltage on the capacitor at t = 3ms. That v then becomes the trailing constant v_o in the integral equation for the next chunk of time. The next chunk of time starts at t = 3ms and goes up to T.

This method is called a "piece-wise" solution. You break up time into pieces and solve each one separately. It is common to use this method when input signals are jumpy or don't have an easy integral.

That v_o in the capacitor's integral i-v equation is pretty cool. In one little number it captures**everything**that has ever happened to that capacitor all the way from - infinity up to that moment in time.

These articles walk through the piece-wise solution process,

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/a/ee-capacitor-equation-in-action

or this updated and revised version,

http://spinningnumbers.org/a/capacitor-iv-equation-in-action.html(1 vote)

## Video transcript

- [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.