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### 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

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# RC step response setup (1 of 3)

The RC step response is an exponential curve. We set up the circuit and create the differential equation we need to solve. We introduce the technique of Natural response + Forced response. Created by Willy McAllister.

## Want to join the conversation?

- It's ordinary because you have one parameter, according to documentation for RC natural response. First order refers to the highest derivative in an equation being first degree only.(4 votes)
- I think there is a chain of sign errors there. For the direction of i shouldn't Vr be equal to -IR ? Also for the same sense it should be I = -C(dV/dt) which ended up cancelling the first error :)(2 votes)
- I think the signs are correct. The labeling of the resistor respects the Sign Convention for Passive Components. That is, the current arrow points into the plus sign of the Vr voltage label. With this arrangement of current arrow and voltage label, Ohm's Law does not have a negative sign. The same holds true for the capacitor, the current arrow points into the + sign of v(t). For this reason, the i-v equation for the capacitor also does not have a minus sign.(3 votes)

- 0:50You said this is where we need calculus skills. Can you please provide the link to learn calculus required to solve this? No more no less.(1 vote)
- Suggested preparation in Calculus is listed in the Calc section of this article: https://www.khanacademy.org/science/electrical-engineering/introduction-to-ee/modal/a/ee-preparing-to-study-electrical-engineering

Also, before taking on this RC Step Response be sure you go through the article or videos on RC Natural Response.

Another resource is my web site, a continuation of this work I did at KA. See https://spinningnumbers.org/t/topic-natural-and-forced-response.html for revised and updated articles on Step and Natural Response.(2 votes)

- could you explain more this statement @5:51"This is what the circuit is being driven to by that step response input voltage source."(1 vote)
- I wanted to ask about the point where you said that the voltage is not same for capacitor and resistor in step response but current is... while in natural response we assumed current to be different for each component and we use Ir for resistor and Ic for capacitor but here use same current I but different voltage. can you plz explain it that why is it different in N.R and S.R?(1 vote)
- The source of your confusion is the RC natural response circuit. The circuit we draw for RC natural response is just two components, R and C. This super simple circuit has a really odd and confusing property that most textbooks ignore. It turns out to be oddly challenging to label the voltage polarity and current direction while
*at the same time*respecting the Sign Convention for Passive Components.

I have a particular way I teach this, which involves defining I_R and I_C in opposite directions. (They have the same magnitude. I_R = -I_C) This is a "trick" that allows me to use the Sign Convention (current flowing*into*the positive voltage polarity) and not have to introduce a - sign into Ohm's Law or the capacitor's I-V equation. This awkward definition of two currents means Ohm's Law is what you think, v = IR and the capacitor's equation is I = C dv/dt. Neither equation has a negative sign. I like that from a teaching/learning standpoint. Then we combine the two I-V equations into a KCL equation at the upper node, I_R + I_C = 0 we end up getting the right differential equation for the natural response.

If instead I had defined just a single current, I, it would be going "backwards" through one of the components, and I would be forced to introduce a - sign into the I-V equation of either R or C. I find that distasteful for the teacher and learner.(1 vote)

- Hi.

If you look at the GPU or CPU loading pattern, There is a GPU/ CPU load which could be 6A/ 1us (forcing function di/dt).

Can you please show the derivation when the function is a current load with di/dt = 6A/ 1 uS.

Thank You.(1 vote)- I'm not sure i understand the circuit you have described. It may be somewhat similar to this video: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/v/ee-capacitor-integrates-current.

6A/us is a rather huge current slope. In 2 usec the current is up to 12 amperes.(0 votes)

## Video transcript

- [Voiceover] In the last video
we looked at this RC circuit and we gave it a step
input with a step source, a step from V nought up to V S, with a sharp change right
here at T equals zero. And we sort of took an intuitive guess at what this voltage looks like here. That's V and T. And in this video we're going to develop a precise expression for what happens here to this voltage across the capacitor. And as we said, this is a
very very common circuit, there's millions and millions
of these types of circuits inside every computer, so it really pays to understand this one quite well. Because this is a function of time, because we have a function of time here, this is going to end up
needing some techniques from calculus to really
figure out properly. So this is going to be a place where our calculus skills
are really necessary. And I'm also going to
assume that you've listened to the video on RC natural response where we worked out the expression for the natural response when
there was no voltage here, where this was zero. The natural response looks like V equals V not E to the minus T over RC. And there's a video on Khan
Academy about natural response and there's also an
article you can read too. So let's get back here,
let's analyze this fourmulae. And one of the things we can do is, well, first thing we always do is
we always label the quantities that we want in here, and
I'll put, there's the current, and here we'll call this voltage of R, and if I've used a
lowercase letter V or I, that means that quantity
is changing with time and if I use a big, a
capital letter V here, that means that those values
don't change with time. So that's why notation. And when we analyze the
circuit, a good thing to do is just to look at what
are the compound equations for this thing. So let's talk about Ohm's
Law for this resistor here. We know that V R equals I times R Or another way that we can write that is I equals V R over R So the other expression that I could write for the capacitor is the
capacitor of I equation which is I equals C DV DT and this means the slope of the voltage or the rate of change of the voltage times C is equal to the current. Alright, so far so good. I have one I. There's actually only
one I in this circuit. So this I here is equal to this one here, that's a good relationship for us to know. Oh and we also have two
different V variables, we have regular V that's
across the capacitor and we have V sub R. So let me see if I can write V sub R in terms of V. And one thing I can
tell from this circuit, this point right here, this is V S. This point right here this is V C. So I can write V R equals V S minus V C. Alright, that's going
to be useful. Alright. So now what I'm going to do is I'm going to set these two I's equal to each other and at the same time I'll plug in this expression V S minus V C for V R up here. It will go in right here. So we can write C DV DT equals one over R times V R V R is V S minus V C. I shouldn't write V C, I'll just write V. I'll take that out there because we don't need that subscript. And multiply through by
one over R et on, we get C DV DT equals one over R V S minus one over R V and we'll get C DV DT minus one over R times V equals V S Alright so now we have, this
is our differential equation. It's a differential equation because it has derivatives in it. In particular this is called a non-homogeneous ordinary differential equation. It's ordinary because
there's a first derivative, the first derivative
has a power of one here, that's the ordinary. It's a differential equation
because it has a derivative and it's called non-homogeneous because this side over here, this is not V or a derivative of V. So this equation is sort of mixed up, it's non-homogeneous. When we did the natural response analysis, this term right here was zero in that equation, so we were
able to solve this rapidly. So we're going to develop
a new strategy here for solving this kind of equation this kind of non-homogeneous ODE. So this equation is a little tricky to solve because this
is not zero over here. We have a mixture of things going on here. This side of the equation represents the forcing function. This is what the circuit
is being driven to by that step response
input voltage source. And this side here we're
going to get tied up with the initial conditions that that capacitor has, so there's sort of two
things going on here, that make this a little more complicated than usual to solve. And so what our strategy is,
we take on a strategy here. We break the problem down into two pieces. This is always what engineers do when they're faced with something that is a little
complicated, is you try to break it into bits and pieces. So what we're going to
do is, our strategy is to figure out what we
call the natural response. And we're going to figure
out something called the forced response. And then we're going to add those together with a plus sign, and that's going to be equal to what we
call the total response. So this is what the circuit really does. Alright so we got two new words here. One is the forced response,
and one's the total response. And if we remember before,
the natural response was what the circuit did
when nothing was forcing it. We just put in some initial
energy, some charge, and let the circuit sort of sag toward wherever it was going. That's the natural response. The forced response is going to be what the input source
forces the circuit to do. And when we take those two
things and add them together we'll get what the circuit
is going to do altogether. And this is another good
application of super position. We're going to take two
things that the circuit does and analyze them separately and then add them
together, superimpose them, to get the total response. So they way we do this, the strategy is for the natural response, basically we take for
the initial conditions, that's the charge on the capacitor, we'll take whatever we get. We'll take the initial Q
and the initial V W, V not. And when we look at the inputs, we'll set the inputs to zero. So that's how we analyze,
that's the conditions for the natural response. And when we take the forced response we do the opposite, what we do is we set the initial conditions to zero, simplify the circuit down and we use the inputs that
are actually applying. In this case, it was V of S. And then you get the total response, we add those together, with a plus sign. So then for the total response, you get the initial conditions,
which is Q and V not, and you get V S, all together. So that's our strategy. And by turning it into
two simpler problems, basically we make this
differential equation a little bit simpler to solve two times than trying to solve it
all together at once. We'll stop here now and in the next video, we'll follow through on the strategy of combining the forced response
with the natural response.