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# Capacitor i-v equations

The capacitor current-voltage equation in derivative form and integral form. Created by Willy McAllister.

## Want to join the conversation?

• Really enjoying this subject matter. I have been stopping periodically to get other source material on how capacitors work and why they are important in electronics. I appreciate the need to break things down to the math, but, it seems that a more practical explanation of what is actually happening would enhance the lesson plan. For example, I assume a capacitor is effectively a charge container. I gather the equations tell us things like, how much charge a given capacitor can store, perhaps how it can receive a charge, how it can be discharged, etc. For my purposes, jumping into the equations before i have some conceptual view of what the component is supposed to be doing and how, matters greatly. • why is the integral from -infinity to T rather than 0 to T?? • Good question. The right side of the integral equation is the term "cv". This product is equivalent to the amount of charge stored in the capacitor, q = cv. The integral's lower bound is t = -infinity to make sure we account for all the charge that has ever flowed into or out of the capacitor.

We often assign a moment where t=0, but that is always specific to the problem at hand. At t=0 we don't necessarily know how much charge is on the capacitor. So the most general way to express the equation is to go all the way back in time as far as possible to keep track of all the charge the capacitor has ever seen.

Now, that said, for many problems we often pick a time to assign as t=0. We pick a time when we know (or define) the charge on the capacitor to be q = cv = 0. When we do this, we can change the lower bound on the integral from t = -inf to t=0. If the capacitor is empty at t=0 ("empty" means zero charge stored), then everything that ever happened in the past to this capacitor has integrated to zero, and we can leave the past out of the bounds on the integral. This allows us to concentrate on the time interval of interest, from t=0 to t=T.
• I'm sorry, but do you have time to re-address the "t vs. tau" question? Your lessons are excellent btw, and I am a khan contributor, but I am heartily confused by t and tau. As to when t=0, wasn't the only time when t was actually zero was at minus infinity? When does t become zero AFTER minus infinity? Is charge gained and then lost over the period from minus inf to zero? How do we know the answer to any of this? And finally, is the "trick" actually parametrization? it seems similar, but not quite. I don't understand the description of t as "continuously running time", and"Big-T" as " the length of the current pulse. " Is big-T the "tau"? Please forgive my ignorance, but I am utterly perplexed. • Beyond , everything is confusing, when he transforms from t to T. Can anybody explain to me that? • From the author:I can understand how this switch from t to T happens quickly without a lot of explanation. Introducing the greek tau variable is a calculus trick.

There are two time variables in this video, little-t is continuously running time. Big-T is the length of the current pulse.

At the equation is written as v(big-T) where big-T is the argument to v and it is also the upper limit on the integral symbol.

At I change my mind about how I want to use the variable names. From now on I want to use little-t as the length of the pulse. This is strictly a matter of style and taste on my part. I want the answer to come out v(little-t) just because I thing it looks prettier. So the equation at now starts out v(little-t) = ...
In this version of the equation, little-t becomes the upper limit on the integral. Now I've substituted little-t for Big-T.

But there's one more thing I have to do before it's ok to proceed. Before, I was using little-t inside the integral, as the dt term. Since little-t now has another job, I change i(t) dt to i(tau) d-tau.

All of this is just a symbol change. There is no math happening.
• at dq/dt is the change of q of capacitor over a infinitesimal change of time.but how are we sure that it is exactly the same as the current that passes through it.i mean it is reasonable that if the current is high then the change of the charge of capacitor over a infinitesimal change of time(dq/dt of capacitor)is high.but why are we sure that they are the same. • Hi, I'm just stuck on part of the math

How come when you do the integral, on the one side you use a definite integral between negative infinity and T, but then on the other side of the equation you do an indefinite integral? Shouldn't they both be the same type of integral?

Thanks • begins "find v in terms of i"
shows an equation with an integral on each side. There are no limits shown.
By we are down to one integral. The other integral has vanished and become just 'v'.
the bounds are added to the integral, making it clear it's a definite integral. It has been all along.

(admission: I'm not always a purist when it comes to integral notation indefinite/definite.)

There is some informality in how this is presented. The point is to see how the voltage now depends on the whole history of the current. It is similar to how the water level in a bucket (voltage) depends on the entire history of how water has been poured in and poured out.
• At , the author integrates the right hand side without adding an arbitrary constant. Is this ever taken care of? In particular, it should be "Cv + d" instead of merely "Cv," where d is some arbitrary constant, right? • In the integral form of the capacitor i-v equation the arbitrary constant turns out to be equivalent to the starting voltage on the capacitor.

The initial version of the equation has a lower limit of t = -infinity. When you don't see an arbitrary constant at the end of the equation, that means we are assuming the capacitor was manufactured and first shipped with no charge on it (the constant is 0).

The i-v equation tells us that the voltage on the capacitor right now depends on the entire past life of the capacitor, all the way back to its birth. That's not so convenient to work with, so we start the time clock ticking at some arbitrary time we call t = 0. We can pick any t = 0 we want as long as it is a moment when we happen to know the voltage on the capacitor. The i-v equation changes its lower limit to t = 0 and that known voltage becomes Vo, the constant at the end of the equation.

It is amusing (or cosmically deep) that the entire life of the capacitor all the way from t = -infinity up until t = 0 is completely captured in the value of Vo.
• What exactly is an integral?
(1 vote) • Hey
Can I have the the link for the explanation on what are integrals?
Thanks
(1 vote)  