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# 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.

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• 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.
• 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.

Regards,

APD
• Can someone please tell me how the integral is just t?
• 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
• At , 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.
• 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" at .
(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
• 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).
• sorry, but how is that an equation of a line? y=mx+b, there is not +b? How can a beginner identify that?
• 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.
• 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.
• 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))
• 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.
• 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,