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RC natural response - derivation

We derive the natural response of an RC circuit and discover it has an exponential form. Created by Willy McAllister.

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• At about I am confused by the direction of i_c.

Is this the direction of the current that charged the capacitor (before switching?).

I don't understand how one can then apply KCL at . It seems to me that i_r should = i_c in size and direction.
• The direction of i_c is arbitrarily designated according to the passive sign convention; see https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/v/ee-passive-sign-convention for more on the passive sign convention, because it's really important to the application of all these equations.

As for KCL, the span between the two components still counts as a node, and so the (two) branches still have their currents sum to zero. The issue you're running into is the desire to have all the arrows drawn in the "right" orientation before you've analyzed the circuit. You are correct, the current is the same throughout the node; that's why he invoked KCL. But it is perfectly valid, or even (as in this case) useful to draw arrows in orientations you know to be impossible, and just get a negative result for some (meaning it's in the opposite direction of how you drew it).

So, yes, the current in both ends of the node is necessarily the same amount, directed to the left. But because we drew i_c as pointing to the right, opposite the direction of i_r, it necessarily must equal the NEGATIVE of i_r. Had we drawn both pointing to the left, they would both equal +[amount], or if we had drawn them both to the right they would both be -[amount], but the all the relationships remain the same; we just lose the passive sign convention.
• At , why it is told that -(t/RC) has to have no units before we evaluate? What is the reason behind this thing?
• The term -(t/RC) is in the exponent. Exponents have to be pure numbers with no dimensions. That's because exponents are really a shorthand notation for a multiplication operation. If I write 2^3 it means multiply 2 by itself 3 times. If I write something like 3^(4 ohms) where there are dimensions in the exponent, I can't make sense of it: multiply 3 by itself "4 ohms" times?
• Solving the ODE:

dv/dt + v/RC = 0
dv/dt = -v/RC
dv/v = -dt/RC
∫dv/v = -∫dt/RC
ln(v) = -t/RC + k
v = e^(-t/RC + k)
= K e^(-t/RC)
v(t) = v_0 e^(-t/RC)
• What law or property allows dropping the constant K from the exponent e^(-t/RC + k) to get Ke^(-t/RC)?
• at you use the formula iC = C dv/dt

silly question but what is 'd'?
• Hello Doug,

This is the notation of calculus - d stands for delta or change. You would think of this as "a change in voltage per unit time."

In English the equation reads The current on the capacitor equals the product of capacitance and the speed at which the voltage is changing.

If the voltage is changing quickly the capacitor will have a high current.

Regards,

APD
• at (probably a stupid question), why don't you also account for current from the capacitor?
• We account for the current in the capacitor back at where we develop the original differential equation.
• At , I get confused. You state that according to KCL,

i_R + i_C = 0,

but I thought that is only the case when they are flowing at the same time. In here i_C stops flowing once you turn the switch, right? So how do you know if i_C and I_R are the same magnitude in opposite directions? Why isn't it possible for i_R to be of a different magnitude than i_C?
• In the previous video there was a battery and switch connected to the R and C. The battery provided some charge that flowed through R and got stored on the plates of C. As charge was being added to the capacitor, its voltage increased, (q = Cv) or (v = q/C). At some time in the past, the voltage on the capacitor became equal to the battery voltage. When that happened, you have 0 volts across the resistor, which means v = iR or 0 = iR, telling us i = 0. So just before we flip the switch, capacitor current = resistor current = 0.

Now we throw the switch and the battery basically vanishes. The circuit becomes just R and C connected at both ends, and there is some charge stored on the capacitor. That's where this video starts. There is only one loop in the circuit, which means there is only one current. Just before the switch is flipped, the current is 0. Just after the switch flips, charge starts flowing out of the capacitor and into the resistor. Moving charge is a current, and it is flowing in both R and C.

I use a notation trick that maybe causes some confusion. The point of this video is to show you how to derive an equation for what happens next. Looking ahead, I planned out how to get a nice looking equation that didn't have too many minus signs. With this foresight, I labeled the one current with two names i_R and i_C. I did this so I could use the i-v equations for R (Ohm's Law) and C (i = Cdv/dt) in the upcoming steps.

Please let me know if this helps you understand what's going on here.
• In your prior video (RC natural response) you show the current flowing in the opposite direction out of the capacitor as that shown in this video. In the last video you said you were going to " say more about why you chose that current direction later" , BUT you didn't. Also, you failed to explain why you show current flowing in 2 (opposite) directions in the same node from the capacitor source.
• How would this circuit change if the current went through the capacitor first, then the resistor?
(1 vote)
• This RC circuit is symmetric and there is a single loop. That means there is only one current, and it flows in both R and C at the same time. There isn't a sense that current flows in the resistor "first". It flows everywhere all at once.
• So question, around the variable e was introduced, what does it represent? K was the initial voltage. S became time/sec. I don't know where e came from or what it represents.
(1 vote)
• Admittedly my Calculus is horribly rusty (I took it almost 30 years ago). At the @ mark, you say that "the S comes down..." How does that happen? I guess I'm missing a step in solving the equation that brings S from an exponent into part of the constant. I want to say it is e^s would be S(e^s-1)dx, but I'm not sure if that's correct.
(1 vote)
• I think you are remembering the derivative of x^n
as in

d x^n/dx = n x^(x-1)

We are doing something different, where the independent variable (t) is up in the exponent and the base is the natural number "e".

The derivative of e^t (with no "s" coefficient) is

de^t/dt = e^t
(the exponential is its own derivative)

If we slip in a coefficient s the answer becomes,

d e^st/dt = s e^st

This happens where I said "comes down".

Sal does a proof of the derivative of e^x here: