RC time constant problem from a book - Page 1
Forum section: electronics / beginners

author: KaTo
author_rank: Newbie
timestamp: March 04, 2023, 09:30:14 pm
content: Hi, I am trying to teach myself electronics. I am learning about RC time constant at the moment and I have this problem I am not sure about.It's problem 22-43. As I understand it - R2 should not play roll in capacitor charging speed, because all of the charging current comes through R1. R1 and R2 are a voltage divider that sets final voltage of the cap to 30V and R2 discharges the cap when switch is open. So time constant should be t=R1*C=1ms. So it should take 5ms to fully charge the cap. But book gives 3ms as the answer. I built the circuit and measured it with scope and it really charges with 3ms. What am I missing? 3ms means that time constant is 3ms/5=0.6ms. That means total resistance should be R=t/C=6k. 10k||15k=6k, but they are not parallel.Would be cool, if someone could explain me, what I am doing wrong.

author: Benta
author_rank: Super Contributor
timestamp: March 04, 2023, 09:36:08 pm
content: I'll give you a hint:Take V1, R1, R2 and think "Thevenin Equivalent".

author: TimFox
author_rank: Super Contributor
timestamp: March 04, 2023, 09:38:04 pm
content: Also, it's a trick question: the capacitor will not "fully charge" in a finite amount of time, only approach an asymptote.

author: Benta
author_rank: Super Contributor
timestamp: March 04, 2023, 09:41:02 pm
content: Well, the title is "Critical Thinking", I hope it applies.

author: IanB
author_rank: Super Contributor
timestamp: March 04, 2023, 11:14:26 pm
content: Suppose I have a voltage source Vs with a source resistance Rs. Then if I apply a short circuit to the source, the current Is that will flow is given by Is = Vs / Rs.Now consider the voltage divider arrangement. The charging voltage seen by the capacitor is Vc = 30 V. If we now replace the capacitor with a short circuit, what is the current Ic that will flow? Therefore, what will be the equivalent source resistance Rc as "seen" by the capacitor?

author: golden_labels
author_rank: Super Contributor
timestamp: March 04, 2023, 11:54:17 pm
content: IanB provided one way of looking at the situation and calculating the result. Let me rearrange the schematic just a tiny bit, so you could see R2from a yet different perspective.

author: ledtester
author_rank: Super Contributor
timestamp: March 05, 2023, 12:17:11 am
content: Here is the explanation of a very similar problem:https://www.allaboutcircuits.com/textbook/direct-current/chpt-16/complex-circuits/And a video:Parallel RC CIrcuit Using Thevenin Equivalent -- Physics Ninjahttps://youtu.be/7lUOEpthkpU

author: KaTo
author_rank: Newbie
timestamp: March 05, 2023, 10:08:16 am
content: Ok, with Thevenin Equivalent it is quite simple to solve. I don't know why I didn't think of that.IanB's explanation makes sense also. Basically, C sees 30V, but max charging current through R1 is 5mA, which is more than 30mA, that would be if it was just series circuit with R1, C and 30V. That's why the cap charges faster. And when C sees 30V and charging current is 5mA then yeah, R must be 6k. I think i get it now.Thank you very much for your help, everyone.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: March 05, 2023, 10:32:58 am
content: What does it even mean when a capacitor is "fully charged"?For me it means that it's charged to its maximum rated voltage, but that is not even mentioned.

author: MrAl
author_rank: Super Contributor
timestamp: March 05, 2023, 05:21:36 pm
content: Quote from: KaTo on March 04, 2023, 09:30:14 pmHi,First, i like the other replies here too i just want to add a little on how to simply understand this circuit and also comment on what it means to "fully charge" a capacitor.A simple way to understand this is probably akin to the Thevenin/Norton equivalents, but easier to picture in your mind i think.First, the circuit voltage is 50 volts but that is not what can get to the capacitor because there is a voltage divider R1 and R2. The maximum voltage that can "get to" the capacitor is thus modified by the voltage divider and that is:VCmax=50*R2/(R1+R2)That's the voltage divider voltage.Next, the resistance that causes the capacitor to charge is no longer R1 (10K) it is the PARALLEL combination of R1 and R2:Re=R1*R2/(R1+R2)So now you can calculate the entire response by using VCmax and Re rather than 50v and 10k by thinking of the circuit with a voltage source of just VCmax and a charging resistance of Re and the capacitor value stays the same. If you know how to do a one resistor one capacitor circuit with a DC voltage source then now you know how to do this circuit also.To calculate the time constant, now you know Re and the capacitor value is still C so you just multiply as if you had just one resistor and one cap.The phrase "fully charged" is subjective and not quantitative so there could be many meanings. In this case though it should mean when the capacitor is charged to the voltage of VCmax (see above). That would be considered fully charged.There is a slight catch to this though as someone mentioned. That is, to be exact, the cap never charges up to the maximum voltage but it gets SO close that we usually refer to it as "fully charged". It just depends how accurate you want to be with this.For example, to charge to within 1 percent of the full voltage it takes about 4.6 time constants. This is also sometimes approximated as 5 time constants.To charge to within 0.1 percent takes 6.9 time constants.To charge to within 0.0001 percent takes 13.8 time constants.To charge to within 1e-7 percent takes 20.7 time constants.To charge to within 1e-10 percent takes 27.6 time constants.At this point it makes no sense to wait any longer as capacitor leakage will be the main effect to consider now. That will not allow the cap to be charged to the full voltage even if we wait a lifetime. However, in the ideal case it still never charges up completely, in theory. For example, to charge to within 1e-99 percent of max would take 233 time constants.That's not the end either, you can go on and on with this. The time is:t=-ln(k)*RCwhereRC is the RC time constant,k is the level of the charge, 1 percent is 1/100, 0.1 percent is 0.1/100, etc.,t is the time it takes to get to that level,ln(k) is the natural log of k.So you can do this math to see how this works.I should mention one additional theoretical point. That is, the cap is considered "fully charged" after we allow time to approach infinity, but that's pure theory.Ok so now you know how the two resistor one capacitor circuit works. Next, you should look at how to calculate the time constant and the voltage across the cap when the cap has some INITIAL voltage across it. Up to now the capacitor has had no voltage across it at the start, but next you want to think of the cap as having some initial voltage first.For example, say the charge voltage was 50 volts as before, but now the cap already has 25 volts across it at the start at t=0. Calculate the time constant and the voltage across the capacitor at time t or after a very long time has passed.