# Capacitors in sereis charging & discharging time?

Discussion in 'General Electronics Chat' started by samy555, Oct 1, 2013.

1. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
I know that when two caps are connected in sereis to a 12 volt battery as shown below, the 12 volts are divided between the two caps in reverse.

and voltage will remain fixed value on each capacitor for a long time

When I connect a 100KΩ resistor in parallrl to the lower (10u) capacitor, I noticed that the voltage on this capacitor decreases, and at the same time the voltage on the upper capacitor is increasing.

When I connect a 100KΩ resistor in parallrl to the upper (22u) capacitor, I noticed that the voltage on the upper capacitor decreases, and at the same time the voltage on the lower capacitor is increasing.

My questions are:
(1) Why is voltage decreasing on the capacitor connected to the 100K resistor?

(2) In fig.2,,, At what time (t =??) will the voltage at the lower capacitor (10u) = 3 volt??
thanks

2. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
The resistor provides a discharge path for the capacitor. But the total voltage across both capacitors (i.e., the sum of the voltages) is fixed at 12V due to the battery.

What do you know about first order RC circuits?

3. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
for this circuit

can you please give me the formula to calculate at what time the voltage in the lower cap = 3 volts

thanks

Last edited: Oct 3, 2013

Nov 30, 2010
16,705
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5. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
The circuit isn't showing up. It looks like you are trying to pull an image from a cache. I've had similar problems. The best way is to just re-upload the image from scratch.

What is your level of circuit analysis? Are you able to work with derivatives and integrals, or just with cookbook equations, or something in between? That will determine how I approach explaining things.

6. ### LvW Active Member

Jun 13, 2013
674
100
Samy 555, did you ever hear about the "two capacitor problem"?
If a series connection of two IDEAL (lossless) capacitors is connected to a dc voltage the divider ratio is UNDEFINED. (By the way: That is the reason all simulation programs do not allow such a combination - all nodes must have a dc connection to ground).

That means, for real capacitors the voltage divider ratio is determined by the resistive losses of the capacitors.
Therefore my question: Did you measure/simulate the mentioned results with real or ideal capacitors? What about the input resistance of the dc meter?

7. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
It's undefined unless initial conditions are specified. In the real world, it's a bad idea because parasitics dominate the initial conditions, which pretty much by definition are transient.

The simulator he is using appears to apply initial conditions of the capacitors uncharged. If the meters are being modeled as having 10MΩ resistance, then if he lets the simulation run long enough (say 20 to 30 minutes simulated time), then he should see the readings equalize.

This looks very much like a homework or lab exercise, especially since the OP is so focused on one specific question -- at what time will the voltage at the junction be 3V.

So let's work with that.

Q1) In terms of Vcc (the 12V) and Vc (the junction between the caps), what is the instantaneous current in the 10kΩ resistor?

Q2) Consider the end of the resistor connected to Vc. Where does the current in Q1 have to go?

Q3) For a given amount of current over a given amount of time, how much does the voltage across a capacitor change by?

Let's work with this set of questions for now.

8. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Thank you all for trying to help
I have reviewed many books and sites the net
All those sources are talking about one capacitor connected with one resistor in series.
the following image is found in most sources:

In my circuit:

My question was: at what time will the voltage on the 10u cap decrease to 3 volts?
I need a formula
thanks

9. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
I'm not going to just give you a formula. For one thing, it's pretty obvious this is homework and I'm not going to work your homework for you. I will help YOU figure out how to work YOUR homework. Second, if I just give you a formula that applies to this particular circuit, you will have learned nothing and if you were asked a similar question regarding the final circuit (the one with the resistor across the top cap) you would be right back here asking for someone to give you a formula for that.

If your approach is going to be to look for sites that have exactly the problem you are trying to work, save yourself a lot of time and grief and go into some other field that is far removed from this one, because you aren't going to go anywhere. If you want to stay in this field, you need to learn how to apply fundamental concepts (not memorized equations) to problems that you haven't seen because, in the real world, people aren't going to pay to to solve problems that already have an answer.

Now, on several occasions I tried to get you to describe your level of understanding (or at least your presumed level of understanding) and you've largely ignored it. But in this last post you give a strong hint based on the material you quoted. So I am going to assume that you are familiar with integrals and derivatives and the relationships between voltage and current in resistors and capacitors in terms of integrals and derivatives.

10. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
I swear I'm not a student, I finished my studies a long time ago
Yes, I'm good in calculus, and I know a lot in electronics
But I'm not an expert, and I find it difficult to deal with real electronic circuits

You're right, and I'm convinced your opinion, and your words evidence you a wonderful teacher
Let's start
i(10k) = Vc/R = Vcc[1-exp(-t/RC)]/R

To ground
Sorry, I did not understand the question well
I'll be waiting for your response, and in the meantime will be reviewing some books
Thank you very much

11. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
Close, but not quite. First, you have two capacitors, so you need to be sure you clearly identify which is which. So let's call C1 the top cap and C2 the bottom cap.

You are correct that

i(10kΩ) = Vc2/R.

However, we can't just throw the same capacitance into the classic first order equation because that is based on the current in the resistor being the same as the current in the capacitor. In this case it's not, because the current in the resistor is made up of two components -- one from the lower capacitor as it discharges and one from the upper capacitor as it charges further!

Taking the component currents to all be flowing from top to bottom, we therefore have

i(C1) = dv(C1/dt)
i(C2) = dv(C2/dt)
i(R) = v(C2)/R

These are all directly from the defining relations for a capacitance and resistance.

Q4) What does KCL require about the relationship between these three currents?

Q5) What does KVL require about the relationship between Vs (the 12V supply) and v(C1) and v(c2)?

Q6) Using the results from Q4 and Q5, can you write a differential equation that only involves i(R) and it's integrals/derivatives?