Natural response of an RC circuit with two capacitators.

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Discussion Overview

The discussion revolves around analyzing the natural response of an RC circuit containing two capacitors after switches are opened. Participants explore the voltage v0(t) for t > 0, using various methods including node-voltage analysis and step-response approaches. The conversation includes theoretical considerations and mathematical reasoning related to capacitor discharge and equilibrium states.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • The original poster (OP) calculates the current io(t) and attempts to determine v0(t) based on the voltage across capacitors and resistors after the switches open.
  • Some participants suggest treating the problem as a step-response due to the change in voltage/current supply to the capacitors.
  • One participant asserts that the response should be treated as a zero input response, indicating that the given answer is incorrect based on their analysis of capacitor discharge over time.
  • Another participant questions the discharge path for the capacitors, noting that switches prevent discharge through the resistor.
  • There is a contention regarding whether the capacitors can reach an equilibrium voltage without discharging, with some arguing they will stabilize at a certain voltage while others claim they will discharge completely.
  • One participant claims that the solution provided in the book is correct and suggests a method for finding the voltage equation using the correct capacitance value.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the behavior of the capacitors and the validity of the provided solution. There is no consensus on the correct approach to analyze the circuit or the final voltage response.

Contextual Notes

Participants reference various assumptions about ideal capacitors, the impact of resistors on discharge, and the conditions under which the circuit operates, but these assumptions remain unresolved and are subject to differing interpretations.

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Homework Statement


I'm supposed to find the voltage v0(t) for t > 0 for the following circuit:
http://img516.imageshack.us/img516/5858/circuitta7.jpg

Homework Equations


The first part of the exercise was to find the current io(t). I did this by using the node-voltage method to find the voltage across the 8k resistor and the 10k resistor to find the voltages across the two capacitators. Then I figured since, once the switches open, they'll be in series, I find R*C by finding the total C which was 100 nF.

Then I found the voltage drop across Io which then became 120e^-5000t - 100e^-5000t = 20e^-5000t and divided this by the resistance to find the current, which would be 10e^-5000t mA.


The Attempt at a Solution


That's when things got ugly. I'm thinking that the voltage at v0 will have to be the voltage-drop between Io and the 300nF capacitator. Fair enough. If the current in I0 is 10e^-5000t mA I figure the voltage will have to be 20e^-5000t V. Then I'm thinking the voltage across the 300nF supply will have to be 100e^-5000t V (since the voltage at the 10k resistor was 100v before the switches were opened) which makes me believe the voltage at v0 should be -80e^-5000t v.

However, they gave us the correct answers beforehand, and I know it's supposed to be:
-6.67e^-5000t + 106.67 v.
This has me somewhat stumped. If this is correct, that means even when t -> infinity there will be a voltage of 106.67 v over v0. But then I'm thinking, since these really are capacitators, won't the 150nF capacitator, being the one with the highest voltage, somehow charge the other one? Or what?

I'm thinking some kind of equilibrium here is involved, but I'm having real trouble seeing where those numbers are coming from. A push in the right direction would truly be helpful!
 
Last edited by a moderator:
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I've done some more thinking to the problem, should I treat this as some kind of step-response? Considering the 300nF capacitator goes from one voltage/current supply to a new one, that would explain how I'm supposed to get that extra voltage in the equation. I tried calculating as a step-response by using the foruma Vc = IsR + (V0 - IsR)e^(-t/(RC)) but to no appearant avail, maybe because according to my book i'd normally have to transform into a norton-equivalent before I could use this formula, and since I can't find anything on doing anything similar when another capacitator is the voltage supply, I should probably just drop that train of thoughts.
 
You must treat it as a zero input response.
[tex]v_c(t) = v_c(0).e^{-\frac{t}{RC}}[/tex]
This means that the answer they gave you is wrong.
Analyzing it physically: The two capacitors are previously charged. As time goes by they discharge into the resistor and the energy they had stored is lost as heat. After an infinite time all energy is spent and the capacitors are discharged.
 
"As time goes by they discharge into the resistor..."
Am I missing something? What resistor will the caps discharge through? Since there's switches on either side, ideal caps would come to some kind of equalibrium and remain there for infinity. Right?
 
dlgoff said:
"As time goes by they discharge into the resistor..."
Am I missing something? What resistor will the caps discharge through? Since there's switches on either side, ideal caps would come to some kind of equalibrium and remain there for infinity. Right?

The 2k resistor through which [tex]i_0[/tex] flows.
 
Nah. There's no path to ground(-) inorder for them to discharge; once the switches are open.
 
dlgoff said:
Nah. There's no path to ground(-) inorder for them to discharge; once the switches are open.

So, you are saying that the two capacitors and the 2k resistor don't make a closed loop?
 
Everyone who has posted in this thread is wrong, but in different ways. Of course the resistor discharges some of the power; that's what resistors do. But they only do it when there is a voltage drop over them.

The capacitors don't entirely discharge. They reach an equilibrium voltage of 106.67v, at which point there is no longer a current running through the circuit.

The OP was right to combine the caps to find the tau, but you need to use 300e-9 F to find the equation for V(t). You need Io so that you can differentiate the equation Vf + Ae^-5000t and set it equal to Io/C and find A.

The solution given by the book is absolutely correct.
 
Last edited:
Aha, I was thinking something like this might happen if there's no voltage drop across the resistor. It was just so hard to analyze physically.
Thanks loads for the help!
 
  • #10
CEL said:
So, you are saying that the two capacitors and the 2k resistor don't make a closed loop?
No. They do make a closed loop. But like I said:
ideal caps would come to some kind of equalibrium and remain there for infinity.
 

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