Three capacitors RC circuit questions

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Homework Help Overview

The discussion revolves around a circuit problem involving three capacitors, three resistors, a battery, and two switches. The initial conditions include all capacitors being uncharged with the switches open. The problem is divided into two parts: determining the charge on a specific capacitor after a long time with the switches closed and calculating the time it takes for that charge to drop to a certain percentage after one switch is reopened.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the method for calculating the charge on capacitor C2 after the switches are closed, with one participant expressing uncertainty about their reasoning. Another participant confirms the method used for part A is acceptable.
  • For part B, questions arise regarding the time constant and the configuration of resistors once a switch is reopened, prompting further exploration of circuit isolation and parallel arrangements.

Discussion Status

The discussion is ongoing with participants validating each other's approaches for part A. There is a focus on clarifying the implications of opening switch S2 for part B, with some guidance provided regarding the circuit's configuration.

Contextual Notes

Participants are navigating assumptions about circuit behavior when switches are opened and closed, as well as the implications for calculating time constants in the context of capacitor discharge.

syhpui2
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Homework Statement



Three resistors, three capacitors, a battery and two switches are connected in the circuit shown below. The values of all circuit elements are given in the figure. Originally, the switches S1 and S2 are open (as shown) and all of the capacitors are uncharged. At time t = 0, both switches are closed.

http://i.imgur.com/BOq2c.png
BOq2c.png



Part A

What is the charge Q2 on capacitor C2 a very long time after the switches are closed?
(a) Q2 = 0 μC
(b) Q2 = 33 μC
(c) Q2 = 90 μC
(d) Q2 = 180 μC
(e) Q2 = 270 μC (Correct Answer)



Part B

After a very long time with both switches in the closed position, switch S2 is reopened. How long (t1/e) does it take for the charge on capacitor C2 to drop to 1/e (36.8%) of its fully-charged value (i.e. of the value it had just before S2 was reopened)?
(a) t1/e = 1200 μsec
(b) t1/e = 1500 μsec (Correct Answer)
(c) t1/e = 3000 μsec
(d) t1/e = 3600 μsec
(e) t1/e = 4800 μsec



Homework Equations



KVL,KCL

The Attempt at a Solution



For part A, what I tried is
Voltage across is 18 X ¾ (R3/ (R1+R3))= 27/2
(Because Q=CV and in this case Ic=0 so no current on R2?)
I get right answer, just not sure if I am thinking correctly.

For part B,
I used

Q= Q(0)e^-(t/tau)

However, I am not sure how do I find time constant in this case.

THX!
 
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Your part A method is fine.

For part B, once switch S2 is opened the right hand portion of the circuit is isolated from the left hand portion. So it's just two parallel capacitors and a resistor. What does that suggest to you?
 
gneill said:
Your part A method is fine.

For part B, once switch S2 is opened the right hand portion of the circuit is isolated from the left hand portion. So it's just two parallel capacitors and a resistor. What does that suggest to you?

How about R1 and R3 in this case?
Are they in parallel?
Thanks
 
syhpui2 said:
How about R1 and R3 in this case?
Are they in parallel?
Thanks

No! With switch S2 open they are isolated from each other (no complete circuit).
 

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