RC Circuit question

  • Thread starter valarking
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  • #1
valarking
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Homework Statement



All capacitors of the open circuit in fig. 3 are discharged when, at time t=0, the switch S1 is closed. At some point later, at time t = t1, the switchS2 is also closed. What is the charge Q1(t2) on the capacitor C1 at time t = t2 > t1?

http://www.vkgfx.com/misc/fig3.jpg [Broken]


Homework Equations



Charging capacitor:
[tex]q = C\epsilon(1 - e^{\frac{-t}{RC}})[/tex]
Discharging capacitor:
[tex]q = {Q_0}e^{\frac{-t}{RC}}[/tex]


The Attempt at a Solution



Ok, I said that when S1 was closed at t1, the resistors arein parallel, giving q at t1 equal to:
[tex]{C_1}\epsilon(1 - e^{\frac{-t}{(R_1+R_2+R_3+R_4)(C_1+C_2)}})[/tex]

Now the thing I need help with is determining the equivalent resistance when S2 is closed. The best I can think of is that only R2 and R4 are acting. That would give:
[tex]{C_1}\epsilon(1 - e^{\frac{-t}{(R_1+R_2+R_3+R_4)(C_1+C_2)}})e^{\frac{-t}{(R_2+R_4)(C_1+C_2)}}[/tex]

Can anyone help me determine the resistance in this last step? I'm out of ideas.
Thanks!
 
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Answers and Replies

  • #2
rl.bhat
Homework Helper
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When S2 is closed, you are creating a short circuit. Hence no charging will take place. But initially charged capacitors will discharge through R3 and R4.
 
  • #3
valarking
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So would the time constant for the discharging period be (R3+R4)(C1+C2)?
 
  • #4
rl.bhat
Homework Helper
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Yes.
 
  • #5
valarking
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Ok, that would give:
[tex]{C_1}\epsilon(1 - e^{\frac{-t}{(R_1+R_2+R_3+R_4)(C_1+C_2)}})e^{\frac{-t}{(R_3+R_4)(C_1+C_2)}}[/tex]
as the answer then.

Thanks for your help.
 
  • #6
rl.bhat
Homework Helper
4,433
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Ok, that would give:
[tex]{C_1}\epsilon(1 - e^{\frac{-t}{(R_1+R_2+R_3+R_4)(C_1+C_2)}})e^{\frac{-t}{(R_3+R_4)(C_1+C_2)}}[/tex]
as the answer then.

Thanks for your help.
The above expression is the combined charge on C1 and C2 at time t2. Bur in the problem you require charge on C1 only.
 
  • #7
valarking
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I thought that both capacitors were used in RC, the time constant, and that C1 was used by itself only when dealing with Qf, or C*epsilon.

Basically, I was fairly certain that the time constant is the same regardless of which capacitor you are looking at. Is this not the case?
 
  • #8
rl.bhat
Homework Helper
4,433
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Across C1 and C2 potential difference is the same. During discharge, time constant for C1 is C1(R3+R4). At the instant of discharge charges on C1 and C2 are not the same.
 
  • #9
valarking
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Ok, I see what you're saying, and that sounds right.

Thanks again for the help.
 

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