1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A little different RC circuit problem! Help!

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data
    First off, thanks for taking a look at my problem guys. I have been given this question as extra credit for a calc-physics II class and at the moment am struggling with pulling the pieces together. At first glance I didn't have to much trouble with it but when I go back to it to write it up, it begins to make less and less sense. We didn't really do a whole lot of RC circuit stuff in this summer term which is why I think in part he gave us this question, so I'm really just here to get some confirmation, some help and a little advice where ever possible.

    So here it goes:

    Find the time constant and the explicit time dependence for all currents. Assume that the capacitor is initially uncharged when switch S is closed. Make plots of the currents clearly indicating any features characteristic of the system.

    You can view the diagram/full question here: http://www.scribd.com/doc/59234592/Problem-2

    2. Relevant equations

    We've got the charging/discharging formulas for RC circuits.
    tao = RC
    Series/parallel relationships.


    3. The attempt at a solution

    The plots are not an issue for me, the jist of it would be that at t=0, the capacitor drains all the current from the circuit, and the current exponentially decreases to infinity. At the same time the charge will have an exponential increase to a plateau.

    My real issue stems from being unable to identify with reducing the circuit down into an effective resistance. When trying to calculate the time constants, for example where it goes from Battery -> R1 -> R2 -> battery, it is clear that R2 and C are in parallel, but at the same time R1 is in series to R2C. So I am thinking do we use the 1/jwc (resistance of a capacitor. We haven't actually learned this yet but I've read it online) and reduce the circuit into an effective resistance as you would any other regular R or C circuits. Basically what it comes down to is if someone could help me with one of the loops, I think I could take it from there. I am just unsure what exactly is our resistance and what exactly is our capacitance and how does the effective circuits come into play.

    Again I am new to RC circuits in general but have a solid understanding of your basic R and C circuits.

    Thanks in advance for any help provided, I'm under a pretty tight deadline and hope to hear from someone soon :)
     
  2. jcsd
  3. Jul 18, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi juicec, welcome to Physics Forums.

    What circuit laws have you learned? Have you done any work with Thevenin or Norton equivalent circuits?
     
  4. Jul 18, 2011 #3
    To my knowledge we haven't learned either, but its possible the names just weren't brought up.

    As far as circuit laws... Not sure exactly what this entails but, when in series additive, when in parallel inversely proportional, etc.

    I think I'm simply over thinking this problem and I feel confident I can do it, just need a little kick to get started on the right track.
     
  5. Jul 18, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    The tricky bit is going to be establishing the resistance to use for the time constant without resorting to solving the circuit's differential equation ("bone knives and bearskins" :smile:). Or is this what you are expected to do?
     
  6. Jul 18, 2011 #5
    I actually took differential equations but no we are not doing that in this class. From my best guess we are to reduce the circuit into the effective resistance and move on from there.

    Finding tao = RC sounds pretty simple and straight forward but I am having a hard time getting started.
     
  7. Jul 18, 2011 #6

    gneill

    User Avatar

    Staff: Mentor

    Okay. Well, to cut to the chase, one method of finding the equivalent resistance for a circuit is to "suppress" all sources (voltage and current sources) and then use the usual resistance combination formulas on the remaining network. To suppress a voltage source it is replaced in the circuit by a short circuit -- a wire. To suppress a current source, simply remove it entirely from the circuit.

    In the present case you're looking for the resistance that the capacitor "sees" when it looks back towards the voltage source. So (temporarily) remove the capacitor, suppress the voltage source, and determine the resistance presented at the connections where the capacitor was.
     
  8. Jul 18, 2011 #7
    OK, so just to let you know off the get go, I don't overly enjoy physics -_- Although I tend to do well in it, I do whats required and never go beyond. With that said, I am in physics II and haven't explored much past it.

    So I can relate a bit better, isolating the example of Battery --> R1 --> R2 --> Battery:

    If we wanted to find RC, you're suggested to remove the capacitor from the equation, find the effective resistance and reestablish the capacitor afterwords? I'm not sure whats going on here.
     
  9. Jul 18, 2011 #8

    gneill

    User Avatar

    Staff: Mentor

    The time constant for the circuit will depend in some fashion on BOTH resistances. In order to establish what that dependence is, we need to find the equivalent resistance that the capacitor "sees" .

    Perhaps a slightly different way of looking at would help. Suppose that the capacitor had finished charging up (some long time after the switch closure). Now suddenly the voltage supply were removed and replaced by wire. The capacitor would begin to discharge. What would be the time constant for that circuit?
     
  10. Jul 18, 2011 #9
    Using the same diagram, the capacitor would just discharge through R2 and C.

    So the time constant when discharging would simply be tao = R2C ?
     
  11. Jul 18, 2011 #10
    Oh your not saying open the switch, so the capacitor also discharges across R1, eh, again confused.
     
  12. Jul 18, 2011 #11

    gneill

    User Avatar

    Staff: Mentor

    Correct :smile:

    The capacitor discharges through BOTH resistors. What's the equivalent resistance that the capacitor sees?
     
  13. Jul 18, 2011 #12
    Well I'm not to sure, don't we separate into different loops?

    So it sees R1 in one situation and R2 in another, I am not sure how to do the effective resistance in this situation.
     
  14. Jul 18, 2011 #13

    gneill

    User Avatar

    Staff: Mentor

    Would you say that the resistors are in series or in parallel?

    attachment.php?attachmentid=37299&stc=1&d=1311046440.gif
     

    Attached Files:

  15. Jul 18, 2011 #14
    well this I think is my problem, and perhaps after you help me out here I will be good to go. I thought that R2 and C are in parallel, but R1 and R2C are in series.

    So perhaps I just answered my own question, the effective resistance is therefore in series? But the current still undergoes a junction, but then again capacitor holds no resistance... hm.
     
  16. Jul 18, 2011 #15

    gneill

    User Avatar

    Staff: Mentor

    Let me redraw the circuit one more time. Are the resistors in series or in parallel? What will the capacitor see?

    attachment.php?attachmentid=37300&stc=1&d=1311047202.gif
     

    Attached Files:

  17. Jul 18, 2011 #16
    lol I feel like we are beating this to death here, but still so foggy :(

    The resisters are now in parallel, the capacitor will have its current split across both?
     
  18. Jul 18, 2011 #17

    gneill

    User Avatar

    Staff: Mentor

    Yes :smile:
    Note that all the connections are still the same -- it's not a different circuit. Nothing changed but the way the circuit was drawn in order to make the relationship between R1 and R2 more obvious.

    Yes :smile:

    So the equivalent resistance that the capacitor "sees" is R1 in parallel with R2. That's the resistance that will go into the time constant.
     
  19. Jul 18, 2011 #18
    OK so with resisters in parallel we simply find: R(eff) = R1 + R2.

    So for the time constant of discharging with the battery removed at 100% charge,

    tao = (R1 + R2)(C) ?
     
  20. Jul 18, 2011 #19

    gneill

    User Avatar

    Staff: Mentor

    R1 + R2 is for resistors in series (they add up). What's the formula for combining resistors in parallel?
     
  21. Jul 18, 2011 #20
    Woops...

    So its, R(eff) = (1/R1 + 1/R2)^-1(C)

    which in turn gives us ((R2+R1)/R1R2)^-1(C)

    which in turn gives us [(R1R2)/(R1+R2)](C)?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A little different RC circuit problem! Help!
  1. RC Circuit Help (Replies: 1)

  2. RC Circuit Help (Replies: 3)

Loading...