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Help analyzing an RC circuit?

  1. Feb 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Ok guys, so this is a slightly long question, but bear with me because I think I figured out most of it. I just need help towards the end.

    So I have a circuit with a current source in parallel with a capacitor and also in parallel with a resistor (see attached image). This is what the problem states:

    a) Give the differential equation between t = t0 and t = t1. During this time, the switch is open and I(t) = 0. The initial value of V(t), the voltage across the resistor, is 150 V.

    b) Use a Laplace transform to solve the differential equation in part a and give the time constant.

    c) If V(t) were to decay to 100 V at t = t1, calculate the time constant during this interval where t1 - t0 = .500 sec.

    d) Now, consider a different interval, when 0<t<t0. During this time the switch is closed and and I(t) = I0u(t). Give the differential equation of V(t) during this interval knowing that V(0) = 100 V and I0 = 5. Write the expression of V(t) in terms of the unknown resistance R.

    e) Calculate R from part d knowing that V(t) increases from 100 V to 150 V within a .2 sec time interval (i.e., t0 = .2 s).


    2. Relevant equations

    The only equations I've used so far:
    1) I = C*dVc/dt
    2) V=IR

    I may also need C = Q/V? (not sure)

    3. The attempt at a solution

    Ok so part a using the current law I think its just C*dV(t)/dt + V(t)/R = 0. This wasn't bad.

    For b using the Laplace in this scenario wasn't hard, I just plugged in the initial value from part a and I got V(t) = 150e^(-t/RC). So the time constant is RC then.

    For c, this where I may be skeptical, but I think what I did makes sense. I did 150 - 100 = 50 V. So I set 50 equal to the equation in part b and I plugged in .5 sec for t in the equation. So I got the time constant, RC, equal to .455.

    Parts d and e are where I'm stuck. If I(t) = I1 + I2 (currents through capacitor and resistor), then I think the differential equation should be 5u(t) = CdV(t)/dt + V(t)/R. Is this the correct equation? Because when I took the Laplace I was getting answers that seemed a bit complicated (the V(0) was ruining everything). The part that I have no idea about is how to write V(t) in terms of R. Without this I have no hope of getting e. I can't find a way to write it without also having C in there. But Im supposed to get C later on so that has to be wrong. Can anyone help?? Should I be thinking about the behavior of the circuit at steady state? What about a transfer function? I guess I just dont know what concept I should be using to get R.


    Edit: Woops, Im not sure what section this should go in...sorry Im new to this.
     

    Attached Files:

    Last edited: Feb 1, 2014
  2. jcsd
  3. Feb 1, 2014 #2

    gneill

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    Staff: Mentor

    Hi madgorillaz15, Welcome to Physics Forums.

    Okay. You didn't show your work via Laplace to get there, so I can't comment on your process, but that is a good result for V(t) for this part.

    Careful there. The problem states that the voltage drops from 150 down to 100 volts over the given time interval. So V(t1) = 100V, not 50V. You derived an equation for V(t), not ΔV(t).

    You could write the final equation by inspection knowing the initial and final states and that those states will be "connected" by an exponential decay. But to derive that via the differential equation is a bit more involved.

    Can you show some detail of your work in attempting to derive the Laplace form of the differential equation?

    (My own preference for attacking the problem would be to work directly in the Laplace domain, modelling the initial capacitor voltage as a separate voltage source, using the Laplace domain version of capacitor impedance, and writing a node equation... but you may have your own preferred method)

    If you solve the differential equation you should find that, except for the exponent term in which ##\tau = RC## appears, only R and constants appear in the magnitude portion of the equation.

    No, it's fine in this section.
     
  4. Feb 1, 2014 #3
    Hey gneill, thanks for choosing to help me. I've been stuck on this for way too long.

    OK so for part c, tbh Im still not sure why taking the difference is wrong, but I did what you said and set the equation from part a equal to 100 V, and this time I got the time constant to be 1.23. Pretty straight forward math.

    Now for part d I'll tell you what I did. I know that 5u(t) = CdV(t)/dt + V(t)/R is true because of the current law. After taking the Laplace and dividing by C I get 5/Cs = sV(s) - 100 + V(s)/R. Solving for V(s) I got V(s) = (5/C)(1/s(s+1/RC)) + 100/(s+1/RC). The second term will just become 100e^-t/RC. For the first term I just multiplied (1/RC)/(1/RC), and then I took the inverse and I got it to be 5R(1-e^-t/RC). So V(t) = 5R(1-e^-t/RC) + 100e^-t/RC. So I guess this might be right or wrong (not sure honestly), but my problem is that I dont know how to use this to write an expression purely in terms of R, and then use the information in part e to find the unknown R value.

    Do you know where I should go from here, or what concept I should employ? I think the method you proposed is interesting, I think I remember seeing something like that from when I took an earlier circuits class, but Im not sure. Could you show me what you're talking about, if that would help? Thanks a lot.
     
  5. Feb 1, 2014 #4

    gneill

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    Staff: Mentor

    That looks better. What you did before was not correct because your expression was for V(t), the voltage with respect to time, and they gave you a particular voltage for a particular time. You plugged in a change of voltage, not the voltage at that time.

    You mean:

    5/Cs = sV(s) - 100 + V(s)/(RC)

    if you're dividing through all terms by C. I presume that the missing "C" on the last term was a typo, since you arrived at a correct equation for V(s) afterwards.

    First, you can collect together the exponentials, since they have identical exponents. Next, I think you're safe in assuming that that there is a time constant ##\tau = RC## from part (c) that you can use for the "RC" term in the exponent. In part e you're again given a final voltage and time (since the switch closes). Write the equation and plug in the now known values for V, ##\tau##, and t. Solve for R1.

    In the Laplace domain the capacitor has impedance 1/(sC). A capacitor with an initial voltage can be modeled as an uncharged capacitor of the same value in series with a voltage source of the same value as that initial voltage. Your equivalent circuit becomes:

    attachment.php?attachmentid=66227&stc=1&d=1391314224.gif

    Write the node equation for V(s) and solve it for V(s). Then you'll have your Laplace form for the voltage, ready to take the inverse transform to find v(t).
     

    Attached Files:

  6. Feb 1, 2014 #5
    ...wow, that was much easier than I thought it would be. The answer was in front of me the whole time.

    OK, so part c I think I'm understanding the difference. When I was subtracting before, that wasn't actually the voltage at that point in time, I think.

    For part d, yeah, I forgot the C. So I did what you said. I plugged in 150 for V(t), .2 for t, and 1.23 for RC, and I got R = 29.8 ohms (part e). That really makes sense to me now. RC is a constant. It doesnt change. That's what I need to take away from this.

    I love the way you modeled that other circuit. I've never seen that before though, and I'm having trouble getting an equation that matches the other one I made. So following the current law, shouldn't the equation be 5/s + (100/s -V(s))sC1 = V(s)/R1? But this doesnt make any sense because this would give me a different time constant?

    Also, I have one final question. How would I go about finding the Transfer function? The question asks me to find the Laplace transform of the transform function, but I always thought that the transfer function was input voltage over output voltage. Do I have sufficient information to answer this question?

    Edit: i realize my equation would be different based on which direction you draw the currents by the way.
     
  7. Feb 2, 2014 #6

    gneill

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    Staff: Mentor

    That's the idea, but I'm seeing a different value for R. When you equation is rearranged it can look something like:

    ##V(t) = 5 R + (100 - 5 R)e^{-t/\tau}##

    Presumably that was where you started before plugging in your values? Maybe you can show more of your working for isolating R?

     
  8. Feb 2, 2014 #7
     
  9. Feb 2, 2014 #8

    gneill

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    Staff: Mentor

    That's still not the value I'm seeing for R. Can you lay out your work here?

    Hmm. I'm not conversant with Simulink. But so far you've got the V(s) for the circuit after the switch closes. You should be able to manipulate it into a standard form. I'm not certain how Simulink deals with shifted time periods and transfer functions since the derivation used assumes time starts at t=0 when the switch closes, but in reality there was other circuit action taking place before the switch closure, so that portion needs a separate transfer function to model it. The Laplace transform for the transfer function for the switch open period should be simple enough to write.

    I suppose you might get away with changing the input to something like Iou(t-t1) to simulate switching the current on at a shifted time.
     
  10. Feb 2, 2014 #9
    Yeah, I think I put something wrong into my calculator because I couldnt reproduce that answer. Anyways, e^(-.2/1.23) = .849, so the equation becomes 150 = 5R +(100-5R)(.849)

    So 150 = 5R + 84.9 -4.245R, so 65.1 = .755R, so should R = 86.2 ohms?

    About the other part, I want to make sure I'm understanding you correctly. Are you saying that I should have a transfer function for when the switch is open, and input that before I insert the transfer function for when the switch is closed? And to do that, I should try inserting something like you suggested, I0u(t-t1)?
     
  11. Feb 2, 2014 #10

    gneill

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    Staff: Mentor

    Much better! I'm seeing 86.7926 to four decimal places, no rounding or truncating of intermediate values in the computation, so you can verify your calculations. It's a good idea to keep extra decimal places though all intermediate steps and only round the final result. This keeps rounding and truncation errors from growing into the significant digits.

    What I'm trying to get across is that, due to the action of the switch, the circuit behavior overall is nonlinear. The circuit actually changes when the switch does. So you need your model to reflect this, either by doing the simulation in two distinct parts or by incorporating the change in the math of one model. The simplest way I can think of doing this with one transfer function is to model the input function accordingly. I0u(t-t1) should produce a step function that is delayed by amount t1. As I said, I'm not familiar with how Simulink implements its modelling, so perhaps this isn't the way to go. I'll have to leave that for you to investigate, or perhaps others here will have an idea for you .
     
  12. Feb 2, 2014 #11
    Nope, you helped A LOT more than you realize! Thanks a bunch, I learned a lot and feel so much more comfortable now that I'm out of this jam. I especially really liked that other model of the circuit you presented, I think I'll be using that line of thinking in the future. I'll update the thread and let you know about what happens with the transfer function/modeling, after I do some more investigation myself.
     
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