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Homework Help: RCL circuit containing only a capacitor, inductor, and generator

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A series RCL circuit contains only a capacitor (C = 8.46 μF), an inductor (L = 6.01 mH), and a generator (peak voltage = 77.4 V, frequency = 2.45 x 103 Hz). When t = 0 s, the instantaneous value of the voltage is zero, and it rises to a maximum one-quarter of a period later. (a) Find the instantaneous value of the voltage across the capacitor/inductor combination when t = 4.32 x 10^-4 s. (b) What is the instantaneous value of the current when t = 4.32 x 10^-4 s?

    2. Relevant equations
    v(t) = V*sin(2pi*ft)
    z= square root(R^2 + (Xl-Xc)^2)

    3. The attempt at a solution
    v(t)=(77.4)*sin(2pi*2450*4.32*10^-4) = 8.96 V

    z=square root(0^2 + (.00601-8.46x10^-6)^2) = .00600154

    I=V/Z = 8.96/.00600154 = 1492.95 A

    Niether 8.96 or 1492.95 are correct answers.. Am I doing this completely wrong? Any help will be greatly appreciated, this problem is due soon and I'm freaking out! :[
  2. jcsd
  3. Mar 2, 2009 #2


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    Better run (a) through your calculator again - I don't get the same answer you have when I key it in. Be sure your calculator is set to radians or change the 2pi to 360.

    The z calc seems wild to me. For the inductor I have
    X = 2(pi)fL = 2*3.14*2450*6.01*10^-3 = 92.5
    Last edited: Mar 2, 2009
  4. Mar 2, 2009 #3
    Wow, the problem was that my calc wasn't in radians... Okay so I calculated part a and got 27.77 V, and that answer is correct.. :]

    I'm not really sure how to do part b (since my first try was obviously way off lol), and I can't figure out how your calculation for the inductor fits in with the calculation of the current in part b.
  5. Mar 2, 2009 #4
    Do I plug in the 92.5 for X of L in my Z equation and then solve for I=V/Z?

    z=square root(0^2 + ((92.5-(8.46x10^-6))^2) = 92.49

    I=V/Z = 27.77/92.49 = .3 A
    Last edited: Mar 2, 2009
  6. Mar 2, 2009 #5


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    You have to find the combined reactance of the capacitor and the inductor. I only did the inductor. Then use I = V/Z.

    You will get a complex number as the answer - how do you evaluate that at the given time? I forget how to do that - hope you will be able to teach me!
  7. Mar 2, 2009 #6
    I have absolutely no idea! :[ lol is what I did on my post above even correct?

    Does anyone else out there know how to evaluate the number at the given time?
  8. Mar 2, 2009 #7


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    Looks like you express V as [tex] \ \tilde{V} = 77.4*e^{j(\omega t )} [/tex]
    and the total X as the number times [tex]e^{j*\pi/2} [/tex]
    in I = V/X
  9. Mar 2, 2009 #8
    I'm lost.. I don't understand how the equation for capacitance relates to the instantaneous value of the current. We haven't really used that equation yet in my class, and I don't really know what numbers to use for the constants either.. :/
  10. Mar 2, 2009 #9


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    As you said in the first post, it is
    The X for the inductor was found with Xl = 2pi*f*L and it works out to 14.5, not 92.5. Next, you need to use Xc = 1/(2*pi*f*C) to find the reactance for the capacitor. Put those together with your z formula.
    Then you use I = V/Z to find the current.
    This approach gives you the amplitude of your current, but not the phase shift due to the reactance. The phase shift should be 90 degrees for a pure reactance so you could multiply your I value by sin(2pi*ft - pi/2) and evaluate at the time you are interested in.

    A better way to do this phase business is to use complex numbers called phasors.
    No doubt you will soon be taking this in your class.

    A better way, if you have taken it in class, is with complex numbers for the quantities.
    Z = R + j(Xl-Xc), where R = 0 (no resistance) and j is the imaginary number sqrt(-1).
  11. Mar 2, 2009 #10
    Thanks so much for that explanation, this is all making a lot more sense now... However, I'm still getting 92.5 for my calculation for Xl. How is it 14.5??
    Last edited: Mar 3, 2009
  12. Mar 3, 2009 #11
    Here are my calculations (using 92.5 for now). The final answer is incorrect still:

    instantaneous value of voltage = 27.77 V

    X of L=2pi(2450)(.00601)=92.5 Ohms
    X of X=1/(2pi(2450)c) = 7.678 Ohms
    Z=square root(0^2 - (92.5 - 7.678)^2) = 84.822
    I0=V0/Z = 27.77/84.822 = 0.32739 A
    I=I0sin(2pi*ft + (pi/2))
    I=(.32739)sin(2pi(2450)(4.34x10^-4) + (pi/2)) = 0.3 A

    Am I confusing I0 with I? Should I instead be calculating I=V/Z as I=77.4/84.822 = 0.91249
    and then solve I=I0sin(2pi*ft + (pi/2)) for I0?
  13. Mar 3, 2009 #12
    Please help! Somebody! This problem is due tonight and I've been struggling with it for 3 days now and can't figure out what I'm doing wrong :[
  14. Mar 3, 2009 #13


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    Yes, we need somebody else to help - I am not familiar with the phase part of this and seem to be making calculator mistakes on the first part. How in the world did you get .006 and me 14.7 for the reactance of the inductor? I did have it right in my first post and then must have got it mixed up with some other question.

    So sorry I wasn't here last night.

    When you are not using the complex numbers, you are calculating the Io.
    so use Io=77.4/84.822 = 0.91249. Then the current vs time function must be something like
    I=0.912*sin(2pi*ft + (pi/2))
    In http://en.wikipedia.org/wiki/Impedence the graph on the right (click on it to enlarge) shows that in and inductive circuit the voltage leads the current by 90 degrees so current should peak at time zero. That +pi/2 phase shift looks correct to me. Evaluating at the specified time I get 0.851 amps.
  15. Mar 3, 2009 #14
    Ok, that's what I'm getting right now too.. But it's still incorrect. :[

    Anyone else wanna take a shot at this?? Thanks so much for your help Delphi!
  16. Mar 3, 2009 #15
    WOW. haha someone answered this on my more recent post. The answer is -.841... the pi/2 had to be subtracted instead of added. !!!
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