RLC Circuit Help!

Discussion in 'Homework Help' started by stupidlogic, Aug 10, 2010.

  1. stupidlogic

    Thread Starter Member

    Aug 10, 2010
    I have a homework assignment working with RLC circuit analysis. Here is what the assignment says:

    1. Go to URL: www.falstad.com/circuit
    2. Open up the RLC Circuit. (It should be the first one that pops up, but if not go to Circuits>Basics>LRC Circuit.)
    3. Change the circuit values to build an
    a. underdamped
    b. critically damped and
    c. overdamped circuit with natural frequency of f=1000 Hz
    4. Plot the natural response in both current and voltage for each of the components R L and C of the second order system when the switch is opened (6 plots total).
    5. Write an equation that matches each of the plots above.


    My first thoughts were to find values of R L and C by using the fact that, ω=2000∏ and that ζ must be <1, =1, and >1 for each of the different situations.

    I also used the following: ω = 1/√[L*C]; ζ = (R/2)*√[C/L]

    Using the tools at the above website and some other sources I was able to design an underdamped circuit with the following conditions:

    R = 10 Ω; L = 100 mH; and C ≈ 253.3 nF
    .: ζ ≈ 0.007921 < 1 and the circuit is underdamped.

    I know that the general form for the solution for an underdamped circuit is:

    x(t) = e^(-ζωt)*[A1 cos(ω*√[1-ζ^2]*t) + A2 sin(ω*√t[1-ζ^2]*t)]

    I plug in what I got for ζ and ω but don't know how to solve for A1 and A2.

    Now the next part is that I'm having trouble coming up with the critically damped and overdamped R L and C values while keeping the 1000 Hz frequency requirement.

    For the critically damped circuit ζ = 1. I thought that I could keep L and C the same as the underdamped circuit (L = 100 mH; and C ≈ 253.3 nF) and use the equation ζ = (R/2)*√[C/L] to solve for a new resistance, R that will make ζ = 1. So I get the equation

    1 = (R/2)*√[253.3n/100m]

    Solve for R,

    R = 2/√[253.3n/100m] ≈ 1256.6 Ω

    I plugged the new R value in the circuit simulator and I can't tell whether or not it's a critically damped circuit by looking at the plots below. It's too small to tell, so I'm wondering if someone can show me how to check that or find a better way to go about doing it.

    Provided that this new circuit is correct where:

    R = 1262.39 Ω; L = 100 mH; and C ≈ 253.3 nF
    .: ζ = 1 = 1 and the circuit is critically damped.

    I know that the general form for the solution for a critically damped circuit is:

    x(t) = B1*e^(-ζωt) + B2*t*e^(-ζωt)

    Again, I plugged in ζ and ω but don't know how to solve for B1 and B2.

    As far as the overdamped circuit goes... I haven't gotten anywhere because I am a bit confused on whether or not I'm on the right track with what I've done so far and want to see what you think.

    I was going to talk to my prof after class, but he wasn't there today... go figure. Thanks for the help in advance.
    Last edited: Aug 10, 2010
  2. stupidlogic

    Thread Starter Member

    Aug 10, 2010
    So I've been working on this problem throughout the afternoon and still haven't solved for the constants A1, A2, B1, and B2 that are apart of the general solutions for an underdamped and critically damped RLC circuit. I did however do the following:

    Since I couldn't see in the simulator whether or not the critically damped circuit was actually critically damped, I decided to solve for new L and C values while keeping the R = 10 Ω to see what I would come up with.

    Again I used the two equations:

    ω = 1/√[L*C] and ζ = (R/2)*√[C/L],
    where ω = 2000∏ and ζ = 1

    I solve both equations for L and set them equal to each other, resulting in:

    L = 1/(2000)^2*C and L = C*(R/2)^2


    1/(2000)^2*C = C*(R/2)^2

    .: C = 1/(R*1000∏)

    Plug in R = 10, and you get C = 1/10000∏ ≈ 31.831 uF
    Go back and plug in R and C you get L ≈ 795.77 uH

    Recap... now,
    R = 10 Ω; L = 795.77 uH; and C ≈ 31.831 uF
    .: ζ = 1 = 1 and the circuit is critically damped.

    Unfortunately I'm still having trouble going back and solving for the initial conditions.

    Anyway, I used the same procedure above to solve for L and C values for an overdamped circuit. My results were:

    R = 10 Ω; L = 79.577 uH; and C ≈ 318.31 uF
    .: ζ = 10 > 1 and the circuit is overdamped.

    I know that the general form for the solution for a overdamped circuit is:

    x(t) = K1*e^(-(ζω-ω√[ζ^2-1])t) + K2*e^(-(ζω+ω√[ζ^2-1])t)

    And now it comes down to K1 and K2... surprise, surprise.

    My problems seems to be dealing more with solving differential equations rather than getting appropriate values of R, L, and C. I got an A when I took Diff EQ, but that was quite some time ago.

    Then again, maybe I'm approaching this the wrong way. Should I look at it from the Diff EQ side first using something like this:

    Ri + (1/C)∫i(x)dx + Vc(to) + L(di/dt) = Vs(t)

    for the loop equation of the series RLC circuit.

    Again, thanks for any help.
    Last edited: Aug 10, 2010
  3. stupidlogic

    Thread Starter Member

    Aug 10, 2010
    I appreciate anyone that may have looked at this, but I believe that I figured it out... or at least most of it.

    I am posting 4 pdf's with all my solutions and work shown for anyone that may be interested. The only thing I couldn't figure out is how to derive an equation for the voltage across the inductor. If you know how to do that, let me know please. Thanks and I hope to frequent this site more and more often.
  4. Ghar

    Active Member

    Mar 8, 2010
    If you have an equation for any voltage or the current you can find any of the others.

    If you know the voltage across the resistor, I = V/R
    If you know the voltage across the capacitor, I = CdV/dt

    If you know the current, V = L dI/dt

    If you know the voltages across the resistor and inductor then Vc + VL + VR = 0
  5. hgmjr

    Retired Moderator

    Jan 28, 2005
    Your question is difficult to answer in much detail in a forum setting. I recommend that you supplement the information you obtain here by googling the subject and see what you can find on the web in answer to your question.

  6. stupidlogic

    Thread Starter Member

    Aug 10, 2010
    Thank you Ghar... that actually helped out a lot and I feel kind of dumb that I didn't think of something so basic to get VL. I appreciate the help.

    If anyone is interested I can re-upload my work for anyone to look at. The only difference between my final answers and what I have above is the addition of VL so if there are no requests I'll just leave it.

    Once again thanks.
  7. victorhugo289


    Aug 24, 2010
    Man, that circuit simulator is awesome!
  8. stupidlogic

    Thread Starter Member

    Aug 10, 2010
    Yeah, it is. One of the kids in my class found it and shared it with the prof and he decided to have us do this assignment with it. It's great if you're not sure how one of the circuits in there works, then you can visually see which direction the current is flowing and the voltage, etc. It's great.