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Time Constant for RLC circuit

  1. Dec 19, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-12-19_13-11-40.png

    2. Relevant equations
    I know for RL circuit T = L/R
    For RC circuit it is RC
    But how to go ahead for RLC circuit.


    3. The attempt at a solution
    I calculated for RL as 1/4 = 0.25
    And RC as 1
    Then I added both time constant and got 1.25
    Book answer is 0.5. How to solve this?
    In solved examples they've used RL or RC circuits.
     
  2. jcsd
  3. Dec 19, 2017 #2

    donpacino

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    Gold Member

    You need to stop using equations and start solving problems :)

    Those equations are derived using certain methods. Things will almost never be exactly ideal, you need to learn how to derive the equations for yourself. You need to solve the differential equation, either using the time domain, or laplace to convert to the frequency domain (if you have learned that method, which actually makes these problems very easy to solve).
     
  4. Dec 19, 2017 #3
    Converting to laplace I get
    upload_2017-12-19_14-10-42.png
    The L becomes 1s. The C becomes 4/s
    The resistances are in series so that becomes 4 ohm.
    For t < 0 the current through inductor is 1A as it is short circuit.
    The voltage across Capacitor is 2V
    So now when switch is open
    the inductor will be 1A current source and the capacitor will be 2V voltage source.
    Voltage across capacitor is Ldi/dt + 4(current) = 1di/dt + 4i = 2V(c)

    Using laplace I get
    upload_2017-12-19_14-25-7.png
    Now using kvl I get
    2/s = (1/s) (4/s + 4 + 1s)
    So 2 = 4/s + 4 + 1s
    so 0 = 4/s + 2 + s
    s2 + 2s + 4 = 0
    Now how to proceed?
    Book answer is 0.5
    I'll do guess work
    Laplace is V(s) = [ s^2 + 2s + 4 ] I(s)
    So we do inverse laplace
    I(s) = V(s) / (s^2 + 2s + 4)
    Which is I(t) = V(t) e^[-(1+j1.732)t]
    And time constant is when the e(raised to part) becomes e raised to 1
    So (1 + j1.732)t = 1
    t = inverse and inverse of this is 0.5
    But I got this only cause answer was given. It's like I went backwards.

    Laplace is V(s) = [ s^2 + 2s + 4 ] I(s)
    I didn't understand after this.
    Why not do laplace inverse now?
     

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    Last edited: Dec 19, 2017
  5. Dec 19, 2017 #4

    gneill

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    Staff: Mentor

    One problem. The initial inductor current should not be placed as a fixed, permanent source in your Laplace version of the circuit. That current is driven by the stored energy in the inductor and can be expected to dissipate through losses elsewhere in the circuit (resistors). Instead, it should be represented as a potential rise with the value ##I_oL##, where ##I_o## is the initial inductor current. No "s" is associated with this value as it's not a "real" source, but rather represents stored energy.

    Things are a a bit different for the capacitor where you DO represent the initial charge with a "real" voltage source. Unlike current, charge is conserved. So for the capacitor you do use a source: ##V_o/s##.

    Wikipedia has a reasonable treatment of Laplace transforms, and even provides a list of valid "transformed" circuit elements if you ever need to jog your memory.
     
  6. Dec 19, 2017 #5
    How do I solve this question then? Book says for inductor Laplace is parallel current source divided by 's' n inductor is LS.
     
  7. Dec 19, 2017 #6

    gneill

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    Staff: Mentor

    You have two choices when you convert the inductor to the s-domain. One places a current in parallel with the inductor, the other places a voltage in series with it. Since you're using KVL and want to write a loop equation of potential changes, the series voltage source is the way to go.

    Did you check out the Wikipedia entry that I mentioned? In particular, look at the nice diagram in the subsection,
    "s-domain equivalent circuits and impedances"
     
  8. Dec 19, 2017 #7
    I'm on my mobile now. I'll check wiki n Laplace tomorrow. I hope they give Laplace list in exam. Else its gonna be hard to remember.
     
  9. Dec 20, 2017 #8
    So now i get:
    upload_2017-12-20_15-35-17.png
    I(s) = (1 + 2/s) / (1s + 2 + 2 + 4/s)
    Which gives I(s) = (s + 2) / (s^2 + 4s + 4) = 1/(s + 2)
    I(t) = e^(-2t) ---> By doing inverse laplace.
    So time constant = 1/2 = 0.5

    Got it. Thanks. But why did we take current [ I(s) ] and why not equation for V(s)?
     

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  10. Dec 20, 2017 #9

    gneill

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    Staff: Mentor

    You could have, but with only one loop, KVL is a one-equation simple approach. If you wish to solve via some V(s) instead, pick a reference node and another node in the circuit and write the KCL node equation.
     
  11. Dec 20, 2017 #10

    donpacino

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    Gold Member

    All you needed was the time constant
     
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