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RLC Circuit Second Order Differential Equation

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi there guys im new to this forum and i have a problem with a bit of cw. It's regarding an RLC circuit. I've come up with a picture (attached) that denotes the equation.

    2. Relevant equations

    I know the equation is [tex] L C \frac{d^2 i}{d t^2} + \frac{L}{R} \frac{di}{dt} + i =i_s (t) , t≥
    0 [/tex]

    3. The attempt at a solution

    I do not really know where to start to show that, i've been given the equation and i was able to draw out the circuit but from there I do not really know where to go as I do not know physics of a circuit very well
     

    Attached Files:

  2. jcsd
  3. Mar 20, 2013 #2

    SammyS

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    attachment.php?attachmentid=56909&d=1363800929.png


    Hello SALAAH_BEDDIAF. Welcome to PF !

    (If your question is principally in regards to solving the differential equation, then you have posted this in the correct place. On the other hand, if it's principally in regards to understanding the physics leading to the differential equation, then we need to ask a moderator to move this thread to Introductory Physics.)

    Is the quantity on the right hand side of the equation, the driving current? -- usually a sinusoidal such as [itex]\displaystyle \ i_s(t)=I_0\sin(\omega t)\ [/itex] or [itex]\displaystyle \ i_s(t)=I_0\cos(\omega t)\ [/itex]
     
  4. Mar 20, 2013 #3
    no i'm able to solve the equation, it's just that i'm ahving difficulty showing that the equation equals what i have given
     
  5. Mar 20, 2013 #4

    SammyS

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    Start with
    [itex]\displaystyle i_s(t)=i_1(t)+i(t)+i_2(t)\ .[/itex]​
    This comes from applying Kirchhoff's Current Law.

    Then since the inductor, the resistor and the capacitor are all in parallel, the instantaneous voltage across any one of these devices is equal to the instantaneous voltage across any other one of these devices.
     
  6. Mar 22, 2013 #5
    I think i may have got it? here is what i have
    Current through resistor: [itex] \frac{1}{R} [/itex]

    Current through inductor: [itex] \frac{1}{L} \int i\,dt [/itex]

    Current through Capacitor: [itex] C\frac{dv}{dt} [/itex]

    Applying kirchoff's law i got

    [itex] \frac{1}{R} + \frac{1}{L} \int i\,dt + C\frac{dv}{dt} = 0 [/itex]

    Differentiating wrt t i got

    [itex] \frac{1}{R} \frac{di}{dt} + \frac{i}{L} + C \frac{d^2 i}{dt^2} = 0 [/itex]

    Rearranging to get the equation

    [itex] LC \frac{d^2 i}{dt^2} + \frac{L}{R} \frac{di}{dt} + i = i_s (t) [/itex]

    I'm pretty sure i missed something here, anyone ought to help me out?
     
  7. Mar 22, 2013 #6

    SammyS

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    Some of your expressions are not quite right.

    Let [itex]\displaystyle \ v(t)\ [/itex] be the instantaneous voltage across each component.

    For the resistor, [itex]\displaystyle \ v(t)=i_1 R\ .\ \ [/itex] Notice that in the figure, the current through R is labeled, i1 .

    For the inductor, [itex]\displaystyle \ v(t)=L\frac{di}{dt}\ .\ \ [/itex]

    For the capacitor, [itex]\displaystyle \ i_2=C\frac{dv}{dt}\ .\ \ [/itex]

    Use these to find how i, i1, and i2 are related.
     
  8. Mar 23, 2013 #7
    okay i got [itex] i_1 + i_2 + i = i_s (t) [/itex]
    using the equalities [itex] v = iR = L\frac{di}{dt} = \frac{Q}{C} [/itex]
    I got [itex] i_1 R = L \frac{di}{dt} → i_1 = \frac{L}{R}\frac{di}{dt} [/itex]
    not too sure how to transform [itex] C\frac{dv}{dt} → LC\frac{d^2 i}{dt^2} [/itex]
     
  9. Mar 23, 2013 #8

    SammyS

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    Use [itex]\displaystyle \ v(t)=L\frac{di}{dt}\ \ [/itex] to find the derivative of [itex]\displaystyle \ v(t)\ .\ \ [/itex]

    Then plug that into [itex]\displaystyle \ i_2=C\frac{dv}{dt}\ .\ \ [/itex]
     
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