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Math Help - RLC second order differential equation question

  1. #1
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    RLC second order differential equation question

    How do I set up and solve a second order differential equation to find the voltage across a capacitor knowing R, L, C, the peak to peak voltage source, and the frequency?

    As far as I can tell, it is:

    L*(d^2i/dt^2)+R(di/dt)+(1/C)i=wVcoswt


    But I am unsure about the RHS of the equation and then what to do from there to find the voltage across the capacitor...


    Any help is much appreciated
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  2. #2
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    Re: RLC second order differential equation question

    1. You could "guess" the form of the answer. e.g. i = A sin (mt) + B cos (mt)
    2. Calculate the derivative and second derivative of your guessed i
    3. Substitute for i and its derivatives in the differential equation
    4. Choose A, B and m to make the equation work

    The last bit can be done by choosing special values for t such as t= 0 and/or mt = pi/2.

    Another trick is to solve it with the RHS=0 first (power supply shorted out). Then solve it with the power supply turned on. You can then add these two answers together to form another solution. The two seperate solutions are known as the natural response and the forced responce respectively (and may have different frequencies).
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  3. #3
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    Re: RLC second order differential equation question

    So if I assume the form of the answer is i = A sin (mt) + B cos (mt), with i' = mA cos (mt) - mA sin (mt) and i'' = - Am^2 sin (mt) + Bm^2 cos (mt)
    and then substitute into Li'' + Ri' + (1/C)i = mV cos mt
    to make:

    L(- Am^2 sin (mt) + Bm^2 cos (mt)) + R(mA cos (mt) - mA sin (mt)) + (1/C)(A sin (mt) + B cos (mt)) = mV cos mt

    How do I get from this point to the values of A, B and m?
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  4. #4
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    Re: RLC second order differential equation question

    Take a look here

    RLC circuit - Wikipedia, the free encyclopedia

    Alternatively, if you post your question on the differential equations page you will probably get plenty of replies. I have added some doodles below.

    assume
    i = A sin (mt + phi)
    i' = Am cos (mt + phi)
    i'' = -Am^2 sin (mt + phi)

    Then

    L(- Am^2 sin (mt+phi)) + R(mA cos (mt+phi)) + (1/C)(A sin (mt+phi)) = mV cos mt

    this must be true for any t. So consider t = pi/2/m, then

    L(- Am^2 sin (pi/2+phi)) + R(mA cos (pi/2+phi)) + (1/C)(A sin (pi/2+phi)) = 0
    simplifying:

    (1/C-Lm^2)
    sin (pi/2+phi) + Rm cos (pi/2+phi) = 0

    but sin (pi/2 + phi) = cos (phi) and cos (pi/2 + phi) = -sin(phi) so

    (1/C-Lm^2)cos (phi) = Rm sin (phi)

    or tan(phi) =
    (1/C-Lm^2) / R / m ONE UNKNOWN FOUND!

    Now if we try t = 0 in:
    L(- Am^2 sin (mt+phi)) + R(mA cos (mt+phi)) + (1/C)(A sin (mt+phi)) = mV cos mt

    we get

    L(- Am^2 sin (phi)) + R(mA cos (phi)) + (1/C)(A sin (phi)) = mV
    or

    (1/(mC)-mL) sin (phi) + R cos (phi) = V/A. Perhaps this is useful since we know tan (phi)?


    Now if we try t = -phi/m in:
    L(- Am^2 sin (mt+phi)) + R(mA cos (mt+phi)) + (1/C)(A sin (mt+phi)) = mV cos mt

    we get

    RmA= mV cos m(-phi/m) = mV cos (phi)

    or

    RA= V cos (phi)

    So

    A = V cos (phi) /R and we know tan (phi) from above



    Last edited by Kiwi_Dave; March 30th 2012 at 05:22 PM.
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