Complex RLC circuit

Discussion in 'Homework Help' started by john williamsXT, Jul 14, 2009.

  1. john williamsXT

    Thread Starter Member

    Jul 14, 2009
    10
    0
    This is really a mathematics question.
    I have obtained the following differential equations for a complex RLC circuit.
    <br />
(L_1D^2+RD+\frac{1}{C})i_1+(RD+\frac{1}{C})i_2=-5000\sin{100t}<br />
    <br />
L_1D^2i_1-(R_2D+\frac{1}{C_2})i_2=0<br />
    Eliminating either i1 or i2 leads to a third order differential equation, which looks nasty!
    Anyone got any tips for solving?
    Best regards
    John
     
  2. millwood

    Guest

    eq 1 + eq 2 gives you

    (2L1D2 + RD+1/c)i1=-5000sin100t, from which you can solve for i1.

    once you have i1, you have i2 from the 2nd equation.
     
  3. john williamsXT

    Thread Starter Member

    Jul 14, 2009
    10
    0
    Not quite,Millwood since the first equation has R and C but the second one has R2 and C2!
    Regards
    John
     
  4. millwood

    Guest

    that's true. my bad.
     
  5. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    john williamsXT,

    Rearranging the second equation gives L1*D^2*i1 = R2*D*i2+i2/C2 .
    Substituting into the first term of the first equation and rearranging gives, [(R+R2)*D+1/C +1/C2]*i2 = -5000*sin(100*t) . This is a linear equation of order one, from which an integrating factor can be found and i2 solved. Once you find i2, can i1 be far behind? Looks like lots of algebra.

    Ratch
     
  6. john williamsXT

    Thread Starter Member

    Jul 14, 2009
    10
    0
    Hi there
    After substituting into the first equation the first equation becomes.
    <br />
[(R+R_2)D+\frac{1}{C}+\frac{1}{C_2}]i_2+(RD+\frac{1}{C})i_1=-5000\sin{100t}<br />
    So you still have a term in i1.
    Regards
    John
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    You could post the original circuit - forum members may be able to offer a simpler approach to a solution.
     
  8. john williamsXT

    Thread Starter Member

    Jul 14, 2009
    10
    0
    Actually I have progressed further, as follows.
    As indicated in my original post elimination of i1 or 12 leads to a third order differential equation.
    The homogeneous equation is as follows.
    <br />
[D^3+(\frac{RR_2CC_2+L_1C+L_1C_2}{CC_2L_1(R+R_2})D^2+(\frac{R_2C_2+RC}{CC_2L_1(R+R_2)})D+\frac{1}{CC_2L_1(R+R_2)}]i_1=0<br />
    Now the circuit components have the following values.
    R=2,R2=5,C=2.10(-4),C2=3.10(-4),L1=2
    After substitution of these values I get the following(approximate) equation for i1.
    <br />
(D^3+1191D+2262D+1190476)i_1=0<br />
    This will give a characteristic equation in m of.
    <br />
m^3+1191m^2+2262m+1190476=0<br />
    If correct so far is this now best solved by iteration method?
    Cheers
    John
     
  9. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    john williamsXT,

    You are correct. Sorry for the mistake in algebra.

    The roots of your equation in 'm' are -1189.939821, -.5300895-31.62545812i, -.5300895+31.62545812i .

    Ratch
     
  10. john williamsXT

    Thread Starter Member

    Jul 14, 2009
    10
    0
    Thanks for that Ratch.
    How did you find those roots, by an iteration method?
    Do you think these values are realistic?
    And what about the nonhomogeneous equation?
    Best regards
    John
     
  11. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    john williamsXT,

    I used a computer program. There are zillions of them around than can find the roots of polynomials. So can many hand calculators. http://www.savetman.com/roots.html

    Plug in the roots and see if they satisfy the equation.

    I did not write it, so I cannot vouch for its veracity.

    Ratch
     
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