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Math Help - Electrical Circuit problem?

  1. #1
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    Electrical Circuit problem?

    A simple electrical circuit contains a condenser of capacity C farads, a coil of inductance L henrys, a resistance of R ohms, and a generator which produces an electromotive force E volts, in series. If the current intensity at time t at some point of the circuit is I amperes, the differential equation governing the current I is:

    L d^2I/dt^2 + R dI/dt + 1/C I = dE/dt

    There is also a picture attached, not sure if its important so let me know.

    Find I as a function of t if:

    1. R=0, 1/(LC) = w^2, E= constant
    2. R=0, 1/(LC) = w^2, E= Asin(alpha*t); alpha=constant≠w
    3. R=0, 1/(LC) = w^2, E= Asin(wt)
    4. R=50, L=5, C=9x10^-6, E= constant

    If someone can just get me started, it would be much appreciated. I have utterly no clue what to do here
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  2. #2
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    I've been thinking about this for a while now and am making.. some.. headway

    1. Since R=0 there would be no 1st derivative term and since E is a constant, its derivative would just be 0, it would just be:

     L d^2I/dt^2+1/CI=0

    So would I just continue and solve like a regular 2nd order DE?

     Lr^2+ r/CI = 0

    but then where would the 1/LC come in?

    2. Same thing with R=0 but the right side would have a cosine instead? or an (alpha)cosx?
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  3. #3
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    Quote Originally Posted by Jeffman50 View Post
    A simple electrical circuit contains a condenser of capacity C farads, a coil of inductance L henrys, a resistance of R ohms, and a generator which produces an electromotive force E volts, in series. If the current intensity at time t at some point of the circuit is I amperes, the differential equation governing the current I is:

    L d^2I/dt^2 + R dI/dt + 1/C I = dE/dt

    There is also a picture attached, not sure if its important so let me know.

    Find I as a function of t if:

    1. R=0, 1/(LC) = w^2, E= constant
    2. R=0, 1/(LC) = w^2, E= Asin(alpha*t); alpha=constant≠w
    3. R=0, 1/(LC) = w^2, E= Asin(wt)
    4. R=50, L=5, C=9x10^-6, E= constant

    If someone can just get me started, it would be much appreciated. I have utterly no clue what to do here
    I assume it's actually

    L\,\frac{d^2 I}{dt^2} + R \, \frac{dI}{dt} + \left(\frac{1}{C}\right) I = \frac{dE}{dt}.

    Please use brackets where they're needed or learn better LaTeX...


    Anyway

    1. R = 0, \frac{1}{LC} = \omega ^2, E = \textrm{constant}.

    Since E = \textrm{constant}, \frac{dE}{dt} = 0.

    The DE becomes

    L\,\frac{d^2 I}{dt^2} + \left(\frac{1}{C}\right) I =0

    \frac{d^2 I}{dt^2} + \left(\frac{1}{LC}\right) I = 0

    \frac{d^2 I}{dt^2} + \omega ^2 I = 0.


    Now yes, proceed as you would for a 2nd order linear constant coefficient ODE.

    Characteristic equation:

    m^2 + \omega^2 = 0

    m^2 = -\omega^2

    m = 0 \pm i\omega.

    Since these are complex conjugates, the solution is

    I = e^{0t}[C_1 \cos{(\omega t)} + C_2 \sin{(\omega t)}]

     = C_1\cos{(\omega t)} + C_2\sin{(\omega t)}.

    Do you have any boundary conditions?


    Use similar processes for the next few questions.
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  4. #4
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    Thanks, and yes I messed up with the original equation.. it is (1/C)I

    And no boundary restrictions, but just to be curious what do u mean by that and how would that change the solution?
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  5. #5
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    I've started 2:

    R=0 so no first derivative term
    d(Asin(alpha*t)/dt = A(alpha)cos(alpha t)

    so i have  r^2 + w^2 = (A(alpha) cos (alpha*t))/L

    which goes to

     r= sqrt[(A(alpha) cos (alpha*t))L - w^2]

    are the two roots that whole thing + and -?

    so I= e^(whole thing)pi + e^(- whole thing)pi ?
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  6. #6
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    Quote Originally Posted by Jeffman50 View Post
    I've started 2:

    R=0 so no first derivative term
    d(Asin(alpha*t)/dt = A(alpha)cos(alpha t)

    so i have  r^2 + w^2 = (A(alpha) cos (alpha*t))/L

    which goes to

     r= sqrt[(A(alpha) cos (alpha*t))L - w^2]

    are the two roots that whole thing + and -?

    so I= e^(whole thing)pi + e^(- whole thing)pi ?
    You need to set up a 2nd order linear ODE, solve for the homogenous case, then solve for the particular case.



    So for 2.

    R = 0, \frac{1}{LC} = \omega ^2, E = A\sin{(\alpha t)}, \frac{dE}{dt} = \alpha A \cos{(\alpha t)}.


    So the DE becomes:

    L\,\frac{d^2 I}{dt^2} + \left(\frac{1}{C}\right)I = \alpha A \cos{(\alpha t)}

    \frac{d^2 I}{dt^2} + \left(\frac{1}{LC}\right)I = \frac{\alpha A}{L}\cos{(\alpha t)}.

    \frac{d^2 I}{dt^2} + \omega ^2 I = \frac{\alpha A}{L}\cos{(\alpha t)}.


    Homogeneous case Characteristic Equation:

    m^2 + \omega ^2 I = 0

    m = 0\pm i\omega as seen before.

    So the homogeneous solution is

    I_C = C_1\cos{(\omega t)} + C_2\sin{(\omega t)} as before.


    Particular case - use the method of undertermined coefficients:

    Assume a solution of the form:

    I_P = C_3\cos{(\alpha t)} + C_4\sin{(\alpha t)}.

    It is important to note that \alpha \neq \omega. If it did we have to adjust our assumption slightly.


    Anyway, if I_P = C_3\cos{(\alpha t)} + C_4\sin{(\alpha t)} then

    \frac{dI_P}{dt} = -\alpha C_3\sin{(\alpha t)} + \alpha C_4\cos{(\alpha t)}

    \frac{d^2 I_P}{dt^2} = -\alpha ^2 C_3\cos{(\alpha t)} - \alpha ^2 C_4\sin{(\alpha t)}.


    Our original DE is:

    \frac{d^2 I}{dt^2} + \omega ^2 I = \frac{\alpha A}{L}\cos{(\alpha  t)}.

    Plugging everything into the equation gives:

    -\alpha ^2 C_3\cos{(\alpha t)} - \alpha ^2 C_4\sin{(\alpha t)} + \omega ^2 \left[C_3\cos{(\alpha t)} + C_4\sin{(\alpha t)}\right] =  \frac{\alpha A}{L}\cos{(\alpha  t)}

    -\alpha ^2 C_3\cos{(\alpha t)} - \alpha ^2 C_4\sin{(\alpha t)} + \omega^2 C_3\cos{(\alpha t)} + \omega ^2 C_4\sin{(\alpha t)} = \frac{\alpha A}{L} \cos{(\alpha t)}

    (\omega^2 C_3 - \alpha ^2 C_3)\cos{(\alpha t)} + (\omega ^2 C_4 - \alpha ^2 C_4) \sin{(\alpha t)} = \frac{\alpha A}{L} \cos{(\alpha t)}

    C_3(\omega^2 - \alpha^2) \cos{(\alpha t)} + C_4(\omega ^2 - \alpha ^2)\sin{(\alpha t)} = \frac{\alpha A}{L}\cos{(\alpha t)} + 0\sin{(\alpha t)}.


    By equating like coefficients, we find

    C_3(\omega^2 - \alpha^2) = \frac{\alpha A }{L} and  C_4(\omega^2 - \alpha ^2) = 0

    We can therefore see that C_4 = 0 and C_3 = \frac{\alpha A}{L(\omega^2 - \alpha^2)}.


    So now we have our particular solution:

    I_P = \frac{\alpha A}{L(\omega^2 - \alpha^2)} \cos{(\alpha t)}.



    Our general solution is the sum of the homogeneous solution and the particular solution.

    So I = I_C + I_P

    I = C_1\cos{(\omega t)} + C_2\sin{(\omega t)} + \frac{\alpha A}{L(\omega^2 - \alpha^2)} \cos{(\alpha t)}.



    It also seems that you need to revise solving second order linear ODEs.
    I suggest you read http://www.mathhelpforum.com/math-he...-tutorial.html
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