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Math Help - Deriving RL circuit (integration factor)

  1. #1
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    Deriving RL circuit (integration factor)

    I'm having trouble with this, here's my attempt:
    V=V_{R}+V_{L}
    = IR + L \frac{dI(t)}{dt}

    Now introduce f(t) for the integratino factor, divide by L to get the first order on its own

    f(t)\frac{V}{L} = f(t)\frac{IR}{L} + f(t)\frac{dI(t)}{dt}
    f '(t) = \frac{Rf(t)}{L}
    f(t) = e^{\int \frac{R}{L} dt}
    So far this seems somewhat logical, as the final answer has the minus factor in it.

    Continuing:
    e^{\frac{Rt}{L}}\frac{V}{L} = \frac{d}{dt} [e^{\frac{Rt}{L}} I]
    Integrating this gets me back to
    V = IR
    as theres nothing i can do with the exponentials

    The answer i should be getting is:
    I = I_{0} [1-e^{-\frac{Rt}{L}}]

    Hope you can help me out
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  2. #2
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    Quote Originally Posted by imagemania View Post
    I'm having trouble with this, here's my attempt:
    V=V_{R}+V_{L}
    = IR + L \frac{dI(t)}{dt}

    Now introduce f(t) for the integratino factor, divide by L to get the first order on its own

    f(t)\frac{V}{L} = f(t)\frac{IR}{L} + f(t)\frac{dI(t)}{dt}
    f '(t) = \frac{Rf(t)}{L}
    f(t) = e^{\int \frac{R}{L} dt}
    So far this seems somewhat logical, as the final answer has the minus factor in it.

    Continuing:
    e^{\frac{Rt}{L}}\frac{V}{L} = \frac{d}{dt} [e^{\frac{Rt}{L}} I]
    Integrating this gets me back to
    V = IR
    as theres nothing i can do with the exponentials

    The answer i should be getting is:
    I = I_{0} [1-e^{-\frac{Rt}{L}}]

    Hope you can help me out
    What about the constant of integration? You have completely ignored it.

    It would help if the original question was posted. Assuming a series RL circuit and assuming V does not depend on time:

    5. Application of ODEs: Series RL Circuit
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  3. #3
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    It's not a question it's just something I am trying to derive.

    As for the constant, the first integral should have a constant (though generally in first order integrating factor questions it's saved till the end). In this case it woudl still cancle anyway from what i can see.
    As for the second integral, it would be between 0 and t (i.e. no constant).

    So following that:
    i would get
    \frac{V}{L}e^{\frac{Rt}{L}+c} = \frac{d}{dt}[e^{\frac{Rt}{L}+c}I]

    \frac{V}{R}e^{\frac{Rt}{L}+c} = Ie^{\frac{Rt}{L}+c}

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by imagemania View Post
    It's not a question it's just something I am trying to derive.

    As for the constant, the first integral should have a constant (though generally in first order integrating factor questions it's saved till the end). In this case it woudl still cancle anyway from what i can see.
    As for the second integral, it would be between 0 and t (i.e. no constant).

    So following that:
    i would get
    \frac{V}{L}e^{\frac{Rt}{L}+c} = \frac{d}{dt}[e^{\frac{Rt}{L}+c}I]

    \frac{V}{R}e^{\frac{Rt}{L}+c} = Ie^{\frac{Rt}{L}+c}

    Mr. F was not talking about the constant for that integration.
    V = IR + L \frac{dI}{dt}

    Your integrating factor is correct, so let's jump to
    \frac{d}{dt} \left ( e^{(R/L)t}I \right ) = e^{(R/L)t}

    Integrating:
    e^{(R/L)t}I + C = \frac{V}{R}e^{(R/L)t}

    I = -Ce^{-(R/L)t} + \frac{V}{R}

    Now define I0 = V/R and use the condition that I(t = 0) = 0 A.

    -Dan
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