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Thread: Differential equation and electrical circuits

  1. #1
    Senior Member Vinod's Avatar
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    Differential equation and electrical circuits

    Hello,

    If we solve the differential equaton given a periodic electromotive force $E_0 sin\omega t$, we get the general solution

    $I=\frac {(R*sin\omega t -2*L*cos\omega t)}{4*L^2+R^2}+ C*e^{(-R/L)*t}$
    Now how to verify that this general solution can be written in the form
    $I=c*e^{(-R/L)*t} +\frac {E_0*(sin(\omega t+\phi))}{(\sqrt{R^2+\omega^2*L^2})}$

    where $\phi$, the phase angle is given by $arctan((-\omega*L)/R).$

    Note:- The exponential term approaches 0 as $t\rightarrow \infty$ This implies that current approaches the periodic function.
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  2. #2
    Senior Member Vinod's Avatar
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    Re: Differential equation and electrical circuits

    Hello,
    I think my general solution is not correct. Is there any difference between $E_0 sin(\omega t)$ and only $sin(\omega t)$ Is the equation $L*(dI/dt)+R*I=sin(\omega t)$ same as $L*(dI/dt)+R*I=E_0*sin\omega t?$
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    MHF Contributor MarkFL's Avatar
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    Re: Differential equation and electrical circuits

    Yes, that constant factor will make a difference in the solution.
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  4. #4
    Senior Member Vinod's Avatar
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    Re: Differential equation and electrical circuits

    Quote Originally Posted by MarkFL View Post
    Yes, that constant factor will make a difference in the solution.
    Hello,
    You mean to say my general solution is wrong. Isn't it?
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Differential equation and electrical circuits

    Quote Originally Posted by Vinod View Post
    Hello,
    You mean to say my general solution is wrong. Isn't it?
    No, I was just answering your question about the constant factor. I haven't solved the given ODE. Let's take a look:

    $\displaystyle L\frac{dI}{dt}+RI=E_0\sin(\omega t)$

    Divide through by $\displaystyle L$ to put into standard linear form:

    $\displaystyle \frac{dI}{dt}+\frac{R}{L}I= \frac{E_0}{L}\sin(\omega t)$

    Compute the integrating factor:

    $\displaystyle \mu(t)=\exp\left(\frac{R}{L}\int\,dt\right)= e^{\Large\frac{R}{L}t}$

    Multiplying the ODE by $\displaystyle \mu(t)$ there results:

    $\displaystyle e^{\Large\frac{R}{L}t}\frac{dI}{dt}+\frac{R}{L}e^{ \Large\frac{R}{L}t}I= \frac{E_0}{L}e^{\Large\frac{R}{L}t}\sin(\omega t)$

    Rewrite the LHS as the derivative of a product:

    $\displaystyle \frac{d}{dt}\left(e^{\Large\frac{R}{L}t}I\right)= \frac{E_0}{L}e^{\Large\frac{R}{L}t}\sin(\omega t)$

    Integrate with respect to $\displaystyle t$:

    $\displaystyle e^{\Large\frac{R}{L}t}I= \frac{E_0}{L}\cdot\frac{e^{\Large\frac{R}{L}t}} {\left(\dfrac{R}{L}\right)^2+\omega^2} \left(\frac{R}{L}\sin(\omega t)- \omega\cos(\omega t)\right)+c_1$

    Hence:

    $\displaystyle I(t)= \frac{E_0}{L}\cdot \frac{\dfrac{R}{L}\sin(\omega t)- \omega\cos(\omega t)}{\left(\dfrac{R}{L}\right)^2+ \omega^2} +c_1e^{-\Large\frac{R}{L}t}$
    Thanks from Vinod
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  6. #6
    Senior Member Vinod's Avatar
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    Re: Differential equation and electrical circuits

    Quote Originally Posted by MarkFL View Post
    No, I was just answering your question about the constant factor. I haven't solved the given ODE. Let's take a look:

    $\displaystyle L\frac{dI}{dt}+RI=E_0\sin(\omega t)$

    Divide through by $\displaystyle L$ to put into standard linear form:

    $\displaystyle \frac{dI}{dt}+\frac{R}{L}I= \frac{E_0}{L}\sin(\omega t)$

    Compute the integrating factor:

    $\displaystyle \mu(t)=\exp\left(\frac{R}{L}\int\,dt\right)= e^{\Large\frac{R}{L}t}$

    Multiplying the ODE by $\displaystyle \mu(t)$ there results:

    $\displaystyle e^{\Large\frac{R}{L}t}\frac{dI}{dt}+\frac{R}{L}e^{ \Large\frac{R}{L}t}I= \frac{E_0}{L}e^{\Large\frac{R}{L}t}\sin(\omega t)$

    Rewrite the LHS as the derivative of a product:

    $\displaystyle \frac{d}{dt}\left(e^{\Large\frac{R}{L}t}I\right)= \frac{E_0}{L}e^{\Large\frac{R}{L}t}\sin(\omega t)$

    Integrate with respect to $\displaystyle t$:

    $\displaystyle e^{\Large\frac{R}{L}t}I= \frac{E_0}{L}\cdot\frac{e^{\Large\frac{R}{L}t}} {\left(\dfrac{R}{L}\right)^2+\omega^2} \left(\frac{R}{L}\sin(\omega t)- \omega\cos(\omega t)\right)+c_1$

    Hence:

    $\displaystyle I(t)= \frac{E_0}{L}\cdot \frac{\dfrac{R}{L}\sin(\omega t)- \omega\cos(\omega t)}{\left(\dfrac{R}{L}\right)^2+ \omega^2} +c_1e^{-\Large\frac{R}{L}t}$
    Hello,
    Now how to verify that this general solution can be written in the form given in my post #1?
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: Differential equation and electrical circuits

    Quote Originally Posted by Vinod View Post
    Hello,
    Now how to verify that this general solution can be written in the form given in my post #1?
    Your sinusoidal term has some numbers where I would expect parameters to be. However, the second form does not, and it is arrived at by using a linear combination identity. This allows us to write:

    $\displaystyle \frac{R}{L}\sin(\omega t)-\omega\cos(\omega t)= \sqrt{\left(\frac{R}{L}\right)^2+\omega^2} \sin\left(\omega t-\arctan\left(\frac{L\omega}{R}\right)\right)$

    And so, our general solution then becomes:

    $\displaystyle I(t)= \frac{E_0}{L}\cdot \frac{\sqrt{\left(\dfrac{R}{L}\right)^2+ \omega^2} \sin\left(\omega t- \arctan\left(\dfrac{L\omega}{R}\right)\right)} {\left(\dfrac{R}{L}\right)^2+ \omega^2} +c_1e^{-\Large\frac{R}{L}t}$

    And this may be simplified to:

    $\displaystyle I(t)=\frac{E_0\sin\left(\omega t-\arctan\left(\dfrac{L\omega}{R}\right)\right)}{ \sqrt{R^2+(L\omega)^2}}+ c_1e^{-\Large\frac{R}{L}t}$

    This is equivalent to the second form you posted.
    Thanks from Vinod
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