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Thread: Differential equation for electrical circuit

  1. #1
    Senior Member Vinod's Avatar
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    Differential equation for electrical circuit

    $L\frac{dI}{dt}+RI=E$

    In this equation, I is the current, R is the resistance, L is the inductance, and E is the electromotive force (voltage).

    A) Solve the differential equation given a constant voltage $E_0$

    B)Use the result of Exercise A) to find the equation for the current if I(0)=0,$E_0$=110volt, R=550 ohms, and L=4 henrys.
    When does the current reach 90% of its limiting value?

    Answer:

    I am trying to answer these questions. But if any member knows the answer he may reply.
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    Member Walagaster's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by Vinod View Post
    $L\frac{dI}{dt}+RI=E$

    In this equation, I is the current, R is the resistance, L is the inductance, and E is the electromotive force (voltage).

    A) Solve the differential equation given a constant voltage $E_0$

    B)Use the result of Exercise A) to find the equation for the current if I(0)=0,$E_0$=110volt, R=550 ohms, and L=4 henrys.
    When does the current reach 90% of its limiting value?

    Answer:

    I am trying to answer these questions. But if any member knows the answer he may reply.
    Divide the equation by L and use the integrating factor method to solve the DE. Have you studied that?
    Thanks from topsquark and Vinod
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    Re: Differential equation for electrical circuit

    From $\displaystyle L\frac{dI}{dt}+ RI= E$ we have $\frac{dI}{dt}= \frac{E- RI}{L}$. We can separate that as $L\frac{dI}{E- RI}= dt$.

    Let u= E- RI so that $du= -RdI$, $dI= -\frac{1}{R}du$. Now we can write the equation as $-\frac{L}{R}\frac{du}{u}= dt$, $\frac{du}{u}= -\frac{R}{L}dt$.

    Integrate both sides to get u as a function of t. Then go back to I.
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    Senior Member Vinod's Avatar
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    Re: Differential equation for electrical circuit

    Hi,
    $Ie^{\frac{Rt}{L}}=\frac1L\int e^{\frac{Rt}{L}}*E_0$

    What to do next? What is the value of $E_0$ ?
    Last edited by Vinod; May 16th 2018 at 11:21 PM.
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    Senior Member Vinod's Avatar
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    Re: Differential equation for electrical circuit

    Hello, By integrating both sides of equation, I got $ln u= \frac{-Rt}{L}$.So E-RI=$\frac{-Rt}{L}$ Is this a general solution to this problem?
    Last edited by Vinod; May 17th 2018 at 12:15 AM.
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    Senior Member Vinod's Avatar
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    Re: Differential equation for electrical circuit

    Hi,
    Please read $e^{\frac{-Rt}{L}}=E-RI$
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    Member Walagaster's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by Vinod View Post
    Hi,
    $Ie^{\frac{Rt}{L}}=\frac1L\int e^{\frac{Rt}{L}}*E_0$

    What to do next? What is the value of $E_0$ ?
    Vinod, you need to quote the post to which you are responding, and then you need to show your steps. I assume this is in response to my post #2. But you have some arithmetic error and the constant of integration should be added, not multiplied. What happened to the $E$? Then you need to solve for $I$. Please check your work and show your steps.
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    Senior Member Vinod's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by Walagaster View Post
    Vinod, you need to quote the post to which you are responding, and then you need to show your steps. I assume this is in response to my post #2. But you have some arithmetic error and the constant of integration should be added, not multiplied. What happened to the $E$? Then you need to solve for $I$. Please check your work and show your steps.
    Hello,

    Sorry for delay in replying your post #7. I devided the equation by L and the integrating factor is $e^{(R/L)*t}$.Now what is the general solution?
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    MHF Contributor MarkFL's Avatar
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    Re: Differential equation for electrical circuit

    You are given:

    $\displaystyle L\frac{dI}{dt}+RI=E$

    So, you divided through by $\displaystyle L$ to obtain an equation in standard linear form:

    $\displaystyle \frac{dI}{dt}+\frac{R}{L}I=\frac{E}{L}$

    You computed the integrating factor:

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

    Multiplying through by the integrating factor, we obtain:

    $\displaystyle e^{\Large\frac{R}{L}t}\frac{dI}{dt}+\frac{R}{L}e^{ \Large\frac{R}{L}t}I=\frac{E}{L}e^{\Large\frac{R}{ L}t}$

    You should now see that the LHS can be rewritten as the derivative of a product...can you write it in that form?
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    Senior Member Vinod's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by MarkFL View Post
    You are given:

    $\displaystyle L\frac{dI}{dt}+RI=E$

    So, you divided through by $\displaystyle L$ to obtain an equation in standard linear form:

    $\displaystyle \frac{dI}{dt}+\frac{R}{L}I=\frac{E}{L}$

    You computed the integrating factor:

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

    Multiplying through by the integrating factor, we obtain:

    $\displaystyle e^{\Large\frac{R}{L}t}\frac{dI}{dt}+\frac{R}{L}e^{ \Large\frac{R}{L}t}I=\frac{E}{L}e^{\Large\frac{R}{ L}t}$

    You should now see that the LHS can be rewritten as the derivative of a product...can you write it in that form?
    Hello,
    So the general solution is $I=\frac{1}{L*e^{(R*t)/L}}*(E*L*e^{(R*t)/L})/R$
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    MHF Contributor MarkFL's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by Vinod View Post
    Hello,
    So the general solution is $I=\frac{1}{L*e^{(R*t)/L}}*(E*L*e^{(R*t)/L})/R$
    Continuing where I left off, we can rewrite the ODE as:

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

    Integrating with respect to $\displaystyle t$, we find:

    $\displaystyle e^{\Large\frac{R}{L}t}I=\frac{E}{R}e^{\Large\frac{ R}{ L}t}+c_1$

    Hence:

    $\displaystyle I(t)=\frac{E}{R}+c_1e^{-\Large\frac{R}{L}t}$

    Now, given $\displaystyle I(0)=0$, we then find:

    $\displaystyle I(0)=\frac{E}{R}+c_1=0\implies c_1=-\frac{E}{R}$

    And so the solution to the given IVP is:

    $\displaystyle I(t)=\frac{E}{R}-\frac{E}{R}e^{-\Large\frac{R}{L}t}=\frac{E}{R}\left(1-e^{-\Large\frac{R}{L}t}\right)$

    Can you determine the limiting value?
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    Senior Member Vinod's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by MarkFL View Post
    Continuing where I left off, we can rewrite the ODE as:

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

    Integrating with respect to $\displaystyle t$, we find:

    $\displaystyle e^{\Large\frac{R}{L}t}I=\frac{E}{R}e^{\Large\frac{ R}{ L}t}+c_1$

    Hence:

    $\displaystyle I(t)=\frac{E}{R}+c_1e^{-\Large\frac{R}{L}t}$

    Now, given $\displaystyle I(0)=0$, we then find:

    $\displaystyle I(0)=\frac{E}{R}+c_1=0\implies c_1=-\frac{E}{R}$

    And so the solution to the given IVP is:

    $\displaystyle I(t)=\frac{E}{R}-\frac{E}{R}e^{-\Large\frac{R}{L}t}=\frac{E}{R}\left(1-e^{-\Large\frac{R}{L}t}\right)$

    Can you determine the limiting value?
    Hello,
    I am very sorry for not adding constant of integration to my general solution. By the way, Current's limiting value is $\frac{t*E}{L}$. If we plug in E=110 volts and L=4 Henrys into this limiting value equation assuming t=1, we get 0.2. Now when does the current reach 90% of its limiting value 0.2? How to answer that question?
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    Member Walagaster's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by Vinod View Post
    Hello,
    I am very sorry for not adding constant of integration to my general solution. By the way, Current's limiting value is $\frac{t*E}{L}$.
    That is wrong. Please show your work instead of just giving answers or guesses.
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    MHF Contributor MarkFL's Avatar
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    Re: Differential equation for electrical circuit

    The limiting value is:

    $\displaystyle \lim_{t\to\infty}I(t)=\lim_{t\to\infty}\left(\frac {E}{R}\left(1-e^{-\Large\frac{R}{L}t}\right)\right)=\,?$
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  15. #15
    Senior Member Vinod's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by Walagaster View Post
    That is wrong. Please show your work instead of just giving answers or guesses.
    Hello,
    I am very sorry for posting wrong answer as well as making some arithmetic error. By the way, Current's limiting value is $\frac{E}{R}$ assuming $t\rightarrow \infty$ If we plug in E=110 volts, and R=550 ohms, we get 0.2. Now when does current reaches its limiting value 0.2?
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