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Thread: Differential equation for electrical circuit

  1. #16
    MHF Contributor MarkFL'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 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?
    Well, we need to solve:

    $\displaystyle I(t)=\frac{9}{10}\cdot\frac{E}{R}$

    $\displaystyle \frac{E}{R}\left(1-e^{-\Large\frac{R}{L}t}\right)=\frac{9}{10}\cdot\frac{ E}{R}$

    $\displaystyle 1-e^{-\Large\frac{R}{L}t}=\frac{9}{10}$

    Can you proceed to find $\displaystyle t$?
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  2. #17
    Senior Member Vinod's Avatar
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    Re: Differential equation for electrical circuit

    Quote Originally Posted by MarkFL View Post
    Well, we need to solve:

    $\displaystyle I(t)=\frac{9}{10}\cdot\frac{E}{R}$

    $\displaystyle \frac{E}{R}\left(1-e^{-\Large\frac{R}{L}t}\right)=\frac{9}{10}\cdot\frac{ E}{R}$

    $\displaystyle 1-e^{-\Large\frac{R}{L}t}=\frac{9}{10}$

    Can you proceed to find $\displaystyle t$?
    Hello,
    So,$e^{(-R/L)*t}=0.1$. By plugging in the value of R=550ohms and L=4 Henrys we get t=5.19387222998E58. I think the answer is in seconds.Isn't it?
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  3. #18
    MHF Contributor MarkFL's Avatar
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    Re: Differential equation for electrical circuit

    Yes, we have:

    $\displaystyle e^{-\Large\frac{R}{L}t}=\frac{1}{10}$

    Or:

    $\displaystyle e^{\Large\frac{R}{L}t}=10$

    This implies (converting from exponential to logarithmic form):

    $\displaystyle \frac{R}{L}t=\ln(10)$

    $\displaystyle t=\frac{L}{R}\ln(10)$
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