# Thread: Differential equation for electrical circuit

1. ## Re: Differential equation for electrical circuit

Originally Posted by Vinod
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 $\displaystyle t$?

2. ## Re: Differential equation for electrical circuit

Originally Posted by MarkFL
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?

3. ## 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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