# Thread: Differential equation for electrical circuit

1. ## 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.

2. ## Re: Differential equation for electrical circuit Originally Posted by Vinod $L\frac{dI}{dt}+RI=E$

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?

3. ## 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.

4. ## 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 $E_0$ ?

5. ## 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?

6. ## Re: Differential equation for electrical circuit

Hi,
Please read $e^{\frac{-Rt}{L}}=E-RI$

7. ## Re: Differential equation for electrical circuit Originally Posted by Vinod 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.

8. ## Re: Differential equation for electrical circuit Originally Posted by Walagaster 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?

9. ## 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?

10. ## Re: Differential equation for electrical circuit Originally Posted by MarkFL 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 $I=\frac{1}{L*e^{(R*t)/L}}*(E*L*e^{(R*t)/L})/R$

11. ## Re: Differential equation for electrical circuit Originally Posted by Vinod 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?

12. ## Re: Differential equation for electrical circuit Originally Posted by MarkFL 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?

13. ## Re: Differential equation for electrical circuit Originally Posted by Vinod 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.

14. ## Re: Differential equation for electrical circuit

The limiting value is:

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

15. ## Re: Differential equation for electrical circuit Originally Posted by Walagaster 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?

### precalculus with limits 7.1 page 481 textbook page

Click on a term to search for related topics.