# Thread: Differential equation and electrical circuits

1. ## 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.

2. ## 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 $L*(dI/dt)+R*I=E_0*sin\omega t?$

3. ## Re: Differential equation and electrical circuits

Yes, that constant factor will make a difference in the solution.

4. ## Re: Differential equation and electrical circuits

Originally Posted by MarkFL
Yes, that constant factor will make a difference in the solution.
Hello,
You mean to say my general solution is wrong. Isn't it?

5. ## Re: Differential equation and electrical circuits

Originally Posted by Vinod
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:

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

6. ## Re: Differential equation and electrical circuits

Originally Posted by MarkFL
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?

7. ## Re: Differential equation and electrical circuits

Originally Posted by Vinod
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.