# Thread: Setting up system of differential equations, circuits.

1. ## Setting up system of differential equations, circuits.

Okay, here is what it looks like:

The problem is stated as:

Set up a system of differential equations with initial conditions to describe the currents $\displaystyle i_2$ and $\displaystyle i_3$ in the electrical network. Assume $\displaystyle E$ is 100V, $\displaystyle R_1$ is 20 ohms, $\displaystyle R_2$ is 40 ohms, L_1 is 0.01 H and $\displaystyle L_2$ is 0.02 H. Currents are initially zero.

My work:
$\displaystyle E(t) = L_1\frac{di_1}{dt}+i_1R_1$

$\displaystyle E(t) = L_2\frac{di_2}{dt}+i_2R_2$

Kirchhoff's Law

$\displaystyle E(t) = L_1\frac{d(i_2+i_3)}{dt}+(i_2+i_3)R_1$

$\displaystyle E(t) = L_2\frac{di_2}{dt}+i_2R_2$

After this, I can take it from there...but am I setting it up right? How do you set it up?

2. $\displaystyle Sorry, \: I \: can't \: see \: i_3 \: in \: this \: picture.$

3. There's no $\displaystyle i_3$ in the picture.

A law is though $\displaystyle i_1=i_2+i_3$.

4. Originally Posted by A Beautiful Mind
There's no $\displaystyle i_3$ in the picture.

A law is though $\displaystyle i_1=i_2+i_3$.
That makes no sense. Are you defining i3 via i1 = i2 + i3? According to your circuit, i = i1 + i2 ...?

5. No, I don't think so. See, I'm confused about this because there's no $\displaystyle i_3$ to work with in the circuit. I've looked closely at the one in my homework and it's plainly not there. The only thing I would've figured to work would've been a substitution of $\displaystyle i_1=i_2+i_3$because we're supposed to "set up a system of differential equations with initial conditions to describe the currents and in the electrical network". The only thing I read in the book pertaining to some kind of substitution.

6. Originally Posted by A Beautiful Mind
No, I don't think so. See, I'm confused about this because there's no $\displaystyle i_3$ to work with in the circuit. I've looked closely at the one in my homework and it's plainly not there. The only thing I would've figured to work would've been a substitution of $\displaystyle i_1=i_2+i_3$because we're supposed to "set up a system of differential equations with initial conditions to describe the currents and in the electrical network". The only thing I read in the book pertaining to some kind of substitution.
Why are you insisting on having an i3 where there is none? The relatinship you are expected to use is i = i1 + i2. I think you need to go back and try to understand Kirchoff's circuit laws a bit better before attempting to set up this question.

7. Okay, I figured that would be it, but why is there an i_3 mentioned in the question? That part is throwing me off. Just a typo you think or no?

8. Originally Posted by A Beautiful Mind
Okay, I figured that would be it, but why is there an i_3 mentioned in the question? That part is throwing me off. Just a typo you think or no?
did you label the current yourself in the diagram or it's as written in the original question?

9. There was a picture of it. I copied the thing and made sure it was right.

10. That was an unfortunate error on my part.
Rename them, 1st, 2nd and 3rd instead of 0th, 1st and 2nd..
So i becomes i_1, 1_1 becomes i_2 and i_2 becomes i_3.
Sorry.
Dr.

At 04:19 AM 9/26/2010, you wrote:
>Professor,
>
>I am a bit confused. In number 7, it says "Set up a system of differential
>equations with initial conditions to describe the currents i_2 and i_3 in the
>electrical network pictured." In the picture, there isn't an i_3, there is
>only
>an i, i_1, and i_2.
>
>Thanks.
>
So, what about my equations I wrote now?

$\displaystyle E(t) = L_1\frac{di_1}{dt}+i_2R_1$

$\displaystyle E(t) = L_2\frac{di_3}{dt}+i_3R_2$

Kirchhoff's Law

$\displaystyle E(t) = L_1\frac{d(i_2+i_3)}{dt}+(i_2)R_1$

$\displaystyle E(t) = L_2\frac{di_2}{dt}+i_3R_2$

11. Originally Posted by A Beautiful Mind
So, what about my equations I wrote now?

$\displaystyle E(t) = L_1\frac{di_1}{dt}+i_2R_1$

$\displaystyle E(t) = L_2\frac{di_3}{dt}+i_3R_2$

Kirchhoff's Law

$\displaystyle E(t) = L_1\frac{d(i_2+i_3)}{dt}+(i_2)R_1$

$\displaystyle E(t) = L_2\frac{di_2}{dt}+i_3R_2$
I agree with your first equation bot not with the second. There are two loops involved in this circuit, the small box on the left and the whole box(perimeter of the circuit). It doesn't quite make sense when you consider the right box as a loop since there is no emf cell providing the potential difference in there. (Review Kirchoff's second law.)

12. Thanks, the way you explained it directly to me helped a lot and I got the right answer.

13. It doesn't quite make sense when you consider the right box as a loop since there is no emf cell providing the potential difference in there.
Not sure I agree. In mesh current analysis, you do that sort of thing all the time. In fact, in the wiki to which I've just linked, there's an example of a mesh current going around a planar loop that has no voltage source.

14. Originally Posted by Ackbeet
Not sure I agree. In mesh current analysis, you do that sort of thing all the time. In fact, in the wiki to which I've just linked, there's an example of a mesh current going around a planar loop that has no voltage source.
Thanks for pointing that out, Ackbeet. On second thought, it can be considered as a loop to set up a 3rd equation:

0.02 d(I2)/dt + 40 I2 - 20 I3 =0

But how did Beautiful mind arrive at the correct answer? Perhaps the 2 equations are sufficient to set up the differential equation. What do you think, Ackbeet?

15. Based on the original drawing, and if I were doing mesh analysis, I would set up $\displaystyle i_{1}$ as the mesh current going clockwise in the left panel, and $\displaystyle i_{2}$ as the mesh current going clockwise in the right panel. In addition, you'd need to assign the positive and negative polarities to the voltage source. I would say the positive terminal is on top, and the negative on the bottom. Using the passive sign convention, my equation would look like this:

$\displaystyle \displaystyle{-E(t)+L_{1}\,\frac{di_{1}}{dt}+(i_{1}-i_{2})R_{1}=0,}$ and

$\displaystyle \displaystyle{(i_{2}-i_{1})R_{1}+L_{2}\,\frac{di_{2}}{dt}+i_{2}R_{2}=0. }$

You can certainly do KVL around the two loops you mentioned, though. You should get the same results either way.