Setting up system of differential equations, circuits.

**A Beautiful Mind** · September 25th 2010, 11:53 AM

Okay, here is what it looks like:

Setting up system of differential equations, circuits.-circuits.png

The problem is stated as:

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

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

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

Kirchhoff's Law

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

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

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

**zzzoak** · September 25th 2010, 03:39 PM

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

**A Beautiful Mind** · September 25th 2010, 08:03 PM

There's no $i_3$ in the picture.

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

**mr fantastic** · September 25th 2010, 09:11 PM

Originally Posted by A Beautiful Mind

There's no $i_3$ in the picture.

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

That makes no sense. Are you defining i3 via i1 = i2 + i3? According to your circuit, i = i1 + i2 ...?

**A Beautiful Mind** · September 25th 2010, 09:17 PM

No, I don't think so. See, I'm confused about this because there's no $i_3$ 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 $i_1=i_2+i_3$ 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.

**mr fantastic** · September 25th 2010, 09:35 PM

Originally Posted by A Beautiful Mind

No, I don't think so. See, I'm confused about this because there's no $i_3$ 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 $i_1=i_2+i_3$ 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.

**A Beautiful Mind** · September 25th 2010, 09:41 PM

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?

**mathaddict** · September 25th 2010, 10:31 PM

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?

**A Beautiful Mind** · September 26th 2010, 06:39 AM

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

**A Beautiful Mind** · September 26th 2010, 10:38 AM

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?

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

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

Kirchhoff's Law

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

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

**mathaddict** · September 27th 2010, 05:07 AM

Originally Posted by A Beautiful Mind

So, what about my equations I wrote now?

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

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

Kirchhoff's Law

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

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

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.)

**A Beautiful Mind** · September 30th 2010, 07:59 PM

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

**Ackbeet** · October 1st 2010, 01:40 AM

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.

**mathaddict** · October 1st 2010, 05:44 AM

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?

**Ackbeet** · October 1st 2010, 06:01 AM

Based on the original drawing, and if I were doing mesh analysis, I would set up $i_{1}$ as the mesh current going clockwise in the left panel, and $i_{2}$ 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:

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

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

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

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