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 and in the electrical network. Assume is 100V, is 20 ohms, is 40 ohms, L_1 is 0.01 H and is 0.02 H. Currents are initially zero.
My work:
Kirchhoff's Law
After this, I can take it from there...but am I setting it up right? How do you set it up?
No, I don't think so. See, I'm confused about this because there's no 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 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.
So, what about my equations I wrote now?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.
>
Kirchhoff's Law
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.)
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.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.
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?
Based on the original drawing, and if I were doing mesh analysis, I would set up as the mesh current going clockwise in the left panel, and 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:
and
You can certainly do KVL around the two loops you mentioned, though. You should get the same results either way.