Results 1 to 15 of 15

Math Help - Setting up system of differential equations, circuits.

  1. #1
    Member
    Joined
    Mar 2008
    From
    A forrest
    Posts
    162
    Awards
    1

    Setting up system of differential equations, circuits.

    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:
    <br />
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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    <br />
Sorry, \: I \: can't \: see \: i_3 \: in \: this \: picture.  <br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    From
    A forrest
    Posts
    162
    Awards
    1
    There's no i_3 in the picture.

    A law is though i_1=i_2+i_3.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by A Beautiful Mind View Post
    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 ...?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2008
    From
    A forrest
    Posts
    162
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by A Beautiful Mind View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Mar 2008
    From
    A forrest
    Posts
    162
    Awards
    1
    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by A Beautiful Mind View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Mar 2008
    From
    A forrest
    Posts
    162
    Awards
    1
    There was a picture of it. I copied the thing and made sure it was right.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Mar 2008
    From
    A forrest
    Posts
    162
    Awards
    1
    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?

    <br />
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
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by A Beautiful Mind View Post
    So, what about my equations I wrote now?

    <br />
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.)
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Mar 2008
    From
    A forrest
    Posts
    162
    Awards
    1
    Thanks, the way you explained it directly to me helped a lot and I got the right answer.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    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.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by Ackbeet View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Setting up a system of equations for a matrix
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 29th 2010, 04:06 AM
  2. setting up a system of equations
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 31st 2009, 04:09 PM
  3. Setting up a system of equations! + Matrix solving
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 21st 2009, 08:31 PM
  4. Replies: 2
    Last Post: February 19th 2009, 11:48 PM
  5. system of differential equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 14th 2008, 09:40 AM

Search Tags


/mathhelpforum @mathhelpforum