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Kirchhoff's Law Problem

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Using Kirchhoff's rules, calculate the current in R1 with the directions indicated in the figure below.

    http://img11.imageshack.us/img11/6018/sbpic2822.png [Broken]

    Assume that R1 = 2.00 kΩ, R2 = 3.00 kΩ, R3 = 4.00 kΩ, E1 = 75.0 V, E2 = 70.0 V and E3 = 85.0 V.

    2. Relevant equations

    Kirchhoff's Rules of voltage loops.

    3. The attempt at a solution

    The problem I'm having is establishing what the voltage loops are here. My book doesn't really explain what to do in the case of multiple batteries and branches...

    Here's what I have so far:

    I2 = I3 + I1 at the junction points.

    Left Loop: E2 - E1 - I1R1 - I2R2 = 0

    Right Loop: E2 - E3 - I3R3 - I2R2 = 0

    I assumed that since the current goes in the opposite direction on the other side, I change it from counter clockwise on the left to clockwise on the right.

    Can anyone please tell me how to properly do this problem? It would be much appreciated. Thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 23, 2009 #2
    I think you are doing fine.
    Solve for I1 and if it turns out negative you know what that means.
     
  4. Feb 23, 2009 #3
    I don't think it works. I've tried solving it a few times, but I keep getting the answer wrong.
     
  5. Feb 23, 2009 #4
    The idea is to match the number of independent equations to the number of unknowns. There is a great deal of flexibility (re. ambiguity) in doing this. What you wrote looks okay but read the following paragraph for a more expected solution. You can solve for I1,I2,I3 but beating the equations together (elimination) or formulating the equations as matrices (the system is linear) and doing a matrix inversion; which is more generic.
    The typical (and probably expected) solution in this case would be to choose two loop currents that are independent; say the left hand loop and right hand loop. Then assign names to the loop currents I1,I2; leaving I3 out. Usually you would make the loop currents consistent; say clockwise. Write the two loop equations and solve. Don't assign names to the loop voltages ; of course the loop equations include the sources but they are constants.

    The ambiguity is in not knowing what "independence" is until after the equation is solved. I have found a way, using exterior products, that clarifies the situation. You can throw the equations and variables together any which way, take successive exterior products and all is answered. The operations are simple but not taught early in college; typically after a BA. That's a shame because it really makes these problems simpler. The alternative is talking about determinants and minors; which I find to be horribly tedious, ambiguous, and unmotivated.
    Having said that, they will usually give you simple things like your problem where the loops are crisp and you just pick the obvious ones. In more complicated cases, non-planar graphs or old-fashioned telephone hybrids, the exterior products are life savers.

    Gotta go!
    Ray
     
  6. Feb 23, 2009 #5
    I didn't really understand a lot of that; I'm only first year Engineering :p. Is it right to assume though that since each loop is going a different direction that I should change from clockwise in one to counter-clockwise in the other?
     
  7. Feb 23, 2009 #6
    It is more conventional to assign loop currents consistently; say all clockwise. In EE it really doesn't make any difference because the the variables will just have a different sign. Speaking from experience consistency makes the work easier to check and easier on the teacher; that counts for something. Keep in mind that this is just a set of linear equations dictated by the Kirchhoff's laws. Any way you put them together they will generate the right answers if the number of unknowns is matched by the number of independent equations. Have faith and try obvious (and hopefully consistent) approaches.
    If you want me to I can describe "dimensional analysis" that allows you to check for simple algebra errors; it's not hard but you probably haven't been taught it. I actually haven't seen it stated in EE books; but it's a life-saver when you move on to linear AC analysis and a quick check in DC analysis. In hindsight it's will seem simple minded.

    Ray
     
  8. Feb 23, 2009 #7
    I have learned dimensional analysis. I can't see how it would help much at this point though. I know that everything in my equations is in the proper units. Like I said though, I'm only in first year engineering, so I probably don't really know what I'm talking about :p.

    Anyway, here is my solution so far. Perhaps someone can point out where I went wrong?

    I2 = I3 + I1 at the junction points.

    Left Loop: E2 - E1 - I1R1 - I2R2 = 0

    - Solve for I1

    I1 = [tex]\frac{ (E2 - E1 - I2R2)}{R1}[/tex]

    Right Loop: E2 - E3 - I3R3 - I2R2 = 0

    - Solve for I2

    I2 = [tex]\frac{ (E2 - E3 - I3R3)}{R2}[/tex]

    - Substitute for I3 (I3 = I2 - I1)

    I2 = [tex]\frac{ (E2 - E3 - I2R3 + I1R3)}{R2}[/tex]

    - Consolidate I2 to one side

    I2(1 +[tex]\frac{R3}{R2}[/tex])= [tex]\frac{ (E2 - E3 + I1R3)}{R2}[/tex]

    I2 = [tex]\frac{(E2 - E3 + I1R3)}{(R2 + R3)}[/tex]

    - Substitute I2 into equation for I1

    I1 = [tex]\frac{ (E2 - E1 - (\frac{(E2 - E3 + I1R3)}{(R2 + R3)})R2)}{R1}[/tex]

    - Consolidate I1 to one side

    I1(1- [tex]\frac{R3R2}{R1(R2 + R3)})[/tex] = [tex]\frac{ (E2 - E1 - (\frac{(E2 - E3)}{(R2 + R3)})R2)}{R1}[/tex]

    - Solve for I1

    I1 = [tex]\frac{ (E2 - E1 - (\frac{(E2 - E3)}{(R2 + R3)})R2)}{R1(1- \frac{R3R2}{R1(R2 + R3)})}[/tex]

    After getting through that most likely convoluted mess, I come up with approximately 0.00500 A, which seems a little bit ridiculously small to me.
     
  9. Feb 23, 2009 #8

    Delphi51

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    Homework Helper

    I started with your two equations for the voltages around the loop and used your I3 = I2 - I1 to eliminate I3. This left
    2*I1 + 3*I2 = 5
    -4*I1 + 7*I2 = -15
    Too lazy to solve the system of equations, I went to
    http://www.analyzemath.com/Calculators/Calculator_syst_eq.html
    and plugged the 6 numbers in. It said I1 = 3.077 and I2 = -.3846
     
  10. Feb 23, 2009 #9
    Neither of those are correct apparently :(. Perhaps someone can look over my two equations in the beginning more closely to see what I did wrong?

    I'm starting to get desperate here. I might just have to get a circuit simulation program to do it for me and then work backwards from that :p.
     
  11. Feb 23, 2009 #10

    Delphi51

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    Checked again - and found an error.
    2*I1 + 3*I2 = 5
    should be -5 instead of 5. Makes a big difference - now I1 = 0.385 and I2 = -1.92

    There was one thing I didn't know when I checked those equations. What is the polarity of the battery symbols and is it conventional positive current or electron current?

    It seems likely we are using positive current, and didn't Benjamin Franklin get the battery symbol backwards so the little end is negative? If so, we have the signs on all the batteries wrong.
     
  12. Feb 23, 2009 #11
    Those answers are unfortunately not correct either. I got the same ones earlier.

    The circuit schematic is showing everything just as it is. Those battery symbols are just as they are on any circuit schematic; Long end = negative terminal, short = positive. So it's kind of weird since their are negative terminals connected directly to each other :p.

    Thanks anyway though, I appreciate your help.
     
  13. Feb 23, 2009 #12

    Delphi51

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    Check http://en.wikipedia.org/wiki/Electronic_symbols
    the short end is negative!

    checking very quickly I think this just means the numbers on the right have their signs changed so
    2*I1 + 3*I2 = 5
    -4*I1 + 7*I2 = 15
    which yields I1 = 0.385, I2 = 1.92
     
  14. Feb 23, 2009 #13
    Jesus ********** Christ

    As Basil Fawlty memorably said to Manuel:
    Getting the pigeon out of the watertank was
    "Not a Proposition from Wittgenstein"
     
    Last edited: Feb 23, 2009
  15. Feb 23, 2009 #14

    Delphi51

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    Indeed, you would think that would have been corrected after more than 200 years!
     
  16. Feb 23, 2009 #15
    Uh.. yea, I don't know what that's about.

    I assumed the short end was the positive terminal due to the fact that it is on actual batteries. I haven't done much of this electricity stuff in a while :p.
     
  17. Feb 23, 2009 #16
    Now that the problem is solved; there are alternative methods.
    Using Thévenin's theorem and superpostion gives direct results:)
    http://en.wikipedia.org/wiki/Superposition_theorem
    In addition there is a diagram method that gives explicit control of what the designer considers dependent and independent variables.
    Signal Flow Graphs
    http://en.wikipedia.org/wiki/Signal-flow_graph
    I haven't read the Wikipedia article in detail. This degree of control in this technique allows extremely sharp analysis.
    Always another way to skin a cat:)


    Ray
     
  18. Feb 23, 2009 #17
    The problem is still not solved. Delphi51's answers still aren't correct.
     
  19. Feb 23, 2009 #18
    Ok, I recalculated everything using a circuit simulating program and it turns out that Delphi51's initial answers of 0.385 and -1.92 are correct, but should be in milliamps, which I find quite strange. The circuit really should be moving in a different direction, but it's according to the diagram... so ya :p.

    Thanks for your help guys!

    EDIT: Just realized it's milliamps because Delphi51 used kOhms rather than ohms.
     
    Last edited: Feb 23, 2009
  20. Feb 23, 2009 #19

    Delphi51

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    Small correction - I used ohms when I should have used kOhms, so that accounts for the mA.

    Interesting, it appears the "correct" answers are obtained from taking the small end of the battery as positive after all.

    Well, I enjoyed it - hope you did, Ownaginatious.
     
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