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Network Analysis/Loop Rule Problem

  1. Jun 16, 2014 #1
    1. The problem statement, all variables and given/known data

    This is out of a book on Excel, where we are solving simultaneous equations. However, how do we find the equations for this network in the first place? (image at bottom)




    2. Relevant equations

    ∑ I= 0

    ∑ V =0


    3. The attempt at a solution

    i1 + i2 + i3 = 2

    i6 + i7 +i10 = 3
     

    Attached Files:

    Last edited: Jun 16, 2014
  2. jcsd
  3. Jun 16, 2014 #2

    LOL, no attempt at solution no help...

    First start making equations at various junctions using Kirchoff's Junction Rule.
    No, I will not start making them myself... -_-
     
  4. Jun 16, 2014 #3
    oops! sorry, did not mean to add all that. Will edit in a second.

    ok, so for top point junction, i1 + i2 +i3 = 2

    and i6 + i10 + i 7 = 3 for bottom left junction.

    Am stuck on the rest.
     
  5. Jun 16, 2014 #4
    You need 4 more equations of KJL. Come on, its current inwards = current outwards. Make them.

    Other is i1+i6=i4...

    Make 3 more...
     
  6. Jun 16, 2014 #5
    Hmmm,

    so for bottom right junction:

    i10 + i9 = i8

    And then top right junction:

    i5 = i3 + i9

    And for the Center:

    Wait, what to do for center? Is the current flowing in all dribbling out into the ground?

    Guessing, by conservation of energy:

    i4 + i5 + i8 + i7 = 5
     
  7. Jun 17, 2014 #6
    Hmmm, no.

    That thing is connected to the earth. So at center, sum of all currents equals zero.

    Now you have 6 equations but 10 variables. Applying Kirchoff's Voltage rule for 4 loops will give you 4 more equations and consequently you'll have 10 variables and 10 equations...
     
  8. Jun 19, 2014 #7
    Uh oh. Okay.

    SO, then for the center, i4 + i5 +i8 +i7 = 0

    And then for the loops, assuming current is flowing clockwise:

    Top left loop:

    (+10)i1 -(40)i2 + (10)i4 = 0

    Top right loop:

    (+40)i2 - (5)i3 - (20)i5 = 0

    Bottom right loop:

    (-10)i9 -(10)i8 +(20)i5 = 0

    Center loop:

    (+10)i8 + (5)i10 -(50)i7 = 0

    Bottom left Loop:

    (-10)i4 + (50)i7 - (2)i6 = 0
     
  9. Jun 19, 2014 #8

    gneill

    User Avatar

    Staff: Mentor

    That's not right. It must be assumed that the currents flowing into the ground node will sum to the external currents entering at the nodes where currents are shown entering the circuit. Otherwise there will be a net buildup of charge on the circuit coming from the external sources, and that can't be right.

    If you like, imagine 2 A and 3 A current sources connected between ground and those nodes where the 2 A and 3A currents are shown "coming in from nowhere".
     
  10. Jun 19, 2014 #9
    Oh man this is hard :D

    Yes I thought as much.

    Thus: Ground node equation :

    i4 + i5 + i8 + i7 = 5

    However, this one is not mentioned in the answers. So it may be redundant.

    AND, question, is it possible for the current to flow backwards into the source?
     
  11. Jun 19, 2014 #10

    gneill

    User Avatar

    Staff: Mentor

    Yup! But don't forget i2.

    You should assume that the external currents shown are correct in their directions. The other branch currents are unknowns so their actual directions are not necessarily as assumed in the diagram. {I presume that you are meant to solve for the individual branch currents}.

    You might find it convenient to write Node Equations (see: Nodal Analysis) to first determine the node potentials with respect to the ground node. There will be five equations in five unknowns. Then use the node potentials to determine the individual branch currents (Ohm's Law).
     
  12. Jun 19, 2014 #11
    Yes.

    Revised Ground Node eq: i2+ i5 + i8 + i7 + i4 = 5 amps

    Moving on to system of equations:

    For the nodes/corners of the pentagon:

    i1 + i6 = i4

    So

    i1 + i6 - i4 = 0 ----Eq 1

    2= i1 + i2 + i3

    So

    i1 + i2 + i3 = 2 ----Eq 2

    i3 = i5 + i9

    So

    i3 - i5 - i9 = 0 ----Eq 3

    i9 + i10 = i8

    so

    i9 + i10 - i8 = 0 ---- Eq 4

    3 = i10 + i7 + i6

    so

    i10 + i7 + i6 = 3 ---- Eq 5

    For the loops, using V = IR, and ∑ V = 0 :

    (-2)(i6) +(-10)(i4) + (50)(i7) = 0 ----Eq 6

    (10)(i1) + (-40)(i2) + (10)(i4) = 0 ----Eq 7

    (-5)(i3) + (-20)(i5) + (40)(i2) = 0 ----Eq 8

    (20)(i5) + (-10)(i9) + (-10)(i8) = 0 ---- Eq 9

    (10)(i8) + (5)(i10) + (-50)(i7) = 0 ----Eq 10


    Only problem now is, a few of my equations, namely Eq 6 - Eq 10 are backwards. Attached below are the answers that I am trying to match, although I am not sure how to modify mine with out messing it up.
     

    Attached Files:

  13. Jun 19, 2014 #12

    gneill

    User Avatar

    Staff: Mentor

    I've looked (quickly) over your loop equations, and they appear to be fine. I'm not sure what you mean by some of them being "backwards". Keep in mind that you may multiply an entire equation through by -1 and it will still be the same equation...
     
  14. Jun 19, 2014 #13
    Bah. Right on. Yay that means it is all ready for matrix solving.
    Thank you guys!
     
  15. Jun 20, 2014 #14
    Thanks for the catch Gneill ! :redface:
     
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