Network Analysis/Loop Rule Problem

  • Thread starter hagobarcos
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  • #1
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Homework Statement



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)




Homework Equations



∑ I= 0

∑ V =0


The Attempt at a Solution



i1 + i2 + i3 = 2

i6 + i7 +i10 = 3
 

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Answers and Replies

  • #2
785
15

Homework Statement



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)




Homework Equations



∑ I= 0

∑ V =0


The Attempt at a Solution



i1 + i2 + i3 = 2

i6 + i7 +i10 = 3

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution



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... -_-
 
  • #3
34
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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.
 
  • #4
785
15
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.

You need 4 more equations of KJL. Come on, its current inwards = current outwards. Make them.

Other is i1+i6=i4...

Make 3 more...
 
  • #5
34
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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
 
  • #6
785
15
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

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...
 
  • #7
34
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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
 
  • #8
gneill
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Hmmm, no.

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

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".
 
  • #9
34
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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?
 
  • #10
gneill
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Oh man this is hard :D

Yes I thought as much.

Thus: Ground node equation :

i4 + i5 + i8 + i7 = 5
Yup! But don't forget i2.

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?

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).
 
  • #11
34
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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.
 

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  • #12
gneill
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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...
 
  • #13
34
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Bah. Right on. Yay that means it is all ready for matrix solving.
Thank you guys!
 
  • #14
785
15
Thanks for the catch Gneill ! :redface:
 

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