Multiple Power Source Circuit Analysis

  • #1
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Homework Statement


Compute all the currents labeled using the mesh loop method.
FZRNPRn.png


Homework Equations


Parallel Resistors
1/Rt=1/R1+1/R2...+1/Ri
Series Resistors
R1+R2...+Ri

The Attempt at a Solution



My attempt at the kvl and kcl analysis, I think I did this wrong because I can't solve for the variables[/B]
v1 - (I1 - I6 - I3)*r1 - (-I6 - I3 + I1)*r6 - (-I3 - I6 + I1)*r3 - v3=0
v2 - (I2 - I5 - I3)*r2 - (-I5 - I3 + I2)*r5 - (-I3 - I5 + I2)*r3 - v3=0
v1 - (I1 + I4 - I2)*r1 - (I4 - I2 + I1)*r4 - (-I2 + I1 + I4)*r2 - v2=0
I4 = I1 - I2
I6 = I3 - I1
I5 = I3 - I1

This is my first time doing this so I'm sure I did this wrong. Any help is appreciated!
 

Answers and Replies

  • #2
phinds
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Your equations are WAY too complex, bringing in terms that don't belong.

Just take the first one, for example, and your inclusion of I1 and I6 as multipliers for R1 when only I3 is flowing through R1

They are all like that.
 
  • #3
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Thanks for the reply.
Ok, then I'm confused.

So for the left area it goes:
V3-I3*R3-I3*R6-I3*R1-V1=0

So then what happens to I1 and I6?
 
  • #4
phinds
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There is a current designator that shows the current through R3. It is I3. There IS no other current going through that branch (R3)

There is a current designator that shows the current through R6. It is I6. There IS no other current going through that branch (R6)

You are making things more complicated than they need to be.
 
  • #5
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Ok. Trying to simplify this and knowing that the current through each resistors is ONLY it's corresponding current number.

V3-I3*R3-I6*R6+I1*R1-V1=0
V3-I3*R3-I5*R5+I2*R2-V2=0
V1-I1*R1-I4*R4+I2*R2-V2=0

Am I at least on the right track?
 
  • #6
phinds
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I haven't checked in detail but yes, that's exactly the kind of thing you want to see
 
  • #7
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Ok thanks for your help!
 
  • #8
phinds
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Glad to help. I love these easy ones :smile:
 
  • #9
gneill
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Mesh loop analysis is based upon mesh currents. Each loop is assigned a "brand new" current of its own which circulates within that loop alone. KVL equations are written for each loop using these new currents and solved.

Fig1.gif


Thus in the above we define mesh currents ##i_1##, ##i_2##, and ##i_3##.

Note that more than one mesh current can flow in a given component (the ones that border two loops). In the above circuit R1 borders loops 1 and 3, so that it conducts ##i_1## and ##i_3## (flowing in opposite directions).

After solving for the mesh currents you can then go back and form any desired current in the circuit by noting which mesh currents contribute to them. For example, in the above circuit ##I_3 = i_1 - i_2##, and ##I_5 = -i_2##.

There's a very good reason for defining and solving for mesh currents in this way. One can write the loop equations in matrix form easily by inspection -- there's a very simple procedure. It's a virtually foolproof way to obtain the circuit equations quickly and without error.
 
  • #10
phinds
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OOPS ... gneill obviously has it right. I was so focused on one thing that I lost track of another. o:)

EDIT: aseyals, in case it's not clear, it's not that the equations you wrote were wrong, and that's what I was focusing on ... helping you get them right, it's that they are right but not helpful. They involve too many variables because, as gneil pointed out, they are not loop current equations.
 
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