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Homework Help: Node analysis

  1. Sep 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine the voltage across all elements

    2. Relevant equations

    3. The attempt at a solution
    I'm not sure which of the nodes I should use to take my KVL from. The a and b are labeled, but should I combine the set of 2-6-6 resistors and label that as an essential node? Should I have four essential nodes?

    Attached Files:

  2. jcsd
  3. Sep 18, 2015 #2


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    You can use KVL ( loops ) or KCL ( nodes ).

    So if you use KVL, you must sketch three closed loops ( with a positive direction ), like the two loops, S1 and S2 here:

    Kirchhoff says that the sum of voltage changes along a closed loop = 0

    So as for S1, you could write:

    ε1 - S1*R1 - S1*R2 + S2*R2 = 0

    Write another equation as for S2.

    Having solved the equations, you find: i2 = S1 - S2 , and so on.
    Last edited: Sep 18, 2015
  4. Sep 18, 2015 #3


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    There are olny unknown three nodes (four if you include ground). at two of the nodes you'll need to do a supernode if you use KCLs

    Also you say you need to do node analysis. If that's the case follow Hesche's advice. If you need to do NODAL ANALYSIS as your method, then use KCLs.
  5. Sep 23, 2015 #4
    It says I must use the node-voltage method. I'm confused what to do when there's a voltage source already at the node. Am I suppose to take 3 different super nodes?
  6. Sep 23, 2015 #5


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    Staff: Mentor

    I don't see any supernode situations here. Just three essential nodes:
    The ground (reference) node was already provided.

    You should be able to write node equations for the three nodes.
  7. Sep 23, 2015 #6
    Okay, I made two formulas with node 1) and node 2)

    Node 1)
    (v1-12)/2 + (v1-v3)/6 + (v1-v2)/2

    Node 2) (v2-v1)/2 + (v2-v3)/6 + (v2-12)/2

    Node 3) seems to be (v3-v1)/6+12/6+(v3-v2)/6


    I set v1 as node 1, v2 as node 2, and v3 as node 3.

    Is there a way I can verify my answer by myself?
  8. Sep 23, 2015 #7


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    Staff: Mentor

    Check that middle term. It should involve v3.
    Usually one would insert the values and confirm consistency, say by verifying that KCL is satisfied at each node using the voltages solved for, or, using another method entirely to see if you get the same results (mesh analysis perhaps).
  9. Sep 23, 2015 #8
    I'm trying to solve for all the currents in the ciruit and prove power delivered=power absorbed

    (v1-12)/2 =-1 a
    (v1-v3)/6 =1 a
    (v1-v2)/2 = 0 a(?)

    (v2-v1)/2 =0 a
    (v2-v3)/6=1 a
    (v2-12)/2= -1 a

    (v3-v1)/6=-1 a
    12/6=2 a
    (v3-v2)/6=-1 a

    I have two values of zero for the resistor between v1 and v2.

    power delivered = 12*1a + 12*1a + 12*2a = 48 W
    power absorbed = 1^2*2ohm + 1a^2*6ohm + 1^2a*6ohm + 2a^2*6ohm+1^2*2ohm +2a^2*2ohm = 48 W

    I reuploaded the picture with all the values labeled , does this look correct?

    Attached Files:

  10. Sep 23, 2015 #9


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    Staff: Mentor

    Zero volts and zero current for the resistor between nodes 1 and 2 is to be expected by the symmetry of the circuit. Symmetry should also tell you something about the directions of certain current pairs.

    You still haven't fixed that middle term in the node 3 equation. Fix it and re-solve.
  11. Sep 23, 2015 #10
    (v1-12)/2 + (v1-v3)/6 + (v1-v2)/2
    (v2-v1)/2 + (v2-v3)/6 + (v2-12)/2


    I also tried solving another way,

    (v1-12)/2 + (v1-v3)/6 + (v1-v2)/2
    (v2-v1)/2 + (v2-v3)/6 + (v2-12)/2

    v1=48/5, v2=48/5, v3=12/5

    I think I did something wrong, I'm suppose to be calculating the current for the entire circuit so I can solve for power absorbed=power delivered

    power delivered = 12*1a + 12*1a + 12*1a = 36 W
    power absorbed = ? I'm not sure how to get the currents/voltages for the individual pieces. According to my equations most of them will be zero now, since they are all 12-12.
  12. Sep 24, 2015 #11


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    Staff: Mentor

    The (v3-12)/6 term is not correct; be careful with the polarity of the source in that leg! It's not the same as that of the other voltage sources. (I see that you've fixed that in your next attempt below)
    Those values look good :smile:
    The individual terms in your node equations represent the currents in the branches that they were written for. So you can plug in your v1, v2, v3 values to those terms and pull out the individual branch currents. Use those currents to calculate the powers produced or consumed by each of the components.
  13. Sep 25, 2015 #12


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    Have you considered simplifying the circuit using symmetry? Sorry if this has already been discussed. The internet connection from this hospital bed isn't great so I type in haste.
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