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AC circuit analysis -- mesh and nodal

  1. Jan 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine, using the values given in TABLE A, the current I in the circuit of FIGURE 2 by:
    (a) mesh analysis
    (b) nodal analysis.

    Any help verify for part (a)? And to help me get started with (b) as V3 is driving me nuts.

    2. Relevant equations

    3. The attempt at a solution
    For (a) my equations are
    120=2I1+(I1-I2)(-j5) ---Loop I1
    (-j5)(I2-I1)+(-j5)(I2-I4)+(j4)(I2-I3)=0 ---Loop I2
    (j4)(I3-I2)+(4)(I3)=j120 ---- Loop I3
    -14.14+j14.14=(-j5)(I4-I2) ---Loop I4

    I1=23.542+j17.5695 A
    I2=16.5142+j2.98628 A
    I3=21.764+j24.7503 A
    I4=13.6862+j5.81428 A

    Attached Files:

  2. jcsd
  3. Jan 11, 2015 #2


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    Staff: Mentor

    Note that V3 creates a fixed potential difference across Z2. This means for loop 2 you can forget about Z2 and just use that known potential difference in its loop equation. Since you're looking for current ##I##, you won't even need to include loop 4 in your analysis (or if you like, exchange the positions of V3 and Z2 on the diagram. Since their node connections remain the same as before, the circuit operation does not change at all and the current through Z2 becomes obvious...).
  4. Jan 11, 2015 #3
    Thankyou for your kind assistance, I have little confidence with circuits. I did have a suspicion that the current through Z2 would be I=V3/Z2. I will attempt another solution.
  5. Jan 11, 2015 #4
    Hello, I swapped Z2 with V3 and came up with the following newer equations.;
    Loop 1 --- 120-(2)I1-(I1-I2)(-j5)=0
    Loop2 --- (-j5)(I2-I1)+(j4)(I2-I3)-14.14+j14.14=0
    Loop3 --- (j4)(I3-I2)+(4)I3-j120=0

    I2= 14.1367-j3.77088
    I3= 23.9538+j20.1829

    Not 100% percent sure how what to do with those loop currents to get current through Z4.

    Part (b)(Updated PDF file detailing labelled nodes.)
    V20-V30=14.14+j14.14 (supernode)
    and ((120-V20)/2)+(0-V20)/-j5)+(0-V30)/j4)+(j120-V30)/4=0
    So..., V20= 86.3761+j45.7584 and V30=72.2361+j31.6184

    Current I is V20/Z4= -9.15168+j17.2752 A

    Am I alright with my approach so far?

    Attached Files:

  6. Jan 11, 2015 #5


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    Staff: Mentor

    You need to be a bit careful with the voltage supplies when you walk the loop. If you're summing potential drops (as positive numbers) then passing through a voltage supply from + to - counts as a positive drop. If you're summing potential rises, then each potential change through a resistance in the direction of the current is a negative value, and passing through a voltage source from + to - counts as a negative change, too.

    So take another look at your Loop 3 equation. It looks like you're summing drops as positive values as you "walk" clockwise around the loop, but then you make the drop across V2 negative...

    For Loop 2, I don't understand how your V3 components ended up with different signs.

    Your nodal analysis looks fine to me. You should be able to compare results between mesh and nodal if you determine ##I## by both methods.
  7. Jan 12, 2015 #6
    Ok, for loop analysis I have, Current I is I1 - I2 = (16.8119-22.8792i)-(25.9636-40.1544i)=-9.1517+j17.2752

    Mesh analysis= Current I is V20/Z4= -9.15168+j17.2752 A

    Sounds good to me dont you think?
  8. Jan 12, 2015 #7


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    Staff: Mentor

    it not only sounds good, but it looks good too :)
  9. Jan 12, 2015 #8
    Thanks for your help!:)
  10. Apr 1, 2015 #9
    Hi Folks, Could anyone help please? I'm trying to find the same loop currents as His_Tonyness and this is what I have so far

    LOOP 1 V1=Z1(i1)+Z4(i1-i2) = 120+J0 = i1(2-J5)-i2(0-J5)
    LOOP 2 -V3=Z5(i2-i3)+Z4(i2-i1) = -14.14+J14.14 = -i1(0-J5)+i2(0-J1)-i3(0+j4)
    LOOP 3 -V2=Z5(i3-i2)+Z3(i3) = - 0+J120 = -i2(0+J4)+i3(4+J4)

    When I plug these into a spreadsheet they don't work out when putting them back into the equation......

  11. May 21, 2015 #10
    Hi Guys, i've got the same as above^^; but am not sure on how to proceed with my mesh analysis - how do i solve the system of equations?
  12. May 21, 2015 #11


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    Staff: Mentor

    Exactly the same? Because I've spotted a sign error in justadaftspark's second loop equation's numerics (the symbolic version is fine).

    Once that's sorted out, use whatever technique you're familiar with to solve the three equations in three unknowns.
  13. May 21, 2015 #12
    No not exactly, i've factored the equation differently. but no worry! in the mean time i'd had a chat with my work colleague who's on the same course and between us we've managed to figure it out. :) thanks!
  14. May 21, 2015 #13
    Now we're working on part b and are stumped. not sure where to begin with our reference node?
  15. May 21, 2015 #14


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    Staff: Mentor

    You'll have to show more effort there. Have you identified the essential nodes at least?
  16. May 21, 2015 #15
    Sorry, i thought this forum was for guidance? or am i mistaken? I'm not looking the the whole answer obviously, just a bit of guidance would be appreciated.

    The two nodes connecting Z4, Z2 and Z3 are both essential, along with the lowermost. but they all appear to have four branches. normally we'd have gone for the node with the most branches? Or do we not include the branch to V3?
  17. May 21, 2015 #16


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    Staff: Mentor

    The forum is indeed for guidance. That does not include handing out answers, even bit by bit. The idea is to guide you towards solving the problem by your own efforts. By the forum rules, "I'm stumped" or "I don't know where to begin" are not counted as showing effort. Hints and suggestions are given once you've demonstrated your effort and ideas.
    Choosing a reference node is sometimes a bit of an art if you have a particular potential or current to solve for and a particular choice of reference node happens to make that simpler. But for the most part you choose a node with the most connections as you say. When number of connections isn't a deciding factor then you need to go by instinct which develops with practice. Often the person who designed the circuit makes the choice obvious by drawing the circuit with a practical reference node at the bottom.

    Now for a hint. The V3 source connecting the two upper nodes forces your hand in how you approach this problem using nodal analysis. The two nodes have a fixed potential difference, so they cannot be independent nodes. Check your course notes or text for the Supernode concept.
  18. May 22, 2015 #17
    If you do this, which way is the voltage going? In the original circuit the voltage is going in an anti-clockwise direction (as indicated by the arrow above it) - if we swap V3 and Z2 do we kept the voltage going in an anti-clockwise direction - ie arrowhead pointing towards Z3 - or do we effectively pull the symbol straight down and keep it in the same direction - ie arrowhead pointing towards Z1 and the voltage flowing into the same node?

  19. May 22, 2015 #18
    My incorrect working out so far looks like this:

    Loop 1:

    V1 - Z1I1 - Z4 (I1 - 2) = 0

    V1 - (Z1 + Z4) I1 + Z4I2 = 0

    V1 = (Z1 + Z4) I1 - Z4I2

    120 = (2 + (-j5)) I1 - (-j5)I2

    120 = (2 - j5) I1 - (-j5) I2

    Loop 2:

    -V3 - Z5 (I2 - I3) - Z4 (I2 - I1)

    -V3 + Z4 I1 - (Z5 + Z4) I2 + Z5 I3

    (-14.142 + j14.142) + (-j5) I1 - (j4 + (-j5)) I2 + (j4) I3

    -14.142 + j14.142 = - (-j5) I1 + (-j1) I2 - (j4) I3

    Loop 3:

    -V2 - Z5 (I3 - I2) - Z3 I3

    -V2 - Z5 I2 - (Z5 + Z3) I3

    -j120 + (j4) I2 - (j4 + 4) I3

    -j120 = -(j4) I2 + (4 + j4) I3

    Now i know that's incorrect as when i plug it into my spreadsheet it wont have it - can anyone tell me where it goes astray? I imagine that it's an error with signs somewhere.

    Last edited: May 22, 2015
  20. May 22, 2015 #19


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    Staff: Mentor

    All connections must remain the same in order to preserve the circuit topology. So the voltage source connections must remain as before as well. That is, the polarity arrow must point to the same node as it did before.
  21. May 22, 2015 #20
    That's what i suspected, thanks.

    I see on loop 3 that V2 should have been -V2, but that still isnt correct by the looks of it.

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