AC circuit analysis -- mesh and nodal

  • #1

Homework Statement


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.

Homework Equations




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

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

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Answers and Replies

  • #2
gneill
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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...).
 
  • #3
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...).
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.
 
  • #4
Hello, I swapped Z2 with V3 and came up with the following newer equations.;
(a)
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

Answers;
I1=19.1624+j12.5642
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?
 

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  • #5
gneill
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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.
 
  • #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?
 
  • #7
gneill
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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?
it not only sounds good, but it looks good too :)
 
  • #8
Thanks for your help!:)
 
  • #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......


Thanks
 
  • #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?
 
  • #11
gneill
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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?
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.
 
  • #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!
 
  • #13
Now we're working on part b and are stumped. not sure where to begin with our reference node?
 
  • #14
gneill
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Now we're working on part b and are stumped. not sure where to begin with our reference node?
You'll have to show more effort there. Have you identified the essential nodes at least?
 
  • #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?
 
  • #16
gneill
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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 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.
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?
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.
 
  • #17
Gremlin
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(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...).
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?

Thanks.
 
  • #18
Gremlin
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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.

Thanks.
 
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  • #19
gneill
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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?

Thanks.
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.
 
  • #20
Gremlin
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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.

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.

Thanks.
 
  • #21
gneill
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You have a sign problem with the V3 voltage source components in Loop 2.

You are right when you state that V2 should have been -V2 in the Loop 3 equation.
 
  • #22
Gremlin
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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
I honestly cant see where.

V3 = 20∠45 which is 14.142 + j14.142.

If we start at midnight in the loop you go against V3 so -14.142 + j14.142, then you walk into Z5 so -(0+j4) x (I2 - I3) and finally you walk into Z4 so - (0-j5) x (I2 - I1).

Do we have to include Z2 somehow?
 
  • #23
gneill
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I honestly cant see where.

V3 = 20∠45 which is 14.142 + j14.142.
Yes.
If we start at midnight in the loop you go against V3 so -14.142 + j14.142,
No. Why do you negate only the real term of the complex value? It's all one number!
Do we have to include Z2 somehow?
Nope. It belongs to a separate loop and is not in the path of the current in loop 2.
 
  • #24
Gremlin
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No. Why do you negate only the real term of the complex value? It's all one number!
Because i dont fully understand complex numbers i suppose is the answer. I cant wait to see the back of them. Although i look at part b of this question and i can see that i'm going to have to simplify a simultaneous equation full of them.... great.

Still, even if i change it to -14.142 - j14.142 in my complex matrix spreadsheet it still tells me that it can not solve the equation.
 
  • #25
gneill
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Because i dont fully understand complex numbers i suppose is the answer. I cant wait to see the back of them.
Don't hold your breath! Complex numbers are ubiquitous in science and engineering! :wink:
Although i look at part b of this question and i can see that i'm going to have to simplify a simultaneous equation full of them.... great.
You can always simplify symbolically first. Plug in the complex values at the end.
Still, even if i change it to -14.142 - j14.142 in my complex matrix spreadsheet it still tells me that it can not solve the equation.
Then you must have other issues entering the values for the solver. The equations that you presented look fine if you've made the corrections that I pointed out. Failing all else, show us the nine entries of the 3 x 3 impedance matrix.[/quote]
 
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