AC circuit analysis -- mesh and nodal

[Hndstudent]

Surely it would just be some algebra? Your v20 only appears once in the expression so you should be able to move its term to one side of the equation, then proceed as usual (straightforward if tedious complex number work).

That answer was obviously very wrong. however, this is where I am at but i am still getting an incorrect answer. can you advise me on where I am going wrong please

-v20(0.75+j0.05)+(67.071+j30)=0
(67.071+j30)=v20(0.75+j0.05)

divide both sides by (o.75+j0.05);

[(67.071+j30)/(0.75+j0.05)]=v20

I= v20/z4

but I'm not getting the same answer as in part A

 

[gneill]

It's difficult to jump into the middle of a derivation that uses numbers rather than symbols. Can you back up a bit and tell me what v20 is (which node it is in the circuit) and show your initial node equation? (I'm presuming that you're using a supernode thanks to the presence of V3).

 

[Hndstudent]

It's difficult to jump into the middle of a derivation that uses numbers rather than symbols. Can you back up a bit and tell me what v20 is (which node it is in the circuit) and show your initial node equation? (I'm presuming that you're using a supernode thanks to the presence of V3).

hi gneill, i started off with

((v1-v20)/z1) + (-v20/z4) + ((-20-v3)/z5) + (v2-(v20-v3)/z3)=0

thanks

 

[gneill]

Okay, so you're summing currents into the supernode. I think there's a sign issue with the third term. I'm seeing

$\frac{0 - (v20 - v3)}{z5} = \frac{-(v20 - v3)}{z5} = -\frac{v20 + v3}{z5}$

 

[Hndstudent]

Okay, so you're summing currents into the supernode. I think there's a sign issue with the third term. I'm seeing

$\frac{0 - (v20 - v3)}{z5} = \frac{-(v20 - v3)}{z5} = -\frac{v20 + v3}{z5}$

Hi gneill, when you say there is a sign issue with the 3rd term, are you saying that (-(v20+v3)/z5) is incorrect?

this is what i have so far, and I have tried changing the signs but I keeping on getting a wrong answer.

-v20(1/z1 + 1/z4 + 1/z5 + 1/z3) + ( v1/z1 + v3/z5 + v2/z3 +v3/z3)=0

does this look correct?

thanks

 

[gneill]

Yes, I think that looks good.

 

[Hndstudent]

Yes, I think that looks good.

Im obviously going wrong somewhere when calculating the brackets.

Im ending up with v20= (67.071+j30)/(0.75+j0.05)

is it my numerator or denominator that is incorrect? or both!?

thanks

 

[gneill]

Sign issue in the denominator.

 

[macca67]

Hi Folks,

Sorry to revive an old thread but I am on the same unit and the notes are not helping me much!

What I have so far is -

Mesh 1
V1-(Vz1)-(Vz4)=0
V1-(I1Z1)-((I1-I2)Z4)=0
V1-I1(Z1-Z4)+(I2Z4)=0
120-I1(2+j5)-I2(-j5)=0

Mesh 2
-V3-Vz5+Vz4=0
-V3-Z5(I2-I3)+Z4(I2-I1)=0
-V3-(Z5I2)+(Z5I3)+(Z4I2)-(Z4I1)=0
-V3-(j4I2)+(j4I3)+(j5I2)-(j5I1)=0
-V3-(j1I2)+(j4I3)-(J5I1)=0

Mesh 3
-(Vz3)-V2-(Vz5)=0
-(I3Z3)-V2-(I3-I2)Z5=0
-(I3Z3)-V2-(Z5I3)+(Z5I2)=0
-I3(Z3-Z5)-V2+Z5I2=0 (Z5 term moved to the left end in next step)
(I2j4)-I3(4-j4)-V2=0

When I put the above into my spreadsheet (same excel one as posted earlier) I am not getting the answer I expected. I have done the nodal analysis (next question) but this doesn't fit with that answer.

Can anyone spot any obvious mistakes in my loop walks?

Thanks in advance
Mac

 

[gneill]

Hi Folks,

Sorry to revive an old thread but I am on the same unit and the notes are not helping me much!

What I have so far is -

Mesh 1
V1-(Vz1)-(Vz4)=0
V1-(I1Z1)-((I1-I2)Z4)=0
V1-I1(Z1-Z4)+(I2Z4)=0$~~~~~~~$← Sign issue with the I1 term: Why is Z4 negative?
120-I1(2+j5)-I2(-j5)=0

Mesh 2
-V3-Vz5+Vz4=0$~~~~~~~~$ ← Sign issue with the Vz4 term: Why is Vz5 negative and Vz4 positive?
-V3-Z5(I2-I3)+Z4(I2-I1)=0$~~~~~~~$← Sign issue with the Z4 term
-V3-(Z5I2)+(Z5I3)+(Z4I2)-(Z4I1)=0
-V3-(j4I2)+(j4I3)+(j5I2)-(j5I1)=0
-V3-(j1I2)+(j4I3)-(J5I1)=0

Mesh 3
-(Vz3)-V2-(Vz5)=0
-(I3Z3)-V2-(I3-I2)Z5=0
-(I3Z3)-V2-(Z5I3)+(Z5I2)=0
-I3(Z3-Z5)-V2+Z5I2=0 (Z5 term moved to the left end in next step)
(I2j4)-I3(4-j4)-V2=0

When I put the above into my spreadsheet (same excel one as posted earlier) I am not getting the answer I expected. I have done the nodal analysis (next question) but this doesn't fit with that answer.

Can anyone spot any obvious mistakes in my loop walks?

Thanks in advance
Mac

See the red flagged lines.

 

[macca67]

Hi gneil,

thanks for the reply.

In Mesh one I have tried to justify my thinking but see the error now, thanks.

In mesh two I put Vz4 as a positive as I think the +ve term would be at the top of z4. I started top left of the mesh and walked clockwise, -v3 as I come to the + side first, same for Vz5, for Vz4 I thought the side I would meet first would be -ve, so gave it a +ve in the initial walk round.

In mesh 1 I had Vz4 as +ve at the top, does it need to stay +ve at the top when I walk mesh 2?

 

[gneill]

What you need to consider when you "walk" over a component is whether or not your walk is in the same direction as the current that pertains to the term you are writing.

Before you took your KVL walk you assigned current directions for the mesh currents. In this case they are all clockwise. When you walk over a component in the same direction as a current you will have a potential drop due to that current. When you walk against a current flow you have a potential rise.

Consider mesh 2. The mesh current is clockwise and if you walk the loop clockwise all the resistors encountered will show a potential drop for that mesh current. This includes Z4. So you must find the drops: I2(Z4 + Z5). The border components of the mesh will have the currents of the bordering meshes flowing in the opposite direction to the mesh 2 current and your "walk" will be against the directions of those currents. So for mesh 2 you will see potential rises of I1Z4 and I3Z5.

When in doubt, on the schematic sketch in the potential changes for all the assumed currents. That should help when writing out the equations:

By inspection then:

+(I1)(Z4) - (I2)(Z4 + Z5) + (I3)(Z5) = 0

 

[macca67]

Got to the bottom of it, many thanks for your help!

 

[frenchy59]

does I = -6.84008+j17.3988?
for nodal way

 

[frenchy59]

Sorry made a mistake in calculation so i have :
V20(0.75+J0.05)=67071+J30
V20=91.6872+J33.875
V20/Z4=I
I=91.6872+J33.875/-J5
I=-6.7775+J18.3374

 

[The Electrician]

In post #118 you said you had "got to the bottom of it". Does that mean that you derived a correct solution using mesh analysis? If so, wouldn't you think the value of I would be the same for nodal analysis? So why are you asking if the value you derived using nodal analysis is correct? Why don't you just compare the value you got with nodal analysis to the value you got with mesh analysis?

Do you see why I'm asking these questions? The very fact that you're asking about your nodal solution makes me wonder if you got the correct value from your mesh analysis.

What solution for the currents did you get with mesh analysis? What did you get for the value of I?

You don't have the correct value for V20, and probably not for V30 either although you didn't show what you got.

It makes it easier for those who would help you if you show your work.

 

[frenchy59]

Dear The electrician,
post 118 wasn t mine, i m currently offshore and my pc won t allow me to use the matrix for some reason (administrator issue, i m currently offshore), that s why i was wondering if the result was good or not. but i understand where you coming from i will post the analysis later on, thank you taking the time.

 

[The Electrician]

Sorry I didn't notice that there two different posters. As soon as you post your work, we'll help you out.

 

[macca67]

Sorry made a mistake in calculation so i have :
V20(0.75+J0.05)=67071+J30
V20=91.6872+J33.875
V20/Z4=I
I=91.6872+J33.875/-J5
I=-6.7775+J18.3374

Your final answer is different to mine, looking at V20(0.75+j0.05)=67.071+j30, you have a sign different somewhere in the build up to this.

I don't want to post exactly what as I am new to the forum and I could be wrong and I am not sure what is acceptable to post in regards to solutions.

In the previous post to above you said this was for nodal analysis, but I would imagine I = I, which ever way we calculate it, so compare the results to your mesh analysis should help.

The notes aren't much help are they haha!

 

[macca67]

Also I couldn't get the spreadsheet to work on my work pc, used wolfram instead

 

[The Electrician]

This is a long thread, and correct answers have been given in early postings. So there should be no question as to what the correct answer is, but if recent posters can't get the correct answer and want help to find their error, they will have to post their work.

macca67, did you get the correct answer with both mesh and nodal analysis?

frenchy59, we'll wait for you to post your work.

 

[macca67]

Yeah going on the post on the first page I got it right (Thanks to some coaching by gneill).

 

[Stephen Forster]

Hi I am looking for guidance
I am happy with my 4 loops, however which is the best way to find current I1 I2 and I3 is it by deterimnents 3*3 or by a spreadsheet from what I have seen on this forum?

 

[gneill]

Hi I am looking for guidance
I am happy with my 4 loops, however which is the best way to find current I1 I2 and I3 is it by deterimnents 3*3 or by a spreadsheet from what I have seen on this forum?

I'm not sure that you can declare a "best" method. Use whatever tools you have to hand and are comfortable with.

 

[frenchy59]

ok , i still have problem with spreadsheet but it look
R j I v
2 0 0 -5 5 0 120 0
0 0 0 5 1 -4 -14.14 + j -14.14
0 0 4 0 -4 4 0 -120
But unfortunatly the result i get don t make sens
for nodal analysis i ve got

(V1-V20/Z1)+(-V20/Z4)+(-(V20-V3)/Z5)+(V2-(V20-V3)/Z3)

-V20((1/Z1)+(1/Z4)+(1/Z5)+(1/Z3))+((V1/Z1)+(V3/Z5)+(V2/Z3)+((V3/Z3)=0

33.535+J33.535/J5
I=-6.707+J6.707

 

[frenchy59]

R j I v
2 0 0 -5 5 0 120 0
0 0 0 5 1 -4 -14.14 + j -14.14
0 0 4 0 -4 4 0 -120
LOOK BETTER WITH SPACES

 

[frenchy59]

trying to put my spreadsheet up but when i post it it cancel the spaces sorry

 

[frenchy59]

Can the result I in nodal analysis can be -7.24761+j25.4258
This is not going well!

 

[NascentOxygen]

trying to put my spreadsheet up but when i post it it cancel the spaces sorry

If you place text inside a pair of [CODE] ... [/CODE] tags it will prevent the compression of spaces. (You can find CODE under the + on the menu bar.)

It will then preserve spacing, as here:

1       3.14159          2.71828

A bit messy, but that's the best we can do, I believe.

(Otherwise, of course, you could save as a jpeg or take a screenshot and then attach that image here.)

 

[frenchy59]

Thank you for your help.

 

[David J]

Thanks for the help, for some reason. the spreadsheet given to me just is not calculating the mesh currents. i have tried an internet calculator and it calculates it just fine.

Hi, I am working on this problem at the moment and experiencing the same problem I think you and many others had with the spread sheet. Do you mind if I ask which online calculator you eventually ended up using to get your results ??

 

[David J]

Leave $V_3$ as $V_3$ for now! That way you don't have to carry around any complex digits through the initial simplifications. Don't be in a hurry to plug in numbers, especially complex numbers; That just gives more opportunities to make transcription and sign errors, and makes it more difficult to spot mistakes in the workings.

You should then have:

$\frac{V_1 - V_{20}}{Z_1} + \frac{-V_{20}}{Z_4} + \frac{-(V_{20} - V_3)}{Z_5} + \frac{V_2 - (V_{20} - V_3)}{Z_3} = 0$

which becomes, after distributing the signs:

$\frac{V_1 - V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + \frac{V_3 - V_{20}}{Z_5} + \frac{V_2 - V_{20} + V_3}{Z_3} = 0$

Can you carry on from there to isolate $V_{20}$? You should be able to reach a point where you have something like:

$V_{20}[$ some terms $] + [$ more terms $] = 0$

Then you can work on reducing the "some terms" and "more terms" down to individual complex numbers, since then all the variables in them will be known values.

Some sign errors snuck in. So no, not so good

Hi, I am trying to understand and work my way through this question using the info posted in this forum. Its very useful for me but a quick question. How does $\frac{-(V_{20} - V_3)}{Z_5}$ become $\frac{V_3 - V_{20}}{Z_5}$ ? I can't understand how it went from 20-3 to 3-20

 

[gneill]

Hi, I am trying to understand and work my way through this question using the info posted in this forum. Its very useful for me but a quick question. How does $\frac{-(V_{20} - V_3)}{Z_5}$ become $\frac{V_3 - V_{20}}{Z_5}$ ? I can't understand how it went from 20-3 to 3-20

The sign outside the parentheses gets distributed over the contents of the parentheses (so the signs of the terms in the parentheses get reversed).

 

[David J]

Ok i see this now in my working out, thanks

 

[GeorgeSparks]

Leave $V_3$ as $V_3$ for now! That way you don't have to carry around any complex digits through the initial simplifications. Don't be in a hurry to plug in numbers, especially complex numbers; That just gives more opportunities to make transcription and sign errors, and makes it more difficult to spot mistakes in the workings.

You should then have:

$\frac{V_1 - V_{20}}{Z_1} + \frac{-V_{20}}{Z_4} + \frac{-(V_{20} - V_3)}{Z_5} + \frac{V_2 - (V_{20} - V_3)}{Z_3} = 0$

which becomes, after distributing the signs:

$\frac{V_1 - V_{20}}{Z_1} - \frac{V_{20}}{Z_4} + \frac{V_3 - V_{20}}{Z_5} + \frac{V_2 - V_{20} + V_3}{Z_3} = 0$

Can you carry on from there to isolate $V_{20}$? You should be able to reach a point where you have something like:

$V_{20}[$ some terms $] + [$ more terms $] = 0$

Then you can work on reducing the "some terms" and "more terms" down to individual complex numbers, since then all the variables in them will be known values.

Some sign errors snuck in. So no, not so good

Hi Gneill, can you recommend any good reading material on AC circuit nodal circuit analysis with complex numbers? I have been trawling the net searching for examples incorporating the super node as in part b) but have yet to find anything!
I would like to try and fully understand the method before I go ahead and use it.
Many thanks