AC circuit analysis -- mesh and nodal

[brabbit87]

If you do a KVL "walk" around the second loop in a clockwise fashion, does v3 cause a potential rise or a potential drop?

V3 will cause a potential drop due to assumed current (clockwise) and it passing from + to - on the source.

 

[brabbit87]

I have redrawn the polarity of impedances and voltage sources in the attached image.

Mesh 1 -

v1 - (z1)(I1) - (z4)(i1 - i2) = 0

v1 - (2)(i1) + (5j)(i1) -(5j)i2 = 0

120 = (2 - 5j)i1 + (5j)i2

Mesh 2

-v3 - (z5)(i2 - i3) - (z4)(i2 - i1)

-v3 -(4j)(i2) +(4j)(i3) +(5j)(i2) - (5j)(i1)

-v3 = +(4j)(i2) - (4j)(i3) -(5j)(i2) +(5j)(i1)

-14.14 - 14.14j = + (5j)(i1) - (1j)(i2) -(4j)i3


Mesh 3 -

-v2 - (z5)(i3 - i2) - (z3(i3)

-v2 -(4j)(i3) +(4j)(i2) - (4j)(i3)

-120j = -(4j)(i2) + (4 +4j)(i3)the answers are slightly diffrerent from last time but still no joy in the spreadsheet

 

[brabbit87]

wow... your equation for mesh two is not correct.
are you probing for someone to do just your work for you?
you have been told multiple times that there is a problem with your equation for mesh 2. Its hard to be more clear.

If you do a KVL "walk" around the second loop in a clockwise fashion, does v3 cause a potential rise or a potential drop?

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.

 

[ProfNut]

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.

I had a similar problem with the spreadsheet.

I'm currently working through the nodal analysis..

I have got to..

$V_20(1/Z_1+1/Z_4+1/Z_3+1/Z_5)+(V_1/Z_1-V_3/Z_3-V_3/Z_5)$

is this correct...I'm struggling to simply further. My next step is to solve for values in brackets. Minus second brackets, divide by 1st brackets... I hope that's clear?! am i on the right lines?

 

[brabbit87]

I had a similar problem with the spreadsheet.

I'm currently working through the nodal analysis..

I have got to..

$V_20(1/Z_1+1/Z_4+1/Z_3+1/Z_5)+(V_1/Z_1-V_3/Z_3-V_3/Z_5)$

is this correct...I'm struggling to simply further. My next step is to solve for values in brackets. Minus second brackets, divide by 1st brackets... I hope that's clear?! am i on the right lines?

Like gneil had said to me previous to now. I had mixed my directions up going into the supernode.

I drew current arrows into the supernode and did the 4 equations based on that. And simplified. Bearing in mind you need an initial equation for v30 to substitute into it.

Hope it helps

 

[ProfNut]

Like gneil had said to me previous to now. I had mixed my directions up going into the supernode.

I drew current arrows into the supernode and did the 4 equations based on that. And simplified. Bearing in mind you need an initial equation for v30 to substitute into it.

Hope it helps

My initial equations:

$(V_20-V_1/Z_1)+(V_20/Z_4)+(V_30-V_2/Z_3)+(V_30/V_5)=0$
And;
$V_30=V_20-V_5$

Subsitute;

$(V_20-V_1/Z_1)+(V_20/Z_4)+(V_20-V_5/Z_3)-(V_2/Z_3)+(V_20-V_5/Z_5)-(V_3/z_5)=0$

 

[brabbit87]

My initial equations:

$(V_20-V_1/Z_1)+(V_20/Z_4)+(V_30-V_2/Z_3)+(V_30/V_5)=0$
And;
$V_30=V_20-V_5$

Subsitute;

$(V_20-V_1/Z_1)+(V_20/Z_4)+(V_20-V_5/Z_3)-(V_2/Z_3)+(V_20-V_5/Z_5)-(V_3/z_5)=0$

Can't really see the equation as I'm on the phone app. But you know the current from the mesh analysis. So you can compare and tinker.

 

[Hndstudent]

Hi gneill, I am having a little trouble with the matrix spread sheet when entering my values. I have read through this forum and notice that my answers are the same as that in post #31 and #33 which you say are correct. But when i put these figures into the matrix spread sheet it is incorrect. I noticed in post #73 that you suggested in loop 2 the -jI2 should in fact be a jI2. my calculations for loops two are as below;

-(14.142-j14.142)+(-j5)I1-(j4+(-j5)I2+j4I3

(-14.142-j14.142)=j5I2-jI2-j4I3

Thanks

 

[gneill]

HI Hndstudent. As far as I can see the values that you show entered in the spreadsheet look fine. There should be no reason why it cannot solve it assuming that the spreadsheet itself is properly coded.

I see that the spreadsheet displays det(R) = 0 and det(X) = 0. If the spreadsheet is rejecting the problem because of zero determinants here, then it is incorrect. For while those determinants are indeed zero, it is more important that det(R + jX) be nonzero, which it is.

 

[Hndstudent]

HI Hndstudent. As far as I can see the values that you show entered in the spreadsheet look fine. There should be no reason why it cannot solve it assuming that the spreadsheet itself is properly coded.

I see that the spreadsheet displays det(R) = 0 and det(X) = 0. If the spreadsheet is rejecting the problem because of zero determinants here, then it is incorrect. For while those determinants are indeed zero, it is more important that det(R + jX) be nonzero, which it is.

I gneill, the spreadsheet is off the Universities Blackboard. Would you recommend that I contact them and inform them that there is a problem with it? Or is there something that i can do to fix it?

Thanks

 

[gneill]

I gneill, the spreadsheet is off the Universities Blackboard. Would you recommend that I contact them and inform them that there is a problem with it? Or is there something that i can do to fix it?

Thanks

I'm not familiar with the spreadsheet so I don't know how its implemented, or what Excel skill set would be required to debug it. It may be doing the heavy lifting in macros. It may be best to contact the University to see if they've had other reports on it.

Myself I use MathCad to solve these sorts of things.

 

[Hndstudent]

I'm not familiar with the spreadsheet so I don't know how its implemented, or what Excel skill set would be required to debug it. It may be doing the heavy lifting in macros. It may be best to contact the University to see if they've had other reports on it.

Myself I use MathCad to solve these sorts of things.

Ok thanks gneil. Whilst i have been waiting for the university to reply regarding the matrix, I was working through my equations for each loop again and for loop1 I have;
v1-z1I1-z4(I1-I2)=0

if z1=2 , z4=-j5 and v1=120

then 120=(2+j5)I1-j5I2 (is this not correct?)

as before I had 120=(2-j5)I1-j5I2.

Thanks

 

[gneill]

v1-z1I1-z4(I1-I2)=0

$120 - 2 I_1 - (-j5)(I_1 - I_2) = 0$
$120 - (2 - j 5) I_1 - j5 I_2 = 0$
$120 = (2 - j5) I_1 + j5 I_2$

 

[Hndstudent]

$120 - 2 I_1 - (-j5)(I_1 - I_2) = 0$
$120 - (2 - j 5) I_1 - j5 I_2 = 0$
$120 = (2 - j5) I_1 + j5 I_2$

Cheers gneil. The university confirmed what you said about the spread sheet and it eventually worked :)

however, for my next problem on 2b, I have factorised and multiplied by conjugates to end up with the following;

v20(-0.1667-j0.028)+(17.896+j7.670)=0

Im just not sure how to get v20 on its own?

Thanks

 

[gneill]

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).

 

[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