Engineering AC circuit analysis -- mesh and nodal

[Gremlin]

You're ignoring gneill's advice from post #43, where he said:

"Leave V3 as V3 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."

Is there any further you can go without starting to put the figures in?

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

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

 

[The Electrician]

Is there any further you can go without starting to put the figures in?

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

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

You can't do this:
$- \frac{V_{20}}{Z_1} - \frac{V_{20}}{Z_4} - \frac{V_{20}}{Z_5}- \frac{V_{20}}{Z_3}$ becomes $- \frac{V_{20}}{Z_1 + Z_3 + Z_4 + Z_5}$

Try this with numbers to see if these two expressions are the same.
Is $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}$
the same as
$\frac{1}{3+4+5+6}$ ?

Work it out on your calculator. Then review how to combine fractions in algebra.

 

[The Electrician]

Gremlin, it might be worth your while to review this:
http://www.mathsisfun.com/least-common-denominator.html

 

[Gremlin]

Thanks, i shall have a look.

 

[Gremlin]

Is there any further you can go without starting to put the figures in?

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

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

I've looked at that.

To be honest if i was dividing with real numbers as in the above link i wouldn't have a problem (i should have stopped and thought about adding the denominators in post #51), collapsing the 4 fractions with complex numbers however...

The last hint of an idea that i have is collapsing the 4 equations into 2 equations.

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

$\frac{V_{20}}{2} x2 = \frac{2V_{20}}{4}$

$-\frac{2V_{20}}{4} - \frac{V_{20}}{4} = - \frac{3V_{20}}{4}$

$\frac{V_{20}}{-j5} x4 = \frac{4V_{20}}{-j20}$

$\frac{V_{20}}{j4} x5 = \frac{5V_{20}}{j20}$

$- \frac{4V_{20}}{-j20} - \frac{5V_{20}}{j20} = Math Error(!)$

 

[donpacino]

Is there any further you can go without starting to put the figures in?

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

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

yes. example

$\frac{a}{b}+\frac{c}{d}=0$
$\frac{ad+bc}{bd}=0$

In many cases you want to get it in the form of 1 numerator over 1 denominator.

 

[Gremlin]

Looked hard at this tonight and back at my notes and I've nearly cracked it. Hopefully tomorrow i can put it to bed.

Thanks for all of your help.

 

[MrBondx]

1. 
   Hi guys this is my work for part a. What am I doing wrong. Please help thanks
   
   -120 – 2i1 – (i1 – i2 ( -j5) = 0-120 – 2i1 –(-i1j5 + i2j5 = 0-120 – 2i1 + i1j5 – i2j5 = 0i1(-2+j5) – i2(j5) = 120
   (14.142 + j14.142) – (i2 – i3)(j4) + (i2 – i1)(-j5) = 0(14.142 + j14.142) – i2j4 + i3j4 – i2j5 + i1j5 = 0(14.142 + j14.142) = i2j4 - i3j4 + i2j5 - i1j5 (14.142 + j14.142) = i2j4 - i3j4 + i2j5 - i1j5(14.142 + j14.142) = i2(j4+j5) - i3j4 - i1j5
   -4i3 + j120 +(i3 –i2)(j4) = 0-4i3 + j120 +(i3j4 –i2j4) = 0-4i3 + j120 + i3j4 – i2j4 = 0j120 = 4i3 –i3 – i3j4 + i2j4j120 =i3(4-j4) + i2(j4)
   

 

[MrBondx]

I have brought it down to this but my signs still seem different from other posts on this thread. I don't understand why because I have tried to follow loops round properly.-120 - i1(2 - j5) + i2(-j5) = 0-i1(2 - j5) + i2(-j5) = 120 ……………………………………………… 1i1(-j5) – i2 (j4 – j5) + (14.142 + j14.142) + i3(j4)i1(-j5) – i2 (-j) + i3(j4) = - 14.142 - j14.142 …………….. 2i2(j4) – i3(4 + j4) + j120 i2(j4) – i3(4 + j4) = -j120 ……………………………………………. 3

 

[The Electrician]

I have brought it down to this but my signs still seem different from other posts on this thread. I don't understand why because I have tried to follow loops round properly.

Read post #5

 

[MrBondx]

Read post #5

Thanks

 

[MrBondx]

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

In the first expression isn't the last part supposed to be (V4 - V3) / Z3 . How do we lose V4 to only end up with V2 and V3

 

[MrBondx]

In the first expression isn't the last part supposed to be (V4 - V3) / Z3 . How do we lose V4 to only end up with V2 and V3

Ok am sorted v4 is same as v2, just made sense

 

[gneill]

In the first expression isn't the last part supposed to be (V4 - V3) / Z3 . How do we lose V4 to only end up with V2 and V3

I don't recall any V4 being defined. There was a V₄₀, which was at the junction of Z₃ and V₂. As such, V₄₀ is dropped in favor of using V₂ which has a given value. There was also a voltage V₃₀ defined to be the potential of the node at the top of Z₅. But thanks to the supernode, it is replaced with V₂₀ - V₃.

 

[gneill]

Ok am sorted v4 is same as v2, just made sense

Hah! I guess I took too long preparing my post

 

[MrBondx]

I don't recall any V4 being defined. There was a V₄₀, which was at the junction of Z₃ and V₂. As such, V₄₀ is dropped in favor of using V₂ which has a given value. There was also a voltage V₃₀ defined to be the potential of the node at the top of Z₅. But thanks to the supernode, it is replaced with V₂₀ - V₃.

View attachment 84374

Thanks gneill by V4 I meant V40 will correct that.

 

[MrBondx]

Hah! I guess I took too long preparing my post

In the first expression isn't the last part supposed to be (V4 - V3) / Z3 . How do we lose V4 to only end up with V2 and V3

I'm coming up with

60 + j30 = V20(0.75 + j0.05)

V20 = 76.99 + j45.13

I = V20 / Z4

= (76.99 + j45.13) / -j5

= j15.398 - 9.026

if I compare to part a my j value seems off, where am I going wrong?

 

[Gremlin]

I'm coming up with

60 + j30 = V20(0.75 + j0.05)

V20 = 76.99 + j45.13

I = V20 / Z4

= (76.99 + j45.13) / -j5

= j15.398 - 9.026

if I compare to part a my j value seems off, where am I going wrong?

Your 60 + j30 should be 67.07 + j30 - you're missing a V₃ / Z₅ and a V₃ / Z₃.

 

[MrBondx]

Your 60 + j30 should be 67.07 + j30 - you're missing a V₃ / Z₅ and a V₃ / Z₃.

Your 60 + j30 should be 67.07 + j30 - you're missing a V₃ / Z₅ and a V₃ / Z₃.

It was a sign error in my calculation, that was helpful thanks.

Finally done with this question, thanks yall

 

[ProfNut]

Hi I am stuck on part a of this question:

the 3 mesh equations i have are:

Mesh1:
2I₁-j5I₁+j5I₂=120
Mesh2:
j5I₁-j1I₂-j4I₃=-14.14+14.14
Mesh:3
-j4I₂+4I₃+j4I₃=-120

I have tried to use the matrix spread sheet to find I₁,I₂ & I₃. But my answers seem well out.

9.11E+00+j1.23E+01
4.18E+00-j8.04E+00
-8.89E+00-j1.69E+01

Any help would be much appreciated.

 

[donpacino]

Hi I am stuck on part a of this question:

the 3 mesh equations i have are:

Mesh1:
2I₁-j5I₁+j5I₂=120
Mesh2:
j5I_1-j1I2-j4I₃=-14.14+14.14
Mesh:3
-j4I₂+4I₃+j4I₃=-120

I have tried to use the matrix spread sheet to find I₁,I₂ & I₃. But my answers seem well out.

9.11E+00+j1.23E+01
4.18E+00-j8.04E+00
-8.89E+00-j1.69E+01

Any help would be much appreciated.

what is the matrix spreadsheet??
can you show us what you input to the matrix

 

[ProfNut]

 

[gneill]

Verify the signs and magnitudes for the circled entries.

 

[volvot]

copy that,Thank you for your help.I will try to do as you said.thank you.

 

[ProfNut]

Thanks got it. Now to the Nodal analysis!

 

[brabbit87]

im struggling abit with the nodal analysis. i think i have simplified the equation appropriately.

Equation with substitution for v30: ( v10 - v20/ z1)) + ( v20/ z4) + ( v20 - v3 / z5) + (v2 - v20 + v3/ z3) = 0

i have split the fractions to start isolating v20(v1 / z1) - (v20 / z1) + (v20/z4) + (v20/z5) - (v3/z5) - (v3/z5) + (v2/z3) - (v20/z3) + (v3/z3) = 0

So

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

v20( -(1/2) + (1/-5j ) + (1/4j ) - (1/4) ) + (120/2) - (14.14 + 14.14j /4j) + (120j /4) + (14.14 + 14.14j /4) = 0

v20( - 0.5 + 0.2j + 0.25j - 0.25) + ( 60 - ( 3.535 - 3.535j) + (30j) + (3.535 + 3.535i) = 0

v20( - 0.5 + 0.2j + 0.25j - 0.25) + ( 60 - 3.535 + 3.535j + 30j + 3.535 + 3.535i = 0

am i going the right way about this? The double negatives in the mesh analysis started to boil my head so thought finding the nodal analysis would be more simple, and then i could compare the two answers for the current through z4.

 

[gneill]

im struggling abit with the nodal analysis. i think i have simplified the equation appropriately.

Equation with substitution for v30: ( v10 - v20/ z1)) + ( v20/ z4) + ( v20 - v3 / z5) + (v2 - v20 + v3/ z3) = 0

It looks to me like you're mixing current directions. You want to either sum all currents flowing into the node, or all currents flowing out of the node, and set the result to zero. Don't mix directions.

For example, your first term: ( v10 - v20/ z1)) {which I assume to mean ( (v10 - v20)/ z1)): watch your parentheses to group operations appropriately!} represents a current flowing into the supernode at V20 through impedance Z1. But your second term, ( v20/ z4), represents a current flowing OUT of the supernode via Z4.

Check each of your terms and assure that they all represent currents flowing in the same manner, either into or out of the supernode, and not both.

 

[brabbit87]

It looks to me like you're mixing current directions. You want to either sum all currents flowing into the node, or all currents flowing out of the node, and set the result to zero. Don't mix directions.

For example, your first term: ( v10 - v20/ z1)) {which I assume to mean ( (v10 - v20)/ z1)): watch your parentheses to group operations appropriately!} represents a current flowing into the supernode at V20 through impedance Z1. But your second term, ( v20/ z4), represents a current flowing OUT of the supernode via Z4.

Check each of your terms and assure that they all represent currents flowing in the same manner, either into or out of the supernode, and not both.

Thankou. Maybe being abit too careless. Will make another attempt.

 

[brabbit87]

All finished nodal analysis and answer is correct. Had another attempt at the mesh analysis, unfortunately the matrix spreadsheet will not calculate I1,2,3. I am lost here, i have repeated the same 3 equations.

KVL walk clockwise around the 3 loops

Mesh 1 -

v1 - (z1)i1 - (z4)(i1 - i2) = 0
v1 = (z1)i1 + (z4)(i1 - i2)
v1 = (2)i1 + (-5j)i1 + (5j)i2

120 = (2 - 5j)i1 + (5j)i2 ....(1)

mesh 2 -

-v3 + (z5)(i2 - i3) + z4(i2 - i1) = 0
v3 = (4j)i2 - (4j)i3 (-5j)i2 + (5j)i1
v3 = (5j)i1 + (4j - 5j)i2 (-4j)i3
14.14 + 14.14j = (5j)i1 + (-1j)i2 + (-4j)i3 ...(2)

Mesh 3 -

-v2 + z5(i3 - i2) + z3(i3)
v2 = + (-4j)i2 + (4 + 4j)i3
120j = (-4j)i2 + (4 + 4j)i3 ...(3)

 

[gneill]

Check the sign of v3 in your mesh 2 equation. You appear to be writing a sum of potential drops, so...

 

[brabbit87]

Check the sign of v3 in your mesh 2 equation. You appear to be writing a sum of potential drops, so...

I'm not so sure. What is wrong with v3 in my mesh 2 equation.

Assuming clockwise current of circuit. It would be a negative voltage with positive voltage drops across the impedances?

 

[donpacino]

I'm not so sure. What is wrong with v3 in my mesh 2 equation.

Assuming clockwise current of circuit. It would be a negative voltage with positive voltage drops across the impedances?

you need to define which way your currents go...

 

[brabbit87]

you need to define which way your currents go...

I have defined which way my current goes in each mesh, clockwise. Or are referring to something else. I may have possibly missed?

Apologies if i am being stupid with something

 

[donpacino]

I have defined which way my current goes in each mesh, clockwise. Or are referring to something else. I may have possibly missed?

Apologies if i am being stupid with something

np. I just did not know... if that is the case then one of the Z terms in your mesh equations has the wrong sign...

 

[brabbit87]

Check the sign of v3 in your mesh 2 equation. You appear to be writing a sum of potential drops, so...

I apologise gneill, however i am still unsure.

Is it just the 2nd mesh equation that i have incorrectly done?

 

[donpacino]

I apologise gneill, however i am still unsure.

Is it just the 2nd mesh equation that i have incorrectly done?

yes

 

[brabbit87]

yes

My initial equation for the mesh 2, is that correct?

Regards

 

[donpacino]

My initial equation for the mesh 2, is that correct?

Regards

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.

 

[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.

No, really. I am genuinely not looking for that. I am just struggling with this question.

 

[gneill]

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

 

[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;
<br /> 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;
<br /> 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