# Nodal Analysis

1. Dec 9, 2016

### David J

1. The problem statement, all variables and given/known data
This particular question has been raised in past posts and I have studied those posts but unable to take anything from them. Basically the question comprises 2 parts. Part A requires us to find the current through "I" using mesh analysis. By studying some of the posts on here I managed to identify that value as -9.2 +j17.30 A. Part B of the question requires me to find the same solution by using Nodal analysis. I have attached a copy of the question (2b), nodal analysis.

2. Relevant equations.
I have attached the question as I cannot post it in LATEX as it contains a circuit diagram.

3. The attempt at a solution.

I have attached my attempt "so far" I say so far because I have been working on this question for a long time now and really need to know if I am on the right track or not. I have not fully completed the solution, its only just started. I have seen other posts for this same question and everyone has the same opinion that the notes and information provided by the university are not good. I have attached the only example of a nodal problem provided by the university. It is very basic and does not go into complex numbers so I have tried to adapt this example to answer the question by using the methods explained in the example but trying to input complex numbers. As you will see I dont think what I have created so far it is correct and I would really appreciate any help to:-

First of all identify if the method I am trying to use will actually work for this question?
Identify the errors I have made.

The attachments may not seem to contain an awful lot of working out but to date, I have lost count of the number of times I have started then stopped because I am not getting the correct answer. I have been working on this for some time but getting no where so I have decided to turn create a new post to ask for guidance.

Any help would be appreciated. I know the rules of these forums. I am not looking for answers just some clues or help to get started and to understand it better. This has proven to be a very difficult subject to teach myself and I am not doing very well at the moment

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2. Dec 9, 2016

### The Electrician

At the very end where you say "I will multiply 120(V1) x 4 and j120(V2) x 2"

This is an algebra mistake. You must multiply everything on the left side of your equation by the same value.

3. Dec 10, 2016

### David J

I found the LCM to be 4.

The equation is as follows:-

$\left ( \frac {120-v20} 2\right ) - \frac {v20} {-j5} - \frac {v30} {j4} + \left ( \frac {j120-v30} 4 \right ) = 0$

I tried multiplying everything by 4. As an example:-

$\left ( \frac {120} 2\right) 4=240$
To find the equivalent v20 i tried $\left (\frac 1 2\right) 4 = 2 v20$ so I have $240 - 2 v20$

1: I then looked at the complex number fraction $\frac {v20} {-j5}$. I can use $\left ( \frac 1 {-j5} \right ) 4 = j0.8 v20$

2: I then looked at the complex number fraction $\frac {v30} {j4}$. I can use $\left ( \frac 1 {j4} \right ) 4 = -j v30$

finally I looked at the last part of the equation, $\left ( \frac {j120-v30} 4\right )$

$\left (\frac {j120} 4\right) 4 = j120$, it did not change.

Same happens with v30, it remains as v30.

So I have ended up with a slightly different equation as follows

$(240 - 2 v20) - (j0.8 v20) - (-j v30) + (j120-v30) = 0$

There is a "minus times minus" in this equation $(j0.8 v20) - (-j v30)$

I will change this to $(j0.8 v20) + (j v30)$

So I end up with $(240 - 2 v20) - (j0.8 v20) + (j v30) + (j120-v30) = 0$

As far as I can see I have multiplied everything by the LCM which is 4. I have tried to describe as best as I can my thinking when trying to do this.
The only thing which has confused me is that I am mixing "j" numbers with real numbers and I am not sure if that works in this case.
The 2 examples of this are steps 1 and 2 where I talked about mixing complex fractions ???

I will try to work on with this but if possible could you prompt me on any further errors I have made.

Thanks

4. Dec 10, 2016

### The Electrician

Of course you can "mix" real numbers with "j numbers". The complex numbers are just an extension of the real numbers, so to speak. Complex numbers can have both a real part and an imaginary part, or they can have only a real part (this would be a "real" number), or they can have only an imaginary part (this would be what you are calling a "j number)". These are all complex numbers, and you can treat them just like a pure real number when solving equations.

You only need one more equation. The example problem you show has a constraint equation for the supernode; it is V30-V20=2

You need the equivalent for your given problem. Then you will have two equations that you can solve for V20 and V30.

Give it a try.

Last edited: Dec 10, 2016
5. Dec 10, 2016

### David J

I think I maybe slowly getting to understand this but I have just realized that the example I am using is trying to find the current flowing into the 2V battery. The example text clearly states " the current flowing into the -ve terminal"

My problem requires the current flowing into Z4 (or in the case of the example, the 4 ohm resistor)

Also, the example constraint equation is looking at $v30 -v20$ based on the polarity and requirement of the example question.

I am wondering if my equation should read as :-

$(2v20 - 240)$ and $(v30-j120)$

and the constraint equation to read $v20 - v30 =14.14+14.14j$ (opposite to the example)

apart from this I now have $(2+0.8j)v20 + (1+j)v30 = 240+j120$

I just need to work out the above equation but I will sleep on it tonight ( I am in UK)

If possible any advice on any errors you can see would be appreciated.

Thanks

6. Dec 10, 2016

### The Electrician

Your example wanted the current in the source, but your given problem wants the current in Z4. If you know the voltage V20, then you can easily find the current in Z4.

Don't change your first equation. What I told you was to find the "equivalent" constraint equation; you noticed the polarity direction was different from the example and got the right result, so you're good there.

If you'll solve the constraint for V30 and substitute that in the first equation, you should end up with an equation containing only V20, which you should be able to solve. Then the current in Z4 is trivial to find.

7. Dec 11, 2016

### David J

Good Morning

I dont fully understand the meaning of the constraint equation, however, after doing a lot of research I think its something that is fixed, something we know for definite. So in this case I need to find that equation for this problem.

I still have $( 2 + 0.8j)v20 + (1 + j)v30 = 240+j120$

If I change this around to find v30 I get $(240+120j) - (2+0.8j) v20 = (1+j) v30$

$(240+120j) - (2+0.8j) v20 = 238 - 120j$

If I then $\frac {238-120j} {1 + j} = 59 - 179j$

This could mean $v30 = 59 - 179j$

If I put this into the original equation then I get

$(240 - 2v20) - (.8j v20) + (j (59-179j)) + (120j - (59-179j))$

which works out as $(240 - 2v20) - (.8jv20) = 120 + 358j$

$120 + 240 = 360$ so $360 + 358j = 2v20 + .8j v20$ so $2 + .8j v20 = 360 +358j$

$\frac {360 + 358j} {2 + .8j} = 216.90 +92.24j$

If I then try to use this to find the current "i" by using $\frac {216.90 + 92.24j} {-j5} = -18.448 +43.38j$

I need $-9.2 + 17.30j$

The interesting thing I find about this is that $\frac {-18.448 + 43.38j} {2} = -9.224 + 21.69j$

I dont think this is a coincidence. I think something is wrong with the equation below.

$( 2 + 0.8j)v20 + (1 + j)v30 = 240+j120$

as an example if i half $216.90 +92.24j$ i get $108.45 + 46.12j$

Divide this by $-5j$ i get a value closer again to the elusive $-9.2 + 17.30j$

This time it works out as $-9.224 + 21.69j$

Can you see any errors in these values or can you identify the area I have gone astray ??

much appreciated

8. Dec 11, 2016

### The Electrician

There is a voltage source, v3, connected between v20 and v30. Its presence forces a relationship between v20 and v30. That relationship is given by the equation v20 - v30 = v3

What you must do is to solve the constraint equation for v30 and substitute that result into your first equation (with the sign error fixed). Then you get an equation you can solve for v20.

9. Dec 11, 2016

### David J

Good Morning

I see your point regarding the sign error. Thanks for pointing that out and I am currently re working this out based on that change.

The second error regarding the part in red, this should have read as $238 + 119.2j$

These 2 errors will have affected my end result so I am working on this at the moment

Thanks again for the advice with this

10. Dec 11, 2016

### David J

Hello again

I tried again but still I cannot get the correct answer. I will show you the calculations I have made.

Starting with the change of the sign in $(2+.8j)v20+(1-j)v30=240+120j$

so $(240+120j)-(2+.8j)v20=(1-j)v30$

so $(240+120j)-(2+.8j)v20=238+119.2j)$

so $\frac {238+119.2j}{1-j}=59.4+178.6j$

so $v30 = 59.4+178.6j$

Have i solved the constraint equation for V30, answer being $59.4+178.6$ ?

I would like to think this is correct but I am not so sure as it does not fit into the rest of the equation.

This makes me think something is wrong further back in the working out, possibly the first equation.

I am also now feeling that using the working out from the initial example (i posted) is in now way relevant to this problem and that explains why a lot of the people who have posted about this question in the past have commented on the lack of help from the university.

Going back to the equation in hand and my value for v30 ($59.4+178.6j$

If I input this into the original equation below

$(240-2 v20)-(0.8jv20)+(j(59.4+178.6j))+(120j-(59.4+178.6j))=0$

$(j(59.4+178.6j))+(120j-(59.4+178.6j))=-238+.8j$

$(240-2 v20)-(.8jv20)=-238+.8j$

I have calculated the further answers on paper but they dont work out so I wont post on here. I am still wrong some where.

I thought I was close to solving this but I seem to have hit a brick wall tonight.

As usual any advice more than appreciated. I am now thinking the original equation should be changed.

Maybe I should have started with $\left( \frac{v20-120}{2}\right)-\frac{v20}{-j5}-\frac{v30}{j4}+\left(\frac{v30-j120}{4}\right)=0$

11. Dec 11, 2016

### The Electrician

12. Dec 12, 2016

### David J

Good Morning
Not sure what you mean by this "what happened to v20" At this stage of the equation I was looking for a value

$(240+120j)-(2+.8j)v20=238+119.2j$ What happened to the v20

$(240+120j)-(2+.8j)v20=238+119.2j$
$(240+120j)-(238+119.2j)v30=(2+.8j) v20$
so $(2+.8j) v20 = -117.20+120j$

$\frac{-117.20+120j}{2+.8j}=-29.83+71.93j$

If i then take the constraint equation which is $v20-v30=v3$
move $v30$ to the left as in $v30=v3+v20$
$(14.14+14.14j)+(-29.83+71.93j)$ should equal $v30$ which i calculated earlier to be $59.4+178.6j$
This equation equates to $-15.69+86.07j$ which is not correct and does not agree with my earlier calculation.
However, I do now have the following:-

$v3 = 14.14+14.14j$
$v20 = -29.83+71.93j$
$v30 = -15.69+86.07j$

$v20 - v30 = v3$ so $(-29.83+71.93j) - (-15.69+86.07j) = -14.14-14.14j$

These values do not work out. I have minus values when they should be plus.

What I have discovered is that the value of $-117.20+120j$ may have some meaning in all of this. I changed this to $117.20+120j$

$\frac{117.20+120j}{2+.8j}=71.21+31.52j$

I then used $71.21+31.52j$ as v20

I put this into $v30 = v3 + v20$

$v30 = (14.14 + 14.14j) + (71.21+31.52j)$ so $v30 = 85.35+45.66j$

$v3 = 14.14+14.14j$
$v20 = 71.21+31.52j$
$v30 = 85.35+45.66j$

When i plug these values into various equations they work out correctly but in the opposite way. For example

$v30 - v20 = v3$ It should be $v20 - v30 = v3$

$\frac {85.35+45.66j}{-5j} = -9.132+17.07j$ It should be $v20$ that gives this value, not $v30$

[FONT=PT Sans, san-serif]I appear to have stumbled across the correct answers (but in the wrong state) by accident but this now makes me think something has been crossed in the very beginning of this post. Do I need to start again? Have I missed something? is my interpretation of the constraint equation correct ??

Confused is an understatement but I think I am close now. Please advise.

Thanks again
[/FONT]

13. Dec 12, 2016

### The Electrician

Here's what you would get if you do so. I used Mathematica to produce this image and a special symbol is used for j; that should be easy to follow:

14. Dec 12, 2016

### David J

I thought i had followed this advice but when I check back I have not followed it correctly, due to not fully understanding. I did touch briefly on this subject in post 5
I notice you have described $v30$ as $10\sqrt2(1+j)$

I am trying to recap on this now, going back through each step from the beginning to see where I went wrong or where I mis understood, using that attachment you sent as a guide.
In my first equation i had $(2+.8j)v20 +(1+j)v30$ $(1+j)$ was wrong. It should have been $(1-j)$ Can I ask how ?? because I am trying to understand this and I cant see it.

15. Dec 12, 2016

### The Electrician

You eventually solved the constraint equation correctly in post #12, but you didn't substitute the results of that into the first equation. You substituted something else that you derived from manipulations of the first equation itself.

"I notice you have described v30 as10√2(1+j)"

It's V3 described that way.

You made a sign error when you went from the equation above to the final one: (2+0.8j)v20+(1-j)v30=240+j120
Go through the steps very carefully.

16. Dec 12, 2016

### David J

I see it now.

during $(240-2v20)-(j0.8v20)-(-jv30)+(j120-v30)$

I decided to assume that "minus take away minus" will equal a plus so i did this

$(j0.8v20)+(j v30)$

What I did not take into account is that $-j$ is the same as $1-j$

so it should have been like this $(j0.8v20)+(1-j)$

I guess this is because we are working with complex numbers ???

I also got confused when i saw there were 2 $v30`s$ in the latter part of the equation

$(jv30)+(j120-v30)$

I would like to thank you for your help with this. This is post 16. I have a far better understanding of this than when i started on post 1.

Many thanks for your help with this

best regards for the coming month

17. Dec 12, 2016

### The Electrician

You're welcome. Just watch out for those signs at mid-term time!