Mesh and nodal analysis don't quite match, but why?

In summary: Thanks for the help - I've been stuck on this for quite some time now!In summary, the conversation is about solving a circuit using mesh and nodal analysis. The user has completed the mesh analysis and is having trouble getting the same results using nodal analysis. They have used WolframAlpha to assist with calculations and have noticed discrepancies in the results. After some troubleshooting, it appears that WolframAlpha may have miscalculated due to the signs of the complex terms being reversed. The user has tried different inputs and has found that using the entire complex number instead of just the imaginary part gives the correct results.
  • #1
Numbskull
54
1

Homework Statement


[/B]
Find I by mesh and nodal analysis

Q2(a) Base drawing & values.jpg


Homework Equations

The Attempt at a Solution



##V_{10}## is defined at the junction of ##Z_1##, ##Z_2## and ##Z_4##.
##V_{20}## is defined at the junction of ##Z_2##, ##Z_3## and ##Z_5##.
##0V## is defined at the junction of ##V_1##, ##Z_4##, ##Z_5## and ##V_2##.

I have completed the mesh analysis without too much difficulty (and those answers are correct according to other threads with this question). I thought I was on the right track with the nodal analysis but the answers are close enough to be wrong, but not far enough away to look like a major error somewhere.

I have used Wolfram to do the heavy lifting, and have attached the PDF of the input and the result for ##V_{10}##. When I calculate the current through ##Z_4## (again using Wolfram entering values in complex form), I get the result of ##-8.7792 + j16.5853## when converted to polar is ##18.765## Amps compared to ##19.59## for the mesh analysis, (although the angle is off too).

I've spend hours and hours trying to figure this out, but can't quite get the answers to agree. I've also been through the process of trying to identify compound rounding errors but I don't think it's that. I would appreciate some help or pointers :cry:
 

Attachments

  • Wolfram Q2 (b).pdf
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  • #2
To me it looks like WolframAlpha has mucked up the signs of the two complex terms in your node equation. For example,
$$\frac{v10}{-5j} = \frac{v10}{-5j}\cdot\frac{j}{j} = \frac{v10}{5}j$$
They have made the term negative by writing it as
$$- \frac{v10}{5}j$$
 
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  • #3
Thank you for the tip! I've played with brackets to no effect. What does appear to work is to express the input as the entire complex number, i.e. ##0-j5## instead of just ##-j5##.

However, although I now have the correct magnitude, the angle is 180 degrees out, which suggests to me that the signs of the real and imaginary components are reversed? Would that be a correct supposition?
 
  • #4
Numbskull said:
Thank you for the tip! I've played with brackets to no effect. What does appear to work is to express the input as the entire complex number, i.e. ##0-j5## instead of just ##-j5##.

However, although I now have the correct magnitude, the angle is 180 degrees out, which suggests to me that the signs of the real and imaginary components are reversed? Would that be a correct supposition?
That could be it. You'll be the best judge though. Look at the complex values for your "good" results via mesh analysis and compare with the recent values returned by Wolfram. I can tell you that my own results for v10 and v20 have all components positive.
 
  • #5
Looking at your Wolfram Alpha solution, you have in the very first line (v10/-5i) which Wolfram turns into a fraction with i in the numerator.

You say you've played with brackets, but you don't show us what you did.

I typed (v10/(-5i)) into Wolfram Alpha and got an expression with i in the denominator as it should be. Did you try that?
 
  • #6
The Electrician said:
Looking at your Wolfram Alpha solution, you have in the very first line (v10/-5i) which Wolfram turns into a fraction with i in the numerator.

You say you've played with brackets, but you don't show us what you did.

I typed (v10/(-5i)) into Wolfram Alpha and got an expression with i in the denominator as it should be. Did you try that?
It looks like I used the complete term and brackets to obtain the correct input. I'm a little unsure on equivalent terms, so the second example below looked like the one which is correct to me, but is the first entry also correct? Actually looking at it, they both give the same answer, so they must be...

bracket_example.jpg


bracket_example_fullterm.jpg
 

Related to Mesh and nodal analysis don't quite match, but why?

1. Why are mesh and nodal analysis not always equivalent?

Mesh and nodal analysis are two different methods used to solve electric circuits. While they both use Kirchhoff's laws, they approach the problem from different perspectives and therefore may not always yield the same results.

2. Which method should I use, mesh or nodal analysis?

The choice between mesh and nodal analysis depends on the circuit and the information available. In general, nodal analysis is better suited for circuits with many parallel elements, while mesh analysis is more effective for circuits with many series elements.

3. Can I use both mesh and nodal analysis on the same circuit?

Yes, you can use both methods on the same circuit, but it may not be necessary. In some cases, one method may be more efficient or provide more accurate results than the other. It is important to understand the differences between the methods and choose the appropriate one for the given circuit.

4. What are some limitations of mesh and nodal analysis?

Mesh and nodal analysis assume linear components and do not take into account non-linear behavior, such as diodes or transistors. They also do not account for mutual inductance in circuits with multiple loops. Additionally, these methods may become more complex and time-consuming for circuits with a large number of elements.

5. Are there any other methods for solving electric circuits?

Yes, there are other methods such as Thevenin's and Norton's theorems, superposition, and source transformation. These methods can be useful for simplifying complex circuits or for solving specific types of problems. It is important to have a good understanding of all these methods to choose the most appropriate one for a given situation.

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