Nodal analysis: simple confusion

In summary, the 80+20 part in the denominator is when you add up the voltage sources and resistance values along the path from node 1 to the reference node. The current through that branch is the sum of the two resistances.
  • #1
Marchese_alex
54
0
My question about this excercise is the 80+20 part in the denominator. Why and when do I do that? I tried doing it but got the wrong answer
 

Attachments

  • 20121015_143232.jpg
    20121015_143232.jpg
    27.5 KB · Views: 431
  • 20121015_143346.jpg
    20121015_143346.jpg
    13.8 KB · Views: 379
Physics news on Phys.org
  • #2
There is a single series-connected path from node 1 through the two resistors and the voltage source to the reference node. So the total resistance along that path is the sum of the two resistances. That makes the current through that branch (V1 - 24V)/(80Ω + 20Ω).

There's no need to introduce extra nodes along a series branch; just add up the voltage sources and resistance values.
 
  • #3
Ok!

How about that +125 in node 3. I had it as V3-125. Why is it positive
 

Attachments

  • 20121015_154642.jpg
    20121015_154642.jpg
    17 KB · Views: 385
  • #4
Marchese_alex said:
Ok!

How about that +125 in node 3. I had it as V3-125. Why is it positive

It's positive because of the polarity of the 125V source along the path from node 3 to the reference node. The potential at its "-" terminal is effectively 125V below the reference potential. Since the potential of node 3 is measured with respect to the reference node, this makes the total potential across the resistor V3 + 125V.

When in doubt, do a "KVL walk" along the path and add up the various potential changes. Starting at the reference node:

-125V + i3*R = v3

So the current is (v3 + 125V)/R
 
  • #5
When i multiply to reduce the fractions, which one should i aim for? Any of the deno?
 
  • #6
Marchese_alex said:
When i multiply to reduce the fractions, which one should i aim for? Any of the deno?

In what situation? Sounds like an exercise in fractions and algebra.
 
  • #7
gneill said:
In what situation? Sounds like an exercise in fractions and algebra.

Can i use any denominator to simplify the eacuation
 
  • #8
Surely it depends upon the equation and what the goal of the simplification is. You'll have to be more specific.
 
  • #9
I have a doubt when asked to find the current at each resistor. Sometimes the book uses the same ecuaction established in the nodal analysis. For example, (V1-128)/5 + V1/60 etc. is my ecuation after the analysis. Well, some times they use it like it was established I=(V1-128)/5 but sometimes they change it like so, I=(128-V1)/5... .??
 
  • #10
A sign change usually means that it was altered in order to match the assumed direction of the current being found.

When you write the equations for nodal analysis it's usually the case that you just assume, for example, that all currents are flowing out of each node (or into each node). This is done in order to make writing the equations a straightforward procedure where you don't have to think about "actual" current directions beforehand.

However, it is also often the case that the circuit drawing will have an indication of what direction should be assumed for a given current when you report the value. The match up can be done by adjusting the sign of of the appropriate nodal analysis term to correspond to the requested direction assumption.
 
  • #11
gneill said:
A sign change usually means that it was altered in order to match the assumed direction of the current being found.

When you write the equations for nodal analysis it's usually the case that you just assume, for example, that all currents are flowing out of each node (or into each node). This is done in order to make writing the equations a straightforward procedure where you don't have to think about "actual" current directions beforehand.

However, it is also often the case that the circuit drawing will have an indication of what direction should be assumed for a given current when you report the value. The match up can be done by adjusting the sign of of the appropriate nodal analysis term to correspond to the requested direction assumption.

Awesome! Thanks
 

1. What is nodal analysis and how is it used in science?

Nodal analysis is a method for solving circuit problems in electrical engineering. It involves creating a system of equations based on Kirchhoff's Current Law and solving for the unknown node voltages. In science, nodal analysis is used to analyze and predict the behavior of electrical circuits.

2. What is Kirchhoff's Current Law and how does it relate to nodal analysis?

Kirchhoff's Current Law states that the sum of currents entering a node in an electrical circuit must equal the sum of currents leaving that node. This law is essential to nodal analysis because it allows us to write equations based on the conservation of current at each node.

3. Can nodal analysis be used for complex circuits?

Yes, nodal analysis can be used for both simple and complex circuits. However, for more complex circuits, the number of equations and variables may increase, making the calculation process more tedious. In such cases, other circuit analysis methods may be more efficient.

4. What are the advantages of using nodal analysis?

Nodal analysis allows for a systematic and organized approach to solving circuit problems. It also provides a clear understanding of the behavior of a circuit and can be applied to both DC and AC circuits. Additionally, nodal analysis can handle non-linear components, making it a versatile tool for circuit analysis.

5. Are there any limitations to nodal analysis?

While nodal analysis is a powerful tool, it does have some limitations. It cannot be used for circuits with dependent sources, and it may not be the most efficient method for solving complex circuits with a large number of nodes and components. Additionally, nodal analysis assumes ideal conditions and does not account for real-world factors such as resistance and capacitance.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
1
Views
249
  • Engineering and Comp Sci Homework Help
Replies
4
Views
529
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
900
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
Back
Top