Node Analysis Help: Troubleshoot Your Homework

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Discussion Overview

The discussion revolves around troubleshooting a homework problem related to node analysis in electrical circuits. Participants are examining equations derived from Kirchhoff's laws, focusing on the calculations of node voltages and the correct application of current directions.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents their equations for node analysis but expresses uncertainty about their results, stating that their calculated voltages do not match the expected answers.
  • Another participant points out potential inconsistencies in the sign conventions used in the equations, suggesting that current direction affects the signs in the equations.
  • There are comments regarding misplaced parentheses in the equations, indicating a need for careful notation in mathematical expressions.
  • One participant provides a reduction of the equations and arrives at a different set of values for the node voltages, suggesting that their approach yields simpler results.
  • Another participant challenges the reduction process, indicating that there may be an error in the results presented by others, specifically questioning the relationship between V1 and V2.
  • A later reply corrects an earlier mistake regarding the addition of a constant in the equations, leading to a different conclusion about the correct values for V1 and V2.

Areas of Agreement / Disagreement

Participants express differing views on the correct application of current direction and the resulting equations. There is no consensus on the correct values for V1 and V2, as multiple approaches and interpretations of the equations are presented.

Contextual Notes

Some participants note that their equations may depend on specific assumptions about current direction, which could lead to different interpretations of the results. There are also unresolved mathematical steps that may affect the final answers.

asdf12312
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Homework Statement


2zibyc9.png


Homework Equations


The Attempt at a Solution



At node 1: (V1-10)/5 -2 + (V1-V2)/4 = 0 --> 9/20V1 - V2/4 = 2
At node 2: (V2-V1)/4 + (V2 - 4*I)/6 + V2/2 = 0
and i use also I= V2/2

i get V1=5.83V and V2=2.5V but its not the rite answer. what am i diong wrong??
 
Last edited:
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I've only looked at your equation for node 1 but you don't appear to have been consistent with the sign.

If I define current into the node as +ve I get..

(10-V1)/4 + 2A + (V2-V1)/4 = 0
 
Also you should be more careful with your parens -- they are misplaced in at least 2 places...
 
CWatters said:
I've only looked at your equation for node 1 but you don't appear to have been consistent with the sign.

If I define current into the node as +ve I get..

(10-V1)/4 + 2A + (V2-V1)/4 = 0

its the same eqtn as mine...you just have eq current going into node but i have current going out. so, i use -2A because current going into node. if it was gonig out it would be +2
 
Last edited by a moderator:
asdf12312 said:

Homework Statement


2zibyc9.png


Homework Equations


The Attempt at a Solution



At node 1: (V1-10)/5 -2 + (V1-V2)/4 = 0 --> 9/20V1 - V2/4 = 2
At node 2: (V2-V1)/4 + (V2 - 4*I)/6 + V2/2 = 0
and i use also I= V2/2

i get V1=5.83V and V2=2.5V but its not the rite answer. what am i diong wrong??

asdf12312 said:
its the same eqtn as mine...you just have eq current going into node but i have current going out. so, i use -2A because current going into node. if it was gonig out it would be +2

Thanks for fixing the parens in your original post.

Could you show your work on the simultaneous equations? I agree with your fixed equations, but I get different (simpler) answers...
 
i always meant it like that but i got lazy with parenths.

equation 1 i reduced to:
(9/20)V1 - V2/4 = 2
9V1-5V2=40

equation 2, substitute I=(V2/2):
(V2-V1)/4 + (V2-4(V2/2))/6 + V2/2 = 0
V2/4 - V2/6 + V2/2 - V1/4 = 0
(7/12)V2 = V1/4
V1 = (7/3)V2

substitute V1 into eq 1:
9((7/3)V2)-5V2=40
21V2-5V2=40
16V2=40

V2=2.5V
V1= (7/3)(2.5)= 5.833V
 
Last edited:
asdf12312 said:
i always meant it like that but i got lazy with parenths.

equation 1 i reduced to:
(9/20)V1 - V2/4 = 2
9V1-5V2=40

equation 2, substitute I=(V2/2):
(V2-V1)/4 + (V2-4(V2/2))/6 + V2/2 = 0
V2/4 - V2/6 + V2/2 - V1/4 = 0
(7/12)V2 = V1/4
V1 = (7/3)V2

substitute V1 into eq 1:
9((7/3)V2)-5V2=40
21V2-5V2=40
16V2=40

V2=2.5V
V1= (7/3)(2.5)= 5.833V

Can you show how you got that reduction for equation 1? There is an error in your result. You will get V1 = X*V2 for the result...
 
(V1-10)/5 -2 + (v1-v2)/4 = 0

V1/5 -2 -2 + v1/4 -v2/4 = 0
(9/20)V1 - V2/4 = 4
9V1 - 5V2=80

hence
16V2=80
V2=5V
V1=11.6V

that is the rite answer. and sorry i made an error, forgot to add that 2 to my 1st eq >.>
 
asdf12312 said:
(V1-10)/5 -2 + (v1-v2)/4 = 0

V1/5 -2 -2 + v1/4 -v2/4 = 0
(9/20)V1 - V2/4 = 4
9V1 - 5V2=80

hence
16V2=80
V2=5V
V1=11.6V

that is the rite answer. and sorry i made an error, forgot to add that 2 to my 1st eq >.>

Great! :smile:
 
  • #10
asdf12312 said:
its the same eqtn as mine...you just have eq current going into node but i have current going out. so, i use -2A because current going into node. if it was gonig out it would be +2

Perhaps I made a mistake but last night I thought you had written +2A not -2A.
 

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