Node Analysis Help: Troubleshoot Your Homework

[asdf12312]

Homework Statement

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??

 

[CWatters]

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

 

[berkeman]

Also you should be more careful with your parens -- they are misplaced in at least 2 places...

 

[asdf12312]

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

 

[berkeman]

Homework Statement

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??

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...

 

[asdf12312]

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

 

[berkeman]

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...

 

[asdf12312]

(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 >.>

 

[berkeman]

(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!

 

[CWatters]

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.