Nodal Analysis

  • Thread starter Rozenwyn
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  • #1
Rozenwyn
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http://img209.imageshack.us/img209/8136/423ky7.jpg [Broken]

I have trouble getting the correct answers.

I tried:
At node V1 [tex]\ i_1 + \frac{v_2-v_1}{5} = \frac{v_1}{20} \ \longrightarrow[/tex] Solve for [tex]i_1[/tex]
ok let's try.
[tex] i_1 + \frac{15-4}{5} = \frac{4}{20}[/tex]
[tex] i_1 = \frac{1}{5} - \frac{11}{5} [/tex]
[tex] i_1 = \frac{-10}{5} = -2A[/tex]

@Cornea: Indeed, the equations seem to be correct. *bangs head to the table.* Can't believe a sign error could waste 2 hrs of my life. Hmmm, need more sleep ... more sleep.

Then;

Ar node V2 [tex]\ \frac{v_2-v_1}{5} + \frac{v_2-v_3}{15} = i_2 \ \longrightarrow[/tex] Solve for [tex]i_2[/tex]

When I solve for [tex]i_1, \ i_2[/tex] I get wrong answers.
 
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Answers and Replies

  • #2
Corneo
321
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Check you math. Your equations are correct.
 
  • #3
LeBrad
214
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From your own equations:

i1 = v1/20 + v1/5 - v2/5 = 4/20 + 4/5 - 15/5 = 1/5 + 4/5 -4 = 1 -3 = -2

i2 = 15/5 - 4/5 + 15/15 - 18/15 = 3 - 12/15 + 1 - 18/15 = 4 - 2 = 2
 
  • #4
eetwonder
9
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There are two easy nodes that I used for the question.
First, the node on the top left conner below -2A,I got:
i_1=(v_1-v_2)/5+(v_1-0/20), assuming the -2A flowing into the node while the other two flowing out of the node.There's a point to note .When we assume the current flowing from the left to the right, we are using the concept that the potential on the left is higher than that on the right .(referring to the definition of convectional voltage)
Same thing with equation two, where I utilize the button node in the middle . The 2A is flowing out while two other currents are flowing into the node, i_2=(v_1-0/20)+(v_3-0/10)

Solve above two equations, I got i_1=-2A and i_2=2A---#
 

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