Using node-voltage analytical technique

  • #1
120
0

Homework Statement


Calculate the voltage at each of the extraordinary nodes (V1 – V3), and the current through each of the six resistors.

http://i.imgur.com/VbZQH.png


Homework Equations



KCL: ƩIn = 0
V=ir
I=[itex]A^{-1}[/itex]B
[itex]\Delta[/itex] = a11C11 + a12C12 + a13C13

[itex]A^{-1}[/itex] = [itex]\frac{adjA}{\Delta}[/itex]
adjA = [itex][C_{jk}]^{T}[/itex]


The Attempt at a Solution





For I1+I2+I3=0:

[itex]\frac{V1-4}{0.2}[/itex] + [itex]\frac{V1-V2}{0.5}[/itex] + [itex]\frac{V1-2}{0.1}[/itex] = 0

For I4+I5+I6=0:

[itex]\frac{V2-V1}{0.5}[/itex] + [itex]\frac{V2-V3}{0.5}[/itex] + [itex]\frac{V2}{0.1}[/itex] = 0

For I7+I8+I9=0:

-[itex]\frac{V1-4}{0.2}[/itex] + [itex]\frac{V3-V2}{0.5}[/itex] + [itex]\frac{V3-3}{0.1}[/itex] = 0


([itex]\frac{1}{0.2}[/itex] + [itex]\frac{1}{0.5}[/itex] + [itex]\frac{1}{0.1}[/itex])V1 - ([itex]\frac{1}{0.2}[/itex] + [itex]\frac{1}{0.5}[/itex])V2 + 0V3 = 40

([itex]\frac{1}{0.5}[/itex]+[itex]\frac{1}{0.5}[/itex]+[itex]\frac{1}{0.1}[/itex])V2 - ([itex]\frac{1}{0.5}[/itex])V1 + (-[itex]\frac{1}{0.5}[/itex])V3 = 0

(-[itex]\frac{1}{0.2}[/itex])V1 + (-[itex]\frac{1}{0.5}[/itex])V2 + [itex]\frac{1}{0.5}[/itex] + [itex]\frac{1}{0.1}[/itex])V3 = 10

And then I simplified that to:

17V1 - 7V2 + 0V3 = 40
-2V1 + 14V2 -2V3 = 0
-5V1 - 2V2 + 12V3 = 10


My book says I should use Cramer's rule or matrix inversion to solve this. I chose to use matrix inversion because I thought solving a simultaneous equation would take a while.


I ended up with I1 = 2.627, I2 = 2/3, I3 = 2.039

That's obviously wrong since the sum should equal zero. I'm not sure what I did wrong, I just followed the steps in my book.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,857
1,655
Cannot tell without your working.
 
  • #3
The Electrician
Gold Member
1,283
169

The Attempt at a Solution





For I1+I2+I3=0:

[itex]\frac{V1-4}{0.2}[/itex] + [itex]\frac{V1-V2}{0.5}[/itex] + [itex]\frac{V1-2}{0.1}[/itex] = 0

For I4+I5+I6=0:

[itex]\frac{V2-V1}{0.5}[/itex] + [itex]\frac{V2-V3}{0.5}[/itex] + [itex]\frac{V2}{0.1}[/itex] = 0

For I7+I8+I9=0:

-[itex]\frac{V1-4}{0.2}[/itex] + [itex]\frac{V3-V2}{0.5}[/itex] + [itex]\frac{V3-3}{0.1}[/itex] = 0

You must include the effect of V3 in this equation, as shown:

For I1+I2+I3=0:

[itex]\frac{V1-4-V3}{0.2}[/itex] + [itex]\frac{V1-V2}{0.5}[/itex] + [itex]\frac{V1-2}{0.1}[/itex] = 0

And this one:

For I7+I8+I9=0:

-[itex]\frac{V1-4-V3}{0.2}[/itex] + [itex]\frac{V3-V2}{0.5}[/itex] + [itex]\frac{V3-3}{0.1}[/itex] = 0
 
  • #4
120
0
Okay, I redid my work and I'm still getting the wrong answer. In my previous post I mistakenly wrote my answers as current values instead of voltages sinceI'm calculating the voltages at the extraordinary nodes.

17V1 - 2V2 - 2V3 = 40
-2V1 + 14V2 -2V3 = 0
-2V1 - 2V2 + 17V3 = 50

After doing the matrix inversion, I ended up with V1 = 2.85556V, V2= 0.891089V and V3 = 3.3819V.

I built the circuit on my computer and I got v1 = 2.8693V, v2 = 0.625V , v3 = 1.5057V
EDIT:

I messed up on my calculations above. I'll re-do them again.
 
Last edited:
  • #5
The Electrician
Gold Member
1,283
169
Okay, I redid my work and I'm still getting the wrong answer. In my previous post I mistakenly wrote my answers as current values instead of voltages sinceI'm calculating the voltages at the extraordinary nodes.

17V1 - 2V2 - 2V3 = 40
-2V1 + 14V2 -2V3 = 0
-2V1 - 2V2 + 17V3 = 50

After doing the matrix inversion, I ended up with V1 = 2.85556V, V2= 0.891089V and V3 = 3.3819V.

I built the circuit on my computer and I got v1 = 2.8693V, v2 = 0.625V , v3 = 1.5057V
EDIT:

I messed up on my calculations above. I'll re-do them again.

You haven't done your algebra right. You should have:

17V1 - 2V2 - 5V3 = 40
-2V1 + 14V2 -2V3 = 0
-5V1 - 2V2 + 17V3 = 10
 
  • #6
120
0
Yeah, I realized that after I posted. I got the correct values now, thanks for the help.

What confused me was the top part of the circuit, specifically the 4V source between the two extraordinary nodes. When I look at circuit diagrams I usually make the mistake of assuming the corners are all nodes. So I treated the corner right above V2 as a node. I make a lot of careless mistakes.
 

Related Threads on Using node-voltage analytical technique

Replies
3
Views
3K
  • Last Post
Replies
2
Views
497
Replies
2
Views
714
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
2
Views
3K
Replies
7
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
823
Replies
6
Views
4K
Top