# Using node-voltage analytical technique

1. Sep 24, 2012

### november1992

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

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

3. The attempt at a solution

For I1+I2+I3=0:

$\frac{V1-4}{0.2}$ + $\frac{V1-V2}{0.5}$ + $\frac{V1-2}{0.1}$ = 0

For I4+I5+I6=0:

$\frac{V2-V1}{0.5}$ + $\frac{V2-V3}{0.5}$ + $\frac{V2}{0.1}$ = 0

For I7+I8+I9=0:

-$\frac{V1-4}{0.2}$ + $\frac{V3-V2}{0.5}$ + $\frac{V3-3}{0.1}$ = 0

($\frac{1}{0.2}$ + $\frac{1}{0.5}$ + $\frac{1}{0.1}$)V1 - ($\frac{1}{0.2}$ + $\frac{1}{0.5}$)V2 + 0V3 = 40

($\frac{1}{0.5}$+$\frac{1}{0.5}$+$\frac{1}{0.1}$)V2 - ($\frac{1}{0.5}$)V1 + (-$\frac{1}{0.5}$)V3 = 0

(-$\frac{1}{0.2}$)V1 + (-$\frac{1}{0.5}$)V2 + $\frac{1}{0.5}$ + $\frac{1}{0.1}$)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.

2. Sep 25, 2012

### Simon Bridge

3. Sep 25, 2012

### The Electrician

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

For I1+I2+I3=0:

$\frac{V1-4-V3}{0.2}$ + $\frac{V1-V2}{0.5}$ + $\frac{V1-2}{0.1}$ = 0

And this one:

For I7+I8+I9=0:

-$\frac{V1-4-V3}{0.2}$ + $\frac{V3-V2}{0.5}$ + $\frac{V3-3}{0.1}$ = 0

4. Sep 25, 2012

### november1992

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: Sep 25, 2012
5. Sep 25, 2012

### The Electrician

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

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

6. Sep 25, 2012

### november1992

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.