# Node Voltage Analysis

1. Aug 10, 2011

### ohdrayray

1. The problem statement, all variables and given/known data
Solve for V1, V2, V3 and V4 (in decimals) using node voltage analysis method for the following:

2. Relevant equations:
Node Voltage Analysis

3. The attempt at a solution:
For Node #1:

$V_{1}$ = 6 V

For Node #4:

$V_{4}$ = 16 V

For Node #2:

$\frac{V_{2}-V_{3}}{3300}$ + $\frac{V_{2}}{1000}$ + $\frac{V_{2}-6}{1500}$ = 0

0.00197$V_{2}$ - 0.000303$V_{3}$= 0.004 --> equation 1

For Node #3:

$\frac{V_{3}-V_{2}}{3300}$ + $\frac{V_{3}}{4700}$ + $\frac{V_{3}-16}{2200}$ = 0

-0.000303$V_{2}$+0.00097$V_{3}$ = 0.007273

$V_{3}$= 0.3124$V_{2}$ + 7.4979 --> equation 2

Solve for $V_{2}$ by substituting $V_{3}$ into equation 1:

0.00197$V_{2}$-0.000303(0.3124$V_{2}$ + 7.4979) = 0.004

0.001875$V_{2}$ - 0.006272 = 0

$V_{2}$ = 3.3451 V

Solve for $V_{3}$ by substituting $V_{2}$ into equation 2:

$V_{3}$= 0.3124(3.3451) + 7.4979

$V_{3}$ = 8.5428 V

I mainly just wanted to know if my equations were right, I think that I've done it correctly, but at the same time I'm not sure, haha. Thank you in advance!

2. Aug 10, 2011

### Staff: Mentor

Hi ohdrayray, welcome to Physics Forums.

Your formulae and methodology are fine. The results are good to two decimal places.