Node Voltage Analysis: Solve for V1, V2, V3 and V4

[ohdrayray]

Homework Statement

Solve for V1, V2, V3 and V4 (in decimals) using node voltage analysis method for the following:

Homework Equations

:
Node Voltage Analysis

The Attempt at a Solution

:
For Node #1:

[itex]V_{1}[/itex] = 6 V


For Node #4:

[itex]V_{4}[/itex] = 16 V


For Node #2:

[itex]\frac{V_{2}-V_{3}}{3300}[/itex] + [itex]\frac{V_{2}}{1000}[/itex] + [itex]\frac{V_{2}-6}{1500}[/itex] = 0

0.00197[itex]V_{2}[/itex] - 0.000303[itex]V_{3} [/itex]= 0.004 --> equation 1


For Node #3:

[itex]\frac{V_{3}-V_{2}}{3300}[/itex] + [itex]\frac{V_{3}}{4700}[/itex] + [itex]\frac{V_{3}-16}{2200}[/itex] = 0

-0.000303[itex]V_{2}[/itex]+0.00097[itex]V_{3}[/itex] = 0.007273

[itex]V_{3} [/itex]= 0.3124[itex]V_{2}[/itex] + 7.4979 --> equation 2


Solve for [itex]V_{2}[/itex] by substituting [itex]V_{3}[/itex] into equation 1:

0.00197[itex]V_{2}[/itex]-0.000303(0.3124[itex]V_{2}[/itex] + 7.4979) = 0.004

0.001875[itex]V_{2}[/itex] - 0.006272 = 0

[itex]V_{2}[/itex] = 3.3451 V


Solve for [itex]V_{3}[/itex] by substituting [itex]V_{2}[/itex] into equation 2:

[itex]V_{3} [/itex]= 0.3124(3.3451) + 7.4979

[itex]V_{3}[/itex] = 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!

 

[gneill]

Hi ohdrayray, welcome to Physics Forums.

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