- #1

- 6

- 0

## 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!