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

ohdrayray
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Homework Statement


Solve for V1, V2, V3 and V4 (in decimals) using node voltage analysis method for the following:
108csn4.png



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!
 
on Phys.org
Hi ohdrayray, welcome to Physics Forums.

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

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