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

AI Thread Summary
The discussion focuses on solving for node voltages V1, V2, V3, and V4 using the node voltage analysis method. The values determined are V1 = 6 V, V4 = 16 V, V2 = 3.35 V, and V3 = 8.54 V. The equations derived for nodes 2 and 3 are confirmed to be correct, leading to accurate results. The methodology and calculations presented are validated, with results acceptable to two decimal places. The participant expresses uncertainty about their work but receives confirmation of its accuracy.
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:

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

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


For Node #3:

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

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

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


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

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

0.001875V_{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!
 
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Hi ohdrayray, welcome to Physics Forums.

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