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Node Voltage Analysis

  1. Aug 10, 2011 #1
    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:
    108csn4.png


    2. Relevant equations:
    Node Voltage Analysis

    3. 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!
     
  2. jcsd
  3. Aug 10, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi ohdrayray, welcome to Physics Forums.

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