Nodal Analysis/Superposition Problem

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Homework Statement



***NOTE: SEE 3RD POST FOR PROBLEM STATEMENT!***

Homework Equations



N/A.

The Attempt at a Solution



Okay, I have all my work for both my superposition attempt and my nodal analysis attempt and the [tex]V_{o}[/tex] I found are not the same.

This leads me to believe I've made a mistake(if not several) in either my superposition attempt, nodal analysis attempt or both.

There is a typo on the first page of my nodal analysis attempt for i4 it says,

[tex]i_{4} = \frac{V_{B} - V_{3}}{3R}[/tex]

when it should be,

[tex]i_{4} = \frac{V_{3} - V_{B}}{3R}[/tex]

I used the correct i4 throughout all my calculations.

Any mistakes you find, tips/suggestions/comments are greatly appreciated.

Thanks again!


***This post contains the my work for the superposition attempt (see figures)***
 
Attachments
  • Handwritten nodal analysis on graph paper: a circuit with resistors 2R, R, 2R, 3R and source V1, nodes A and B, KCL equations, and solved result V_B = 12V1/43.
    A2Q1A1P4.jpg
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  • Handwritten circuit derivation: top, a ladder network of resistors labelled 2R, R, R, 2R, 2R with source V2 and output Vo; below, a simplified circuit with 2R, 2R, R and source V2, followed by equations 3R//2R = 6R^2/5R = 6R/5, 2R + 6/5R = 16R/5, i0 = 5V2/16 and Vo = 2R(2/5 i0), so Vo = V2/4.
    A2Q1A1P5.jpg
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  • Handwritten circuit analysis on graph paper: a resistor network with 2R, R and 2R resistors feeding a dependent voltage source V3 = 4R io, with the equation Vo = 8V1/4? + V2/4 + V3/2 written below.
    A2Q1A1P6.jpg
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***This post contains the my work for the nodal analysis attempt (see figures)***
 
Attachments
  • Handwritten nodal analysis on grid paper: a circuit schematic with resistors 2R, R, R, 3R and nodes A, B, source voltages V1, V2, V3 and output V0, followed by KCL equations at nodes A and B and simplified node equations.
    A2Q1A1.jpg
    30.9 KB · Views: 442
  • Handwritten nodal analysis derivation on grid paper: using equation 1, V_A = V1/4 + V_B/2 is substituted into a node equation, then V_B is solved as (4V3 + 6V2 + 3V1)/16.
    A2Q1A1P2.jpg
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  • Handwritten derivation on graph paper of an output voltage V0 in terms of V1, V2 and V3, using i4 = (V3 - VB)/3R, with the boxed result V0 = 12V3 - 2V2 - V1 over 8, equivalently 3/2 V3 - 1/4 V2 - 1/8 V1.
    A2Q1A1P3.jpg
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Problem Statement see figure attached.
 
Attachments
  • Handwritten problem with a resistor ladder circuit: 2R, R and R across the top, 2R in series with sources v1, v2 and v3 down each branch, output vo at the right, and Req1, Req2 and Req3 marked. Problem text above asks for vo and Req.
    A2Q1.jpg
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