Solve V_A in Circuit with Nodal Analysis

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Discussion Overview

The discussion focuses on using nodal analysis to determine the voltage V_A in a given electrical circuit. Participants explore the application of Kirchhoff's laws and the setup of equations necessary for solving the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • The initial setup of currents and application of Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) is presented by the first participant.
  • Equations for currents I_1, I_2, and I_3 are proposed, with some participants questioning the correctness of these equations.
  • A matrix representation of the equations is created, leading to a calculated value for V_A of 22.34 V.
  • One participant points out a potential error in the KCL equation for V_3, specifically regarding the omission of a constant value.
  • The first participant acknowledges the correction and thanks the contributor for their input.

Areas of Agreement / Disagreement

Participants generally agree on the soundness of the equations presented, although there is a specific disagreement regarding the accuracy of the KCL equation for V_3, which is later corrected by the original poster.

Contextual Notes

The discussion includes a potential oversight in the formulation of the KCL equation for V_3, which may affect the overall solution. The reliance on specific circuit parameters and the interpretation of the circuit diagram are also implicit in the discussion.

VinnyCee
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Homework Statement



Use nodal analysis to find V_A in the circuit below.

http://img180.imageshack.us/img180/7004/chapter3problem236vy.jpg

Homework Equations



KVL, KCL, v = i R, super-node?

The Attempt at a Solution



I made a few currents and showed the obvious KVL loop.

http://img201.imageshack.us/img201/4716/chapter3problem23part22nf.jpg

V_1\,=\,V_A <---- Right?

I_1\,=\,\frac{V_1\,-\,0}{2\Omega}

I_2\,=\,\frac{V_3\,-\,0}{16\Omega}

I_3\,=\,\frac{V_1\,-\,V_2}{4\Omega} <------Right?I know KCL for V_1:

\frac{30\,-\,V_1}{1\Omega}\,=\,\frac{V_1\,-\,0}{2\Omega}\,+\,\frac{V_1\,-\,V_2}{4\Omega}

7\,V_1\,-\,V_2\,=\,120I know KCL for V_3:

\frac{V_1\,-\,V_2}{4\Omega}\,+\,3\,=\,\frac{V_3\,-\,0}{16\Omega}

4\,V_1\,-\,4\,V_2\,-\,V_3\,=\,-48For KVL1:

-V_1\,+\,4\,I_3\,+2\,V_A\,+\,V_3\,=\,0

2\,V_1\,-\,V_2\,+\,V_3\,=\,0Now, putting those three EQs into a matrix and rref:

\left[\begin{array}{cccc}7&amp;-1&amp;0&amp;120\\4&amp;-4&amp;-1&amp;-48\\2&amp;-1&amp;1&amp;0\end{array}\right]\,\,\longrightarrow\,\,\left[\begin{array}{cccc}1&amp;0&amp;0&amp;\frac{648}{29}\\0&amp;1&amp;0&amp;\frac{1056}{29}\\0&amp;0&amp;1&amp;-\frac{240}{29}\end{array}\right]

This gives me V_1\,=\,V_A equal to 22.34 V.

Thanks for the help mjsd!
 
Last edited by a moderator:
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all your equations look sound to me..
 
oh..hang on... 2nd line for KCL for V3 is wrong where is the 48 gone?
 
Corrected, thanks much!
 

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