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CIRCUIT ANALYSIS: 3 resistrs, I src, V src, V.C.V.S. in series w/ resistor - Find V_A

  • Engineering
  • Thread starter VinnyCee
  • Start date
489
0
1. Homework Statement

Use nodal analysis to find [itex]V_A[/itex] in the circuit below.

http://img180.imageshack.us/img180/7004/chapter3problem236vy.jpg [Broken]


2. Homework Equations

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


3. The Attempt at a Solution

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

http://img201.imageshack.us/img201/4716/chapter3problem23part22nf.jpg [Broken]

[tex]V_1\,=\,V_A[/tex] <---- Right?

[tex]I_1\,=\,\frac{V_1\,-\,0}{2\Omega}[/tex]

[tex]I_2\,=\,\frac{V_3\,-\,0}{16\Omega}[/tex]

[tex]I_3\,=\,\frac{V_1\,-\,V_2}{4\Omega}[/tex] <------Right?


I know KCL for [itex]V_1[/itex]:

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

[tex]7\,V_1\,-\,V_2\,=\,120[/tex]


I know KCL for [itex]V_3[/itex]:

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

[tex]4\,V_1\,-\,4\,V_2\,-\,V_3\,=\,-48[/tex]


For KVL1:

[tex]-V_1\,+\,4\,I_3\,+2\,V_A\,+\,V_3\,=\,0[/tex]

[tex]2\,V_1\,-\,V_2\,+\,V_3\,=\,0[/tex]


Now, putting those three EQs into a matrix and rref:

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

This gives me [itex]V_1\,=\,V_A[/itex] equal to 22.34 V.

Thanks for the help mjsd!
 
Last edited by a moderator:

Answers and Replies

mjsd
Homework Helper
725
3
all your equations look sound to me..
 
mjsd
Homework Helper
725
3
oh..hang on.... 2nd line for KCL for V3 is wrong where is the 48 gone?
 
489
0
Corrected, thanks much!
 

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