Solve V_A in Circuit with Nodal Analysis

[VinnyCee]

Homework Statement

Use nodal analysis to find [itex]V_A[/itex] 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

[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!

 

[mjsd]

all your equations look sound to me..

 

[mjsd]

oh..hang on... 2nd line for KCL for V3 is wrong where is the 48 gone?

 

[VinnyCee]

Corrected, thanks much!