View Single Post
VinnyCee
#1
Jan22-07, 07:06 PM
P: 492
1. The problem statement, all variables and given/known data

Determine voltages [itex]V_1[/itex] through [itex]V_3[/itex] in the circuit below.




2. Relevant equations

KCL, V = iR


3. The attempt at a solution

So I added some variables to represent currents and a super-node. The variables are in red and the super-node in light-blue.



[tex]V_A\,=\,V_2[/tex]

[tex]V_3\,=\,13\,V[/tex]

[tex]I_1\,=\,\frac{V_1\,-\,V_3}{\frac{1}{2}\Omega}\,=\,2\,V_1\,-\,2\,V_3[/tex]

[tex]I_2\,=\,\frac{V_1\,-\,0}{1\Omega}\,=\,V_1[/tex]

[tex]I_3\,=\,\frac{V_2\,-\,0}{\frac{1}{4}\Omega}\,=\,4\,V_2[/tex]

[tex]I_4\,=\,\frac{V_2\,-\,V_3}{\frac{1}{8}\Omega}\,=\,8\,V_2\,-\,8\,V_3[/tex]

Now I use KCL at the super-node:

[tex]I_1\,+\,I_2\,+\,I_3\,+I_4\,=\,2\,A[/tex]

[tex](2\,V_1\,-\,2\,V_3)\,+\,(V_1)\,+\,(4\,V_2)\,+\,(8\,V_2\,-\,8\,V_3)\,=\,2\,A[/tex]

[tex]3\,V_1\,+\,12\,V_2\,-\,10\,V_3\,=\,2\,A[/tex]

[tex]3\,V_1\,+\,12\,V_2\,-\,10(13\,V)\,=\,2\,A[/tex]

[tex]3\,V_1\,+\,12\,V_2\,=\,132[/tex]

And get the voltage equation from inside the super-node:

[tex]V_1\,-\,V_2\,=\,2\,V_A[/tex]

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

[tex]V_1\,-\,3\,V_2\,=\,0[/tex]

Now put into a matrix and rref to get [itex]V_1[/itex] and [itex]V_2[/itex]:

[tex]\left[\begin{array}{ccc}3&12&132\\1&-3&0\end{array}\right]\,\,\longrightarrow\,\,\left[\begin{array}{ccc}1&0&\frac{132}{7}\\0&1&\frac{44}{7}\end{array}\right][/tex]

So I get these for [itex]V_1[/itex] through [itex]V_3[/itex]:

[tex]V_1\,=\,\frac{132}{7}\,V\,\approx\,18.86\,V[/tex]

[tex]V_2\,=\,\frac{44}{7}\,V\,\approx\,6.286\,V[/tex]

[tex]V_3\,=\,13\,V[/tex]

Does this look right?
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100