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

• Engineering
1. Homework Statement

Use nodal analysis to find $V_A$ 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]

$$V_1\,=\,V_A$$ <---- 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 $V_1$:

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

$$7\,V_1\,-\,V_2\,=\,120$$

I know KCL for $V_3$:

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

$$4\,V_1\,-\,4\,V_2\,-\,V_3\,=\,-48$$

For KVL1:

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

$$2\,V_1\,-\,V_2\,+\,V_3\,=\,0$$

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

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

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

Thanks for the help mjsd!

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mjsd
Homework Helper
all your equations look sound to me..

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

Corrected, thanks much!