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

1. Jan 22, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

Use nodal analysis to find $V_A$ in the circuit below.

2. Relevant 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.

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

Last edited by a moderator: Apr 22, 2017 at 3:18 PM
2. Jan 22, 2007

### mjsd

all your equations look sound to me..

3. Jan 22, 2007

### mjsd

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

4. Jan 22, 2007

### VinnyCee

Corrected, thanks much!