Performing Node analysis with more

Question

Following a previous question problem regarding the node analysis with dependent source in a circuit I made some extensions that I would appreciate some validation if someone know.

I have the following circuit as shown in picture. I think that node equation would be: \$ -10 + \frac{V_{0}}{4} + \frac{V_{0} - 2 i_{1}}{6} = 0 \$ , with \$i_{1} = \frac{V_{0}}{6}\$. Am I correct?

Additionally I added another resistor a 3 Ω in parallel.I think that now I have 2 node equations:

\$-10 + \frac{V_{1}}{4}\ +\frac{V_{1}-V_{2}}{2i_{1}}=0\$

\$\frac{V_{2}-V_{1}}{2i_{1}} + \frac{V_{2}}{6} + \frac{V_{2}}{3}=0\$

with \$i_{1}\$ this time to be with current division (since 3 and 6 Ω resistors are in parallel): \$i_{1} = \frac{3}{9}*10 = 30/9\$ Can someone correct me if I am wrong? I would appreciate it.

Answer 1

I'm really impressed, looking at the chat log. I thought I'd add a few ways I might have written this out for the first case you worked out on your own:

Let's first keep the KVL constraint you mention:

Then:

$$\begin{align*} &\rlap{\text{The currents sinking in the resistors must equal the source:}}\\\\ &\frac{V_o}{4\:\Omega}+\frac{V_x=V_o+2\cdot i_1}{6\:\Omega}=10\:\text{A}\\\\ &\rlap{\text{But }}\quad\quad i_1=\frac{V_x}{6\:\Omega}=\frac{V_o+2\cdot i_1}{6\:\Omega},\text{ or, solving for }i_1, \text{ find } i_1=\frac14 V_o:\\\\ &\frac{V_o}{4\:\Omega}+\frac{V_o+2\cdot \frac14 V_o}{6\:\Omega}=10\:\text{A} \end{align*}$$

That's it. This solves out as \$V_o=20\:\text{V}\$. And from there you can work out \$i_1\$ and also \$V_x\$.

Or just start this way, instead:

$$\begin{align*} &&i_1&=\frac{V_x}{6\:\Omega}=\frac{V_o+2\cdot i_1}{6\:\Omega}\\\\ &&\therefore i_1&=\frac14 V_o\\\\ &&\therefore V_o&=4\cdot i_1\\\\ \rlap{\text{KCL:}}\\\\ &&\frac{V_o}{4\:\Omega}+i_1&=10\:\text{A}\\\\ &&\frac{4\cdot i_1}{4\:\Omega}+i_1&=10\:\text{A}\\\\ &&i_1+i_1&=10\:\text{A}\\\\ &&i_1&=5\:\text{A} \end{align*}$$

From that you know right away that \$V_x=30\:\text{V}\$ and that \$V_o\$ is \$2\cdot i_1\$ or \$10\:\text{V}\$ less, so \$V_o=20\:\text{V}\$.

And, finally, you could have just assigned a new unknown current variable, through the dependent source, and developed two KCL equations using that unknown current, plus your constraint equation, and solved three equations in three unknowns to get the same results, as well.

Lots of ways of going. Just wanted to point this out to help stretch and exercise your thinking processes when looking at a problem.