Basically KCL says that the same amount of current entering a node must leave it. So for instance, if we label the R1/R3/5A as V1 and the R2/R3/3A as V2:

Node 1's equation can be described as:

\$ \dfrac{V1}{2} + \dfrac{V1 - V2}{8} = -5A \$

Node 2's equation can be given as:

\$ \dfrac{V2}{4} + \dfrac{V2 - V1}{8} = -3A \$

Rearranging, you have:

\$ \dfrac{5V1 - V2}{8} = (0.625 * V1) - (0.125 * V2) = -5A \$

and

\$ \dfrac{3V2 - V1}{8} = (0.375 * V2) - (0.125 * V1) = -3A \$

Solving the simultaneous equations gives us:

\$ V1 = -10.28V, V2 = -11.42V \$

AFAIK, you can always solve any linear circuit the 'brute force' way using nodal analysis:

- Write Kirchoff's Current Equations on all nodes except ground
- For every circuit component, (i.e. resistors, capacitors etc.), write down their behaviour (for instance, ohm's law for a resistance, i = c dV/dt for a capacitance and so on)
- At this point, we'll have a handful of equations with us. We can also try to eliminate as many equations from them as possible using any info we have; however in the end, we need to be left with
**N** simultaneous equations in **N** unknowns. Solve them and we'll get all the node voltages and branch currents.

Coming to the circuit above, let's define the current through V2 as I2, and the ones through R_n as I_n. Let me also call the node at the top as V_a, the node between the CCCS and R5 as V_c, the one between the R_7 and R_8 as V_b and the node in the middle as V_e. Now, writing Kirchoff's Current Law on these nodes will leave us with
$$
I_2 = I_7 + I_5\\
I_7 = I_8 + I_x\\
I_x + I_5 = I_6
$$ respectively.
Writing down the 'behaviour' of R6, R5, R7, R8, V2, V3 and the CCCS will respectively yield
$$
V_E = I_6 R_6 \\
V_A - V_C = I_5 R_5 \\
V_A - V_B = I_7 R_7 \\
V_B = I_8 R_8 \\
V_A = V_2 \\
V_B = V_E + 0.7\\
I_5 = 180 I_x
$$

That's 10 linear equations in 10 unknowns. Solve them, and we'll find all I_x as 88.18uAmps...

Of course, 10 equations is a bit too much to solve by hand (I generally use Gauss-Jordan elimination to do this part), but as far as I've seen, this method works in situations where the usual 'text-book' approach using nodal and mesh analyses fail. Furthermore, we don't have to deal with the painful Thevenin equivalent/Super-mesh workarounds here...

On the downside, I'm not quite sure if this approach works with **every possible** circuit (so far I haven't seen any where it fails), so any negative feedback on this part is welcome :)

## Best Answer

Enough time has passed, so I'll write up something for this now.

Just a couple of notes that I'll be discussing shortly, but want to bring up front right now:

Let's start by selecting a convenient ground reference (the bottom wire of your schematic) and re-draw the schematic:

^{simulate this circuit – Schematic created using CircuitLab}I've color-coded each box to make them stand out more distinctly, but I've also included a name for each box below to help those who may not see color differences as well as I do.

First thing to note is that I've simplified the diagram, somewhat, by removing all the busing of wire around for your bottom wire in your schematic. So long as you keep in mind that these ground connections are still connected to each other, the advantage here is that there's some distracting lines removed now and it's a little bit easier to "think about."

Looking at BOX 1, the reason you can eliminate (short across with a wire) \$V_1\$ is that the current won't change no matter what the voltage value happens to be. Whatever the value of \$V_a\$ may happen to be, the only thing \$V_1\$ does is change the voltage drop across \$I_1\$ and the current source \$I_1\$ doesn't care about that. It still provides the same current, regardless of its voltage drop. So \$V_1\$ is completely irrelevant for the purposes of analyzing \$V_a\$ or \$V_b\$. You can lose \$V_1\$ and what happens is that the voltage drop across \$I_1\$ changes by just that much. So what? Just remove it.

Note that you wrote that you are:

So, I'm just telling you here and now that it should NOT confuse you. Just get rid of it. However, in the interest of showing you how it doesn't matter I'll keep it in place for now. You probably need more convincing, not just hand-waving about it. So let's keep it in BOX 1 for now.

(The same arguments would go for BOX 3. But again, let's leave things as they are.)

BOX 2 and BOX 4 are each voltage sources in series with a resistor. These could technically be replaced with a Norton equivalent, with current sources and a parallel resistor. At times, that approach may help simplify the analysis. This might actually be one of those cases, too -- but again I think I'll just leave things as they are.

Okay. So that's it for my hand-waving. Outside of that, all I've really done here is re-draw the schematic. So now let's analyze it.

Here are the nodal equations to solve (out-flowing currents on the left side, in-flowing currents on the right side):

$$\begin{align*} \frac{V_a}{R_1}+\frac{V_a}{R_2}&=\frac{V_2}{R_1}+\frac{V_b}{R_2}+I_1\\\\ \frac{V_b}{R_2}+\frac{V_b}{R_3}+I_2&=\frac{V_a}{R_2}+\frac{V_3}{R_3} \end{align*}$$

Note that in the above pair of equations (the two unknowns are \$V_a\$ and \$V_b\$), neither \$V_1\$ nor \$R_4\$ show up. As I said, they are irrelevant for the reasons I talked about. You can see that \$I_1\$ shows up in the first equation as an in-flowing current. Take careful note that this in-flowing current is completely unaffected by \$V_1\$. You can also see that \$I_2\$ shows up in the second equation as an out-flowing current. And again take careful note that this out-flowing current is completely unaffected by \$R_4\$.

The above equations can be re-arranged into a more standard form for solution:

$$\begin{align*} \frac{V_a}{1\:\Omega}+\frac{V_a}{3\:\Omega}&=\frac{-3\:\text{V}}{1\:\Omega}+\frac{V_b}{3\:\Omega}+3\:\text{A}\\\\ V_a\cdot\left(1+\frac13\right)+V_b\cdot\left(-\frac13\right)&=0\:\text{A}\\\\ &\text{and,}\\\\ \frac{V_b}{3\:\Omega}+\frac{V_b}{2\:\Omega}+3\:\text{A}&=\frac{V_a}{3\:\Omega}+\frac{+2\:\text{V}}{2\:\Omega}\\\\ V_a\cdot\left(-\frac13\right)+V_b\cdot\left(\frac13+\frac12\right)&=-2\:\text{A} \end{align*}$$

At this point, you should find it very easy to work out that \$V_a=-\frac23\:\text{V}\$ and \$V_b=-\frac83\:\text{V}\$. Hence, the voltage across \$R_2\$ has the magnitude of \$V_a-V_b=+2\:\text{V}\$ and from this it should be trivial to calculate the indicated current you want, \$I_{a\:b}\$.

In case it helps, here's a highly simplified equivalent schematic:

^{simulate this circuit}Note that in the above case, I didn't even bother with showing anything at all on one side of the current sources shown there. There's no point. It doesn't matter. It's enough to know the value of the current sources, and direction.

There's more that can be done, now. Let's first take my above comment, where I don't even

"bother with showing anything at all on one side of the current sources", and now make a small change expanding on that idea:^{simulate this circuit}As I said, it doesn't matter. So it's just fine to hook up the other end of any current source to any voltage source I want. In this case, I just arranged it in a way that should make the next step obvious.

At this point, we can convert what now appears to be obvious Norton sources into Thevenin sources. You can treat the left side consisting of \$I_1\$ and \$R_1\$ as a Norton source and convert it to a Thevenin source, instead. You can similarly also treat the right side consisting of \$I_2\$ and \$R_3\$ as yet another Norton source and convert it to a Thevenin source, too. The result is this:

^{simulate this circuit}Now the current is trivial to compute:

$$I_{a\:b}=\frac{0\:\text{V}-\left(-4\:\text{V}\right)}{1\:\Omega+3\:\Omega+2\:\Omega}=\frac{4\:\text{V}}{6\:\Omega}=\frac{2}{3}\:\text{A}$$

So, as you can see there are a number of ways to get the same answer. And, in fact, one way doesn't even require the simultaneous solution of two equations with two unknowns. It just becomes a nearly trivial series circuit to solve with Ohm's Law.