# Using Nodal Analysis Method

I had solved easier circuits problem using nodal analysis but this one it has supernodes on nodes 2 and 3 which I find it difficult solving. thus I need help as I am trying my very best to solve.

i think we had two supernode and applying kCL see below. • I think we had two supernodes. – Surdz Sep 21 '17 at 3:29

I get confused reading what you did. Seriously, you are making this way too complicated.

You should have used the included schematic editor. A picture is nice, but the editor makes it easy to number things. I'd have appreciated the addition, anyway. You might want to consider the time of those from whom you ask for help.

Here's your circuit, below. Actually, twice. The left side is your schematic. The right side is where I added a ground node. I wanted to emphasize the fact that you can do that anywhere you decide to do it. You get to pick any node you want and call it zero! Since you brought up super nodes and seem to be making a big deal about them, I decided to pick the spot I chose in order to break you of some of that worry. It causes an extra node to "show up," but for now I think it is more important that you get broken of so much worry about super nodes. simulate this circuit – Schematic created using CircuitLab

(I added an unknown current because, well, we don't know that current. It's $I_X$, for now. I picked the direction, arbitrarily. It doesn't matter, so long as I apply it consistently.)

Here's a very quick run-down on what we know, plus the nodal equations:

\begin{align*} V_1&=-8\:\textrm{V}\\\\ V_X&=V_3+6\:\textrm{V}\\\\ \frac{V_2}{R_1}+\frac{V_2}{R_2}+\frac{V_2}{R_5}&=\frac{V_1}{R_1}+\frac{V_X}{R_2}+\frac{V_3}{R_5}\tag{V_2 node}\\\\ \frac{V_3}{R_5}+\frac{V_3}{R_4}+I_X&=\frac{V_2}{R_5}+\frac{V_1}{R_4}\tag{V_3 node}\\\\ \frac{V_X}{R_2}+\frac{V_X}{R_3}&=I_X+\frac{V_2}{R_2}+\frac{0\:\textrm{V}}{R_3}\tag{V_X node} \end{align*}

The first two are more than obvious. If I have to explain them, we have a different problem entirely.

The remaining three should be seen as "currents spilling outward from the node" on the left side which must be equal to the "currents spilling inward into the node" on the right side. If you have a current that points outward, add it to the left side. If you have a current that points inward, it goes on the right side.

Move from node to node and just write equations really fast.

Very easy mental picture that keeps everything straight.

A few very minor substitutions leaves us with:

\begin{align*} \frac{V_2}{R_1}+\frac{V_2}{R_2}+\frac{V_2}{R_5}&=\frac{-8\:\textrm{V}}{R_1}+\frac{V_3+6\:\textrm{V}}{R_2}+\frac{V_3}{R_5}\tag{V_2 node}\\\\ \frac{V_3}{R_5}+\frac{V_3}{R_4}+I_X&=\frac{V_2}{R_5}+\frac{-8\:\textrm{V}}{R_4}\tag{V_3 node}\\\\ \frac{V_3+6\:\textrm{V}}{R_2}+\frac{V_3+6\:\textrm{V}}{R_3}&=I_X+\frac{V_2}{R_2}+\frac{0\:\textrm{V}}{R_3}\tag{V_X super node} \end{align*}

Which is, of course, three equations and three unknowns and easily solvable.

Once you know $V_3$, then you also know $V_X$. And knowing $V_X$ you can now compute the current you want as $i=\frac{-V_X}{R_3}$.

I could pick any of the nodes as ground and label the rest and achieve the same thing. If you look at the substitutions I did, you can see the so-called "super nodes" appearing because of those substitutions. But it's all just so much bookkeeping, really. Nothing magical.