# Superposition Theorem/Nodal Analysis

I'm having a little trouble understanding this worked solution.

My understanding of superposition is that we "turn off" every independent power source except the one we plan on analysing, then we superimpose all the individual results to get the real result. But it's in the nodal analysis when we leave the current source on and turn the others off that's getting me.

The solution implies two node voltages when to me there are 3 junctions/nodes. They seem to just follow the current through the 8 ohm and the leftmost 4 ohm resistors all the way to the ground. Is this because there is no branch between the ground node and the other node voltages?

Also, is there a reason other than time saving that we use mesh analysis for the voltage sources and node analysis for the current source?

Sorry for all the questions, but when we remove the current source, I've never understood how the remaining open circuit works. Do we simply disregard the wires that remain in figure (b)? If yes, why do they leave it there?

One last thing I am little shaky on choosing at which junction the make a reference node, is there a reason the solution chose the one it did?

Thanks so much!

In figure (c), there are indeed 3 nodes: one in the middle (v1, common node of the two 4$\Omega$ and the 3$\Omega$ resistor), one on "top" of the current source (v2), and the one connected to the "bottom" of the source. They decided to make it the reference node, valued 0V for easier calculations. They could have chosen any other node for this, but generally you tend to choose the one on the bottom of your schematic. In most cases, this provides you the easiest solution.
Regarding your question on how they solved (c): From the top of your source, there is 3A flowing out. That has two ways to go from there: through the 8$\Omega$ resistor or through the other branch. The sum of these two, obviously, is 3A as well. The solution ignores the fact that the second branch also divides later on, because the current flowing through the rightmost 4$\Omega$ resistor equals to the sum of currents flowing through the lefmost 4$\Omega$ resistor and the 3$\Omega$ one. If you look carefully, the voltage drop v2-v1 is through the rightmost, not the leftmost 4$\Omega$ resistor.
• Indeed, there is no component between that junction and the reference point – that means they are the same. Any piece of continuous, uninterrupted wire has the same potential its whole length long. (In reality, wire has a small resistance, and thus some voltage drop across it, but you can ignore it in a case like this.) v2/8 is just (v2-0)/8 simplified, so yes, the voltage drop across the 8$\Omega$ resistor is from v2 to the reference node. – hryghr Aug 19 '14 at 9:07