# (Superposition/nodal analysis Problem) This solution makes no sense to me

#### jshrimp

Joined Nov 2, 2019
5
I asked my teacher, and they said they checked and insist there is absolutely nothing wrong with the solution. Can someone point out the mistake I am making?

The second set of equations in particular, make no sense. Relating to V2.
How do you get V2/2=6+(4ix-V2)/8 from that second diagram?
In that second diagram, the solution is equating Ix with V2/2. Is that correct?
Then from that equation, you get v2/2 = 6 + (2v2 – v2)/8 which apparently leads to v2 = -16 (according to the solution). Except that it doesn't. It leads to V2 = 16.

Then it says that ix=Vx/2. I don't understand where that came from. If I assume ground is to the right of the 4ix voltage source, the equation I come up with for ix is ix=(4ix+Vx)/2 -> ix = -Vx/2

If I disregard the solution, and make my own equations, my final answer is always Ix=13.333A, instead of -13.333A.
Can someone tell me what I'm doing wrong?

First set of equations:
V1/8 + (V1-(-4ix))/2 =4

ix=(v1+4ix)/2 -> ix=-v1/2

V1 = (-32/3)

Second set of equations:
(4ix+V2)/2 = 6 + (4ix-4ix-V2)/8

ix = (v2+4ix)/2 -> ix= -v2/2

V2=-16

----
Vx = V1+V2 = -26.67...V

ix=(4ix+Vx)/2

ix=-Vx/2=-(-26.67)/2=13.33A

#### RBR1317

Joined Nov 13, 2010
674
I think you have the right answer. I get the same using nodal analysis. The circuit has one supernode (yellow), so there is only one node equation. And there is one constraint equation because of the dependent source. I have not substituted the results back into the circuit for verification; but I trust nodal analysis.

#### StayatHomeElectronics

Joined Sep 25, 2008
1,070
Not so sure I like the way that V2 is defined in the solutions, it is defined as the node voltage to ground but then also as the voltage over the resistor. It can not be both.

• jshrimp

#### RBR1317

Joined Nov 13, 2010
674
I asked my teacher, and they said they checked and insist there is absolutely nothing wrong with the solution.
If there is nothing wrong with the given solution, what happens when you substitute a negative value for Ix back into the circuit?