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

Thread Starter

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
 

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RBR1317

Joined Nov 13, 2010
555
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.
 

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Thread Starter

jshrimp

Joined Nov 2, 2019
5
If there is nothing wrong with the given solution, what happens when you substitute a negative value for Ix back into the circuit?
Yeah, after doing that I see that there's just no way for Ix to be -13.33A in this circuit. Thanks for verifying. Wish my school would rewrite the solution.
 
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