\[ 0V2+V3(0.11)-V4(0.267)=1.8 \]Hello I have this problem as an extra HW problem that isnt worth any points and doesnt have a solution posted. So I figured I try and solve and use Multisim as a checker and I dont know if im solving this correctly.

The problem:

1.)From my understanding since we have dependent sources I need to find Voc(open circuit) and Isc(short circuit) of i2 to find the Rth(thevenin equivalent resistence) of circuit, correct?
2.) I tried to solve for Voc, my work:
I tried to upload a pic of my handwritten work but keep getting an error..... but my equations(edit,nevermind,it worked finally):

Supernode:
\[ 2Va=V4-V2 ->Va = 12Ia \]
\[ 2(12Ia)=V4-V2 -> Ia = 12/15 - V4(1/15) \]
\[ -V2+0V3+2.6V4=19.2 \]
Nodes V2 & V4 for supernode:
\[ V2((1/6)+(1/9))-V3(1/9)-(12/6)+V4(1/15)-12/15+5A=0 \]
\[ V2(0.28)-V3(0.11)+V4(0.067)=-2.2 \]
Node V3:
\[ V2(0)+V3(0.11)-V4(0.267)=1.8 \]
Solving with linear algebra:
V2 = 5.3V -> which is my Voc
V3= 39.24V
V4= 9.42

But in MultiSim I get a Voc of 12V.....Did I do something wrong??



Thanks

 

The circuit lends itself to Mesh analysis with an upper left loop(say I1), upper right loop(say I2) and the large bottom loop (I3). Only the upper left loop (contains the 12ohm resistor, dependent voltage supply and 6 ohm resistor) requires an equation. The other two loops are trivial, I2=5A, and I3=4Ia. Then knowing that Va=12I1 you get everything in your equation in terms of I1. Finally using KCL at the center node as showing I3 = (Voc-12)/6ohm you can find Voc=12V.