I have to find the current across V2 using Thevenin's theorem.

Below in the attachments, draft2.jpg is the circuit and draft5.jpg is the Thevenin equivalent.

R2 and R3 are in series and give a 6 Ohm resistance.
R1 is shorted out. R4 is in parallel with R(23) so 1/Rt = 1/R(23) + 1/R4
1/6 + 1/6 = 2/6 which gives Rt the total resistance equal to 3 Ohm.

Then I write the nodal voltage equations:

Va/2 + (Va-V1)/4 = 0
Vb/6 + 4 + (Vb-V1)/2 = 0
(V1-Va)/4 - 4 + (V1-Vb)/2 = 0

which give:

3*Va - V1 = 0
4*Vb - 3*V1 = -24
-Va - 2*Vb + 3*V1 = 16

and solving the system/matrix for Va, Vb and V1 gives approximately:
Va = 1.1
Vb = -3.4
V1 = 3.4

The Thevenin voltage equivalent should be Va-Vb = 1.1 - (-3.4) = 4.5 Volts

So the current over V2 should be: I = (Vt + V2)/Rt = (4.5 + 8)/3 = 4.17 Amps

But when I do the Ltspice simulation the current is shown to be -6 A.

Can you please tell me what I'm doing wrong?

 

Looking at your diagram we already know the voltage at pony A (Va = 12V)
So the only thing you need to do is to write the nodal equation for Vb and V1

Vb/6 + 4 + (Vb-V1)/2 = 0
(V1-Va)/4 - 4 + (V1-Vb)/2 = 0

Va = 12V then solve for Vb and V1
http://www.wolframalpha.com/input/?i=x/6+++4+++(x-y)/2+=+0,+(y-12)/4+-+4+++(y-x)/2+=+0

 

Looking at your diagram we already know the voltage at pony A (Va = 12V)
So the only thing you need to do is to write the nodal equation for Vb and V1


Va = 12V then solve for Vb and V1
http://www.wolframalpha.com/input/?i=x/6+++4+++(x-y)/2+=+0,+(y-12)/4+-+4+++(y-x)/2+=+0

Thanks very much, I got it.

One last thing: how would the equations be if V1 swapped places with ground?

 

One last thing: how would the equations be if V1 swapped places with ground?

Do you mean this version?
Now we have one supernode so the equation will look like this

For Vb

\(\frac{Vb}{2}+\frac{Vb-V2}{6}+4 = 0 \)

And for V2 + Va

\(\frac{Vb-V2}{6} = \frac{Va}{4} \)

And

\(Va = V2+12V\)

 

Do you mean this version?
Now we have one supernode so the equation will look like this

For Vb

\(\frac{Vb}{2}+\frac{Vb-V2}{6}+4 = 0 \)

And for V2 + Va

\(\frac{Vb-V2}{6} = \frac{Va}{4} \)

And

\(Va = V2+12V\)

Yes, this version.
Thank you so much! I got it!