hi guys

For this question I need to fined V1,V2 and V3 using nodal analysis.
The resister values are in S... which I'm guessing is admittance (my lecturer used Y for admittance in his notes ).

so here is the cct

for the analysis I am unsure as to the voltage between V1 and V2... I'm assuming its just 2Vo.

I'm realy not sure if I constructed my equations correctly...

here is what I have for my equations.
s = 1/R
At node V1
eq1) 2(V1 -V2) +2 + 2Vo = 0

at node V2
eq2) 4V2 -2Vo + 8(V2 - V3) = 0

node 3 = 13V
Vo = 1(V2)

substituting V3 = 13 and Vo = V2
eq1) 2(V1 - 13) + 2 + V1 + 2V2 = 0
3*V1 + 2*V2 = 24

eq2) 4*V2 + 8*V2 - 8*V3 - 2*V2 = 0
10*V2 = 8*13
V2 = 10.4V

subbing V2 into eq1
V1 = 1.1V


Thanks guys

 

In this circuit you have super-node.
So you treat V1 and V2 as a big (super) node.
So we have

\( V1 - V2 = 2*Vo = 2*V2\)

and one equation for the supernode

\( 2A + I4 = I1 + I2 + I3\)

\( 2A + 8S*(13V - V2) = 1S*V1+ 2S*(V1 - 13V) + 4S*V2 \)

And as for admittance we use Y but the unit is Siemens

And if I use this site to solve the equations
http://www.wolframalpha.com/input/?i=+2+%2B+8%2813+-+V2%29+%3D+2%28V1+-+13%29+%2B+V1+%2B+4V2%2C++V1+-+V2+%3D+2V2

I get V1 = 132/7 ; V2 = 44/7

 

Unfortunately your equations are not correct.

You seem to be treating the voltage controlled voltage source as a current source.

Keep in mind that V2=Vo.

So it's worth noting that V1=V2+2Vo=3Vo

I would combine Node 2 (LHS of equation) & Node 1 (RHS of equation) as

8(V3-V2)-4V2=V1+2(V1-V3)-2

Then substitute V1=3Vo, V2=Vo & V3=13 and solve for Vo

 

so I get
2 + 10V3 = 21Vo

subbing V3 = 13V
V1 = 18.857
V2 6.285

correct?

 

Your answers check out.

I would give some attention to Jony130's post.

You will need to understand how to deal with supernodes as they crop up from time to time in circuit analysis problems - so it would be worth doing some reading on that topic.

 

excellent
thanks guys