Hi

In general can someone tell me if i have a circuit like the ones below can i made them a supernode?


P.S
​

 

Yes. IF The current that enters the super node will be the same as the current leaving, then solving the individual nodes is not required, and one can use that as a super node.

 

ok the following circuit has the exact circuit i have described above. I am having trouble trying to get the correct answer which is Vo = 22.34.


This is what i have done

1st equation using KCL

528=24V1 + V2

2nd Equation using KVL

-4i + 2Vo + V2 = V1

where i = V1 - 2Vo-V2

where V1 = Vo

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

7V1+5V2=0

V1=-5V2/7

sub back into 1st equation and we get

528 = -120V2/7 + V2

3696 = -113V2

V2 = -32.707

V1 = (-5*-32.707)/7

So V1=Vo=23.36volts

 

ok the following circuit has the exact circuit i have described above. I am having trouble trying to get the correct answer which is Vo = 22.34.

View attachment 30969
This is what i have done

1st equation using KCL

528=24V1 + V2

2nd Equation using KVL

-4i + 2Vo + V2 = V1

where i = V1 - 2Vo-V2

where V1 = Vo

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

7V1+5V2=0

V1=-5V2/7

sub back into 1st equation and we get

528 = -120V2/7 + V2

3696 = -113V2

V2 = -32.707

V1 = (-5*-32.707)/7

So V1=Vo=23.36volts

I'm not seeing how you get your equations.

Writing a super node at the top with the 4Ω and the dependent voltage source(labeling left node Vo, right node Vx),

\(\text{KCL:}\quad 3 = \frac{3}{2}V_{o} - 30 + \frac{1}{16}V_{x}\)

But from the supernode,

\(V_{o} - V_{x} = 4 \left( \frac{-1}{2}V_{o} + 30 - V_{o} \right) + 2V_{o}\)

\(\Rightarrow V_{x} = 5V_{o} -120\)

Into my KCL,

\(33 = \frac{3}{2}V_{o} + \frac{1}{16}\left( 5V_{o} - 120 \right) \)

\(\Rightarrow Vo = \frac{81}{2} \times \frac{16}{29}\)

Hopefully that clears up your confusion.

Next time tell us what nodes you are labeling what, your equations aren't clear at all.