i had make 2 equation out for nodal analysis
eq 1 : (V1-V3)/12 + V1/5 = (V3-V2)/8 + 6
eq 2 : V1 - V2 = 12

but i had no idea for equation 3, any1 help pls.

for mesh analysis, i need to comfirm the supermesh equation
by apply supermesh, 6A is open circuit.
i obtained the below equation:
-12 + 5I2 + 8(I1+I3) = 5Vx

below 2 loop are I1 and I2 with anticlockwise, I3 is upper loop with clockwise.
i not sure is it = 5Vx or = -5Vx?

 

Firstly are you going for nodal analysis or mesh analysis?

 

both nodal and mesh for the same circuit.
i need to find out the Io.

 

You need to specify that you are assigning a value of 0V to V4. Don't assume people will make the same assignment you do.

Use units properly -- most mistakes you make will mess up the units and let you catch them right away.

Your third equation is KCL applied at Node V3.

In your mesh analysis, what you do mean that the 6A current source is open circuit -- are you doing superposition?

 

by applying supermesh on the bottom part of circuit
the 6A will be short circuit and form an equation.

 

How will applying a supermesh short out a 6A current source?

How will shorting it out "form an equation"?

And you need to make up your mind. First you say it is open and then you say it is shorted out. Which is it?

Hint: Neither.

 

ooppss.... sry
is open circuit for 6A not short....
wrong mentioned

 

Okay, so how does using a supermesh result in a current source becoming open circuited?

Please draw the resulting circuit and the equations that result from it.

 

1 is the ori circuit and 2 is the circuit open 6A
below is my 3 equation, pls correct me if wrong

-12 + 5I2 + 8 (I1 + I3) = 5Vx
12I3 + 8(I3 + I1) = 12
I2 = I1 + 6

in here Vx = 5I2
and in the nodal analysis Vx = V1/5

am i any mistake here???

 

Clearly your two circuits are NOT the same. So after solving for the second circuit (the one with the current source removed), how do you plan to proceed to get the solution for the original circuit?

It can be done. What you are basically doing is the first part of superposition and the next part would be to turn off the independent voltage source and turn on the current source. But that doesn't seem to be what you are attempting to do here.

It appears that you are confusing how you apply the concept of a supermesh with the concept of how you apply superposition.

When you apply a supermesh, you don't make any changes to the circuit. You simply perform KVL around a loop formed by combining two or more meshes. But you don't ignore the branches that are internal to the supermesh -- you use them to establish the relationships between the mesh currents that make up the supermesh.