Use superposition to solve for vs in the circuit

[sugz]

Homework Statement

Use superposition to solve for vx in the circuit of attached document titled "Problem".

Homework Equations

Vx=V1+V2, where V1 is the voltage obtained using superposition by removing 6A source and V2 is the voltage obtained using superposition by removing 4A source.

The Attempt at a Solution

Solutions for the problem by me is shown in attached document titled "hand_solutions".

It seems as though my answer is incorrect as I obtained 5.33V whereas the solution manual (also attached as titled "solution_manual") shows the answer as -26.67. Can anyone tell me where I went wrong? It seems as though my steps are correct.

 

[gneill]

You moved the reference point (ground) between your solution for v1 and v2. v2 is not the same as v_x for that case.

 

[sugz]

Hi gneill, thank you very much for your reply!
Does the reference point (ground) have to be kept consistent through v1 and v2? I made the assumption I could place them wherever since they v1 and v2 circuits were variations of the first circuit. I think I understand it now as I should choose my ground on the initial circuit given in the problem and then use superposition and remove the independent sources and keep the ground consistent. Is this right?

 

[gneill]

Hi gneill, thank you very much for your reply!
Does the reference point (ground) have to be kept consistent through v1 and v2? I made the assumption I could place them wherever since they v1 and v2 circuits were variations of the first circuit. I think I understand it now as I should choose my ground on the initial circuit given in the problem and then use superposition and remove the independent sources and keep the ground consistent. Is this right?

You can move your reference point or even change analysis methods altogether as long as you actually determine the right quantity When you moved the reference point in your second part you placed the controlled voltage source between the 8 Ohm resistor and that reference . So it's now in the path from v2 through the 8 Ohm resistor to the reference. You'd have to take that into account to determine vx.