To solve for Vth, can I do a simply KVL on the outer loop or do I need to do a mesh/loop analysis?

 

There is no shortcut. The analysis of this circuit will need to take all components into account.

hgmjr

 

If I were you, I would use superposition and voltage/current division. That's a whole lot faster than nodal/loop analysis. If you replace the current source with an "open", then you can solve for Vo in terms of the 10V source. Then, put the current source back, and short out the voltage source. Solve for Vo in terms of the current source alone. Then, add the two results. Once you get the hang of it, you can solve circuits like these in a fraction of the time it would normally take with nodal/loop analysis.

 

I have been trying this problem many different ways. My I can only see solving for Vth via mesh/loop analysis then possibly cramer's rule. However, I am very out of practice. This is my work - can you guys check to see if I am doing anything wrong or if I am even doing anything right lol..

thanks

 

I have been trying this problem many different ways. My I can only see solving for Vth via mesh/loop analysis then possibly cramer's rule. However, I am very out of practice. This is my work - can you guys check to see if I am doing anything wrong or if I am even doing anything right lol..

thanks

Your first equation for mesh 1 is:

-5(3) - 1(Inct) = 0

but this should be:

-5(3) - 1(Inct) = (voltage across 3A source)

The sum of the voltages across the two resistors isn't zero.

The equation for mesh 1 should be simply:

I1 = -3

Substitute that into your equation for mesh 2 and you should be able to get I2.