The superposition theorem with dependent sources

[DODGEVIPER13]

Homework Statement

Use the superposition principle to determine the value of Ix.

Homework Equations

I1+I2+I3=0 KCL
V1+V2+V3=0 KVL
Ix'+Ix''= answer

The Attempt at a Solution

So I understand how to get Ix' its just standard KVL it is -10+2Ix'+Ix'+2Ix'=0 which solving for Ix' gives 2A. This is the part I don't get so using nodal analysis (V''/2)+((V''-2Ix'')/1)=3 which when I simplify I get 6=3V''-4Ix''. So staying on this part why V''/2 is it this because that gives the current that goes across the 2 ohm resistor and V'' is the voltage on the wire? Also the second part ((V''-2Ix'')/1) is it this because your getting the current going from left to right over the 1 ohm resistor and thus by using the potential over 1 ohm which gives the current flowing from the left side of the resistor to the right. You would subtract off the dependent source part because it flows against the other current, so you get the current that was being solved for which is from left to right? essentially you are doing this I1=I2+I3 where I1=3A. So am I right there? On the next part I am confused Vx''+2Ix''=0 which when solved gives Vx''=-2Ix'' is this because the 3A source is in parallel with the dependent source and so the voltage is the same across both and thus V''=-2Ix''. Or is it because of KVl and so starting from the far left of the circuit I get 2Ix''+Vx''=0. I guess what I am asking is am I doing so form of KVL here or is it just simply relating the dependent source quantity. Anyways when this is all done I get the answer which is 2A.

 

[DODGEVIPER13]

Sorry I meant 1.4 not 2A on the last part

 

[DODGEVIPER13]

Sorry if it is really confusing

 

[DODGEVIPER13]

Oh also is my picture readable I am sorry if it is too small let me know

 

[DODGEVIPER13]

I can re upload or clarify anything just let me know please

 

[NascentOxygen]

I suggest that you follow an identical problem being examined here: https://www.physicsforums.com/showthread.php?t=674396

 

[DODGEVIPER13]

Ok thanks on his problem I get Ix''=-1.75 A is that correct

 

[DODGEVIPER13]

So 3-1.75=1.25

 

[DODGEVIPER13]

Where v''=-3Ix'' but why is this is it because of KVL equations giving me 3Ix''+v''=0 or is it just simply a relation

 

[NascentOxygen]

Ok thanks on his problem I get Ix''=-1.75 A is that correct

Once you have worked out a value for the current, use Kirchoff's Laws to confirm that that value satisfies ∑ currents into a node = 0. If it does, it must be right. Or check your current together with the knowledge that the voltage at the centre node is 20¼ volts.