Superposition for dependent sources and Pspice

[gneill]

The previous Ix =0 which is lower so should I use that however that wouldn't make sense because that would mean negative current.

I believe the problem wanted the largest positive current. .125A is larger than 0A.

 

[DODGEVIPER13]

Ah ok thanks

 

[gneill]

Ok man well if you are not tired I have this problem https://www.physicsforums.com/showthread.php?t=676267 posted in calculus it was suppose to be in engineering but I posted it wrong it has my work for 2

I don't follow the work that you posted there; there are no comments accompanying the equations to explain what you are attempting to do in each step. It would help if you could elucidate your attempt.

 

[DODGEVIPER13]

Ok so don't follow the (1) at the top, it starts at (2). From the first equation with the 3A source removed -8+3ix'+3ix'+2ix'=0 I solve for ix'=1A. Then I remove the 8v source and get ( v''/3)+((v''-2ix'')/3)=4 and relate the dependent source to v'' with v''=-2ix'' solve for ix'' which I find to be -2A and then plug into my formula for Ix=Ix'+Ix'' and I get 1+-2=-1A

 

[gneill]

Ok so don't follow the (1) at the top, it starts at (2). From the first equation with the 3A [4 amp source] source removed -8+3ix'+3ix'+2ix'=0 I solve for ix'=1A.

Okay

Then I remove the 8v source and get ( v''/3)+((v''-2ix'')/3)=4 and relate the dependent source to v'' with v''=-2ix'' solve for ix'' which I find to be -2A and then plug into my formula for Ix=Ix'+Ix'' and I get 1+-2=-1A

Why would you write v''=-2ix'' ? One is a node voltage and the other a dependent supply voltage, and they are separated by a resistor.

What you want is to replace v'' with some function of Ix''. How is Ix'' related to v'' ?

 

[DODGEVIPER13]

Uh I have no idea that how the book did it I assumed sice they were in parallel or something

 

[DODGEVIPER13]

Hmm I was thinking could it be v''/3=-2Ix so that when I sole I get Ix'' to be -(6/7) which when added to 1 A I get (1/7) for Ix

 

[gneill]

Hmm I was thinking could it be v''/3=-2Ix so that when I sole I get Ix'' to be -(6/7) which when added to 1 A I get (1/7) for Ix

Nope. Look at the diagram and where Ix is. How is it related to the node voltage? It has nothing (directly) to do with the dependent source. How would you write Ix if you were performing nodal analysis?

 

[DODGEVIPER13]

Ok so 3Ix+4=I1

 

[DODGEVIPER13]

Oh wait could it be 3ix+4+2ix(1)=0

 

[gneill]

No, you want to relate Ix to the node voltage. What term wold you include in a node equation for the current in the Ix branch?

 

[DODGEVIPER13]

-8+3 Ix=0

 

[gneill]

-8+3 Ix=0

Isn't the 8V source removed (suppressed) at this point?

 

[DODGEVIPER13]

Oh yah good point so could it be v''=-3Ix

 

[gneill]

Oh yah good point so could it be v''=-3Ix

Yup, could be!

 

[DODGEVIPER13]

Heh good because I solved it out and found Ix''=-1.5 and Ix=-.5A

 

[gneill]

Heh good because I solved it out and found Ix''=-1.5 and Ix=-.5A

Yup. That's it.