Superposition for dependent sources and Pspice

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Discussion Overview

The discussion revolves around a homework problem involving dependent sources in circuit analysis, specifically using PSpice and nodal analysis. Participants share their attempts at solving the problems and seek confirmation on their methods and results.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks confirmation on their work and the correct command for a dependent current source in PSpice, noting confusion with the symbols used.
  • Another participant mentions using LTSpice and expresses uncertainty about the syntax for dependent current sources in PSpice.
  • Several participants share their calculations for the maximum current through resistors, discussing the use of inequalities and current divider formulas.
  • One participant describes their approach to a Δ-Y transformation and shares their calculations, but others question the accuracy of their results.
  • There are discussions about the potential drops across resistors and how they relate to the node voltage, with suggestions for applying nodal analysis.
  • Participants express uncertainty about their calculations and seek clarification on the correct application of nodal analysis for the circuit.
  • One participant offers to re-upload their work for further feedback, indicating a desire to correct potential errors in their approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of their calculations, with multiple competing views on the results and methods used. Uncertainty persists regarding the application of nodal analysis and the interpretation of dependent sources.

Contextual Notes

Some participants note discrepancies in their results and express confusion over the correct application of formulas and transformations. There are unresolved issues regarding the interpretation of node voltages and the effects of dependent sources on circuit behavior.

  • #61
DODGEVIPER13 said:
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.
 
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  • #62
Ah ok thanks
 
  • #63
DODGEVIPER13 said:
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.
 
  • #64
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
 
  • #65
DODGEVIPER13 said:
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'' ?
 
  • #66
Uh I have no idea that how the book did it I assumed sice they were in parallel or something
 
  • #67
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
 
  • #68
DODGEVIPER13 said:
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?
 
  • #69
Ok so 3Ix+4=I1
 
  • #70
Oh wait could it be 3ix+4+2ix(1)=0
 
  • #71
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?
 
  • #72
-8+3 Ix=0
 
  • #73
DODGEVIPER13 said:
-8+3 Ix=0

Isn't the 8V source removed (suppressed) at this point?
 
  • #74
Oh yah good point so could it be v''=-3Ix
 
  • #75
DODGEVIPER13 said:
Oh yah good point so could it be v''=-3Ix

Yup, could be!
 
  • #76
Heh good because I solved it out and found Ix''=-1.5 and Ix=-.5A
 
  • #77
DODGEVIPER13 said:
Heh good because I solved it out and found Ix''=-1.5 and Ix=-.5A

Yup. That's it.
 

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