Superposition for dependent sources and Pspice

In summary: For number 4, it looks like you're attempting a Δ-Y transformation (or two?). If so, the numbers don't look right to me. Can you describe the steps you're making?in summary, the student is trying to solve a current problem using a dependent current source but is having difficulty understanding the syntax or format. He has uploaded his work for 3 and 4 and notes that at least one of the current values he found doesn't match values he finds when solving the problem using his own method. For #2, he notes that at least one of the current values doesn't agree with values he finds. For #3, he has a different answer this time. Lastly, for #4, he
  • #36
DODGEVIPER13 said:
I1=.1A and I2=.2A

Nope.

You need to get the equation squared away before leaping to results.
 
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  • #37
Okie dokie so I sense I am getting closer here is my first equation again with modifications -12.5+100I1+25(I1-Ix)=0
 
  • #38
DODGEVIPER13 said:
Okie dokie so I sense I am getting closer here is my first equation again with modifications -12.5+100I1+25(I1-Ix)=0

Do I1 and Ix flow in different directions through the 25Ω resistor?
 
  • #39
ah they both flow in correct so -12.5+100I1+25(I1+Ix)=0
 
  • #40
DODGEVIPER13 said:
ah they both flow in correct so -12.5+100I1+25(I1+Ix)=0

Yes, that's better. That is now an equation relating I1 and Ix. You have a maximal value for I1 you can plug in and find the corresponding Ix. Then you'll have to do a bit more fiddling with the equation to work out the value of Ix corresponding to the maximum allowed current through the 25Ω resistor. Compare the two values of Ix.
 
  • #41
ok so I did the first equation and solved for it I get Ix=0 when I1=.1A due to i=sqrt(p/R). For the second equation -12.5+100(Ix-I2)+25I2=0 is that correct?
 
  • #42
I also found I2=sqrt(P/R)=.2A for the mavimum current through the 25 ohm resistor
 
  • #43
DODGEVIPER13 said:
ok so I did the first equation and solved for it I get Ix=0 when I1=.1A due to i=sqrt(p/R).
Yes, that's fine.
For the second equation -12.5+100(Ix-I2)+25I2=0 is that correct?
How did you arrive at that equation? Why is Ix flowing through the 100Ω resistor? Isn't Ix the mesh current for the right hand mesh?
 
  • #44
ah ok so -12.5+100I2+25I2=0
 
  • #45
oppps incorrect sorry man -12.5+100I2+25Ix lol no Ix in my equation above
 
  • #46
is my equation now correct if I use that equation and using I2=.2A from p=i^2R I will get a negative number so I am not real confident in my answer
 
  • #47
The basic equation for the two cases does not change; the circuit has not changed. Use the same equation in both cases!

-12.5 + 100i1 + 25(i1 + Ix) = 0

In the first case you fix i1 (the current in the left mesh) at 100mA to max-out the power in the 100Ω resistor. In the second case you want to max-out the current in the 25Ω resistor, which you've found to be 200mA. But the current in the 25Ω resistor is i1+Ix, so you write: i1+Ix = 200mA. This gives you a value for i1 to plug into the circuit equation in order to find the corresponding value of Ix.
 
  • #48
Ok so doing that for Ix I get 249.875 ok so this is much higher than 0 Sotheby's 100 ohm resistor is the limiting value
 
  • #49
Or .249 amps
 
  • #50
Can you show your work? That result doesn't look right.
 
  • #51
-12.5+100(200-Ix)+25((200-Ix)+Ix)=0 then -12.5+20000-100Ix+5000=0 solve for Ix I get .249 amps where did I go wrong again
 
  • #52
I1=200-Ix
 
  • #53
DODGEVIPER13 said:
-12.5+100(200-Ix)+25((200-Ix)+Ix)=0 then -12.5+20000-100Ix+5000=0 solve for Ix I get .249 amps where did I go wrong again

The current I1 is not 200mA (or 0.2mA). It's the current through the 25Ω resistor (I1 + Ix) that's .2A . So I1 + Ix = 0.2, or I1 = 0.2 - Ix. That's what you plug in for i1.
 
  • #54
I thought I did I plugged 200-Ix into I1 which is I1+Ix=200 this is in mA
 
  • #55
With the equation being -12.5+100I1+25(I1+Ix)=0
 
  • #56
DODGEVIPER13 said:
I thought I did I plugged 200-Ix into I1 which is I1+Ix=200 this is in mA
Okay, but you have to keep units consistent in the equations. The voltage source is in volts (12.5) so you'd have to scale it accordingly (12500 mV) if you want to work in mA. Sorry if I confused things by writing the current in mA.
 
  • #57
Ah yah ur right gotcha ill redo and post
 
  • #58
Yes woot thank you man re did it 125 mA
 
  • #59
The previous Ix =0 which is lower so should I use that however that wouldn't make sense because that would mean negative current.
 
  • #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.
 
  • #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
 

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