Superposition for dependent sources and Pspice

[DODGEVIPER13]

ok so mesh 1: -12.5+100I1+25(I1-I2)=0

 

[DODGEVIPER13]

Should I have set 25(I1-I2) = Ix

 

[DODGEVIPER13]

I1=.1A and I2=.2A

 

[gneill]

ok so mesh 1: -12.5+100I1+25(I1-I2)=0

That's okay, but you really don't have to introduce another current variable (I2); You've already got that second mesh current as Ix. (Although note that it is a counterclockwise current, so adjust the resulting potential drop across the shared resistor accordingly).

 

[gneill]

Should I have set 25(I1-I2) = Ix

Certainly not. The left hand side is a voltage while the right hand side is a current. Ix is the mesh current; get rid of I2.

 

[gneill]

I1=.1A and I2=.2A

Nope.

You need to get the equation squared away before leaping to results.

 

[DODGEVIPER13]

Okie dokie so I sense I am getting closer here is my first equation again with modifications -12.5+100I1+25(I1-Ix)=0

 

[gneill]

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?

 

[DODGEVIPER13]

ah they both flow in correct so -12.5+100I1+25(I1+Ix)=0

 

[gneill]

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.

 

[DODGEVIPER13]

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?

 

[DODGEVIPER13]

I also found I2=sqrt(P/R)=.2A for the mavimum current through the 25 ohm resistor

 

[gneill]

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?

 

[DODGEVIPER13]

ah ok so -12.5+100I2+25I2=0

 

[DODGEVIPER13]

oppps incorrect sorry man -12.5+100I2+25Ix lol no Ix in my equation above

 

[DODGEVIPER13]

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

 

[gneill]

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.

 

[DODGEVIPER13]

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

 

[DODGEVIPER13]

Or .249 amps

 

[gneill]

Can you show your work? That result doesn't look right.

 

[DODGEVIPER13]

-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

 

[DODGEVIPER13]

I1=200-Ix

 

[gneill]

-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.

 

[DODGEVIPER13]

I thought I did I plugged 200-Ix into I1 which is I1+Ix=200 this is in mA

 

[DODGEVIPER13]

With the equation being -12.5+100I1+25(I1+Ix)=0

 

[gneill]

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.

 

[DODGEVIPER13]

Ah yah ur right gotcha ill redo and post

 

[DODGEVIPER13]

Yes woot thank you man re did it 125 mA

 

[DODGEVIPER13]

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

 

[DODGEVIPER13]

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