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gneill
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DODGEVIPER13 said:I1=.1A and I2=.2A
Nope.
You need to get the equation squared away before leaping to results.
DODGEVIPER13 said:I1=.1A and I2=.2A
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
DODGEVIPER13 said:ah they both flow in correct so -12.5+100I1+25(I1+Ix)=0
Yes, that's fine.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).
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?For the second equation -12.5+100(Ix-I2)+25I2=0 is that correct?
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
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 said:I thought I did I plugged 200-Ix into I1 which is I1+Ix=200 this is in mA
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.
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
OkayDODGEVIPER13 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.
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
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