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Mesh analysis, one source?

 
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Mar16-11, 02:44 PM   #1
 

Mesh analysis, one source?


this question is part of my electronics CW, ive done it before, but briefly. and looking back at all my old notes ive realised that every single one ive done before has 2 sources and only 2 loops, whereas you can see from the image below that this has 3 and its getting me in quite a kerfuffle. help is appreciated

"With the calculated value of R1, determine the current through resistor R5 using mesh analysis. Using this current value, determine the voltage across resistor R5.

You need to clearly indicate the mesh current directions and voltage drop directions. Your answer should include Kirchhoff’s law Voltage (KVL) equations for each loop and method that you used for solving simultaneous equations. "


my attempt at the solution gets stuck at the KVL rule. i am fairly certain im wrong but so far ive gotten this far: left mesh=I1 top right mesh=I2 bottom left =I3

i believe all the loops are clockwise

I1= V=R1*(I1-I2)+R4*(I1-I3)
I2= 0=R5*(I2±I3)+R1*(I2-I1)+R5*I2
I3= 0=R3*I3+R4*(I3-I1)+R5(I3±I2)

can someone please give me a heads up how to keep this going as its really bothering me now.

regards ash kennard

EDIT: R20 was mean to be R2, but multisim wouldnt let me change it for some reason :S

EDIT: i think i have the KVL down now, but the places i have"±" i am confused which sign to have. an input on that for now would be awesome. cheers guys xxx
Attached Thumbnails
mesh analysis.jpg  
 
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Mar17-11, 03:51 AM   #2
 
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The direction of the loops is completely arbitrary. Maintain consistency, and the directions will work themselves out properly once you've performed your analysis.

EDIT: And if all the loops are assumed to have current going clockwise through them, you can see that the currents due to I2 and I3 will be going in opposite directions through R5, and will thus have a negative sign.
 
Mar17-11, 05:27 AM   #3
 
so it is this?

I1= V=R1*(I1-I2)+R4*(I1-I3)
I2= 0=R5*(I2-I3)+R1*(I2-I1)+R5*I2
I3= 0=R3*I3+R4*(I3-I1)+R5(I3-I2)
 
Mar17-11, 07:19 AM   #4
 
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Mesh analysis, one source?


hi ash-kennard!

(try using the X2 icon just above the Reply box )
Quote by ash-kennard View Post
so it is this?

I1= V=R1*(I1-I2)+R4*(I1-I3)
I2= 0=R5*(I2-I3)+R1*(I2-I1)+R5*I2
I3= 0=R3*I3+R4*(I3-I1)+R5(I3-I2)
looks fine … now solve!

(except too many "=" signs and "I"s … perhaps you could write eg "Mesh1:" instead of "I1 =" ? )
 
Mar17-11, 07:29 AM   #5
 
bollocksed the calculation of R1, so heres the new picture haha
Attached Thumbnails
mesh analysis.jpg  
 
Mar17-11, 07:51 AM   #6
 
just a quick on now, if i have
mesh 1:
V=R1*(I1-I2)+R4*(I1-I3)
mesh 2:
0=R5*(I2-I3)+R1*(I2-I1)+R5*I2
mesh 3:
0=R3*I3+R4*(I3-I1)+R5(I3-I2)

rearranged so they can be solved using a matrix will they look like this?
mesh 1:
I1*(R1+R4)-I2*(R1)-I3*(R4)=V
mesh 2:
-I1*(R1)+I2*(R1+R2+R5)-I3*(R5)=0
mesh 3:
-I1*(R4)-I2*(R5)+I3*(R3+R4+R5)=0

is this right? if so i can crack out a matrix i think :)
 
Mar17-11, 08:01 AM   #7
 
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looks ok
 
Mar17-11, 08:15 AM   #8
 
brilliant, thank you very much tiny-tom and matlabdude. ill let you know how i get on
 
Mar17-11, 09:05 AM   #9
 
hi again, had another snag. just need to know what

means before i can substitute it into the rows like in this example

http://upload.wikimedia.org/math/d/e...9c00a80a7b.png
 
Mar17-11, 10:50 AM   #10
 
ignore that last one im going to try a different method haha
 
Mar17-11, 12:23 PM   #11
 
You're not forced to use a matrix to solve. Just solve your first two equations for different unknowns and substitute into the third.
 
Mar17-11, 06:34 PM   #12
 
solved it using the matrix, cheers lads xxx
 
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