Needing help with calculating currents

[astenroo]

Homework Statement

The currents in a DC circuit is to be calculated, and I end up with a system of 5 equations and 5 unknowns. This is where my headache starts.

Homework Equations

(1) I$$_{1}$$=I$$_{2}$$ + I$$_{3}$$
(2) I$$_{2}$$=I$$_{4}$$ + I$$_{5}$$
(3) 6 - 2I$$_{1}$$ - I$$_{3}$$ = 0
(4) 4.5 - 5I$$_{2}$$ - 3I$$_{5}$$ + I$$_{3}$$ = 0
(5) 3I$$_{5}$$ - 3I$$_{4}$$ -3 = 0

The Attempt at a Solution

Now, I have no problem solving equations with two unknowns. But 5?
Anyway, I gave it a try. I tried to substitute (2) in (1) and substituting the result into (3) and then substituting (2) into (4), resulting in a new system with three unknowns

(3') -3I$$_{3}$$ - 2I$$_{4}$$ - 2I$$_{5}$$ = -6
(4') I$$_{3}$$ - 5I$$_{2}$$ - 8I$$_{5}$$ = -4,5
(5') -3I$$_{5}$$ + 3I$$_{5}$$ = 3

I've been on this problem for the last 6 hours and I still haven't figured out how to solve it.
I've even tried using a matrix (never done one before today), and it all goes wrong, since I apparently don't know how to use a matrix :P

 

[phyzguy]

You are on the right track with matrices. Write the current as a vector C=(I1,I2,I3,I4,I5), and then you can write the equations as a matrix equation M * C = B, where B is a constant vector which has the constant coefficients in it. Then you invert the matrix M, and can write the current C as C = Inverse(M) * B. I carried it out and got an answer for the currents using Mathematica. Can you try to write down the matrix M and the constant vector B?

 

[astenroo]

You are on the right track with matrices. Write the current as a vector C=(I1,I2,I3,I4,I5), and then you can write the equations as a matrix equation M * C = B, where B is a constant vector which has the constant coefficients in it. Then you invert the matrix M, and can write the current C as C = Inverse(M) * B. I carried it out and got an answer for the currents using Mathematica. Can you try to write down the matrix M and the constant vector B?

Thanks for the quick reply!
i have to get back on that one as soon as I find instructions on how to set up said matrices... The one I tried was using it on (3'), (4') and (5'). Apparently that is a no go ;)

 

[dulrich]

Matricies are cleaner and if you have the time to figure it out, that is the preferable way to go. However, this "brute-force" way will work. The best part is that if you keep clean notes, simply plug your answers back in and you will know for sure if you have the right answer regardless how you get there.

Except for a couple of typos in your subscripts, your equations look right. If you substitute out I₅ from (3') and (4') using (5'), you will end up with two equations (3'') and (4'') with two unknowns, I₃ and I₄. Solve for those and then plug them back into (5') and you have I₅. Plug these back into (2) then (1) and you'll be done.

 

[astenroo]

Matricies are cleaner and if you have the time to figure it out, that is the preferable way to go. However, this "brute-force" way will work. The best part is that if you keep clean notes, simply plug your answers back in and you will know for sure if you have the right answer regardless how you get there.

Except for a couple of typos in your subscripts, your equations look right. If you substitute out I₅ from (3') and (4') using (5'), you will end up with two equations (3'') and (4'') with two unknowns, I₃ and I₄. Solve for those and then plug them back into (5') and you have I₅. Plug these back into (2) then (1) and you'll be done.

Oh yes! Indeed, it is exactly as solving systems with two equations and 2 unknowns. Now I noticed the typos :) I might stick around using the brute-force method for now, but I will eventually have to learn matrices also :) Thank you for the help.