Solving 3x3 Matrix: Solving 3 Equations-3 Unknowns

[Blu3eyes]

Homework Statement

[PLAIN]http://img220.imageshack.us/img220/8187/questiong.jpg

NOTE: I's are the unknowns, R's and emfs are given

Homework Equations

The Attempt at a Solution

[PLAIN]http://img4.imageshack.us/img4/2180/solutionv.jpg

Using Junction, loop rule to come up with 3x3 matrix, 3 equations-3 unkowns

Junction C:
I1 + I2 + I3 =0 (1)

Loop ABCF:
-[tex]\epsilon1[/tex] - I1r1 - I1r2 +I2r3 - [tex]\epsilon2[/tex]=0 (2)

Loop FCDE:
+[tex]\epsilon2[/tex] - I2r3 - I3r4 +I3r5 + [tex]\epsilon3[/tex]=0 (3)

I'm not so sure about r2 and r4 as to which current should I use.

 

[gneill]

Junction C:
I1 + I2 + I3 =0 (1)

Loop ABCF:
-[tex]\epsilon1[/tex] - I1r1 - I1r2 +I2r3 - [tex]\epsilon2[/tex]=0 (2)

Loop FCDE:
+[tex]\epsilon2[/tex] - I2r3 - I3r4 +I3r5 + [tex]\epsilon3[/tex]=0 (3)

I'm not so sure about r2 and r4 as to which current should I use.

I'm a bit confused by your sign convention. For loop ABCF you used "-" when, in your journey around the loop, you crossed a resistor going in the same direction as the assumed current, and "+" when you crossed a resistor in a direction against the current flow. Thus, for example, the terms -I1r1 and +I2r3. But in your second loop you have - I3r4 +I3r5, when both are being traversed "against the flow".

 

[Blu3eyes]

I'm a bit confused by your sign convention. For loop ABCF you used "-" when, in your journey around the loop, you crossed a resistor going in the same direction as the assumed current, and "+" when you crossed a resistor in a direction against the current flow. Thus, for example, the terms -I1r1 and +I2r3. But in your second loop you have - I3r4 +I3r5, when both are being traversed "against the flow".

My teacher said that we did not need to care about the direction of currents. We could have all the currents going the same direction. Eventually, after solving for I's if you have a negative sign, just switch the direction around.
Also, I was told that electron current enters the resistors with negative and exits with positive. This is I got confused whether the signs are right with R2, and R4.
Should Loop FCDE be like this:
Loop FCDE:
+[tex]\epsilon2[/tex] - I2r3 + I3r4 +I3r5 + [tex]\epsilon3[/tex]=0 (3)

Am I heading the right way or should I do it differently.
By blowing up B,C and C,D then hook the two wires and have 2 more currents

 

[gneill]

The important thing is to be consistent in applying whatever rule you adopt. Your teacher is correct that the directions of the currents will sort themselves out if you made a wrong choice in the initial assignment of their direction. However, the values you get can be wrong if you aren't consistent in the application of your convention within the problem, and particularly for the same current!

You should probably try to get used to using conventional (positive) current rather than electron current. It'll save you a lot of grief in the long run, particularly when communicating with others over a problem, and when interpreting a lot of the mathematics that revolves around circuits and electronic systems.

By the way, you did realize, right, that you could simply add R2 to R1 and R4 to R5 to form two single resistors?

 

[Blu3eyes]

By the way, you did realize, right, that you could simply add R2 to R1 and R4 to R5 to form two single resistors?

This solves everything.
Silly me, I tried doing it the hard way. I was so distracted by the signs and currents.
Thank you so much!