Quote History Quoted:
I'm trying to follow what you're doing, would you mind posting up your work up to this point? Where you got your equations from, and what you were given to begin with, and what you're trying to find?
View Quote
Check the image I posted above. It is a schematic of the circuit. Was there something else you wanted?
The algebraic solution is quite easy to do, bordering on trivial once you recognize it is a balanced bridge.
As I said, though, that's not the point. I'm wondering why I can't use Cramer's Rule - what aspect of this disallows its use, causes the dependence or otherwise causes the method to fail.
Is it the "balanced bridge" that's causing the failure - R5 is actually irrelevant to the circuit causing me to over-specify the order of the matrix? I expected I5 to calculate out as zero but not to have the whole thing bomb.
Loop Equations
1) E1 - I2R2 - I4R4 - IsRs = 0
2) E1 - I1R1 - I3R3 - IsRs = 0
3) I1R1 - I2R2 + I3R3 - I4R4 = 0
4) I1R1 - I2R2 - I5R5 = 0
5) I3R3 -I4R4 + I5R5 = 0
6) E1 - I2R2 - I3R3 - I5R5 - IsRs = 0
The coefficients in the matrices are the values for Ri with polarity. Values for Ri are given in the figure, above.
You asked, why six equations?
In its general form, before you know it is actually a balanced bridge, there are six unknown currents (I1, I2, ...I6). Six unknowns requires six equations. I actually have 7 loop equations but need and used only 6.