For the enlightment of lurkers and neophytes, here is rcowboy's set of "3 loop equations":
-140 I1 + 0 I2 + 35 I3 = -35
-140 I1 -210 I2 + 0 I3 = -147
0 I1 + 210 I2 + 35 I3 = 112
Now dum-dum claims that I1=-0.25, I2= 0.166, I3 = 0 solves this system of equations differing from the correct solution I1 = 0.3, I2=0.5, I3 =0.2 which you can find in my earlier posts or in rcowboy's solution by his first method, thereby showing that there isvsomething badly wrong with the "loop equations". However the fact of the matter is that his stated solutiom, I1=-0.25, I2= 0.166, I3 = 0, is NOT a solution of the given 3 equations, as almost any fool, rcowboy being a notable exception, can see. It does not satisfy even the first equation, yielding the nonsensical result 35 = -35.
On the other hand, I1 = 0.3, I2=0.5, I3 =0.2 is a solution to this system of 3 equations, and that is the correct circuit solution.
Without having to do tedious algebra one can see that I1 = 0.3, I2=0.5, I3 =0.2 is indeed a solution by direct substitution.
On the other hand
det
so one can see that there are in fact infinitely many solutions.
Those solutions will be of the form I1 = 0.3, I2=0.5, I3 =0.2 plus any solution of the associated homogeneous system of equations
-140 I1 + 0 I2 + 35 I3 = 0
-140 I1 -210 I2 + 0 I3 = 0
0 I1 + 210 I2 + 35 I3 = 0
or I1 = (35/140) I3, I2 = (-35/210) I3, I3 arbitrary.
So, what is going on ?
The problem is that the equations are not in fact "loop equations" in the correct sense of linear circuit theory. In linear circuit theory the loop currents flow around their respective loops, in this case the two mesh "windows of the circuit, and the current in a common elenent, R2 for instance is the supereposition of the two loop currents. This gives the implicit relation I2 = I1 + I3 in rcowboy's notation, which adds a constraint and results in the complete system having a unique solution. So the problem is not that one cannot solve the circuit with 2 or 3 loop equations, but rather that the loop equations were not formulated properly. See my loop equation solution above for a proper formulation.
An alternative is to solve by means of nodal equations. See my subsequent post for a demonstration of how that is properly done.
So, basically rcowboy screwed up the formulation of the circuit equations, then further confused the issue by screwing up the solution of an underdetermined system of equations, failing to recognize that the correct solution was included in the solution set of his undetermined system. That should have been a clue that what was missing was an additional needed constraint. But the clue did not surface because he blew some high school algebra.
My mistake was in giving him any credit at all for understanding of circuits and being able to do basic algebra, and therefore foregoing the tedium of checking everything in gory detail, as one would for someone taking the first baby steps in circuit analysis.
You can indeed solve the circuit with just 2 loop equations -- see earlier post where this is done. You can also do it if you throw in a third loop equation, which adds nothing, BUT you have to write correct loop current equations in the first place.
What is true is that if you use nodal currents rather than true loop currents then you cannot find the necessary constraint (here I1 + I3 = I2) just from the fact that the voltage drop around any closed loop is zero. This serves as an illustration of the difference between nodal currents and loop currents.
Note also that the use of loop currents works only for circuits composed of linear elements. Kirchoff's laws apply in the non-linear case, but one is then forced to formulate the circuit description in terms of nodal currents. Superposition of loop currents is not valid in the non-linear case.
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Originally Posted by rcowboy
be the outer loop current and 
be the current in the first mesh. In your notation this means 
, 
and 
, 
and 
