Finding Unknown Currents in a Two-battery Complex Circuit

[gerfken]

Homework Statement

Look at the circuit given in figure 1. Solve for the three currents, I₁,I₂,I₃.

Homework Equations

KVL: Sum of voltages in a closed loop equals zero
KCL: Sum of currents entering a junction equals the sum of currents exiting a junction
Ohm's Law: V=IR
NOTE: Nodal analysis is not allowed for this solution

The Attempt at a Solution

Through KCL, calling I₅ the current leaving the junction on the left-hand side of the circuit and I₄ the current leaving the junction on the right-hand side of the circuit: I₂=I₅+I₃
I₄=I₃+I₁
Then... the KVL equations for each loop:
5I₅-10I₃-8I₄=0
5I₅-12+I₂=0
8I₄-9+I₁=0
-I₂+12-10I₃-9+I₁=0
-I₂+12-5I₅+8I₄-9+I₁=0
-10I₃-9+I₁+5I₅=0
-10I₃-8I₄-I₂+12=0
Okay...so I can't come up with a methodical way of plugging these equations into one another and solving for any unknowns. It just seems...honestly impossible.
We haven't learned any clever tricks involving ignoring one battery (I've seen that in solutions elsewhere), and when the professor mentioned this question he presented it as pretty straightforward...is there something major that I'm missing?
I feel like I have a firm grasp of the mathematics and physics involved but I still can't produce any sort of solution. When I have I've found a mistake somewhere in my math.

 

[phinds]

I think KCL is the hard way to do this one. I'd use KVL. Use the currents as loop currents and just write the equations for the sum of the voltages around each loop and you'll have 3 equations in 3 unknowns, very easy to solve.

 

[ehild]

Homework Statement

Look at the circuit given in figure 1. Solve for the three currents, I₁,I₂,I₃.

Homework Equations

KVL: Sum of voltages in a closed loop equals zero
KCL: Sum of currents entering a junction equals the sum of currents exiting a junction
Ohm's Law: V=IR
NOTE: Nodal analysis is not allowed for this solution

The Attempt at a Solution

Through KCL, calling I₅ the current leaving the junction on the left-hand side of the circuit and I₄ the current leaving the junction on the right-hand side of the circuit: I₂=I₅+I₃
I₄=I₃+I₁

It is correct so far.

There are only 3 independent loops. Write the KVL equation for them, going round along the arrows and following the change of the potential. Note that the potential drops across the resistor if the current flows in the direction of the arrow, and rises if the current flows in the opposite direction.

Then... the KVL equations for each loop:

Well, try again.
If you have the three equation, replace I4 and I5 in terms of I1,I2,I3. You will get three equations with three unknown. It is not so bad!

ehild