Kirchoff's Laws Problem - Power

[KRC]

Homework Statement

At what rate is electrical energy converted to internal energy in the 4.56Ω and the 5.91Ω resistors in the figure?

Homework Equations

Kirchoff's Laws:
ƩI(in) = ƩI(out) Junction Rule

ƩΔV = 0 Loop Rule

The Attempt at a Solution

I have set the top branch current to go from right to left = I(3)
I have set the middle branch current to go from right to left = I(2)
I have set the bottom branch current to go from left to right = I(1)

The left middle junction results in the following I(3) + I(2) = I(1)...Junction#1
Loop #1 = 0 = 6.96V-5.91I(2)-4.56I(3)
Loop #2 = 0 = -9.56I(1)-5.91I(2)+1.86V

I get I1=2.7502
I2=-4.1340
I3=6.8842

The answers I get when using the P=I^(2)*R...do not yield results consistent with the online homework

I have tried this several different ways and the answer according to my homework do not match. Please provide some insight here.

Thank You.

 

[gneill]

I think you're not following your own stipulations for current directions. Once you choose them, if you start writing equations you need to follow through.

When you "pass though" a resistor in the direction of the assumed current flow, you expect a voltage drop. If you pass through against the flow then you get a voltage rise.

Just a thought: you might find a nodal analysis approach to be easy to apply here.