I need to come up with a formula to solve for i2 given the diagram of the circuit attached. I can identify 2 junctions and 3 loops.
2. Relevant equations
*The algebraic sum of the currents into any junction is 0. By convention the currents entering the junction are positive and the currents leaving the junction are negative.
*The algebraic sum of the potential differences in any loop must equal 0.
Junction B: i1-i2+i3=0 (Eq.1)
Junction E: -i1+i2-i3=0 (Eq.2)
Loop ABEF: ε1-i1R1-i2R2=0 (Eq.3)
Loop BCDE: ε2-i3R3-i2R2=0 (Eq.4)
Loop ACDF: ε1-i1R1+i3R3-ε2=0 (Eq.5)
3. The attempt at a solution
Here is where I run into trouble. I am having difficulty using the above equations to find a formula for i2. I am not sure if the equations above have an error in them or if I am heading in the wrong direction. I need to get everything in terms of R1,R2,R3 and ε. In the experiment ε1 is a fixed power supply and ε2 is a variable power supply. I am assuming we can add and subtract them and get:
ε1–1R1+i3R3-ε2=0 → -i1R1+i3R3=0
But I cannot seem to work the equations well enough to get everything in terms of Rx, i2 and ε so like I said I think there is an error above. I am also not very comfortable working with systems of equations.