Basic DC Circuits - Kirchoff's Rules Lab questions

[jamesrb]

Homework Statement

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.

Homework 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)

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 R_x, 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.

 

[carllacan]

Have you tried using Woflram Alpha (or similar programs) to solve them? That could help you to find out if you stated the problem correctly. I can help you if you don't know how to use it.

 

[jamesrb]

I have used Wolfram to solve some simple systems of equations involving x and y. I am not sure where to begin with something like this.

 

[carllacan]

Here is my try at it:

solve[{ c1-c2+c3==0, -c1+c2-c3==0, a1-c1*R1-c2*R2==0, a2-c3*R3-c2*R2==0, a1-c1*R1+c3*R3-a2==0 }]

i had to change the $i$ to $a$ and the ε to $e$ because it confused them with the mathematical constants. The second solution seems to be the valid one, as it is the only one that does not assume 0 for any variable. You can try to rewrite the expression for R3 into one for i2 (which here is a2)