- #1

jamesrb

- 11

- 0

## Homework Statement

I need to come up with a formula to solve for i

_{2}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: i

_{1}-i

_{2}+i

_{3}=0 (Eq.1)

Junction E: -i

_{1}+i

_{2}-i

_{3}=0 (Eq.2)

Loop ABEF: ε

_{1}-i

_{1}R

_{1}-i

_{2}R

_{2}=0 (Eq.3)

Loop BCDE: ε

_{2}-i

_{3}R

_{3}-i

_{2}R

_{2}=0 (Eq.4)

Loop ACDF: ε

_{1}-i

_{1}R

_{1}+i

_{3}R

_{3}-ε

_{2}=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 i

_{2}. 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 R

_{1},R

_{2},R

_{3}and ε. In the experiment ε

_{1}is a fixed power supply and ε

_{2}is a variable power supply. I am assuming we can add and subtract them and get:

ε

_{1}-

_{1}R

_{1}+i

_{3}R

_{3}-ε

_{2}=0 → -i

_{1}R

_{1}+i

_{3}R

_{3}=0

But I cannot seem to work the equations well enough to get everything in terms of R

_{x}, i

_{2}and ε so like I said I think there is an error above. I am also not very comfortable working with systems of equations.