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)

 

[HalfLostAndSearching]

First of all, it is important to note that there is no error in the equations provided. These are known as Kirchoff's Rules and are fundamental principles in circuit analysis.

To solve for i2, we need to use a combination of these equations. Let's start by rearranging Eq. 3 to solve for i2:

i2 = (ε1 - i1R1)/R2 (Eq. 6)

Next, let's substitute this equation into Eq. 1 and Eq. 2:

Junction B: i1 - [(ε1 - i1R1)/R2] + i3 = 0 (Eq. 7)
Junction E: -i1 + [(ε1 - i1R1)/R2] - i3 = 0 (Eq. 8)

Now, we have two equations with two unknowns (i1 and i3). We can solve these equations simultaneously to find the values of i1 and i3. Once we have these values, we can substitute them back into Eq. 6 to find the value of i2.

Alternatively, we can use a combination of Eq. 4 and Eq. 5 to solve for i2. Let's rearrange Eq. 4 to solve for i2:

i2 = (ε2 - i3R3)/R2 (Eq. 9)

Now, let's substitute this equation into Eq. 5:

ε1 - i1R1 + i3R3 - ε2 = 0 (Eq. 10)

Again, we have two equations with two unknowns (i1 and i3). We can solve these equations simultaneously to find the values of i1 and i3. Once we have these values, we can substitute them back into Eq. 9 to find the value of i2.

It is important to note that both methods should give us the same value for i2. If they do not, then there may be an error in your calculations or in the circuit itself.

I hope this helps in solving for i2. Remember, practice makes perfect when it comes to working with systems of equations in circuit analysis. Good luck!