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Multiple Loops and Resistors

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data

    In the figure ε = 15.1 V, R1 = 1560 Ω, R2 = 2130 Ω, and R3 = 4210 Ω. What are the potential differences (in V) (a) VA - VB, (b) VB - VC, (c) VC - VD, and (d) VA - VC? The figure consists of single ideal batter connecting to a diamond shape circuit ABCD. Going clockwise starting with point A, line AB has resistor 1, line AD has resistor 2, line CD also has a resistor 1, and line CA also has a resistor two. A line between CB is between points C and B, and it contains resistor 3.

    ε = 12V, R1 = 2000Ω, R2 = 3000Ω, R3 = 4000Ω

    2. Relevant equations

    ε - i1R2 - i2R2 = 0 (1)
    ε - (i1 - i2)R3 = 0 (2)

    3. The attempt at a solution

    What I did was move around equation (2), so that I would end up with

    i2 = i1 - (ε/R3), and then I would plug that back into equation one, which would help me find the current. The thing is that it doesn't match up with the answers in the book, so I would like to know how to reach the correct answer. i 1-3, in that order, are 0.002625 A, 0.00225 A, and 0.000375 A. Help would be much appreciated. Thanks.
     

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    Last edited: May 21, 2013
  2. jcsd
  3. May 21, 2013 #2

    phinds

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    Uh ... what figure would that be ?
     
  4. May 21, 2013 #3
    Figure Associated With Problem

    Oops. Forgot to upload the drawing. Sorry. It's up now.
     

    Attached Files:

  5. May 21, 2013 #4

    haruspex

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    You haven't defined i1 and i2. No matter how I try to define them I don't understand how you get those equations.
     
  6. May 21, 2013 #5
    Where the Equations Came From

    I'm following along with the solution manual on this one. For this part, the book said,

    "The symmetry of the problem allows us to use i2 as the current in both of the R2
    resistors and i1 for the R1 resistors. We see from the junction rule that i3 = i1 – i2. There
    are only two independent loop rule equations:

    where in the latter equation, a zigzag path through the bridge has been taken. Solving, we
    find i1 = 0.002625 A, i2 = 0.00225 A and i3 = i1 – i2 = 0.000375 A. Therefore, VA – VB =
    i1R1 = 5.25 V."

    I'm just not getting where they are getting their currents from. I've tried looking at the Req, and using that to find the the current by using, i = (V/Req), but it seems as though that didn't work. Of course, I've also used the R for when the current passes through resistors 1, 2, and 3.
     
  7. May 22, 2013 #6

    SammyS

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    It appears that you have two completely different ses of values for the voltage and the resistances.

    Which set are you using for this problem?
     
  8. May 22, 2013 #7

    haruspex

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    OK, so in (1) i1*R2 was a typo for i1*R1; but I still don't see how you get equation (2).
     
  9. May 22, 2013 #8
    Equation (2) comes from the loop rule. I know that the current running through R1 is i1 and the current running through R2 is i2. These two currents add up to the main current i3. The equation would be i3 = i1 + i2. At the moment the currents i1 and i2 run through R3, they add up together to become the main current i3. At least, that's what I believe my understanding of the situation is. Correct me if I'm wrong.
     
  10. May 22, 2013 #9

    haruspex

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    No. Think about the currents flowing into and out of the node at the left of the diamond.
     
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