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Total resistance in the circuit

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data
    It is known that total resistance of the circuit is dependent on the position of the wiper in potentiometer ##R##. Find the biggest and the smallest possible total resistances of the circuit. Potentiometer resistance is ##R=9r##.
    band.png

    2. Relevant equations
    Ohm's law.

    3. The attempt at a solution
    Here's a redrawn image:
    zwixw5.jpg
    where ##R_1+R_2=R##.

    This circuit reminds me of Wheatstone Bridge, so I approached the problem like this:
    ##V_1=\frac{2r+R_1}{3r+R_1}V##

    ##V_2=\frac{r+R_2}{4r+R_2}V##


    I don't know what to do from here.
     
  2. jcsd
  3. May 26, 2015 #2

    Hesch

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    You have no current through the circuit, and that is what you have to manage: Add a voltage-source between A and B ( I suggest 12V ). Assume "r" represents 1Ω.

    So VA = 12V, VB = 0V.

    Now you use Kirchhoffs voltage law ( KVL, 3 loops ) and calculate by algebra the current IAB as R is divided into (x*9r) and (( 9-x)r):
    Minimum current → maximum resistance, and vica versa.

    You have 3 equations but only one unknown ( x ). Reduce the equations algebraically (right word?) and solve x as for I(x) = . . . . . .
     
  4. May 27, 2015 #3
    http://imageshack.com/a/img673/9890/WLcnru.jpg [Broken]
    From Kirchhoff's first rule:
    ##I-I_3-I_1=0##
    ##I_2-I_1-I_5=0##
    ##I_3-I_4-I_5=0##

    From Kirchhoff's second rule:
    ##I_3r+I_4(2r+9rx)=V##
    ##I_3r+I_5r-I_13r=V_1##
    ##I_2(r+(9-x)r)-I_4(2r+9rx)+I_5r=V_2##
     
    Last edited by a moderator: May 7, 2017
  5. May 27, 2015 #4

    SammyS

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    Can you use a Y-Δ transform ? It works nicely with the three resistors on the left. (r, r, and 3r)
     
  6. May 27, 2015 #5
    I've tried Y-Δ transform, but since I am not really familiar with it I don't want to get too much into it. I haven't studied it yet, so it would be better if I solve the problem without it.

    By the way, I've found a simpler way to write the equations:

    ##\frac{V_1}{2r+9rx}+\frac{V_1-V_2}{r}+\frac{V_1-V}{r}=0##

    ##\frac{V_2}{2+(9-x)r}+\frac{V_2-V_1}{r}+\frac{V_2-V}{3r}=0##

    ##R_{total}=\frac{V}{\frac{V_1}{2r+9rx}+\frac{V_2}{r+(9-x)r}}##
     
  7. May 27, 2015 #6

    Hesch

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    What are you doing? 6 equations! Well, you are doing like everybody else: Very complicated. Here is how you do:

    Assume V = 12V, V3 = 0V, r = 1Ω. Draw 3 current-loops:

    La: clockwise in the upper left "window".
    Lb: clockwise in the upper right window.
    Lc: clockwise in the lower window.

    Now use Kirchhoffs voltage law for the La-loop:

    La: -5La + Lb + 3Lc = 0

    Explanation: La is crossing 3r+r+r giving a voltage drop = -5La.
    Lb is crossing r in the La-loop in opposite direction giving a voltage rise = Lb.
    Lc is crossing 3r in the La-loop in opposite direction giving a voltage rise = 3Lc.

    Make two more equation as for the Lb and Lc loops. You now have 3 equations with the unknown La, Lb, Lc and x. The current I through the circuit = Lc.
    The resistance of the circuit = V / I. So now you can determine the resistance R(x).
     
  8. May 27, 2015 #7

    Hesch

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    That's right, but you have at least made one error in the equations ( the last fraction in the first equation ).

    EDIT: No, sorry. It's correct ( just don't like your signs ).
     
  9. May 27, 2015 #8
    I can't find a way how to derive anything useful from my equations. Is something missing?

    Here's the equations using your method:
    Lb: -(12+8x)Lb + (10-x)Lc = 0
    Lc: -(13-x)Lc = 0

    How can I use them? There are too many unknowns to solve them.
     
  10. May 27, 2015 #9

    Hesch

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    First of all when you draw cirkulation-paths with signed direction, you must use the correct sign, otherwise the direction of the path doesn't matter.

    Remember that Lb is the path in the upper right window, so how does 12V come into it? here it is:

    Lb: La - ( 4+8x ) Lb + ( 10 - x ) Lc = 0

    Lc is the circulation path in the lower window and crosses the 12V batteri:

    Lc: 3 La + (10 - x) Lb - ( 13 - x ) Lc = -12

    There are three equations with 4 unknown. You can solve Lc(x) algebraic.

    I(x) = Lc(x).
     
    Last edited: May 27, 2015
  11. May 28, 2015 #10
    I fail to derive anything useful. There are just too many unknowns getting in the way. Are you sure there are only 3 equations? There must be more.
     
  12. May 28, 2015 #11

    Hesch

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    Say you have one equation with two unknown:

    4x + 8y = 10

    You cannot calculate x and y, but you can express x(y):

    x(y) = ( 10 - 8y ) / 4

    Likewise you can express Lc(x) = . . . . . . . having 3 equations and 4 unknown.
     
  13. May 28, 2015 #12

    SammyS

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    Error alert!

    There has been an inconsistency in the use of x throughout this thread.

    If x is a number ranging between 0 and 1, then the resistances for the two "pieces" of the potentiometer are (9r)(x) and (9r)(1-x) .

    If x is a number ranging between 0 and 9, then the resistances for the two "pieces" of the potentiometer are (r)(x) and (r)(9-x) .

    However, using (9r)(x) and (9-x)(r) together is incorrect.
     
  14. May 28, 2015 #13
    (1) La: -5 La + Lb + 3 Lc = 0
    (2) Lb: La - ( 4 + 8x ) Lb + ( 10 - x ) Lc = 0
    (3) Lc: 3 La + (10 - x) Lb - ( 13 - x ) Lc = -12

    I've derived Lb from the first equation, La from the second and put those in the third. This is what I got:

    ##Lc(x)=\frac{(4+8x)Lb+(50-5x)La+12}{53-5x}##

    Did you get the same?
     
  15. May 28, 2015 #14

    SammyS

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    I don't know if it's correct or not.

    but ... It's not the way to solve this set of equations. If you solve Loop a: for La, you should then plug that into the other two loop equations. You then have two equations with only currents, Lb and Lc.
    Solve one of them for Lb & plug that into the other. Then solve for Lc.

    Aside: In my view, it's unfortunate that Hesch labeled these currents as La, Lb, and Lc, rather than the more customary Ia, Ib, Ic except that this default font makes those look bad -- could use ia, ib, and ic -- or change the font Ia, Ib, Ic -- or use LaTeX ##\ I_a,\, I_b,\, I_c\ ##.
     
  16. May 28, 2015 #15

    SammyS

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    Maybe it's time to consider the Y-Δ transformation .

    It's really not that difficult, particularly for the values here.

    Snapshot_Y_delt.jpg
    The delta is the triangle on the left. Those are the resistors in Loop a . Point A is equivalent to point A in your overall circuit. Point B is at the node which you labeled with voltage V1 on some of your diagrams. Point C is at the node which you labeled with voltage V2 on some of your diagrams.

    The idea here is to replace that given set of resistors connected in that way, with a set of resistors in a "Y" configuration, as seen on the right. We need this set of resistors in the Y-configuration to behave exactly as the original set in the Δ-configuration in regards to nodes A, B, and C.

    What this boils down to is that the resistances between the three nodes (taken pair-wise) must be the same for both configurations.

    Starting with the Δ-configuration, find the resistance between A & B, between B & C, and between A & C .

    For example: between A&B: The single resistor, r, is in parallel with the series combination of r and 3r. RAB = (4r2)/(5r) = (4/5)r .

    RBC gives the same result: RBC = (4/5)r .

    For RAC we have 3r in parallel with the series combination of r and r, giving RAB = (3r⋅2r)/(5r) = (6/5)r .

    Now figure out what RA, RB, and RC must be in the Y-configuration on the right. For each pair here, we only have two resistors in series.

    We must have:
    RA + RB = (4/5)r

    RB + RC = (4/5)r

    RA + RC = (6/5)r​

    What do you get for these?
     
  17. May 29, 2015 #16
    I am not really sure what you are asking for. Is this what you want:
    RA = r
    RB = 3r
    RC = r
     
  18. May 29, 2015 #17

    SammyS

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    No.

    Take the triangular (like Δ) configuration on the left. Find the equivalent resistance between points A and B. (Yes, I apologize for using B. It's not the same B as in your overall circuit.) I actually have found this resistance in post #15.

    Do the same thing for the equivalent resistance between A and C, then also for the equivalent resistance between B and C.
     
  19. May 30, 2015 #18
    I think I got this:

    ##R_A+R_B+R_C=\frac{r⋅3r+3r⋅r+r^2}{r+r+3r}=\frac{7}{5}r##

    ##R_A+R_B+R_C-(R_B+R_C)=R_A=\frac{7}{5}r-\frac{4}{5}r=\frac{3}{5}r##

    ##R_A+R_B+R_C-(R_A+R_C)=R_B=\frac{7}{5}r-\frac{6}{5}r=\frac{1}{5}r##

    ##R_A+R_B+R_C-(R_A+R_B)=R_C=\frac{7}{5}r-\frac{4}{5}r=\frac{3}{5}r##

     
  20. May 30, 2015 #19

    gneill

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    Okay, now re-draw your circuit and plug in the new values in the appropriate locations:

    Fig1.gif

    Note that "r" can be treated as a scaling value since every resistance in the circuit is a multiple of it. So you can drop the "r" from the equations while you work out the minimums and maximums without losing generality. Just tack the r's back on at the end!
     
  21. May 30, 2015 #20
    Well I started calculating total resistance of the circuit and got this expression:

    ##R=\frac{25.6+86.4x-81x^2}{12.8}r##

    Is this correct? If so, how should I interpret it?
     
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