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Getting currents in ammeter, with some resistances.

  1. Oct 12, 2013 #1
    1. The problem statement, all variables and given/known data
    ZqCnDhc.gif
    In the figure R1 = 1.29R, the ammeter resistance is zero, and the battery is ideal. What multiple of ε/R gives the current in the ammeter?

    2. Relevant equations
    V=IR, 1/R=1/R_1+1/R_2..(when resistances are parallel)
    etc

    3. The attempt at a solution
    since ammeter doesn't have resistance, two top resistances and two bottom resistances are basically parallel. First, I set the currents coming out from the battery is i, and I was getting R_eq

    R_top: R_1+R=(119/229)R
    R_bottom: R/2
    R_eq=R_top+R_bottom=(467/458)R

    set the voltage as ε

    ε=(467/458)*iR
    ε=V_top+V_bottom=V_top+iR
    (since two bottom resistances are same, so V_bottom=iR)

    (467/458)*iR=V_top+iR
    V_top=(467/458)*iR-iR=(9/458)*iR

    and

    i_topleft=V_top/1.29R=(9/458)iR*1/(1.29R)=(0.015233...)*i
    i_A=current through ammeter
    i_A+i_topleft=i
    i_A=i-(0.015233..)i=(0.9847..)i

    so I got 0.9847.. for the answer, which is wrong. where did I go wrong? I spent like 3 hours on this problem, and still can't get it.
     
  2. jcsd
  3. Oct 12, 2013 #2

    ehild

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    The equivalent top resistance is not 119/229 R.

    It is also wrong that Vbottom=iR.

    It is better to do the derivation symbolically, and substitute the numerical data at the end.

    ehild
     
  4. Oct 12, 2013 #3
    I couldn't managed to pull out the answer, so I did what you said.
    R_t=(R*R1)/(R+R1)
    R_b=R/2

    R_eq=(R*R1)/(R+R1)+R/2

    ε=R_eq*i

    ε=V_t+V_b=V_t+0.5iR
    V_t=R_eq*i-0.5iR
    i_topleft=V_t/(1.29R)

    so
    i_A+i_topleft=i
    i_A=i-i_topleft

    i_A=i-(V_t/(1.29R))=i-((R_eq*i-0.5iR)/(1.29R))=i-(((R*R1)/(R+R1)+R/2*i-0.5iR)/(1.29R))
    R1=R*1.29

    i_A=i-((1/2.29+1/2)i-0.5i)/1.29
    since I won't need i in the answer
    A=1-((1/2.29+1/2)-0.5)/1.29)=0.66148
    http://www.wolframalpha.com/input/?i=1-((1/2.29+1/2)-0.5)/1.29

    which is still wrong. I'll appreciate your correction on my attempt.
     
  5. Oct 12, 2013 #4

    ehild

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    The equation in red is wrong if iAmeans the current through the ammeter.

    ehild
     
  6. Oct 12, 2013 #5
    I just realized I've mixed up my assumptions.

    i_A+i_topleft=0.5i

    is this right?
    and, yes i_A is the current through the ammeter.
     
  7. Oct 12, 2013 #6

    ehild

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    Yes, it is correct.

    ehild
     
  8. Oct 12, 2013 #7
    then it should be

    i_A=0.5i-(((R*R1)/(R+R1)+R/2)*i-0.5iR)/(1.29R))

    substitute R1=1.29R

    i_A=0.5i-(1.06332iR-0.5iR)/(1.29R)
    i_A=0.5i-(0.56332/1.29)i
    i_A=0.5i-0.43668217i
    i_A=0.06331877i


    I=V/R
    i=ε/R

    so this should be the answer, but still..
     
    Last edited: Oct 12, 2013
  9. Oct 12, 2013 #8

    ehild

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    You made it a bit complicated. ((R*R1)/(R+R1)+R/2)*i=ε.

    It is simpler with the topright current.
    iA=i/2-itopright.
    itopright=Vt/r
    Vt=ε-ir/2.
    -->

    iA=i-ε/r.

    i=ε/Req, and Req=r(3r1+r)/(2(r+r1)). ..

    ehild
     
    Last edited: Oct 12, 2013
  10. Oct 12, 2013 #9

    gneill

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    Staff: Mentor

    Question: Have you studied Thevenin equivalents yet?
     
  11. Oct 12, 2013 #10
    it is! Maybe I was too hanging out with R1. Now I got the right answer.
    Even though I got the right answer, I really should trying to get answer with R1, though. So much appreciated for your devotion!
     
  12. Oct 12, 2013 #11
    I don't think I've heard of that. My class just started with Kirchhoff's two rules.
     
  13. Oct 12, 2013 #12

    ehild

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    You need the answer in term of ε/R, but i=ε/Req

    By the way, the derivation is much simpler if you simplify whenever possible.

    Vt= i(R1R)/(R1+R), Itopleft=Vt/R1=iR/(R1+R),
    iA=i/2-iR/(R1+R)= i(R-R1)/(2(R+R1))...

    ehild
     
  14. Oct 12, 2013 #13

    gneill

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    Okay. I was just wondering what level of theory you'd covered. While problems like this one can be solved with basic Kirchcoff's laws, they become easier to tackle when you've covered some additional techniques. I thought perhaps it might have been an exercise intended to evoke such an approach. Never mind, carry on :smile:
     
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