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MCQ on potentiometer

  1. Jul 18, 2009 #1
    1. The problem statement, all variables and given/known data
    In the potentiometer circuit shown,the balanced length l is reduced to l/2,when a resistor of resistance R is connected across the terminals of the cell E.The internal resistance of the cell E is
    (1) R/2 (2)R (3) 2R (4) 3R/2 (5) 3R
    http://img268.imageshack.us/img268/4080/potentiometer.png [Broken]
    2. Relevant equations

    [tex]\Sigma[/tex]E = [tex]\Sigma[/tex]IR

    3. The attempt at a solution

    [tex]\rho[/tex] = potential gadient
    R1 is the resistance of the potentiometer wire

    In the 1st case, E = I*(r+R1) = [tex]\rho[/tex] * l ?
    Now when the resistor R is connected parallel to E,
    V = I* R = [tex]\rho[/tex]*l/2 ?

    I'm not quite sure how I'm supposed to find r from this,interms of R only.
    Hope someone can help.
    Thank you.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 19, 2009 #2

    rl.bhat

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    In the first case, when the galvanometer shows zero deflection at the balancing length, the current is not flowing through the galvanometer because E = ρl....(1)
    In the second case
    V = ER/(R + r) = ρl/2 ...(2)
    Substitute the value of E from eq. (1) in eq.(2) and find the value of r in R
     
  4. Jul 19, 2009 #3
    Thank you so much for replying!
    Now,I get this,
    but I'm not sure I understand how you get this,

    I tried doing it again using the 1st equation you provided
    In the 1st case,
    as you mentioned E= ρl ---------(1)
    (taking I as the current through the external circuit)
    after connecting the resistor in parallel and using Kirchoff's laws for this external circuit,I get
    E = I(R + r) -----------(3)
    therefore (1) = (3) = ρl = I(R + r) -----------(4)
    In the second case,
    V (for the resistor only) = ρl/2 = IR ---------(5)

    Dividing (4) by (5) gives me r as equal to R,which is the same answer I get when I Substitute the value of E from eq. (1) in eq.(2) .
    I would like to know if we're both doing the same thing?Is this also how you get V = ER/(R + r) = ρl/2 ...(2) or have you found the current here?
     
  5. Jul 19, 2009 #4

    rl.bhat

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    You are connecting the positive terminals of two batteries to the left end of the potentiometer wire say A. There will be a potential drop along the wire. When you slide the galvanometer contact along the wire, at one point P the potential difference across AP will be equal to E (emf). So there is no flow of current through E. That leads me to write the eq. (1). This is not equivalent to eq. (3) in your post, because in the first case there is no R in the circuit. When you connect R parallel to E, current will flow through R and r and PD across R is given by the eq. (2) in my post.
     
    Last edited: Jul 19, 2009
  6. Jul 20, 2009 #5

    I'm sorry,I still don't understand why it's wrong to equate (1) and (3)
    http://img198.imageshack.us/img198/4080/potentiometer.png [Broken]

    Before connecting the resistor, as said in your post,
    E= ρl ---------(1)

    Now after connecting the resistor R I've applied Kirchoffs' 2nd law to the outer circuit labelled as ABCD in the above diagram.I get
    [tex]\stackrel{\rightarrow}{ABCD}[/tex]
    E = I*R + I*r
    E = I(R + r) -----------------(3)
    I don't know how to get V = ER/(R + r) from this circuit,without including the current 'I' in it ?

    Since the magnitude of both (1) and (3) is equal to E,I equated them .

    Then for the potential of the resistor R only(i.e. VBC which is equal to IR) )
    Isn't VBC equal to the new balanced length ρl/2,cause it's connected in parallel?
    VBC = ρl/2 = IR -------------(5) ?
     
    Last edited by a moderator: May 4, 2017
  7. Jul 20, 2009 #6

    rl.bhat

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    You can equate (1) and (3). No problem. When I wrote that I was thinking about your first equation E = I(r+R1) in the first post. Even in the third post,after equation (1) you have written (taking I as the current through the external circuit) But there is no current in the external circuit.
    V is nothing but VBC = I*R = E/(R+r)*R = Terminal voltage.
    When you connect this to the potentiometer wire through the galvanometer, deflection in the galvanometer will be zero when PD across AP = VBC
     
    Last edited by a moderator: May 4, 2017
  8. Jul 20, 2009 #7
    Thanks rl.bhat!
    I realise my mistake now.
     
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