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Archived Balanced length of a potentiometer wire

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A battery if emf E0=12v connected across a 4m long uniform wire having resistance 4Ω. The cells of small emf E1=2v and E2=4v having internal resistance 2Ω and 6Ω respectively, are connected as shown in figure below. If galvanometer shows no deflection at the point N, then distance of N from point A is equal to

    a. 1/6 cm
    b. 25 cm
    c. 75cm
    d. 50 cm



    2. Relevant equations
    V=IR
    Kirchoff's laws
    V=kl (v=voltage, k=voltage per unit length, l=length) => for potentiometer

    3. The attempt at a solution
    Since the potentiometer is in balanced condition no current passes through galvanometer.
    So applying V=IR for potentiometer
    12v=(8+4)I
    12=12I
    I=1A

    BY applying V=IR to potentiometer wire
    V=4I=4.1
    V=4v

    Potential drop per unit length
    k=4v/4m K=1 vm-1

    I have no idea to find potential of the points of circuit which is connected to the potentiometer. How to find that. When Kirchoff;s laws applied that potential get 0 and balanced length become 0.

    Applying Kirchoff's Laws
    4v=6I1
    I1=2/3 A

    -2v=2I2
    I2=-1A

    Direction of current is opposite

    Now by V=IR, the potential of circuit become 0.
    How to complete this.

    :confused:
     

    Attached Files:

  2. jcsd
  3. Feb 28, 2016 #2

    cnh1995

    User Avatar
    Homework Helper

    untitled-jpg.48094.jpg
    I believe there is a mistake in the diagram. The middle wire( where i1-i2 is shown) should not be there. It connects one end of galvanometer to 12V directly. If it is there, the galvanomater will read 0 at point A.
    If the middle wire is not there:
    Assigning potentials to various points, let A be at 12V. So, B is at 8V and so on..
    Applying KVL to the loop containing the two sources, it turns out that one end of the galvanometer(fixed end) is at 9.5V potential.
    So, when the other (moving) end is also at 9.5V on the wire, galvanometer will read 0.
    So, starting from point A(12V), N will be 2.5V away i.e. 2.5m away from point A (since potential gradient=1V/m).
    So, the answer should be 250cm, which is none of the options.
     
    Last edited: Feb 28, 2016
  4. Feb 29, 2016 #3
    I agree with CNH to some extent...it looks like a mistake in which case the voltage across the external circuit is 2.5V which equates to a section of wire 2.5m long (any section of wire 2.5m long would 'balance' 2.5V)
    If the diagram is not wrong then there is a short circuit caused by that central connection and the voltage across the circuit is ZERO.
    I think the 'best' answer is 1/6cm..1.3 mm....the wire is 4000mm long and 1.3mm is as close to zero as makes no difference.
    If that is the answer then this is either a stupid question or a question designed to promote discussion.
     
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