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Application of potentiometer






(A) Comparison of EMF's of two cells using potentiometer


  • Consider the circuit arrangement of potentiometer given below used for comparison of emf's of two cells

    Application of potentiometer

  • Positive terminals of two cells of emf's E1 and E2( whose emf are to be compared ) are connected to the terminals A and negative terminals are connected to jockey through a two way key K2 and a galvanometer
  • Now first key K1 is closed to establish a potential difference between the terminals A and B then by closing key Ksub>2 introduce cell of EMF E1 in the circuit and null point junction J1 is dtermined with the help of jockey.If the null point on wire is at length
    l1 from A then
    E1=Kl1
    Where K -> Potential gradient along the length of wire
  • Similarly cell having emf E2 is introduced in the circuit and again null point J2 is determined .If length of this null point from
    A is l2 then
    E2=Kl2
    Therefore
    E1/E2=l1/l2
    This simple relation allows us to find the ratio of E1/E2
  • if the EMF of one cell is known then the EMF of other cell can be known easily



(B) Determination of internal resistance of the cell

  • Potentiometer can also be used to determine the internal resistance of a cell

    Application of potentiometer

  • For this a cell whose internal resistance is to be determined is connected to terminal A of the potentiometer across a resistance box through a key K2
  • First close the key K1 and obtain the null point .Let l1 be the length of this null point from terminal A then
    E=Kl1
  • When key K2 is closed ,the cell sends current through resistance Box (R).If E2 is the terminal
    Potential difference and null point is obtained at length l2(AJ2) then
    V=Kl2
    Thus
    E/V=l1/l2
    But E=I(R+ r) and V=IR
    This gives
    E/V=(r+R)/R
    So (r+R)/R=l1/l2
    giving
    r=R(l1/l2-1)
  • Using above equation we can find internal resistance of any given cell




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    Class 12 Maths Class 12 Physics





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