Friday, January 4, 2013

Comparison of EMF's of Two Cells using potentiometer

A battery of emf ε is connected the end terminal A and B of potentiometer wire with rheostat Rh, ammeter A1 and key K in series. This is called an auxiliary circuit. Two cells of emf ε1 and ε2 are connected as shown in the figure. Positive terminals of the cells connected to A of potentiometer. Negative terminal connected to jockey as shown in figure.
Close the Key K and join the terminals 1 and 3 and find out the ε1 when galvanometer shown no deflection. Suppose no deflection is at l1 of potentiometer wire, then

ε1 = Kl1                                                                                     ……1

Similarly connect 2 and 3 terminal and find out ε2 when there is no deflection in galvanometer. Suppose no deflection is at l2 of potentiometer wire, then
ε1 = Kl2                                                                                     ...…2

Divide 1by 2 we get

ε1/ ε2 = l1/l2


  1. how can we conclude this experiment?
    what is its conclusion??

  2. wow nice explanation liked it!!!
    U must be so talented

  3. nice DMR
    I have also a blog
    can we crossshare each others post