 # 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 • 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 • 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