# 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 E
_{1} and E_{2}( whose emf are to be compared ) are connected to the terminals A and negative terminals are connected to jockey through a two way key K_{2} and a galvanometer

- Now first key K
_{1} is closed to establish a potential difference between the terminals A and B then by closing key Ksub>2 introduce cell of EMF E_{1} in the circuit and null point junction J_{1} is dtermined with the help of jockey.If the null point on wire is at length

l_{1} from A then

E_{1}=Kl_{1}

Where K -> Potential gradient along the length of wire
- Similarly cell having emf E
_{2} is introduced in the circuit and again null point J_{2} is determined .If length of this null point from

A is l_{2} then

E_{2}=Kl_{2}

Therefore

E_{1}/E_{2}=l_{1}/l_{2}

This simple relation allows us to find the ratio of E_{1}/E_{2}

- 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 K
_{2}

- First close the key K
_{1} and obtain the null point .Let l_{1} be the length of this null point from terminal A then

E=Kl_{1}

- When key K
_{2} is closed ,the cell sends current through resistance Box (R).If E_{2} is the terminal

Potential difference and null point is obtained at length l_{2}(AJ_{2}) then

V=Kl_{2}

Thus

E/V=l_{1}/l_{2}

But E=I(R+ r) and V=IR

This gives

E/V=(r+R)/R

So (r+R)/R=l_{1}/l_{2}

giving

r=R(l_{1}/l_{2}-1)

- Using above equation we can find internal resistance of any given cell

Class 12 Maths
Class 12 Physics