Question

Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.

Answer 1

Measurement of internal resistance of a cell using potentiometer:

The cell of emf, E (internal resistance r) is connected across a resistance box (R) through key K₂.

When K₂ is open, balance length is obtained at length AN₁ = l₁

E= Φ l₁…………………………………………………………………………….. (1)

When K₂ is closed:

Let V be the terminal potential difference of cell and the balance is obtained at AN₂ = l₂

∴ V = Φ l₂ ……………………………………………………………………………(2)

From equations (1) and (2), we get

`E/V = (I_1)/(I_2)  ...... (3)`

E = I(r+R)

V =IR

`therefore E/V =(r+R)/R   ..... (4)`

From (3) and (4), we get 

`(R+r)/R= I_1/I_2`

`therefore E/V = (I_1)/(I_2)`

`because r = R (E/V -1)therefore r =R (I_1/I_2 -1)`

We know l₁, l₂ and R, so we can calculate r.