Question

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

Answer 1

Principle: If constant current is flowing through a wire of uniform area of cross-section at constant temperature, the potential drop across- any portion of wire is directly proportional to the length of that portion

 i.e., V ∝ l 

Method: (i) Initially key K is closed and a potential difference is applied across the wire AB. Now rheostat (Rh) is so adjusted that on touching the jockey J at ends A and B of potentiometer wire, the deflection in the galvanometer is on both sides. Suppose that in this position the potential gradient on the wire is k. 

(ii) Now key K₁ is kept open and the position of null deflection is obtained by sliding and pressing the jockey on the wire. Let this position be P₁ and AP₁ = l₁. In this situation the cell is in open circuit, therefore the terminal potential difference will be equal to the emf of cell, 

i.e., emf ε = kl₁ ...(i) 

(iii) Now a suitable resistance R is taken in the resistance box and key K₁ is closed. Again, the position of null point is obtained on the wire by using jockey J. Let this position on wire be P₂ and AP₂ = l₂ .In this situation the cell is in closed circuit, therefore the terminal potential difference (V) of cell will be equal to the potential difference across external resistance R, i.e.,