Question

State the underlying principal of potentiometer. 

Describe briefly, giving the necessary circuit diagram, how a potentiometer is used to measure the internal resistance of a given cell.

Answer 1

Principle of potentiometer: 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 

V ∝l 

Determination of internal resistance of potentiometer. 

(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.,

From this formula r may be calculated.