Question

Describe with the help of a neat circuit diagram how you will determine the internal resistance of a cell by using a potentiometer. Derive the necessary formula.

Explain the use of a potentiometer to determine the internal resistance of a cell.

Answer 1

1. The experimental setup for this method consists of a potentiometer wire AB connected in series with a cell of emf ε, the key K₁, and rheostat as shown in the figure. 
   
2. Terminal A is at a higher potential than Terminal B. A cell of emf ε₁ whose internal resistance r is to be determined is connected to the potentiometer wire through a galvanometer G and the jockey J. 
3. A resistance box R is connected across cell ε₁ through the key K₂. The key K₁ is closed and K₂ is open. 
4. The circuit now consists of cell ε, cell ε₁, and the potentiometer wire. The null point is then obtained.
5. Let l₁ be the length of the potentiometer wire between the null point and point A. This length corresponds to emf ε₁.
   ∴ ε₁ = kl₁ ….(1) 
   where K is the potential gradient of the potentiometer wire which is constant.
6. Now both the keys K₁ and K₂ are closed so that the circuit consists of cell ε, cell ε₁, the resistance box, the galvanometer, and the jockey. Some resistance R is selected from the resistance box and a null point is obtained. 
7. The length of the wire l₂ between the null point and point A is measured. This corresponds to the voltage between the null point and point A.  
   ∴ V = kl₂ ….(2)
   Dividing equation (1) by equation (2),
   ∴ `"ε"_1/"V" = ("k""l"_1)/("k""l"_2) = "l"_1/"l"_2` .....(3)
8. Consider the loop PQSTP,
   ε₁ = IR + Ir and V = IR 
   ∴ `"ε"_1/"V" = ("IR" + "Ir")/("IR") = ("R" + "r")/"R" = "l"_1/"l"_2` ....….[From equation (3)]
   ⇒ r = R`("l"_1/"l"_2 - 1)`
   The above equation is used to determine the internal resistance of the cell.