Question

Describe how a potentiometer is used to compare the EMFs of two cells by connecting the cells individually.

Answer 1

A stable emf E battery is used to create a potential gradient V/L along a potentiometer wire, where V ≡ potential difference across the length L of the wire.
The positive terminals of the cells, whose emf's (E₁ and E₂) are to be compared, are connected to the high potential terminal A. A two-way key connects the negative terminals of the cells to a galvanometer G. The galvanometer's other end is attached to a pencil jockey. The emf E must be greater than the emfs E₁ and E₂.

Comparison of two emf's using a potentiometer by the direct method

The cell with emf E₁ is brought into the circuit by connecting points P and C.

The jockey is tapped along the wire to find the null point D, which is located at a distance of l₁ from A. Then,

E₁ = `l_1("V"/"L")`

w, without changing the potential gradient (i.e., without changing the rheostat setting) point Q (instead of P) is connected to C, bringing the cell with emf E₂ into the circuit. Let its null point D' be at a distance l₂ from A, so that

E₂ = `l_2("V"/"L")`

As a result, `"E"_1/"E"_2` can be calculated by measuring the corresponding null lengths l1 and l2. Using the rheostat, the experiment is repeated for different potential gradients.

Answer 2

1. A potentiometer circuit is set up by connecting a battery of emf (E), with a key (K) and a rheostat such that point A is at a higher potential than point B.
2. The cells whose EMFs are to be compared are connected with their positive terminals at point A and negative terminals to the extreme terminals of a two-way key K₁ and K₂. 
3. The central terminal of the two ways key is connected to a galvanometer. The other end of the galvanometer is connected to a jockey (J).
   
4. Key K is closed and then, key K₁ is closed and key K₂ is kept open. Therefore, the cell of emf E₁ comes into the circuit.  
5. The null point is obtained by touching the jockey at various points on the potentiometer wire AB.
6. Let l₁ be the length of the wire between the null point and point A.
   Here, l₁ corresponds to emf E₁ of the cell. Therefore,
   E₁ = K l₁ ….(1)
   where K is the potential gradient along the potentiometer wire.
7. Now key K₁ is kept open and key K₂ is closed. The cell of emf E₂ now comes in the circuit. Again, the null point is obtained with the help of the Jockey.
8. Let l₂ be the length of the wire between the null point and point A.
   Here l₂ corresponds to the emf E₂ of the cell.
   ∴ E₂ = K l₂ ….(2)
9. Dividing equation (1) by equation (2),
   `"E"_1/"E"_2 = "l"_1/"l"_2`
   Thus, the EMFs of the two cells can be compared and if any one of the EMFs is known, the other can be determined.