Question

Describe how a potentiometer is used to compare the emf's of two cells by the combination method.

Answer 1

A stable emf E battery is used to create a potential gradient `"V"/"L"` along the potentiometer wire,
where V = potential difference across length L of the wire.

The positive terminal of cell 1 is connected to the potentiometer's higher potential terminal A, while the negative terminal is connected to the galvanometer G via the reversing key. The galvanometer's other terminal is connected to a pencil jockey. Cell 2 is connected across the remaining two opposite terminals of the reversing key. The other terminal of the galvanometer is connected to a pencil jockey.

The emf E₁ should be greater than the emf E₂; this can be fine-tuned through trial and error. In positions 1-1, two plugs are inserted into the reversing key. In this case, the two cells help each other, resulting in a net emf of E₁ + E₂.

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

E₁ + E₂ = l₁ (V/L)

Comparison of two emf's using a potentiometer by the combination method (the sum and difference method)

For the same potential gradient (without changing the rheostat setting), the plugs are now inserted into positions 2-2. (instead of 1-1). The emf E₂ then opposes E₁ and the net emf is E₁ - E₂. The new null point D' is, say, a distance l₂ from A and 

E₁ - E₂ = l₂ (V/L)

∴ `("E"_1 + "E"_2)/("E"_1 - "E"_2) = "l"_1/"l"_2`

∴ `"E"_1/"E"_2 = ("l"_1 + "l"_2)/("l"_1 - "l"_2)`

Here, the emf E should be greater than E₁ + E₂. Using the rheostat, the experiment is repeated for different potential gradients.

Answer 2

1. The emfs of two cells can be compared using the sum and difference method.
2. When two cells are connected such that the positive terminal of the first cell is connected to the negative terminal of the second cell, the emf of the two cells are added up and the effective emf of the combination of two cells is E₁ + E₂. This method of connecting two cells is called the sum method.
   
                                   Sum method
3. When two cells are connected such that their negative terminals are together or their positive terminals are connected together, then their emf oppose each other. The effective emf of the combination of two cells is E₁ – E₂ (Considering E₁ > E₂). This method of connecting two cells is called the difference method. 
   
                                 Difference method
4. The circuit for the sum and difference method is connected, as shown in the below figure. When keys K1 and K3 are closed, the cells E1 and E2 are in the sum mode. The null point is obtained using the jockey. 
   
5. Let l₁ be the length of the wire between the null point and point A.
   This corresponds to the emf (E₁ + E₂).
   ∴ E₁ + E₂ = K l₁ ….(1)
   where K is the potential gradient along the potentiometer wire.   
6. Now the keys K₁ and K₃ are kept open and keys K₂ and K₄ are closed. In this case, the two cells are in the difference mode.  
7. Again the null point is obtained. Let l₂ be the length of the wire between the null point and point A.
   This corresponds to emf (E₁ - E₂).
   ∴ E₁ - E₂ = Kl₂ ….(2) 
8. Dividing equation (1) by equation (2),
   `("E"_1 + "E"_2)/("E"_1 - "E"_2) = "l"_1/"l"_2`
   By componendo and dividendo method,
   `(("E"_1 + "E"_2) + ("E"_1 - "E"_2))/(("E"_1 + "E"_2) - ("E"_1 - "E"_2)) = ("l"_1 + "l"_2)/("l"_1 - "l"_2)`
   ∴ `"E"_1/"E"_2 = ("l"_1 + "l"_2)/("l"_1 - "l"_2)`
   Thus, emf of two cells can be compared using sum and difference method.