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Question
Describe how a potentiometer is used to compare the emf's of two cells by the combination method.
Solution 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.
Solution 2
- The emfs of two cells can be compared using the sum and difference method.
- 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 - 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 - 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.
- 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. - 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.
- 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) - 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.
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