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Question
Describe how a potentiometer is used to compare the EMFs of two cells by connecting the cells individually.
Solution 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.
Solution 2
- 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.
- 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.
- 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).
- 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.
- The null point is obtained by touching the jockey at various points on the potentiometer wire AB.
- 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. - 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.
- 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) - 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.
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