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Question
State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf's of two primary cells. Obtain the required expression used for comparing the emfs.
Solution
Working principle of Potentiometer:
When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.
Applications of Potentiometer for comparing emf of two cells:
The following Figure shows an application of the potentiometer to compare the emf of two cells of emf E1 and E2
E1, E2 is the emf of the two cells.
1, 2, 3 forms a two-way key.
When 1 and 3 are connected, E1 is connected to the galvanometer (G).
Jokey is moved to N1, which is at a distance l1 from A, to find the balancing length.
Applying loop rule to AN1G31A,
Φ l1 + 0 − E1 = 0 ..(1)
Where Φ is the potential drop per unit length
Similarly, for E2 balanced against l2 (AN2),
Φ l2 + 0 −E2 = 0 ..(2)
From equations (1) and (2),
`E_1/E_2 = I_1/I_2` .....(3)
Thus we can compare the emfs of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy.
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