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प्रश्न
Describe with the help of a neat circuit diagram how you will determine the internal resistance of a cell by using a potentiometer. Derive the necessary formula.
Explain the use of a potentiometer to determine the internal resistance of a cell.
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उत्तर
- The experimental setup for this method consists of a potentiometer wire AB connected in series with a cell of emf ε, the key K1, and rheostat as shown in the figure.
- Terminal A is at a higher potential than Terminal B. A cell of emf ε1 whose internal resistance r is to be determined is connected to the potentiometer wire through a galvanometer G and the jockey J.
- A resistance box R is connected across cell ε1 through the key K2. The key K1 is closed and K2 is open.
- The circuit now consists of cell ε, cell ε1, and the potentiometer wire. The null point is then obtained.
- Let l1 be the length of the potentiometer wire between the null point and point A. This length corresponds to emf ε1.
∴ ε1 = kl1 ….(1)
where K is the potential gradient of the potentiometer wire which is constant. - Now both the keys K1 and K2 are closed so that the circuit consists of cell ε, cell ε1, the resistance box, the galvanometer, and the jockey. Some resistance R is selected from the resistance box and a null point is obtained.
- The length of the wire l2 between the null point and point A is measured. This corresponds to the voltage between the null point and point A.
∴ V = kl2 ….(2)
Dividing equation (1) by equation (2),
∴ `"ε"_1/"V" = ("k""l"_1)/("k""l"_2) = "l"_1/"l"_2` .....(3) - Consider the loop PQSTP,
ε1 = IR + Ir and V = IR
∴ `"ε"_1/"V" = ("IR" + "Ir")/("IR") = ("R" + "r")/"R" = "l"_1/"l"_2` ....….[From equation (3)]
⇒ r = R`("l"_1/"l"_2 - 1)`
The above equation is used to determine the internal resistance of the cell.
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संबंधित प्रश्न
On what factors does the potential gradient of the wire depend?
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(a) What is the value ε?
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(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
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