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Questions
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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Solution
- 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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