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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.
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
Write two factors by which current sensitivity of a potentiometer can be increased.
On what factors does the potential gradient of the wire depend?
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value ε?
(b) What purpose does the high resistance of 600 kΩ have?
(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?
(f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
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Describe how a potentiometer is used to compare the EMFs of two cells by connecting the cells individually.
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where 'i' is the current in the potentiometer
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In a potentiometer experiment when three cells A, B, C are connected in series the balancing length is found to be 740 cm. If A and B are connected in series, the balancing length is 440 cm and when B and C are connected in series, it is 540 cm. The e.m.f. of A, B, and C cells EA, EB, EC are respectively (in volt) ______
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The conductivity of super - conductor is
Specific resistance of a conductor increase with.
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A particle carrying 8 electron charges starts from rest and is accelerated through a potential difference of 9000 V. Calculate the KE acquired by it in keV.
Draw a neat labelled diagram of Internal resistance of a cell using a potentiometer.
