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प्रश्न
Describe how a potentiometer is used to compare the EMFs of two cells by connecting the cells individually.
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उत्तर १
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.
उत्तर २
- 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.
संबंधित प्रश्न
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?
Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?
In a potentiometer experiment, balancing length is found to be 120 cm for a cell E1 of emf 2V. What will be the balancing length for another cell E2 of emf 1.5V? (No other changes are made in the experiment.)
State the underlying principle of a potentiometer ?
A potentiometer wire of length 1 m has a resistance of 5 Ω. It is connected to a 8 V battery in series with a resistance of 15 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.
Figure shows a long potentiometer wire AB having a constant potential gradient. The null points for the two primary cells of emfs ε1 and ε2 connected in the manner shown are obtained at a distance of l1 = 120 cm and l2 = 300 cm from the end A. Determine (i) ε1/ε2 and (ii) position of null point for the cell ε1 only.
In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs ε1 and ε2 connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε1/ ε2 and (ii) position of null point for the cell ε1.
How is the sensitivity of a potentiometer increased?
Two students ‘X’ and ‘Y’ perform an experiment on potentiometer separately using the circuit given below:
Keeping other parameters unchanged, how will the position of the null point be affected if
(i) ‘X’ increases the value of resistance R in the set-up by keeping the key K1 closed and the Key K2 opens?
(ii) ‘Y’ decreases the value of resistance S in the set-up, while the key K2 remains open and they K1 closed?
Justify.
The net resistance of an ammeter should be small to ensure that _______________ .
The potentiometer wire AB shown in the figure is 40 cm long. Where should the free end of the galvanometer be connected on AB, so that the galvanometer may show zero deflection?
In a potentiometer experiment, the balancing length with a resistance of 2Ω is found to be 100 cm, while that of an unknown resistance is 500 cm. Calculate the value of the unknown resistance.
Draw a labelled circuit diagram of a potentiometer to measure the internal resistance ‘r’ of a cell. Write the working formula (derivation is not required).
Figure below shows two resistors R1 and R2 connected to a battery having an emf of 40V and negligible internal resistance. A voltmeter having a resistance of. 300 Ω is used to measure the potential difference across R1 Find the reading of the voltmeter.
State the uses of a potentiometer.
Distinguish between a potentiometer and a voltmeter.
What are the disadvantages of a potentiometer over a voltmeter?
A cell of e.m.f 1.5V and negligible internal resistance is connected in series with a potential meter of length 10 m and the total resistance of 20 Ω. What resistance should be introduced in the resistance box such that the potential drop across the potentiometer is one microvolt per cm of the wire?
The emf of a standard cell is 1.5V and is balanced by a length of 300 cm of a potentiometer with a 10 m long wire. Find the percentage error in a voltmeter that balances at 350 cm when its reading is 1.8 V.
A potentiometer wire is 4m long and potential difference of 3V is maintained between the ends. The emf of the cell, which balances against a length of 100 cm of the potentiometer wire is ____________.
Two cells having unknown emfs E1 and E2 (E1 > E2) are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell E2 is connected, so as to oppose cell E1, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells `("E"_1/"E"_2)` is ______.
Two cells when connected in series are balanced on 8 m on a potentiometer. If the cells are connected with polarities of one of the cell reversed, they balance on 2 m. The ratio of e.m.f's of the two cells is ____________.
The length of a potentiometer wire is L. A cell of e.m.f E is balanced at length L/3 from the positive end of the wire. If the length of wire increases by L/2, then the same cell will give balance point at length ____________.
A cell of e.m.f. 'E' is connected across a resistance 'R'. The potential difference across the terminals of the cell is 90% ofE. The internal resistance of the cell is ______.
Sensitivity of a given potentiometer can be decreased by ______.
A potentiometer wire is 10 m long and has resistance of 2`Omega`/m. It is connected in series with a battery of e.m.f 3 V and a resistance of 10 `Omega`. The potential gradient along the wire in V/m is ______.
A potentiometer wire of length 100 cm and resistance 3 `Omega` is connected in series with resistance of 8 `Omega` and an accumulator of 4 volt whose internal resistance is 1 `Omega`.
A wire has a length of 2m and a resistance of 10Ω. It is connected in series with a resistance of 990Ω and a cell of e.m.f. 2V. The potential gradient along the wire will be ______
A potentiometer wire has a length of 4m and resistance of 5Ω. It is connected in series with 495 Ω resistance and a cell of e.m.f. 4V. The potential gradient along the wire is ______
If the length of potentiometer wire is increased, then the length of the previously obtained balance point will ______.
Two students X and Y perform potentiometer experiment separately and null point was obtained as shown in diagram. During the experiment, ______.
- X increases the value of R (resistance)
- Y decreases the value of S (resistance)
The position of null point obtained by students X and Y respectively.
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) ______
A potentiometer wire of length 'L' and a resistance 'r' are connected in series with a battery of E.M.F. 'E0' and a resistance 'r1'. A cell of unknown E.M.F, 'E' is balanced at a length 'ℓ' of the potentiometer wire. The unknown E.M.F. E is given by ______
In the potentiometer experiment, the balancing length with a cell E1 of unknown e.m.f. is 'ℓ1' cm. By shunting the cell with resistance R Ω, the balancing length becomes `ℓ_1/2` cm, the internal resistance (r) of a cell is ______
It is observed in a potentiometer experiment that no current passes through the galvanometer when the terminals of the cell are connected across a certain length of the potentiometer wire. On shunting the cell by a 2 Ω resistance, the balancing length is reduced to half. The internal resistance of the cell is ______.
In the experiment of potentiometer, at balance point, there is no current in the ______.
In a potentiometer of 10 wires, the balance point is obtained on the 7th wire. To shift the balance point to 9th wire, we should ______.
In a potentiometer circuit, a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
In an experiment with a potentiometer, VB = 10V. R is adjusted to be 50Ω (Figure). A student wanting to measure voltage E1 of a battery (approx. 8V) finds no null point possible. He then diminishes R to 10Ω and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.
A Daniel cell is balanced on 125 cm lengths of a potentiometer wire. Now the cell is short circuited by a resistance 2 Ω and the balance is obtained at 100 cm. The internal resistance of the Daniel cell is ______.
As a cell age, its internal resistance increases. A voltmeter of resistance 270 Ω connected across an old dry cell reads 1.44 V. However, a potentiometer at the balance point gives a voltage measurement of the cell as 1.5 V. Internal resistance of the cell is ______ Ω.
A cell of internal resistance r is connected across an external resistance nr. Then the ratio of the terminal voltage to the emf of the cell is ______.
The emf of the cell of internal resistance 1.275 Ω balances against a length of 217 cm of a potentiometer wire. Find the balancing length when the cell is shunted by a resistance of 15 Ω.
What will a voltmeter of resistance 200 Ω read when connected across a cell of emf 2 V and internal resistance 2 Ω?
Three identical cells each of emf 'e' are connected in parallel to form a battery. What is the emf of the battery?
The Figure below shows a potentiometer circuit in which the driver cell D has an emf of 6 V and internal resistance of 2 Ω. The potentiometer wire AB is 10 m long and has a resistance of 28 Ω. The series resistance RS is of 2 Ω.
- The current Ip flowing in the potentiometer wire AB when the jockey (J) does not touch the wire AB.
- emf of the cell X if the balancing length AC is 4.5 m.
