Advertisements
Advertisements
प्रश्न
Describe how a potentiometer is used to compare the EMFs of two cells by connecting the cells individually.
Advertisements
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
A potentiometer wire has resistance of per unit length of 0.1 Ω/m. A cell of e.m.f. 1.5 V balances against a 300 cm length of the wire. Find the current in the potentiometer wire.
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?
In the given circuit, with steady current, calculate the potential drop across the capacitor and the charge stored in it.
(i) State the principle on which a potentiometer works. How can a given potentiometer be made more sensitive?
State the underlying principle of a potentiometer ?
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.
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?
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.
When the balance point is obtained in the potentiometer, a current is drawn from ______.
Define a Potentiometer.
What are the disadvantages of a potentiometer?
What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is increased?
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.
A potential drop per unit length along a wire is 5 × 10−3 V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.
What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is decreased?
Describe how a potentiometer is used to compare the emf's of two cells by the combination method.
The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with iL What should be the resistance in the box if it is desired to have a potential drop of 1 µV/mm?
When two cells of emf's E1 and E2 are connected in series so as to assist each other, their balancing length on a potentiometer wire is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emf's of the two cells.
When the null point is obtained in the potentiometer, the current is drawn from the ______
What is the SI unit of potential gradient?
Which of the following instruments is not a direct reading instrument?
In a potentiometer experiment, when the galvanometer shows no deflection, then no current flows through ____________.
The resistivity of potentiometer wire is 40 × 10-8 ohm - metre and its area of cross-section is 8 × 10-6 m2. If 0.2 ampere current is flowing through the wire, the potential gradient of the wire is ______.
A potentiometer is an ideal device for measuring potential difference because ______.
Select the WRONG statement:
If the e.m.f of a cell is not constant in the metre bridge experiment, then the ____________.
Which of the following is true for a potentiometer?
To determine the internal resistance of a cell by using potentiometer, the null point is at 1 m when cell is shunted by 3 Ω resistance and at a length 1.5 m when cell is shunted by 6 Ω resistance. The internal resistance of the cell is ______.
Sensitivity of a given potentiometer can be decreased by ______.
When two cells of e.m.f 1.5 V and 1.1 V connected in series are balanced on a potentiometer, the balancing length is 260 cm. The balancing length, when they are connected in opposition is (in cm) ____________.
In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of 10 `Omega` is connected in parallel with the battery, then it balances on 50 cm length of the potentiometer wire. The internal resistance r of the battery 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`.
In the potentiometer experiment, the balancing length with cell E1 of unknown e.m.f. is ℓ1 cm. By shunting the cell E1 with resistance 'R' which is equal to internal resistance (r) of the cell E1, the balancing length ℓ2 is ______
A potentiometer is used to measure the potential difference between A and B, the null point is obtained at 0.9 m. Now the potential difference between A and C is measured, the null point is obtained at 0.3 m. The ratio `E_2/E_1` is (E1 > E2) ______
The current drawn from the battery in the given network is ______
(Internal resistance of the battery is neglected)
Potentiometer measures the potential difference more accurately than a voltmeter, because ______.
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 ______.
AB is a wire of potentiometer with the increase in the value of resistance R, the shift in the balance point J will be ______.
AB is a wire of potentiometer with the increase in value of resistance R, the shift in the balance point J will be:
Three resistance each of 4Ω are connected to from a triangle. The resistance b / w two terminal is
The value of current I in the network shown in fig.
A wire of resistance R is cut into two equal part. There parts are then connected in parallel. The equivalent resistance of the combination will be
What is the current I in the circuit as show in fig.
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?
Potential difference between the points A and B in the circuit shown is 16 V, then potential difference across 2Ω resistor is ______ V. volt. (VA > VB)
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 ______.
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 ______.
In balanced meter bridge, the resistance of bridge wire is 0.1 Ω cm. Unknown resistance X is connected in left gap and 6 Ω in right gap, null point divides the wire in the ratio 2:3. Find the current drawn from the battery of 5 V having negligible resistance.
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.
In a potentiometer, a cell is balanced against 110 cm when the circuit is open. A cell is balanced at 100 cm when short-circuited through a resistance of 10 Ω. Find the internal resistance of the cell.
