Advertisements
Advertisements
Question
Describe how a potentiometer is used to compare the EMFs of two cells by connecting the cells individually.
Advertisements
Solution 1
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.
Solution 2
- 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.
RELATED QUESTIONS
State the principle of working of a potentiometer.
On what factors does the potential gradient of the wire depend?
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?
State the advantages of potentiometer over voltmeter.
In the given circuit, with steady current, calculate the potential drop across the capacitor and the charge stored in it.
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 a voltmeter should be large to ensure that ______________ .
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 ______.
Why should not the jockey be slided along the potentiometer wire?
Distinguish between a potentiometer and a voltmeter.
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 battery of emf 4 volt and internal resistance 1 Ω is connected in parallel with another battery of emf 1 V and internal resistance 1 Ω (with their like poles connected together). The combination is used to send current through an external resistance of 2 Ω. Calculate the current through the external resistance.
Why is a potentiometer preferred over a voltmeter for measuring emf?
The SI unit of the potential gradient is ______
The instrument which can measure terminal potential difference as well as electromotive force (emf) is ______
What is the SI unit of potential gradient?
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 ____________.
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 ______.
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 ____________.
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 ______.
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 ______.
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 experiment to determine the internal resistance of a cell (E1) using a potentiometer, the resistance drawn from the resistance box is 'R'. The potential difference across the balancing length of the wire is equal to the terminal potential difference (V) of the cell. The value of internal resistance (r) of the cell is ______
If the length of potentiometer wire is increased, then the length of the previously obtained balance point will ______.
A student connected the circuit as shown in the figure to determine the internal resistance of a cell E1 by potentiometer (E > E1). He is unable to obtain the null point because ______.
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, cells of e.m.f. E1 and E2 are connected in series (E1 > E2). the balancing length is 64 cm of the wire. If the polarity of E2 is reversed, the balancing length becomes 32 cm. The ratio `E_1/E_2` is ______
In a potentiometer experiment, for measuring internal resistance of a cell, the balance point has been obtained on the fourth wire. The balance point can be shifted to fifth wire by ______.
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 ______.
The sensitivity of the potentiometer can be increased by ______.
The best instrument for accurate measurement of EMF of a cell is ____________.
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
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.
The instrument among the following which measures the e.m.f of a cell most accurately is ______
While doing an experiment with potentiometer (Figure) it was found that the deflection is one sided and (i) the deflection decreased while moving from one end A of the wire to the end B; (ii) the deflection increased. while the jockey was moved towards the end B.
- Which terminal + or – ve of the cell E1, is connected at X in case (i) and how is E1 related to E?
- Which terminal of the cell E1 is connected at X in case (ii)?
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 ______.
In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ______ cm.
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 potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire x cm long. If the length of the potentiometer wire is increased without changing the cell, the balancing length will ______. (Driving source is not changed)
What is the value of resistance for an ideal voltmeter?
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.
Draw a neat labelled diagram of Internal resistance of a cell using a potentiometer.
State dimension of potential gradient.
Three identical cells each of emf 'e' are connected in parallel to form a battery. What is the emf of the battery?
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.
