Advertisements
Advertisements
प्रश्न
Describe how a potentiometer is used to compare the emf's of two cells by the combination method.
Advertisements
उत्तर १
A stable emf E battery is used to create a potential gradient `"V"/"L"` along the potentiometer wire,
where V = potential difference across length L of the wire.
The positive terminal of cell 1 is connected to the potentiometer's higher potential terminal A, while the negative terminal is connected to the galvanometer G via the reversing key. The galvanometer's other terminal is connected to a pencil jockey. Cell 2 is connected across the remaining two opposite terminals of the reversing key. The other terminal of the galvanometer is connected to a pencil jockey.
The emf E1 should be greater than the emf E2; this can be fine-tuned through trial and error. In positions 1-1, two plugs are inserted into the reversing key. In this case, the two cells help each other, resulting in a net emf of E1 + E2.
The jockey taps along the wire to find null point D. If the null point is located at a distance of l1 from A,
E1 + E2 = l1 (V/L)
Comparison of two emf's using a potentiometer by the combination method (the sum and difference method)
For the same potential gradient (without changing the rheostat setting), the plugs are now inserted into positions 2-2. (instead of 1-1). The emf E2 then opposes E1 and the net emf is E1 - E2. The new null point D' is, say, a distance l2 from A and
E1 - E2 = l2 (V/L)
∴ `("E"_1 + "E"_2)/("E"_1 - "E"_2) = "l"_1/"l"_2`
∴ `"E"_1/"E"_2 = ("l"_1 + "l"_2)/("l"_1 - "l"_2)`
Here, the emf E should be greater than E1 + E2. Using the rheostat, the experiment is repeated for different potential gradients.
उत्तर २
- The emfs of two cells can be compared using the sum and difference method.
- When two cells are connected such that the positive terminal of the first cell is connected to the negative terminal of the second cell, the emf of the two cells are added up and the effective emf of the combination of two cells is E1 + E2. This method of connecting two cells is called the sum method.
Sum method - When two cells are connected such that their negative terminals are together or their positive terminals are connected together, then their emf oppose each other. The effective emf of the combination of two cells is E1 – E2 (Considering E1 > E2). This method of connecting two cells is called the difference method.
Difference method - The circuit for the sum and difference method is connected, as shown in the below figure. When keys K1 and K3 are closed, the cells E1 and E2 are in the sum mode. The null point is obtained using the jockey.
- Let l1 be the length of the wire between the null point and point A.
This corresponds to the emf (E1 + E2).
∴ E1 + E2 = K l1 ….(1)
where K is the potential gradient along the potentiometer wire. - Now the keys K1 and K3 are kept open and keys K2 and K4 are closed. In this case, the two cells are in the difference mode.
- Again the null point is obtained. Let l2 be the length of the wire between the null point and point A.
This corresponds to emf (E1 - E2).
∴ E1 - E2 = Kl2 ….(2) - Dividing equation (1) by equation (2),
`("E"_1 + "E"_2)/("E"_1 - "E"_2) = "l"_1/"l"_2`
By componendo and dividendo method,
`(("E"_1 + "E"_2) + ("E"_1 - "E"_2))/(("E"_1 + "E"_2) - ("E"_1 - "E"_2)) = ("l"_1 + "l"_2)/("l"_1 - "l"_2)`
∴ `"E"_1/"E"_2 = ("l"_1 + "l"_2)/("l"_1 - "l"_2)`
Thus, emf of two cells can be compared using sum and difference method.
APPEARS IN
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 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
SI unit of potential gradient is _______.
(a) V cm
(b) `V/"cm"`
(c) Vm
(d) `V/m`
In the given circuit, with steady current, calculate the potential drop across the capacitor and the charge stored in it.
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.
State the principle of a potentiometer. Define potential gradient. Obtain an expression for potential gradient in terms of resistivity of the potentiometer wire.
State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf's of two primary cells. Obtain the required expression used for comparing the emfs.
Write two possible causes for one sided deflection in a potentiometer experiment.
Would you prefer a voltmeter or a potentiometer to measure the emf of a battery?
The net resistance of an ammeter should be small to ensure that _______________ .
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.
A student uses the circuit diagram of a potentiometer as shown in the figure
(a) for a steady current I passing through the potentiometer wire, he gets a null point for the cell ε1. and not for ε2. Give the reason for this observation and suggest how this difficulty can be resolved.
(b) What is the function of resistance R used in the circuit? How will the change in its value affect the null point?
(c) How can the sensitivity of the potentiometer be increased?
When the balance point is obtained in the potentiometer, a current is drawn from ______.
Describe how a potentiometer is used to compare the EMFs of two cells by connecting the cells individually.
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.
What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is decreased?
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 the null point is obtained in the potentiometer, the current is drawn from the ______
The instrument which can measure terminal potential difference as well as electromotive force (emf) is ______
What are the disadvantages of a potentiometer over a voltmeter?
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 ______.
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)
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 ______.
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) ______
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 ______
A battery is connected with a potentiometer wire. The internal resistance of the battery is negligible. If the length of the potentiometer wire of the same material and radius is doubled then ______.
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 ______.
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.
1°C rise in temperature is observed in a conductor by passing a certain current. If the current is double then the rise in temperature is approximately.
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?
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)?
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.
Two cells of same emf but different internal resistances r1 and r2 are connected in series with a resistance R. The value of resistance R, for which the potential difference across second cell is zero, 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.
A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ε and internal resistance r. A cell C having emt `ε/2` and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in the figure shows no deflection is ______.
What should be the diameter of a soap bubble such that the excess pressure inside it is 51.2 Pa? [Surface tension of soap solution = 3.2 × 10−2 N/m]
Draw a neat labelled diagram of Internal resistance of a cell using a potentiometer.
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.
