Advertisements
Advertisements
प्रश्न
How is potential gradient measured? Explain.
Advertisements
उत्तर १
Consider the following potentiometer, which is made up of a long uniform wire AB of length L and resistance R stretched on a wooden board and connected in series with a cell of stable emf E and internal resistance rand a plug key K.
Let I be the current flowing through the wire when the circuit is closed.
Current through AB, I = `"E"/("R" + "r")`
Potential difference across AB. VAB = IR
∴ `"V"_"AB" = "ER"/("R + r")`
The potential difference (the fall of potential from the high potential end) per unit length of the wire,
`"V"_"AB"/"L" = "ER"/(("R + r")"L")`
As long as E and r remain constant, `"V"_"AB"/"L"` will remain constant. `"V"_"AB"/"L"` is known as a potential gradient along with AB and is denoted by K. As a result, the potential gradient is computed by dividing the potential difference between the ends of the potentiometer wire by the wire's length.
Let P be any point on the wire between A and B and AP = l = length of the wire between A and P.
Then `"V"_"AP" = "Kl"`
∴ `"V"_"AP" prop l` as K is constant in a particular case. Thus, the potential difference across any length of the potentiometer wire is directly proportional to that length. This is the principle of the potentiometer.
उत्तर २
- Potential gradient (K) is defined as a potential difference per unit length of wire.
- It is measured as, `"V"/"L" = "ER"/("L"("R" + "r"))`
where V = Potential difference between two points
L = Length (distance) between two points - Explanation:
- A potentiometer consists of a long wire AC of length L and resistance R having uniform cross-sectional area A.
- A cell of emf E having internal resistance r is connected across AC, as shown in the figure.
- When the circuit is switched on, the current I pass through the wire. Current through AC, I = `"E"/("R" + "r")` ….(1)
- Potential differences across AC,
VAC = IR
VAC = `"ER"/(("R" + "r"))` ….[From equation (1)] - Therefore, the potential difference per unit length of the wire is,
`"V"_"AC"/"L" = "ER"/("L"("R" + "r"))`
As long as E remains constant, `("V"_"AC")/"L"` will remain constant. - `("V"_"AC")/"L"` is known as a potential gradient along with AC and is denoted by K.
The potential gradient can be defined as a potential difference per unit length of wire.
APPEARS IN
संबंधित प्रश्न
Write two factors by which current sensitivity of a potentiometer can be increased.
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 the given circuit in the steady state, obtain the expressions for (a) the potential drop (b) the charge and (c) the energy stored in the capacitor, C.
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.
The net resistance of an ammeter should be small to ensure that _______________ .
The net resistance of a voltmeter should be large to ensure that ______________ .
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.
Draw a labelled circuit diagram of a potentiometer to compare emfs of two cells. Write the working formula (Derivation not required).
State the uses of a potentiometer.
A potentiometer wire has a length of 1.5 m and a resistance of 10 Ω. It is connected in series with the cell of emf 4 Volt and internal resistance 5 Ω. Calculate the potential drop per centimeter of the wire.
Find the equivalent resistance between the terminals of A and B in the network shown in the figure below given that the resistance of each resistor is 10 ohm.
What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is decreased?
Why is a potentiometer preferred over a voltmeter for measuring emf?
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.
The SI unit of the potential gradient is ______
What is the SI unit of potential gradient?
State any one use of a potentiometer.
A voltmeter has a resistance of 100 Ω. What will be its reading when it is connected across a cell of emf 6 V and internal resistance 20 Ω?
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.
The resistance of the potentiometer wire should ideally be ____________.
A potentiometer wire has length L For given cell of emf E, the balancing length is `"L"/3` from 3 the positive end of the wire. If the length of the potentiometer wire is increased by 50%, then for the same cell, the balance point is obtained at length.
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 ____________.
If the e.m.f of a cell is not constant in the metre bridge experiment, then the ____________.
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 ______.
The length of a wire of a potentiometer is 100 cm, and the e.m.f of its standard cell is E volt. It is employed to measure the e.m.f of a battery whose internal resistance is 0.5 `Omega` If the balance point is obtained at `l`= 30 cm from the positive end, the e.m.f of the battery is ____________.
where 'i' is the current in the potentiometer
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 ______.
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 ____________.
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 ______
The current drawn from the battery in the given network is ______
(Internal resistance of the battery is neglected)
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 is 4 m long and a potential difference of 3 V is maintained between the ends. The e.m.f. of the cell which balances against a length of 100 cm of the potentiometer wire is ______
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 ______
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 ______.
In the experiment of potentiometer, at balance point, there is no current in the ______.
The best instrument for accurate measurement of EMF of a cell is ____________.
The instrument among the following which measures the e.m.f of a cell most accurately is ______
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?
Consider a simple circuit shown in figure 
stands for a variable resistance R′. R′ can vary from R0 to infinity. r is internal resistance of the battery (r << R << R0).
- Potential drop across AB is nearly constant as R ′ is varied.
- Current through R′ is nearly a constant as R ′ is varied.
- Current I depends sensitively on R′.
- `I ≥ V/(r + R)` always.
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)?
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.
For the circuit shown, with R1 = 1.0 Ω, R2 = 2.0 Ω, E1 = 2 V, and E2 = E3 = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V) ______.
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.
If you are provided a set of resistances 2Ω, 4Ω, 6Ω and 8Ω. Connect these resistances so as to obtain an equivalent resistance of `46/3`Ω.
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 is the effect of decreasing the current through the potentiometer on the null point?
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.
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.
