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प्रश्न
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?
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उत्तर
(a) Constant emf of the given standard cell, E1 = 1.02 V
Balance point on the wire, l1 = 67.3 cm
A cell of unknown emf, ε, replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm
The relation connecting emf and balance point is,
`"E"_1/"l"_1 = ε/"l"`
ε = `"l"/"l"_1 xx "E"_1`
= `82.3/67.3 xx 1.02`
= 1.247 V
The value of unknown emf is 1.247 V.
(b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
(c) The balance point is not affected by the presence of high resistance.
(d) The point is not affected by the internal resistance of the driver cell.
(e) The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.
(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to ending A. Hence, there would be a large percentage of errors.
The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.
संबंधित प्रश्न
State the principle of working of a potentiometer.
On what factors does the potential gradient of the wire depend?
Would you prefer a voltmeter or a potentiometer to measure the emf of a battery?
How is potential gradient measured? Explain.
If the potential gradient of a wire decreases, then its length ______
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.
Which of the following is true for a potentiometer?
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
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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 ______
What is the current I in the circuit as show in fig.
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.
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.
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 ______.
What will a voltmeter of resistance 200 Ω read when connected across a cell of emf 2 V and internal resistance 2 Ω?
