Home
Class 12
PHYSICS
In order to determine the internal resis...

In order to determine the internal resistance of a primary cell by means of potentiometer the emf of the battery connected across the ends of the potentiometer wire should be

A

equal to the emf of the primary cell

B

smaller than the emf of the primary cell

C

Greater than the emf of the primary cell

D

All the above three options maybe possible

Text Solution

Verified by Experts

The correct Answer is:
C
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • PRACTICAL PHYSICS

    AAKASH SERIES|Exercise LECTURE SHEET LEVEL-II (ADVANCED) Straight Objective Type Questions|19 Videos

  • PRACTICAL PHYSICS

    AAKASH SERIES|Exercise LECTURE SHEET LEVEL-II (ADVANCED) Linked Comprehension Type Questions|4 Videos

  • PHYSICAL WORLD

    AAKASH SERIES|Exercise Exercise -I|10 Videos

  • RAY OPTICS

    AAKASH SERIES|Exercise PROBLEMS ( LEVEL-II)|60 Videos

Similar Questions

Explore conceptually related problems

In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across 0.52 m of the potentiometer wire. If the cell is shunted by a resistance of 5Omega balance is obtained when the cell connected across 0.4m of the wire. Find the internal resistance of the cell.

In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across 0.52 m of the potentiometer wire. If the cell is shunted by a resistance of 5Omega balance is obtained when the cell connected across 0.4m of the wire. Find the internal resistance of the cell.

Knowledge Check

  • The potentiometer wire of length 200 cm has a resistance of 20 Omega . It is connected in series with a resistance 10 Omega and an accumulator of emf 6 V having negligible internal resistance. A source of 2.4 V is balanced against length 1 of the potentiometer wire. Find the length l of the potentiometer wire. Find the length l

    A
    100cm
    B
    120cm
    C
    110cm
    D
    140cm

    • Similar Questions

    Explore conceptually related problems

    There is a potentiometer wire of length 1200 cm and 60 mA current is flowing in it. A battery of emf 5V and internal resistance of 20 ohm is balanced on potentiometer wire with balancing length 1000 cm. The resistance of potentiometer wire is

    Draw a circuit diagram for the determination of the internal resistance of a cell using potentiometer. Derive the formula to be used.

    The length of the potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of emf 2 V and internal resistance 10Omega is balanced on this potentiometer the balance length is found to be 500 cm. The resistance of potentiometer wire will be

    An electron in potentiometer experiences a force 2.4xx10^(-19)N . The length of potentiometer wire is 6m. The emf of the battery connected across the wire is (electronic charge =1.6xx10^(-19)C )

    A potentiometer wire of length 10 m and resistance 10 Omega per meter is connected in serice with a resistance box and a 2 volt battery. If a potential difference of 100 mV is balanced across the whole length of potentiometer wire, then then the resistance introduce introduced in the resistance box will be

    In an experiment to determine the internal resistance of a cell with potentiometer, the balancing length is 165 cm. When a resistance of 5 ohm is joined in parallel with the cell the balancing length is 150 cm. The internal resistance of cell is

    A constant voltage V_(0) is applied to a potentiometer of resistance R connected to an ammeter. A constant resistor r is connected to the sliding contact of the potentiometer and the fixed end of the potentiometer. If the reading of ammeter vary as the sliding contact is moved from one end of the potentiometer to the other. The resistance of ammeter is assumed to be negligible. Find the minimum reading of the Ammeter.

    Home
    Profile