Home
Class 12
PHYSICS
Figure 6.13 shows a 2.0 V potentiometer ...

Figure `6.13` 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`. Whan a resistor of `9.5 Omega` is used in the external circuit of the cell, the balance point shifts to `64.8 cm`, length of the potentiometer. Dentermine the internal resistance of the cell.

Text Solution

Verified by Experts

Here, `l_(1) = 76.3 cm`, `l_(2) = 64.8 cm`, `r =` ?, `r = 95 Omega`
Now, `r = ((l_(1) - l_(2))/(l_(2))) R = ((76.3 - 64.8)/(64.8)) xx 9.5= 1.68 Omega`
Promotional Banner

Topper's Solved these Questions

  • ELECTRICAL MEASURING INSTRUMENTS

    CENGAGE PHYSICS ENGLISH|Exercise Solved Examples|4 Videos

  • ELECTRICAL MEASURING INSTRUMENTS

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 6.1|15 Videos

  • ELECTRIC POTENTIAL

    CENGAGE PHYSICS ENGLISH|Exercise DPP 3.5|15 Videos

  • ELECTROMAGNETIC INDUCTION

    CENGAGE PHYSICS ENGLISH|Exercise compression type|7 Videos

Similar Questions

Explore conceptually related problems

3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Omega is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell is

In a potentiometer arrangement for determining the emf of cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Omega is used in the external circuit of cell , the balance point shifts to 300 cm. Determine the internal resistance of the cell.

A 2.0 V potentiometer is used to determine the internal resistance of 1.5 V cell. The balance point of the cell in the circuit is 75 cm . When a resistor of 10Omega is connected across cel, the balance point sifts to 60 cm . The internal resistance of the cell is

For the given circuit, If internal resistance of cell is 1.5 Omega , then

The internal resistance of a cell is determined by using a potentiometer. In an experimetn, an external resistance of 60 Omega is used across the given cell. When the key is closed, the balance length on the potentiometer decreases form 72 cm to 60 cm. calculate the internal resistance of the cell.

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

In a potentiometer experiment the balancing length with a cell is 560 cm. When an external resistance of 10Omega is connected in parallel to the cell, the balancing length changes by 60 cm. Find the internal resistance of the cell.

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

The adjoining figure shows the connection of potentiometer experiment to determine internal resistance of a leclanche cell. When the cell is an open circuit the balancing length of the potentiometer wire is 3 .4m and on closing the key K_2 the balancing length becomes 17.m. If the resistance R through which current is dream is 10 Omega then the internal resistance of the cell is:

In an experiment with a potentiometer, the null point is obtained at a distance of 60 cm along the wire from the common terminal with a leclanche cell. When a shunt resistance of 1 Omega is connected across the cell, the null point shifts to a distance of 30 cm from the common terminal. what is the internal resistance of the cell?