Explain the use of potentiometer for determining he internal resistance of cell.

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 Omega 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.

Define : Internal resistance

When a balance point is obtained in a potentiometer for finding the internal resistance of a cell, the current through the potentiometer wire is due to

State the uses of a potentiometer.

In an experiment to determine the internal resistance of a cell of emf 1.5 V, the balance point in the open cell condition is 76.3 cm. When a resistor of 9.5 ohm is used in the external circuit of the cell the balance point shifts to 64.8 cm of the potentiometer wire. Determine the internal resistance of the cell.

The formula for internal resistance of a 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

In potentiometer experiment to determine internal resistance of the cell, balance point has been obtained in fourth wire. It can be shifted to 5^("th") wire by

Describe with the help of neat diagram how will you determine the internal resistance of a cell by using potentiometer. Derive the necessary formula.