Determination Of Internal Resistance Of A Cell

Correct diagram for the determination of internal resistance of a primary cell by potentiometer

In the determination of the internal resistance of a cell using a potentiometer, when the cell is shunted by a resistance R and connected in the secondary circuit, the balance length is found to be L_(1) . On doubling the shunt resistance, the balance length is found to increase to L_(2) . The value of the internal resistance is

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

Figure shows a 2.0V potentiometer used for the determination of internal resistance of a 1.5V cell , When the key is not inserted in the plug , the balance point is at 60 cm and in the closed circuit the balance point is at 50 cm. Find the internal resistance of the cell .

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.

Consider the potentiometer circuit for determining the internal resistance of a cell. When switch S is open, the balance point is found to be at 75 cm os the wire. When switch S is closed and value of R is 4 Omega , the balance points shifts to 60 cm . find the internal resistance of cell. .

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.