Determination Of Internal Resistance Of A Cell

In the determination of internal resistance of a cell of e.in.f. 1.5 V with potentiometer, the balancing length is 75 cm in the open circuit and when a resistance of 10 Omega is used the balancing length is found to be 65 cm. Calculate the internal resistance of the cell.

Explain the determination of the internal resistance of a cell using voltmeter.

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

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

Fig. shows a 2.0V potentiometer used for the determination of internal resistance of 1.5V cell- The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Omega 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.

Figure 3.35 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. :

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