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.

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.

Explain the use of potentiometer for determining he internal resistance of 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

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Omega the balance point is found at 40 cm of the wire from the same end. The internal resistance of the cell is

A lead resistance R is connected across a cell of emf E and internal resistance r. If the closed circuit p.d. across the terminals of the cell is V the internal resistance of the cell is.

Two resistances are connected in the two gaps of a meter bridge. The balance point is 20 cm from the zero end. When a resistance 15 Omega is connected in series with the smaller of two resistance, the null point+ shifts to 40 cm . The smaller of the two resistance has the value.

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