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

How will you use potentiometer to determine internal resistances of a cell?

Why is a potentiometer preferred over a voltmeter for determining the e.m.f. of a cell?

In an experiment to determine the internal resistance of a cell, the null point is obtained at 2 cm, when the cell is shnted by a resistance of 5Omega . When the cell is shunted by a resistance of 20Omega , the null point is obtained at 3 cm. Find the internal resistance of the cell.

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.