In a potentiometer experiment , the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of `2Omega`, the balancing length becomes 120 cm. Find the internal resistasnce of the 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.

A cell gives a balance with 85 cm of a potentiometer wire. When the terminals of the cell are shorted through a resistance of 7.5Omega , the balance is obtained at 75 cm. Find the internal resistance of the cell.

In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shuntedd by a 5 Omega resistance, and is at a length of 3 m when the cell is shunted by a 10 Omega resistance. The internal resistance of the cell is, then

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

A cell supplies a current of 0.3 A through a resistance of 7 ohm and a current of 0.9 A through a resistance of 2 ohm. The internal resistance of the cell is

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.

A cell can be balanced against 110 cm and 100 cm of potentiometer wire respectively, when open icrcuited and when short circuited through a resistance of 10 ohm. Find the internal resistance of the cell.

Two cells of emfs E_(1) and (E_(2) (E_(1) gt E_(2)) are connected as shows in figure When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio E_(1)//E_(2)