To Measure Internal Resistance Of A Cell|Questions

Internal resistance of a cell depends on

Determination Of Internal Resistance Of A Cell

In an experiment with potentiometer to measure the internal resistance of a cell when the cell is shunted by 5Omega the null point is obtained at 2m .When the cell is shunted by 10Omega the null point is obtained at 3m .The internal resistance of the cell is

In an experiment with a potentiometer to measure the internal resistance of a cell. when the cell in the secondary circuit is by shounted by 5 Omega , the null point is at 220 cm . When the cell is shunted by 20 Omega the null point is at 300 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

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 shunted by a 4Omega resistance and at 3 m when cell is shunted by a 8Omega resistance. The internal resistance of cell is -

In order to measure internal resistance r1 of a cell emf E, a meter bridge of wire resistance R_0 = 50 Omega , a resistance R_0/2 another cell of emf E/2 and galvanometer G are used in circuit. If null point is founded at l = 72cm. then value of r1

Name the device used for measuring the internal resistance of a secondary cell.