Write equation to measure internal resistance of cell by using potentiometer.

Explain method to measure internal resistance of cell by using potentiometer.

In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted 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

Internal resistance of a cell depends on

What is called as internal resistance of the cell ?

When internal resistance of the cell can be neglected?

show a 2.0V potentiometer used for the determination of internal resistance of a 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.

Write equation of equivalent resistance of n resistors connected in series.

There is a potentiometer wire of length 1200 cm and a 60 mA curreut is flowing in it. A battery of emf 5 V and internal resistance of 20 Omega is balanced on this potentiometer wfre with a balancing length 1000 cm. The resistance of the potentiometer wire is

Assertion: Terminal potential difference of a cell is always less than its emf. Reason: Potential drop on internal resistance of cell increases terminal potential difference.