Figure `6.13` 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`. Whan 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. Dentermine the internal resistance of the cell.

Using a potentiometer how do you determine the internal resistance of a cell?

Potentiometer is used to determine the internal resistance of a cell using potentlometer:- A potentiometer is a device for ________.

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

A battery of emf 10V and internal resistance 3Omega is connected to a resistor.If the current in the circuit is 0.5A,what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

A battery of e.m.f. 10V and internal resistance 3W is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery used across the potentiometer wire, has an emf of 2V and negligible internal resistance. The potentiometer itself is 4m long. When the resistance 'R_(1)' , connected across the given cell has values of (i) infinity (ii) 9.5Omega , the balancing length on the potentiometer wire are found to be 3m and 2.85m respectively. The value of internal resistance of the cell is