In potentiometer method'a cell can be balanced at 480cm. It is balanced at 400 cm when connected through a shunt `5 Omega`. Find the internal resistance of the cell.

In a potentiometer experiment . A cell produces balance in the wire at 240 cm. balancing length change to 120 cm when a resistance 2Omega Is connected parallel to the cell. Find out the internal resistance of the cell.

In a potentimeter experiment , it is found that no current passes through the galvanometer when the termials of the cell are connected across 52 cm of the potentimeter wire . If the cell is shunted by a resistance of 5Omega , a balance is found when the cell is connected across 40 cm of the wire . Find the internal resistance of ht cell .

A length of potentiometer wire of 155 cm balances the emf of a cell in a circuit and a length of 135 cm when the cell has a conductor of resistance 8Omega connected between its terminals . Find the internal resistance of the cell .

The potential difference between the terminals of a cell, in an open circuit was 2.5 volt and when it was connected through a resistance of 5 ohm, the potential difference fell to 1.5v. Find the internal resistance of the cell.

In a potentiometer experiment , a particular cell produces balance in the wire at 560 cm . If a 10Omega resistance is connected with the cell in parallel ,then this balanced lendth change to 412 cm . Find out the internal resistance of the cell .

A current of 2 A flows through a 2 Omega resistor when connected across a battery. The same battery supplied a current of 0.5 A when connected across a 9 Omega resistor. The internal resistance of the battery is

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

In a potentiometric arrangement ,a cell is connected to the potentiometer wire of 60 cm in length to make the deflection zero in galvanometer . Now if the cell is shunted with a 6Omega resistor , a null point is found in 50 cm length of the wire . What is the internal resistance of the cell ?

Draw Wheatstone bridge circuit and write down the balancing condition of it. In a meter bridge the balance point is found to be at 40 cm from one end when the resistor at the end is 10 Omega . Find the resistance at the other side. If the galvanometer and the cell are interchanged at the balance condiotion, would it affect the current through the galvanometer?

In a meter bridge the balance point is found to the be at 40 cm from one end when the resistor at the end is 15Omega , Find the resistance at the other side .