The balancing length for a cell in a potentiometer experiment is 45 cm. In the balanced condition, the current through the potentiometer wire is 2 A. If the total length of the potentiometer wire is 1 m and its total resistance is`60Omega` , the emf of the cell is (in Volts) _________.

When a balance point is obtained in a potentiometer for finding the internal resistance of a cell, the current through the potentiometer wire is due to

In comparing emf's of 2 cells with the helf of potentiometer, at the balance point, the current flowing through the wire is taken from

A cell is balanced at 100cm of a potentiometer wire when the total length of the wire is 400cm .If the length of the potentiometer wire is increased by 100cm ,then the new balancing length for the cell will be

In potentiometer experiment, null point isobtained at a particular point for a cell on potentiometer wire x cm long. If the lengthof the potentiometer wire is increasedwithout changing the cell, the balancing length will (Driving source is not changed)

Calculate the value of unknown potential V for the given potentiometer circuit .The total length 400 cm of potentiometer wire has a resistance of 10 Omega and the balance point in obtained at a length of 240 cm

A constant voltage of 3 V is supplied to a potentiometer wire. A cell of emf 2 V is balanced by a voltage drop across 416 cm of the wire. Find the total length of the potentiometer wire.

A 2 volt battery, a 15 Omega resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance resistance of potentiometer wire is 5 Omega , then the potential gradient of the potentiometer wire is

It is observed in a potentiometer experiment that no current passes through the galvanometer, when the terminals of the potentiometer, wire. On shunting the cell by a 2Omega resistance, the balancing length is reduced to half. The internal resistance of the cell is:-

In an experiment to find the internal resistance of a cell by a potentiometer, a balance was obtained for 50 cm length of the potentiometer wire, with a cell of e.m.f. 2V. When the cell was shunted by a resistance of 2 Omega , the balancing length of the potentiometer wire was 40 cm. What was the internal resistance of the cell ?