A potentiometer wire is 10 meter long and a potential difference of 6 volt is maintained between its ends. Find the e.m.f. of a cell which balances a length of 180 cm of the potentiometer wire.

A potentiometer is an ideal instrument for measuring potential difference or e.m.f. because

A potentiometer wire is 100 cm long hand a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obatined at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emfs is:

A potential drop per unit length along a wire is 5 xx 10^-3 V//m . If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.

A potentiometer wire is 10 m long and has a resistance of 2 Omega//"m". It is connected in series with a battery of e.m.f. 3 V and a resistance of 10Omega . The potential gradient along the wire in V/m is

A potentiometer wire of length 10 m is connected in series with a battery the emf of a cell balances against 250 cm length of wire if length of potentiometer wire is increased by 1 m then new balancing length of wire will be

A potentiometer circuit has been setup for finding. The internal resistance of a given cell. The main battery used a negligible internal resistance. The potentiometer wire itsefl is 4 m long. When the resistance, R , connected across the given cell, has value of (i) Infinity 9.5 Omega , (ii) the 'balancing length' , on the potentiometer wire are found to be 3 m and 2.85 m , respectively. The value of internal resistance of the cell is

Length of potentiometer wire is 600 cm and a driving cell develop 6V potential difference across it. The potential gradient along the wire is?

A potentiometer wire of Length L and a resistance r are connected in series with a battery of e.m.f. E_(0) and a resistance r_(1) . An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by :