A. Potentiometer measures the potential difference more accurately than a voltmeter, because the potentiometer
a. does not draw current from external circuit.
b. has a wire of high resistance.
c. draws a heavy current from external circuit.
d. has a wire of low resistance.
B. With the help of a diagram explain the principle of a potentiometer.

A cell has a potential difference of 6 V in an open circuit, but it falls to 4 V when a current of 2A is drawn from it. Find the internal resistance of the cell.

A potentiometer wire has a length 400 cm and a resistance of 20 Omega . It is connected in series with an external resistance R and a cell of e.m.f. 1.5 V and internal resistance 1 Omega . A source of e.m.f.6 mV is balanced against a length of 30 cm of the potentiometer wire. Find R. According to the principle of potentiometer Vpropl . i.e., V = ilp where l'is the balancing length & p the resistance per cm of the potentiometer wire.

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 2.0 V and a negligible internal resistance.The potentiomter were itself is 4m long.When the resistance, R, connected across the given cell, has values of (i) Infinity (ii)9.5 Omega the'balancing lengths', on the potentiometerwire are found to be 3m and 2.85m, respectively. The value of internal resistance of cell is

The current in a conductor and the potential difference across its ends are measured by an ammeter and a voltmeter.The meters draw negligible current.The ammeter is accurate but the vlotmeter has a zero error(that is ,it does not read zero when no potential difference is applied).Calculate the zero error if the readings for two different conditions are 1.75A, 14.4Vand 2.75a, 22.4V.

Assertion: When a large current passes through the circuit, the fuse wire melts due to Joules heating effect. Reason: Fuse wire has low melting point and high resistance.

Figure 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. When 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 wire. Determine the internal resistance of the cell

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.

A potentiometer wire of length 100 cm has a resistance of 10Omega .It is connected in series with a resistance and a cell of emf 2V having negligible internal resistance.A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. The value of external resistance is

Consider the potentiometer circuit arranged as in figure. The potentiometer wire is 600cm long. At what distance from the point A should the jockey touch the wire to get zero deflection in the galvanometer? b) If the jockey touches the wire at a distance of 560cm from A, What will be the current in the galvanometer?