potentiometer is a device used for measuring emf and internal resistance of a cell. It consists of two circuits one is main circuit in which there is a cell of given emf ε′ and given resistance R which is connected across a wire of length 100cm having resistance r. Another circuit having unknown emf ε and galvanometer.for a given potentiometer
Answers
ANSWER
(a) Constant emf of the given standard cell, E
1
=1.02V
Balance point on the wire, l
1
=67.3cm
A cell of unknown emf, ε,replaced the standard cell. Therefore, new balance point on the wire, l=82.3cm
The relation connecting emf and balance point is,
l
1
E
1
=
l
ε
ε=
l
1
l
×E
1
=
67.3
82.3
×1.02=1.247V
The value of unknown emfis 1.247 V.
(b) The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
(c) The balance point is not affected by the presence of high resistance.
(d) The point is not affected by the internal resistance of the driver cell.
(e) The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.
(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.
The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.