Potentiometer: A potentiometer is a device to measure an unknown emf or potential difference accurately.

Construction: a potentiometer consist of a long wire AB of uniform cross-section, usually 4 to 10 meter long, of constantan or manganin. Usually, 1 m long separate pieces of wire are fixed on a wooden board parallel to each other. The wires are joined in series by thick cooper strips. The end A and B are connected to a battery, a plug key k, and a rheostat R_h. this circuit sends a constant current l through the wire AB, A jockey can slide along the length of the wire.

Principle : the basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-section area and composition, the potential drop across any length of the wire is directly proportional to that length. If we connect a voltmeter between the end A and the jockey J, it reads the potential difference V across the length l of the wire AJ. By ohm’s law,

V = IR = IρL/A

For a wire of uniform cross-section and uniform composition, resistivity ρ and area of cross-section A are constants. Therefore, when a steady current I flows through the wire,

I ρ/A = constant k

Hence, V = kl

Or

V ∝ l

This is the principle of a potentiometer.

Potential gradient: the potential drop per unit length of the potentiometer wire is known as potential gradient. It is given by k = V/l

SI unit of potential gradient = V/m

Cgs unit of potential gradient = V/cm

Determination of resistivity: if r is radius of the wire and l’ its length, then resistivity of the material is

Comparison of emf’s of two cells using potentiometer:

In the case want to compare the emf’s (ɛ₁ - ɛ₂) of two cells, we find the balance points separately for each cell. If l₁ and l₂ are the distance of respective balance points from the ends a, then

ɛ₁ ∝ l₁ and

ɛ₂ ∝ l₂

Determination of internal resistance of a cell: in order to determine the internal resistance r of a cell whose emf is ɛ, the experimental set up is shown in fig. first find the balance point C, when the key k’ is open. In this case, the cell is in the open ckt and

E ∝ pd across AC

ɛ ∝ l₁   or  ɛ ∝ k l₁………………………(1)

again, find the balance point C when the key k is closed and a resistance R is in the circuit. In this case the cell is in the circuit. In this case the cell is in closed circuit and

V = pd across ac₂

 i.e. V ∝ l₁     or      V = K l₂ …………………….(2)

From 1,2,3 we get

This is the internal resistance of the cell.