Question

(a) What is potentiometer? What is its principle?

(b) How will you use potentiometer

(i) to compare e.m.f. s of two cells?

(ii) to determine internal resistance of a cell?

Or

State the working principle of a potentiometer. Explain with the help of a circuit diagram how the e.m.f. of two primary cells are compared by using a potentiometer.

Answer 1

(a) Potentiometer: An instrument which is used for measuring e.m.f. or potential difference accurately is called potentiometer.

A potentiometer consists of a number of segments of wire of uniform area of cross-section stretched on a wooden board between two thick copperstrips. Each segment of wire is 100 cm long. The wire is usually of constantan or magnanin. A metre rod is fixed parallel to its length Fig. a battery connected across the two end terminals sends a current through the wire which is kept constant by using a rheostat. strips. Each segment of wire is 100 cm long. The wire is usually of constantan or magnanin. A metre rod is fixed parallel to its length Fig. a battery connected across the two end terminals sends a current through the wire which is kept constant by using a rheostat.

Principle: The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of wire is directly proportional to the length of that portion.

Let V be potential difference across certain portion of wire whose resistance is R. If I is the current through the wire, then

V = IR

We know that

R = \(\rho \frac{l}{A}\)

where l, A and ρ are length, area of cross-section and specific resistance of the material of wire respectively.

∴ V = Iρ

If constant current is passed through the wire of uniform area of cross-section, then ρ, I and A are constants and we have

V = (constant) l

or V ∝ l

Hence, if a constant current flows through a wire of uniform area of cross-section, then potential drop along the wire is directly proportional to the length of the wire.

(b) (i) Comparison of e.m.f.s of two cells

The circuit diagram for the comparison of e.m.f.s. of two cells is shown in Fig. The auxiliary circuit consists of a battery, an ammeter, rheostat Rh and key K. The positive terminals of two cells E₁ and E₂ whose e.m.f. is to be compared are connected to end A. The negative terminals are connected to two way key. The common terminal of the key is connected to jockey through galvanometer. 

The +ve terminal of auxiliary battery is connected to zero end of the scale. Insert key between 1 and 2 so that cell E₁ is in circuit. Press the jockey and slide it on the wire and find a point P such that galvanometer gives no deflection.

This will happen when potential difference between A and P is equal to e.m.f. of cell. Note length AP = l₁ using principle of potentiometer.

E₁ ∝ l₁ ...........(i)

Remove the plug from 1 and 2, insert it between 2 and 3. Again slide the jockey to get a balance point. If l₂ be the length of wire giving the balance point with the cell of e.m.f. E₂.

∴ E₂ ∝ l₂ .............(ii)

Dividing (i) by (ii) , we get

\(\frac{E_1}{E_2} = \frac{l_1}{l_2}\)

∴ E₁ and E₂ can be compared

(ii) Determination of internal resistance of a cell.

The internal resistance of a cell is given by

r = \(R \frac{(E-V)}{V}\) ..............(iii)

when E = e.m.f. of cell

V = Potential difference

R = External resistance of circuit

The values of E and V can be determined with the help of a potentiometer circuit as shown in Fig. When the key K₁ in the auxiliary circuit is closed, a constant current flows through the wire AB of the potentiometer

Keep the key K₂ open, the jockey J is slided along the wire to get a balance point. Let l₁ be length of wire giving balance point P.

Using principle of potentiometer

E = kl₁

where k is the fall of potential per unit length of the wire AB. Close the key K₂ and take out some resistance R from the resistance box. The jockey is again slided along the potentiometer wire to get balance point.

Let l₂ be length of wire giving the balance point in this case. The potential difference V across the terminals of the cell is given by,

V = kl₂

Substituting for E and V in equation (i), we get

Knowing the values of R, l₁ and l₂ the value of r can be obtained.