To Compare The EMF of Two Given Primary Cells Using Potentiometer

Aim/Aim of Experiment

To Compare the EMF of two given Primary Cells (Daniel and Leclanche cells) using Potentiometer.

Apparatus/Material Required

  • Potentiometer
  • A Leclanche cell
  • A Daniel cell
  • An Ammeter
  • A Voltmeter
  • A Galvanometer
  • A Battery
  • A Rheostat
  • A Resistance box
  • A One way key
  • A Two way key
  • A Jockey
  • A Set Square
  • Connecting wires
  • A piece of Sand paper


We know that a Voltmeter is used to measure only the potential difference (V) between the two terminals of a cell, but using a Potentiometer we can measure the value of emf of a given cell.

When we keep key (K1) closed and (K2) open, let the null point found be lal1,

E1=Klal1 …..(1)

When we keep key (K1) open and (K2) closed, let the null point found be lal2,

E2=Klal2 …..(2)

So, (1)/(2) or E1/E2 = Klal1/Klal2

hence E1/E2 = l1/l2

Where E1 and E2 are the EMF of two given primary cells and l1 and l2 are the corresponding balancing lengths on Potentiometer wire.

Circuit Diagram

Comparison of the emfs of two cells
Comparison of the emfs of two cells


  1. Connect the apparatus as shown in the circuit diagram.
  2. Remove the insulation from the ends of connecting copper wires with the help of Sand paper.
  3. Measure the e.m.f. (E) of the battery and the emfs (E1 and E2 ) of the cells. See that E > E1 and also E > E2.
  4. Connect the positive pole of the battery (a battery of constant e.m.f.) to the zero end (P) of the potentiometer and the negative pole through a one-way key, an ammeter and a low resistance rheostat to the other end (Q) of the potentiometer.
  5. Connect the positive poles of the cells E1 and E2 to the terminal at the zero end (P) and the negative poles to the terminals a and b of the two way key.
  6. Connect the common terminal c of the two-way key through a galvanometer (G) and a resistance box (R.B.) to the jockey J.
  7. Take maximum current from the battery making rheostat resistance zero.
  8. Insert the plug in the one-way key (K) in circuit and also in between the terminals a and c of the two-way key.
  9. Take out a 2,000 ohms plug from the resistance box (R.B.).
  10. Note down the direction of the deflection in the galvanometer by pressing the jockey at zero ends.
  11. Now press the jockey at the other end of the potentiometer wire. If the direction of deflection is opposite to that in the first case, the connections are correct. (If the deflection is in the same direction then either connections are wrong or e.m.f. of the auxiliary battery is less).
  12. Slide the jockey smoothly over the potentiometer wires till you obtain a point where galvanometer shows no deflection.
  13. Put the 2000 Ω plug back into the resistance box and obtain the null point position accurately with the help of a Set square.
  14. Note the length l1 of the wire at the point for the cell E1 Also note the current as indicated by the Ammeter.
  15. Repeat this with E2 by disconnecting the cell E1. Now removing the plug from gap ac of two-way key and connect the cell E2 by inserting plug into gap be of two-way key.
  16. Take out a 2000 ohms plug from resistance box R.B. and slide the jockey along potentiometer wire so as to obtain no deflection position.
  17. Note the length l2 of wire in this position for the cell E2. However, Make sure that Ammeter reading is same as in step 14.
  18. Repeat this observations alternately for each cell again for the same value of current.
  19. By increasing the current and adjusting the rheostat, we get three sets of observations in a similar way.
  20. Record your observations as given below.


  • Range of Voltmeter = 3V.
  • Least count of Voltmeter = 0.05V.
  • E.M.F. of battery eliminator = 4V.
  • E.M.F. of Leclanche cell = 1.4V.
  • E.M.F. of Daniel cell= 1V.
  • Least count of the Ammeter = 0.05A.
  • Zero error of the Ammeter = 0A.

Observation Table for Lengths:

Srl. No. of ObservationsAmmeter Reading (A)Balance Point when E1 is Connected l1 (cm)Balance Point when E2 is Connected l2 (cm)E1/E2=l1/l2
(1)(2)1(3a)2(3b)Mean l1(3c)1(4a)2(4b)Mean l2(4c)(5)


  1. For each observation, find mean l1 and mean l2 and record it in coloum 3c and 4c respectively.
  2. Find E1/E2 for each set, by dividing mean l1 (column 3c) by mean l2 (column 4c).
  3. Find the mean E1/E2.


The ratio of E.M.F.’s E1/E2 ≅ 1.404.


  1. The connections should be neat, clean and tight.
  2. The plugs should be introduced in the keys only when the observations are to be taken.
  3. The positive poles of the battery E and cells E1 and E2 Should, all be connected to the terminal at the zero of the wires.
  4. The jockey key should not be rubbed along the wire. It should touch the wire gently.
  5. The e.m.f. of the battery should be greater than the e.m.f.’s of the either of the two cells.
  6. The ammeter reading should remain constant for a particular set of observation. If necessary, adjust the rheostat for this purpose.
  7. Some high resistance plug should always be taken out from resistance box before the jockey is moved along the wire.

Sources of Error

  1. The auxiliary battery may not be fully charged.
  2. The potentiometer wire may not be of uniform cross-section and material density throughout its length.
  3. End resistances may not be zero.
  4. Other sources of errors same as the previous experiment.

Viva Voice Questions with Answers

1. What do you understand by the e.m.f. of a cell?

Answer: Electromotive force i.e., e.m.f. of a cell is the potential difference across the terminals of the cell when the cell is in an open circuit i.e., when no current is drawn from the cell.

2. What is a potentiometer?

Answer: It is an instrument used to measure potential difference or e.m.f. of a cell.

3. Why is it called a potentiometer?

Answer: Because it measures potential difference between any two points of electric circuits.

4. What is the principle of a potentiometer?

Answer: It works On the principle that for a constant current, fall of potential along a uniform wire is directly proportional to its length.

5. What kind of source of e.m.f. should be used as auxiliary battery?

Answer: The e.m.f. of the source must be steady. A freshly charged accumulator should be used for this purpose.

6. What should be the order of magnitude of the e.m.f. of the auxiliary battery?

Answer: The e.m.f. of the auxiliary battery should be slightly greater than the e.m.f. of the individual cells. (With battery of lesser e.m.f., null point will not be obtained on the potentiometer wire).

7. Why do we not want the balance point to be on the first wire, say?

Answer: The smaller is the balancing length, the greater is the relative uncertainty in its location.

8. What is the merit of a potentiometer over a voltmeter in measurement of e.m.f. of a cell?

Answer: E.M.F. measured by potentiometer is more accurate because the cell is in open circuit, giving no current.

9. What is potential gradient?

Answer: It is the fall of potential per unit length of the potentiometer wire. K =V/l.

10. Why do we use a rheostat in the battery circuit?

Answer: To vary the potential gradient.

Class 12 Physics Practicals:

2 thoughts on “To Compare The EMF of Two Given Primary Cells Using Potentiometer”

Leave a Comment