To compare the EMF of two given primary cells (Daniel and Leclanche cells) with the help of a potentiometer.
- Daniel Cell
- Leclanche Cell
- low resistance Rheostat
- A one-way key
- A two-way key
- Set Square
- Resistance Box
- Connecting wires
- Piece of sandpaper
Using a voltmeter it is possible to measure only the potential difference between the two terminals of a cell, but using a potentiometer we can determine the value of emf of a given cell. where E1 and E2 are EMFs of two cells, l1 and l2 are the balancing lengths when E1 and E2 are connected to the circuit respectively and φ is the potential gradient along the potentiometer wire.
E1 /E2 = φ l1 /φ l 2 = l1 /l 2
- Connect the circuit as shown in the figure.
- With the help of sandpaper, remove the insulation from the ends of connecting copper wire.
- Measure the EMF (E) of the battery and the EMFs (E1 and E2) of the cell and see if E1 > E and E2 > E.
- Connect the positive pole of the battery to the zero end (P) of the potentiometer and the negative pole through the one-way key, low resistance rheostat and the ammeter to the other end of the potentiometer (Q).
- Connect the positive poles of the cells to the terminal at the zero end (P) and the negative poles to the terminals a and b of the two way key.
- Connect the common terminal c of the two-way key through a galvanometer (G) and a resistance box to the jockey J.
- Take maximum current from the battery by making the rheostat resistance zero.
- Insert the plug in the one-way key through the resistance box and the galvanometer to the jockey J.
- Take out 2000 Ω plug form the resistance box.
- Note down the direction of the deflection in the galvanometer by pressing the jockey at zero end.
- Now, press the jockey at the other end of the potentiometer wire. If the deflection is in the opposite direction to that in the first case, the connections are correct.
- Push the jockey smoothly over the potentiometer up to a point where galvanometer shows no deflection.
- Put the 2000 Ω plug back to the resistance box and obtain the null point position accurately with the help of the set square.
- Note the length l1 of the wire for the cell E1.
- Note the current as indicated by the ammeter.
- Disconnect the cell E1 from the plug
- Connect E2 by inserting the plug into gap be of the two-way key.
- take out a 2000 ohms plug from resistance box and slide the jockey along potentiometer wire and obtain no deflection position.
- put 2000 ohms plug back in the RB and obtain null for E2.
- note the length L2 of wire in this position for the cell E2.
- by increasing the current and adjusting the rheostat get three sets of observation.
E.M.F of battery, E =
E.M.F of Leclanche cell, E1 =
E.M.F of Daniel cell, E2 =
Range of voltmeter =
Least count of voltmeter =
Least count of ammeter =
Zero error of ammeter =
Corrected Ammeter Reading
Balance point when E1
(Leclanche cell) in the circuit
Balance point when E2
(Daniel cell) in the circuit
- For each observation, find mean l1 and mean l2 and record it 3c and 4c respectively.
- Find E1/E2, by dividing l1/l2
- Find the mean of E1/E2
The ratio of EMFs, E1/E2 ≅ _____.
1. What is an EMF of a cell?
Ans: Electromotive force is the measurement of the energy that causes the current to flow through a circuit. It is also known as voltage and is measured in volts.
2. What is potentiometer?
Ans: A potentiometer is a three terminal device that is used to measure the potential difference by manually varying the resistance.
3. On what principle does the potentiometer work?
Ans: For a constant current, the fall of potentiometer along a uniform wire is directly proportional to its length.
4. How is the emf of the cell determined for given cells?
Ans: E1/E2 = l1/l2
5. Is the cross section of the potentiometer wire uniform?