The potentiometer- Comparison of emf, Educational Scientific Instruments, Laboratory Glassware instruments.

To compare the emfs of two given primary cells using a potentiometer.


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To compare the emfs of two given primary cells using a potentiometer.

The potentiometer- Comparison of emf:

Theory

Potentiometer

Potentiometer is a device used to compare the e.m.f. (electromotive force) of two cells, to measure the internal resistance of a cell, and potential difference across a resistor. It consists of a long wire of uniform cross-sectional area and of 10 m in length. The material of wire should have a high resistivity and low temperature coefficient.  The wires are stretched parallel to each other on a wooden board. The wires are joined in series by using thick copper strips. A metre scale is also attached on the wooden board.

The potentiometer works on the principle that when a constant current flows through a wire of uniform cross sectional area, potential difference between its two points is directly proportional to the length of the wire between the two points.

Electromotive force (e.m.f) of a cell

Electromotive force (emf) is a measurement of the energy that causes current to flow through a circuit.  It is the energy provided by a cell or battery per coulomb of charge passing through it. It can also be defined as the potential difference across the terminals of a cell, when no current flows through it. Electromotive force is also known as voltage, and it is measured in volts. Electromotive force is not truly a force; rather, it is a measurement of energy per unit charge.

\epsilon =\frac{E}{Q}     , where E is the energy and Q is the charge.

Using a potentiometer, we can determine the emf of a cell by obtaining the balancing length l.  Here, the fall of potential along the length l of the potentiometer wire is equal to the emf of the cell, as no current is being drawn from the cell.

Then,

E\propto l    or,   E=kl ;     where k is the potential gradient along the wire.

Thus it is possible to compare the emf’s of two given cells by measuring the respective balancing lengths l1 and l2.

ie;                                          E_{1}=kl_1           and       E_2=kl_2

or                                                               \frac{E_1}{E_2}=\frac{l_1}{l_2}

Learning Outcomes

  • Students understand the potentiometer apparatus, its parts and how to use it.
  • Students learn the concept of electromotive force in cells.
  • Students are able to construct circuits based on circuit diagrams.
  • Students understand the different components used in the experiment.

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Experiment

Materials Required

  • Potentiometer
  • Daniel cell
  • Leclanche cell
  • Jockey
  • Battery eliminator
  • Resistance box
  • Galvanometer
  • One way key
  • Two way key
  • Rheostat
  • Ammeter
  • Connecting wires

Real Lab Procedure

  • Arrange the required materials on a table and make the connections as per the connection diagram.
  • Tight the plugs of the resistance box.
  • Note the reading on the ammeter.
  • To test the connection, insert plug in the one way key k1 and also in between the terminals a and c of the two way key. Introduce a sufficiently high resistance on the resistance box (R.B). Place the jockey at the two end points of the wire. Press the jockey at both end of the potentiometer wire and note the deflection in galvanometer. If the galvanometer shows opposite deflection, the connections are correct.
  • Now, gently slide the jockey along the potentiometer wire and stop when null point is obtained.
  • Measure the length l1 between this point and the end P of the potentiometer. It is the balancing length for the cell E1.
  • Disconnect the cell E1 by removing the plug from the gap ac of the two way key and connect the cell E2 by inserting plug into the gap bc of the two way key.
  • Again slide the jockey along the potentiometer wire to obtain the null point. Measure the new balancing length l2 for the cell E2 based on this point.
  • Make sure that the reading on the ammeter is constant throughout the observation.
  • Repeat the experiment by increasing the current by adjusting the rheostat and record the observations.
  • Each time, the ratio between the emf’s of the given cells can be calculated using the relation,

frac{E_1}{E_2}=frac{l_1}{l_2}

 Procedure :

  • Select the first primary cell form the drop down list.
  • Select the second primary cell form the drop down list.
  • Select the rheostat resistance using the slider.
  • In the case of other cells, you can select the emf of the first and second cells using the slider.
  • To see the circuit diagram, click on the ‘Show circuit diagram’ check box seen inside the simulator window.
  • Connections can be made as seen in the circuit diagram by clicking and dragging the mouse form one connecting terminal to the other connecting terminal of the devices to be connected.
  • Once all connections are made, click and drag the one way key to insert it into the switch.
  • Drag the plug and insert it in the first gap of the two way key.
  • Slide the jockey along the potentiometer wire to obtain the null point.
  • You can see the first balancing length in the popup.
  • Now drag the plug and insert it in the second gap of the two way key.
  • Slide the jockey along the potentiometer wire to obtain the null point.
  • You can see the second balancing length in the popup.
  • You can compare the emf values of the two cells using the value of first and second resonating lengths.
  • You can repeat the experiment by increasing the current by adjusting the rheostat.
  • To verify your result click on the ‘Show result’ check box.
  • To redo the experiment, click on the ‘Reset’ button.

Observations

No. Ammeter reading (A) Balancing length when frac{E_1}{E_2}=frac{l_1}{l_2}
E1 in the circuit, l1 (cm) E2 in the circuit, l2 (cm)
1      
2      
3      
4      
5      
6      

Calculations

Calculate the ratio of E1 and E2 for each set of l1 and l2. The mean of the calculated values gives the ratio of emf’s of the two given primary cells.

Result

The emf’s of the two given primary cells are compared.

The ratio of emf’s of the two given primary cells, E1/E2 = …………..

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