# Comparision of EMF Using Potentiometer Experiment

Experiment: To compare the e.m.f.’s of two given primary cells by using a potentiometer.

A voltmeter is a device used for measuring the terminal potential difference of a cell. But in the process of measuring, it draws a small current from the cell. A potentiometer is an instrument used for measuring without drawing any current, the potential drop across two points in a circuit, or e.m.f. of a cell. It is also used for comparing the e.m.f s of two given cells.

If E1 and E2 are the e.m.f s of the two given cells and l1 and l2 are the two balancing lengths respectively on the potentiometer then,

E1/E2 = l1/l2

Material Required

A potentiometer with atleast 4 m wire, an ammeter, a galvanometer, a voltmeter, a resistance box, a jockey, sandpaper, a one way key, a Leclanche cell, a Daniell cell, a two way key and rormectine; wires

### How To Perform Experiment

1. Draw the circuit diagram.

2. Make the connections as shown in the circuit. Be sure to remove the insulations from the ends of connecting wires with sandpaper.

3. Make sure that the negative end of the ammeter is connected with the negative terminal of the battery. The positive terminal of the battery should always be connected to the zero end of the potentiometer.

4. Always keep the key, K, open while making connections.

5. Make the resistance in the rheostat minimum by drawing current for full scale deflection of the ammeter from the battery.

6. Close the one-way key (K) and take out a 1000 Ω plug from the resistance box for safety of the galvanometer while first searching approximate position of null point. Insert the plug between the terminals a and c to connect the cell E1, in the circuit. E1, is the Leclanche cell.

7. Now, gently press the jockey at the zero end of the potentiometer and note the deflection in the galvanometer. Press the jockey at the other end of the potentiometer wire. If the two deflections in the galvanometer are in the opposite direction then the connections are correct. In case the two deflections are in same directions, then potential drop across the potentiometer is less than E1. Then current in potentiometer has to be increased.

8. Now, gently slide the jockey over the potentiometer wire till the galvanometer shows no deflection. This position of jockey is approximate position of null point. Put back the 1000 Ω plug in the resistance box and make fine adjustment of null point, since the galvanometer becomes more sensitive with zero resistance in the resistance box. Note this position of null point as first position and now move the jockey 5 cm beyond this position and locate the second position of this null point by sliding back the jockey and take mean of these positions as l1.

9. Take down the ammeter reading and note the length l1 for the cell E1.

10. Disconnect the cell B by removing the plug from gap ac and insert the plug between gap bc to connect the cell E2 and repeat the process.

11. Obtain an accurate position of null point for the second cell E2 also by making resistance zero in the resistance box and note the length l2. Make sure that the ammeter reading remains precise) the same as that for cell E1, when you measured l1.

12. Change the current in the circuit by adjusting the rheostat and obtain at least three sets of observations similarly. If the null point for E1 in first set was on first or second wire, you must decrease the current for subsequent sets so that this null point shifts to 3rd or 4th wire.

### Sources of Error

1. The potentiometer wire may not be of uniform cross-section throughout its length.
2. The e.m.f. of the auxiliary battery may not be constant during the experiment.
3. Contact resistances at the ends of the wires of potentiometer may not be negligible due to rusting and a considerable amount of voltage may drop across them.