EMF of Two Primary Cells Using a Potentiometer
A popular Physics experiment taught in schools is the experiment to compare the EMF of two given primary cells using a potentiometer.
In this article, we will learn how to compare the EMF of two given primary cells using a potentiometer. You will also learn about the potentiometer and its uses. The experiment will then be explained using the theory behind it. Each material required for the experiment is also listed clearly. The procedure is also explained in a simple and easy to understand manner. Refer to the official website of Vedantu or download the app for an elaborate and easy explanation.
The materials which we need for the experiments are a potentiometer along with a Leclanche cell, a Daniel cell, ammeter, voltmeter, galvanometer, battery or battery eliminator, rheostat of low resistance, resistance box, one-way key, two-way key, jockey, set square, then connecting wires and a piece of sandpaper.
What is a Potentiometer?
A potentiometer is a very useful instrument that is used to measure electric potential. It consists of three terminals and a slidable or rotating contact. This acts as a voltage divider in the potentiometer.
They are mostly used in audio equipment to control the volume. They can be also used to control inputs for other electric circuits.
Where, the letter E1 and E2 are the e.m.f. of two cells which are given and l1 and l2 are the corresponding balancing lengths on the wire of the potentiometer.
We will follow the following steps to do the experiments:
First of all, arrange the apparatus as shown in the circuit diagram figure.
Then we need to remove the insulation from the ends of the connecting copper wires with sandpaper.
Next is to measure the e.m.f. That is denoted by E of the battery and the e.m.fs. That is E1 and E2 of the cells. See that E > E1 and also we will notice that E > E2.
Then we need to connect the positive pole of the battery or a battery of constant e.m.f. to the zero end that is denoted as P of the potentiometer and the negative pole through a one-way key. An ammeter and a low resistance rheostat to the other end which is denoted by Q of the potentiometer.
Next, we will connect the positive poles of the cells E1 and E2 to the terminal at the zero end denoted by P and the negative poles to the terminals a and b of the two way key.
Now connect the common terminal that is c of the two-way key through a galvanometer denoted by G and a resistance box denoted by letters R.B. to the jockey J.
We now need to take maximum current from the battery making rheostat resistance zero.
Then we will insert the plug in the one-way key denoted by letter K in the circuit and also in between the terminals a and c of the two-way key.
Now carefully take out a 2,000 ohms plug from the resistance box that is R.B.
Now we will press the jockey at the zero end and then please note the direction of deflection in the galvanometer.
Then press the jockey at the other end of the potentiometer. If the direction of deflection is showing opposite to that in the first case then we can say that the connections are correct. If the deflection is in the same direction itself then either connections are wrong or we can say that the e.m.f. of the auxiliary battery is less.
Then we will slide the jockey gently over the potentiometer wires till we obtain a point where the galvanometer shows no deflection.
Now we need to put the 2000 ohms plug back in the resistance box and obtain the null point position accurately by using a set square.
We need to note that the length l1 of the wire for the cell E1 Also note that the current is as indicated by the ammeter.
Now disconnect the cell which is E1 by removing the plug from gap ac of two-way key and then connect the cell E2 by inserting plug into gap be of two-way key.
Now we need to take out a 2000 ohms plug from resistance box R.B. and slide the jockey along potentiometer wire to obtain no deflection position.
Now put the 2000 ohms plug back in the resistance box that is R.B and obtain an accurate position of the null point for the second cell E2.
We need to note the length l2 of wire in this position for the cell E2. However, we need to make sure that ammeter reading is the same as in step 14.
Now repeat the observations alternately for each cell again for the same value of current which we want.
Then increase the current by adjusting the rheostat and obtain at least three sets of observations in a similar way as mentioned above.
Then record our observations.
For each observation, we will find the mean l1 and mean l2 and record in a column by naming them 3c and 4c.
Then we will find E1/E2 for each set by dividing mean l1 (column 3c) by mean l2 (column 4c).
Now find the mean of E1/E2.
Precautions of the Experiment
The connections which we are making should be neat, clean and tight.
The plugs should be introduced in the keys only when we are doing the experiment and taking the observations.
The poles which are the positive ones of the battery that are E and cells E1 and E2 should all be connected to the terminal at the zero of the wires.
The key of the jockey should not be rubbed along the wire. We need to take care that it touches the wire gently.
The reading of the ammeter should remain constant for a particular set of observations. If necessary we can adjust the rheostat for this purpose as well.
The e.m.f. That is, the battery should be greater than the e.m.f. 's which is either of the two cells.
There are some high resistance plugs also which should always be taken out from the resistance box before the jockey is moved along the wire.
The Source of Error
This could be one reason that the auxiliary battery may not be fully charged.
The potentiometer wire that we are using may not be of uniform cross-section and material density throughout its length.
Then the end resistances may not be zero.
FAQs on To Compare the EMF of Two Given Primary Cells Using Potentiometer Experiment
1. Explain how you can compare the EMF of two cells using a potentiometer?
While using a potentiometer we can determine that which is the emf of a cell by obtaining the balancing length that is denoted by l. Here we can say that the fall of potential along the length l of the potentiometer wire is equal to the emf of the cell. As there is no current which is being drawn from the cell.
2. Explain what is the primary cell in a potentiometer?
The two cells which are the primary cells whose EMFs are to be compared are so connected in the circuit that their terminals which are positive terminals are joined together to the end A of the potentiometer wire AB and their negative terminals are also said to be joined to a galvanometer through a two-way key a, b, c.
3. Explain why the EMF of a cell measured by a potentiometer is accurate?
A voltmeter draws generally current from a cell therefore voltmeter is said to measure terminal potential difference rather than emf. While we can say that a potentiometer at balance does not draw any current from the cell so the cell remains in an open circuit. Hence the potentiometer is said to read the actual value of emf.
4. Explain how we use a potentiometer experiment?
We introduce a sufficiently high resistance on the resistance box which is denoted by R.B. Then we place the jockey at the two endpoints of the wire. Then we press the jockey at both ends of the potentiometer wire and we need to note which is the deflection in the galvanometer. If the galvanometer shows opposite deflection then we can say that the connections are correct.
5. What principle of Physics must be understood well to comprehend the experiment of the potentiometer?
The potentiometer works on the simple principle that the drop of potential across a wire segment of the uniform cross-section is directly proportional to the length of the wire. This is an important principle that must be thoroughly understood to be able to successfully understand the above-mentioned experiment. The potentiometer workings can be understood better using the Vedantu website. Our experts have explained the workings of the potentiometer, its characteristics and applications in a simple manner along with the diagram of the device.