Stepwise Procedure and Key Calculations for EMF Comparison
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.
Theory
E1/E2=l1/l2
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.
The Procedure
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.
Observation
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 Compare the EMF of Two Primary Cells Using a Potentiometer
1. What is the fundamental principle on which a potentiometer works for comparing EMFs?
The working principle of a potentiometer states that for a wire of uniform cross-section and composition carrying a constant current, the potential drop across any length of the wire is directly proportional to that length. This relationship is expressed as V ∝ l. This principle allows for the precise measurement of an unknown EMF by balancing it against a known potential drop along the wire.
2. How do you compare the EMFs of two primary cells using the individual cell method with a potentiometer?
To compare the EMFs of two cells, E₁ and E₂, they are connected one by one in the secondary circuit. First, cell E₁ is connected and the balancing length (l₁) is found, where the galvanometer shows zero deflection. According to the principle, E₁ = K * l₁. Then, cell E₁ is replaced by cell E₂, and the new balancing length (l₂) is found, giving E₂ = K * l₂. By dividing the two equations, we get the ratio of EMFs as E₁/E₂ = l₁/l₂.
3. Why is a potentiometer preferred over a voltmeter for accurately measuring a cell's EMF?
A potentiometer is preferred because it measures the EMF of a cell in an open circuit condition. At the balancing point (null deflection), it draws no current from the cell. In contrast, a voltmeter always draws some current to operate, thereby measuring the terminal potential difference (V = E - Ir), which is always slightly less than the actual EMF (E) of the cell due to its internal resistance (r).
4. What is the importance of the potential gradient in a potentiometer experiment?
The potential gradient (K) is the potential drop per unit length of the potentiometer wire (K = V/L). It determines the sensitivity and range of the potentiometer. A smaller potential gradient leads to a more sensitive instrument, allowing for more precise measurement as a larger balancing length is obtained for a given EMF. The potential gradient must be constant and uniform throughout the experiment for accurate results.
5. What are the essential precautions for the Class 12 experiment on comparing EMFs using a potentiometer?
As per the CBSE 2025-26 practical syllabus, the following precautions are essential:
The EMF of the driver cell in the primary circuit must be greater than the EMFs of the cells being compared in the secondary circuit.
The positive terminals of all cells (driver and the two primary cells) must be connected to the same end of the potentiometer wire.
The jockey should be tapped gently on the wire and not dragged to avoid altering the wire's uniform cross-section.
The current in the primary circuit should be kept constant throughout the experiment by using a rheostat.
6. What happens if the EMF of the driver cell is less than the EMF of the primary cell being measured?
If the EMF of the driver cell is less than the EMF of the primary cell, the potential drop across the entire length of the potentiometer wire will be insufficient to balance the EMF of the primary cell. Consequently, you will not find a null point or balancing length on the wire. The galvanometer will show a deflection in only one direction, no matter where the jockey is placed on the wire.
7. Can an AC source be used instead of a DC battery in the primary circuit of a potentiometer? Explain why.
No, an AC source cannot be used in the primary circuit. A potentiometer works on the principle of a steady potential drop, which requires a constant, unidirectional current provided by a DC source. With an AC source, the magnitude and direction of the current and potential change continuously, making it impossible to find a stable null or balancing point on the wire.
8. What is the formula used to compare EMFs when two cells are connected in the 'sum and difference' method?
In the sum and difference method, the cells are first connected to assist each other (sum), and the balancing length l₁ is found, so E₁ + E₂ = K * l₁. Then, they are connected to oppose each other (difference), and the balancing length l₂ is found, giving E₁ - E₂ = K * l₂. The ratio of EMFs is then calculated using the formula: E₁/E₂ = (l₁ + l₂)/(l₁ - l₂).
