Exams Prep Master
Potentiometer experiment is where the potentiometer will be used for determining and comparing the EMFs of two given primary cells by balancing their length. EMF refers to the Electromotive Force or the energy per unit electric charge that is imparted through an energy source, like an electric generator and battery. Here, energy gets converted from one form to another in the generator or the battery present in the device does work after the electric charge is being transferred to it. By this, one terminal of the device becomes positively charged and the other one becomes negatively charged and the electricity that is gained per unit electric charge is known as the Electromotive Force. A potentiometer is an instrument that is used for measuring an electromotive force by balancing it against the potential difference.
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Key Takeaways: EMF, Electromotive Force, Potentiometer, Electric Charge, Primary Cells, Potential Difference, Electric Generator, Daniel Cell, Leclanche Cell
Aim
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The aim of this experiment is to compare the EMF of two given primary cells namely, Daniel and Leclanche cells, with the help of a potentiometer.
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Apparatus or Material Required
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Before starting, here is the list of materials that will be needed for doing an Experiment are as follows –
- Potentiometer
- Ammeter
- Voltmeter
- Galvanometer
- Daniel Cell
- Leclanche Cell
- Low resistance Rheostat
- One-way key
- Two-way key
- Jockey
- Set Square
- Resistance Box
- Piece of Sandpaper, and
- Connecting Wires
Theory
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We can use Potentiometer to determine the value of emf of a given cell as through voltmeter, we are only able to measure the potential difference between the two terminals of the cell. Let’s take an example where E1 and E2 are the EMFs of two cells l1 and l2 are the balancing lengths where E1 and E2 are connected with the circuit and φ is refers as the potential gradient along the potentiometer wire.
The formula for calculating the emf of the primary cell is given by,
E1 /E2 = φ l1 /φ l 2 = l1 /l 2
Circuit Diagram
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Circuit Diagram of Comparison of the EMF of two Primary Cells
Procedure
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- Arrange the apparatus as shown above in the given Circuit Diagram.
- Remove insulations from the ends of the connecting copper wires with the sandpaper.
- Measure the EMF of the battery and also, the EMF of the cells. Also, have a look that is E>E1 and E>E2.
- Connect the Positive pole of the battery to the zero ends of the Potentiometer and the negative pole through a one-way key, an ammeter, and a low resistance rheostat to the other end of the potentiometer.
- Connect the positive poles of the cells E1 and E2 to the terminal at zero ends 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 and a resistance box to the jockey.
- Take the minimum current from the battery making rheostat resistant zero.
- Insert the plugin the one-way key (K) in the circuit and also in between the terminals a and c of the two-way key.
- Take out a 2,000 ohms plug from the resistance box (R.B.).
- Press the jockey at the zero ends and note the direction of deflection in the galvanometer.
- Press the jockey present at the other end of the potentiometer wire. If the direction of deflection is opposite to that in the first case, the connections are correct. (If the deflection is in the same direction then either connection is wrong or the emf of the auxiliary battery is less).
- Slide the jockey gently over the potentiometer wires till you obtain a point where the galvanometer shows no deflection.
- Put the 2000 ohms plug back in the resistance box and obtain the null point position accurately, using a set square.
- Note the length l1 of the wire for cell E1 and also note the current as indicated by the ammeter.
- Disconnect cell E1 by removing the plug from the gap ac of the two-way key and connect the cell E2 by inserting a plug into the gap be of the two-way key.
- Take out a 2000 ohms plug from resistance box R.B. and slide the jockey along the potentiometer wire to obtain no deflection position.
- Put the 2000 ohms plug back in the resistance box and obtain the accurate position of the null point for second cell E2.
- Note the length l2 of wire in this position for cell E2. However, make sure that the ammeter reading is the same as in step 14.
- Repeat the observations alternately for each cell again for the same value of current.
- Increase the current by adjusting the rheostat and obtain at least three sets of observations similarly.
- By increasing the current and adjusting the rheostat, we get three sets of observations.
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Observations
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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 =
Calculations
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- For each of the observations, find the mean l1 and mean l2 and record it 3c and 4c respectively.
- Find E1/E2 by dividing l1/l2
- At last, find the mean of E1/E2.
Result
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The ratio of EMFs, E1/E2 ≅ _____.
Precautions
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- Connections should be neat and clean and tight.
- The plugs should be introduced in the keys only when the observations have to be taken.
- The positive poles of battery E and cells E1 and E2 should be connected to the terminal at the zero of the wires.
- The jockey key should not be rubbed along the wire. It should touch the wire gently.
- The ammeter reading should remain constant for a particular set of observations. If necessary, adjust the rheostat for this purpose.
- The e.m.f. of the battery should be greater than the e.m.f.’s of either of the two
cells. - Some high resistance plugs should always be taken out from the resistance box before the jockey is moved along with the wire.
Things to Remember
- The only potentiometer can be used for measuring the electric force.
- EMF isn’t the force.
- Connect the poles of the Potentiometer and the rheostat correctly.
- All the circuits must be connected in the right order.
- Voltmeter can only measure the potential difference between two terminal s of the cells.
- Remember, rheostat resistance must be zero at the time of the battery’s current at the maximum.
Also Read:
Sample Questions
Ques: Explain how you can compare the EMF of two cells using a Potentiometer? (2 marks)
Ans: While using the Potentiometer, we can determine that which from the given two cells, is the emf of a cell by obtaining the balancing length that is denoted by I. Here, we can say that the fall of the potentiometer along with the length of the potentiometer wire is equal to the emf of the cell because there is no current which is drawn from the cell.
Ques: Explain what is Primary Cell in a Potentiometer? (2 marks)
Ans: The two cells which are the Primary Cells whose EMFs have to be compared are connected in the circuit that their terminals which are the positive terminals are joined together to the end of the potentiometer wire and their negative terminals are also said to be joined with a galvanometer.
Ques: Why EMF of a Cell that is measured in a Potentiometer is accurate? (2 marks)
Ans: Voltmeter generally measures terminal potential difference rather than an emf. Whereas, the potentiometer at balance does not draw any current from the cell so the cell remains open in an open circuit. Therefore, the potentiometer gives the accurate actual value of the emf.
Ques: Explain how to use the Potentiometer Experiment? (2 marks)
Ans: We introduce a sufficiently high resistance on the resistance box and then, we place the jockey at the two endpoints of the wire. After this, we press the jockey situated at both ends of the potentiometer wire and we need to note which one is the deflection in the galvanometer. If the galvanometer shows opposite deflection, then we can say that connections are correct.
Ques: What is an EMF of a cell? (1 mark)
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.
Ques: What is a potentiometer? (1 mark)
Ans: A potentiometer is a three-terminal device that is used to measure the potential difference by manually varying the resistance.
Ques: On what principle does the potentiometer work? (1 mark)
Ans: For a constant current, the fall of the potentiometer along a uniform wire is directly proportional to its length.
Ques: How is the emf of the cell determined for given cells? (1 mark)
Ans: E1/E2 = l1/l2
Ques: Is the cross-section of the potentiometer wire is uniform? (1 mark)
Ans: No, the cross-section of the potentiometer wire is not uniform.
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