
Exams Prep Master
Potentiometer is a three-terminal resistor that uses sliding or rolling contact to produce an adjustable voltage divider. It comprises a long wire of uniform cross-sectional area and length such that the wire should have high resistivity and a low-temperature coefficient. The potentiometer can be used to determine the internal resistance of a primary cell. The internal resistance of a source is the impedance observed if an electrical power source is a linear electric circuit. According to Thévenin's theorem, such a power source can be represented as an ideal voltage source in series with an impedance, and this impedance is called the internal resistance of the source. This article discusses the experiment to determine the internal resistance of a primary cell using the potentiometer.
| Table of Content |
Key Takeaways: Potentiometer, Resistance, Primary cell, Internal resistance, Electromotive Force (e.m.f), Resistivity, Galvanometer, Ammeter, battery, Voltmeter
Aim
[Click Here for Sample Questions]
The aim of this experiment is to determine internal resistance of a primary cell using a potentiometer.
Also Read:
Materials Required
[Click Here for Sample Questions]
- 1 Galvanometer
- 1 ammeter
- 1 potentiometer
- 1 voltmeter
- 1 battery
- 1 rheostat of low resistance
- A fractional resistance box
- 2 numbers keys (one-way)
- Connecting wires
- A jockey
- 1 high resistance box
- Sandpaper
- 1 Leclanche cell
- 1 set square
Theory
[Click Here for Sample Questions]
The potentiometer is a device that is used to:
- Measure the internal resistance of a cell
- Compare the e.m.f. of two cells
- The potential difference across a resistor
The relation between potential difference, emf, and internal resistance of a cell is expressed by:
\(I = \frac{E}{R+r} \)
or
E = I(R+r)
Therefore,
V = IR = E - Ir
This shows that the value of V is less than E by an amount that is equal to the fall of potential inside the cell due to its internal resistance.
From the above equation:
\(\frac{r}{R} = \frac{E-V}{V} \)
Using a potentiometer, the internal resistance of a cell can be expressed as:
\(r = R\frac{E-V}{V} \)
Where,
r = internal resistance of the cell
E = electromotive force in Volts
I = current in amperes
R = Load resistance of the load in the circuit
Circuit Diagram
[Click Here for Sample Questions]
Here, the internal resistance can be estimated as r = (L1 - L2)/(L2) * R
Where,
l1 and l2- balancing lengths without shunt or with the shunt.
R- shunt resistance in parallel with the given cell
Procedure
[Click Here for Sample Questions]
- Connect different electrical components according to the circuit diagram.
- Using sandpaper, clean the ends of the connecting wires and check the connections. All connections should be tight.
- Check the plugs in the resistance box. They should be tight.
- Check the e.m.f of the cell and battery. It should be more than that of the cell. If it is not then the null point would not be obtained.
- The rheostat resistance can be made minimum by taking maximum current from the battery.
- Check whether the circuit connections are correct. The galvanometer deflections should be in the opposite direction. Insert the key K1 and note the ammeter reading.
- To get the null point on the fourth wire, adjust the rheostat without inserting the key K2.
- Take a small resistance between 1-5 ohm from resistance box R connected in parallel with the cell.
- Slide the jockey and obtain the null point.
- Record the observations.
Observations
[Click Here for Sample Questions]
The least count of voltmeter = ………
Range of voltmeters = ……..
E.M.F of cell = ………
E.M.F of battery = ……
Table for lengths:
| Sr. No. | Corrected ammeter reading (A) | Balance point when E1 (Leclanche cell) in the circuit l1 (cm) | Balance point when E2 (Daniel cell) in the circuit l2 (cm) | (E1/E2) = (l1/l2) |
|---|---|---|---|---|
Calculations
[Click Here for Sample Questions]
- For each set of observations find mean and l2
- Calculate the value of r for each set.
- Take the mean of values of r.
Result
[Click Here for Sample Questions]
The internal resistance of the given cell is …… Ohms.
Precautions
- The e.m.f of the cell should be less than the battery.
- Ensure that the ammeter reading remains constant at least for a single set of readings.
- The current should be passed only while obtaining the null point.
- The rheostat should remain fixed and the cell should remain undisturbed throughout the experiment.
- Do not rub the jockey against the potentiometer wire.
Sources of Error
- The wire of the potentiometer may not be of uniform cross-section.
- The brass strips at the ends may have a finite resistance.
- Emf of the auxiliary battery producing the drop of potential along the wire may not remain constant throughout the experiment.
- The heating of the potentiometer wire by the current can also cause some errors.
- A zero error may occur in the measurement of l, due to the end of the scale not being exactly at the end of the wire.
Things to Remember
- The potentiometer was first invented by Poggendorff.
- The e.m.f of a cell is the potential difference across the terminals of the cell.
- A potentiometer works on the principle that for a constant current, the fall of potential in a wire is directly proportional to the length of the wire.
- Primary cells are those cells that cannot be recharged electrically i.e., the original state of the cell cannot be brought back by passing electrical energy from an external source. For e.g., Daniel Cell, Leclanche Cell, Manganese-alkaline cell, Mercury button cell.
- Secondary cells are those in which the chemical process is reversible and the original chemical state can be brought back supplying electrical energy through an external source. For e.g., alkali cells.
Also Read:
Sample Questions
Ques. What is meant by the e.m.f of a cell? (1 mark)
Ans. e.m.f is referred to as the electromotive force. E.M.F of a cell means the potential difference across the terminals of the cell.
Ques. Why is it called potentiometer? (1 mark)
Ans. It measures the potential difference between any two points in an electric circuit, therefore it is called a potentiometer.
Ques. Mention some practical uses of potentiometers. (4 marks)
Ans. Some practical uses of potentiometers include:
- Audio control: Linear and rotary potentiometers are used in audio equipment for controlling loudness and other audio-related signals.
- Television: Potentiometers are used in televisions to control the picture brightness, color response, and contrast.
- Motion control: Potentiometers are used as position feedback devices known as a servomechanism, that helps create a closed-loop control.
- Transducers: Potentiometers are used in designing displacement transducers as they provide large output signals.
Ques. Define the following terms: a) Potential gradient b) E.M.F. (2 marks)
Ans: a. Potential gradient is the potential difference per unit length of the wire.
- M.F of a cell refers to the potential difference across the terminals of the cell.
Ques. State the principle of a potentiometer. (2 marks)
Ans: The potentiometer works on the principle that for a constant current, the fall of potential in a wire is directly proportional to the length of the wire.
Ques. State the difference between a potentiometer and a voltmeter. (3 marks)
Ans.
| Potentiometer | Voltmeter |
|---|---|
| It measures the emf of the circuit | It measures the end terminal voltage of the circuit. |
| It measures an unknown voltage source with a known voltage | It measures the voltages between any two points of an electrical circuit |
| It has high sensitivity | It has low sensitivity |
| It is based on the null deflection method | It is based on the deflection method |
| While measuring emf, the resistance of the potentiometer becomes infinite | While measuring emf, the resistance of the voltmeter becomes very high but is measurable. |
Ques. A potentiometer wire of length 1m has a resistance of 10Ω. Determine the emf of a primary cell which gives a balance point at 40 cm. (3 marks)
Ans. Given: length of wire (l)= 1m = 100 cm
Resistance (R) = 10 Ω
Emf of a battery (E1) = 6V
R1 = 5Ω
x = 40 cm
Therefore, current (I)= E1/ (R+ R1) = 6/ (10+5) = 6/15 A
VAB = IR = (6/15) × 10 = 60/15 = 4V
Therefore, emf of the primary cell = (VAB/l) × x = (4/100) × 40 = 1.6 V
Ques. Differentiate between emf and terminal voltage. Draw a plot showing variation of terminal voltage (V) versus the current (I). Also, explain how this plot can be used to determine internal resistance of the cell. (3 marks)
Ans.
| Emf | Terminal voltage |
|---|---|
| It indicates the potential difference between two terminals of the cell when no current is flowing through it. | It is the potential difference between the two terminals when current flows through the cell. |
The voltage versus current plot is represented:
The negative slope of the graph gives the internal resistance.
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Check-Out:







Comments