Class 10 CBSE Physics notes|Electricity Notes Pdf download is very crucial and important for class X students who will sit in their board exam very first time this year .For the same scenario the all the notes of this chapter is designed in keep in mind that they will learn and memorise this chapter very easily.
Here’s a full detailed set of Class 10 CBSE Physics notes on the chapter “Electricity“, including explanations, formulas, diagrams, and NCERT questions with answers at the end.
Video Tutorial of Electricity Chapter Class X
Chapter: Electricity – Class 10 Science (Physics)
Class 10 CBSE Physics notes|Electricity Notes Pdf download
CBSE Board – Detailed Notes with Diagrams and NCERT Solutions
This chapter Electricity is very crucial chapter for the students who want to get enroll themselves in engineering or if we specified say so Engineering branches such as Electrical Engineering ,Electronics Engineeering and many others so for the same scrnario this chapter is goinng to play a vital crucial role in their Foundation building so read the fully chapter that we are giving here to boost your basics and concepts.
Lets start with some Baic Definitions and Terminology’
1. Electric Current and Circuit
1. Electric Current and Circuit
Electric Current (I)
Definition: The rate of flow of electric charge
Formula:
I = {Q}/{t}
I= Current (in Amperes),
Q= Charge (in Coulombs),
t= Time (in seconds)
Unit of Current: Ampere (A)
1 Ampere = Flow of 1 Coulomb of charge per 1 second.
2. Electric Potential and Potential Difference
2. Electric Potential and Potential Difference
Electric Potential (V)
Definition: Work done to bring a unit positive charge from infinity to a point in the electric field.
Potential Difference
Definition: Work done in moving a unit charge between two points.OR
The work done to carry a charge from infinity to a definite point is Known as Potential or Voltage.It is also known as Electric Pressure
Formula:
V = {W}/{Q}
V= Potential Difference (Volt),
W = Work Done (Joule),
Q= Charge (Coulomb)
Unit of Potential Difference: Volt (V)
3. Electric Circuit
A closed path in which current flows.
Open circuit: Current does not flow.
Circuit Diagram Symbols:
4. Ohm’s Law
Statement:
The current flowing through a conductor is directly proportional to the potential difference across its ends, provided the temperature remains constant.
Formula:
V = IR
= Potential Difference (V),
= Current (A),
= Resistance (Ω)
Graph of V vs I:
A straight line shows that current is directly proportional to voltage.
5. Resistance (R)
Definition:
The opposition offered by a material to the flow of electric current.
Unit: Ohm (Ω)
Factors affecting Resistance:
R = rho {l}/{A}
Rho= Resistivity,
l= Length,
A= Area of cross-section
6. Resistivity (ρ)
Definition: Resistance of a unit length and unit area of a material.Resistivity only depends upon the material of the conductor no matter what is the length,size and cross section area is.
Unit: Ohm metre (Ω·m)
Depends on: Nature of material only.
7. Series and Parallel Circuits
Series Combination:
When the two or more than two Resistance are connected to each other in such a way that the first end of the resistor is on voltage source end and another end is connected with the other Resistor and the goes on, so such kind of arrangement is called the Series Combination Of Resistance.
In the series sequence of Resistance Current Reamins the same while Voltage divided in per phase of the load as per distribution.
Total Resistance:
R = R1 + R2 + R3
In the series combination of the Resistance Voltage get Divided while the current remains Same or similiar.
Class 10 CBSE Physics notes|Electricity Notes Pdf download
Parallel Combination:
Total Resistance:
\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} +……
Current: Divided.
Diagram – Series vs Parallel:
8. Heating Effect of Electric Current
Formula:
H = I^2Rt
H= Heat produced (Joule),
I = Current (A),
R = Resistance (Ω),
t = Time (s)
Applications:
Electric heater, iron, bulb, fuse
9. Electric Power
9. Electric Power
Class 10 CBSE Physics notes|Electricity Notes Pdf download
Formulae:
P = VI = I^2R = {V^2}/{R}
P= Power (Watt)
Unit: Watt (W),
1 kilowatt (kW) = 1000 W
Electrical Energy:
Electrical Energy:
\text{Energy} = P times t
Commercial Unit: kilowatt-hour (kWh)
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ Joules
AC(Alternating) & DC(Direct Current)
Alternating Current – Alternating Current is that kind of electricity current in which electricity flows sinusoidely in both direction simultaneously. it means it flows positive direcion in a half cycle and it flows negative direction in seconf half.
Direct Current – Direct current is that kind of Electricity in which Charges flow in unidirectional movevement of the Oscillation. It means it flows in one direction only in a time
Direct current has no frequency at all beacause of no variation in cycle .Thats why we say that Direct current has Zero frequency
Comparison Of AC & DC – Comperative Study of Alternating Current & Direct Current
| Factors | Alternating Current | Direct Current |
|---|---|---|
| 1. Flow of current | Flow of current changes direction periodically | Flow of current is in one constant direction |
| 2. Frequency | Frequency is 50 Hz (India), 60 Hz (USA) | Frequency is zero |
| 3. Generation | Produced by generators and power stations | Produced by batteries, solar cells, etc. |
| 4. Transmission | Can be easily transmitted over long distances | Difficult to transmit over long distances |
| 5. Voltage | Voltage can be increased or decreased using a transformer | Voltage can’t be changed easily |
| 6. Uses | Used in homes, industries, and appliances | Used in mobile phones, laptops, vehicles |
| 7. Safety | Safer at low voltages but more dangerous at high voltages | Generally safer and more stable |
| 8. Efficiency | More efficient for long-distance power supply | Less efficient for long-distance supply |
| 9. Losses | Can cause more heat and losses in small devices | Ideal for low-voltage, precision devices |
| 10. Waveform | Represented by a sine wave in diagrams | Represented by a straight line in diagrams |
NCERT Exercise Questions with Answers
NCERT Exercise Important Questions with Answers for Class 10 CBSE Physics Electricity Chapter
Q1. What does an electric circuit mean?
Answer: A closed conducting path through which current can flow is called an electric circuit.
Q2. Define the unit of current.
1 Ampere is the current when 1 Coulomb of charge flows through a conductor in 1 second.
Q3. Calculate the number of electrons constituting one coulomb of charge.
Answer: We know that Q = n.e Or n= Q/e
Charge of one electron =
n ={1}/{1.6 \times 10^{-19}} = 6.25 \times 10^{18} { electrons}
6.25 ×1018 Electrons.
Q4. Name a device that helps to maintain a potential difference across a conductor.
Answer: Cell or battery.
Q5. What is meant by saying that potential difference between two points is 1V?
It means 1 Joule of work is done to move 1 Coulomb of charge between those two points.
Q6. How much energy is given to each coulomb of charge passing through a 6V battery?
W = V times Q = 6 times 1 = 6 { Joules}
W = V .Q
Q1. What does an electric circuit mean?
Answer: A closed conducting path through which current can flow is called an electric circuit.
Q2. Define the unit of current.
1 Ampere is the current when 1 Coulomb of charge flows through a conductor in 1 second.
Q3. Calculate the number of electrons constituting one coulomb of charge.
Answer:
Charge of one electron =n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} \text{ electrons}
Q4. Name a device that helps to maintain a potential difference across a conductor.
Answer: Cell or battery.
Q5. What is meant by saying that potential difference between two points is 1V?
Answer:
It means 1 Joule of work is done to move 1 Coulomb of charge between those two points.Q6. How much energy is given to each coulomb of charge passing through a 6V battery?
Answer:
W = V \times Q = 6 \times 1 = 6 \text{ Joules}
Q7. On what factors does the resistance of a conductor depend?
Answer:
Length of the conductor
Area of cross-section
Nature of material
Temperature
Q8. Why are coils of electric toasters made of alloy rather than pure metal?
Answer:
Alloys have higher resistivity and do not oxidize easily at high temperatures.Q9. Why is tungsten used for filament of electric lamps?
Answer:
Because it has high melting point and high resistivity.Q10. Calculate the resistance of 1 km long copper wire of cross-section .
Given: ρ =
Answer:R = \rho \frac{l}{A} = \frac{1.7 \times 10^{-8} \times 1000}{0.5 \times 10^{-6}} = 34 \, \Omega
Q12. What are the advantages of connecting electrical devices in parallel?
Answer:
Each device gets same voltage.
Devices can operate independently.
Total resistance decreases.
Q1. What does an electric circuit mean?Class 10 CBSE Physics notes|Electricity Notes Pdf download
Answer: A closed conducting path through which current can flow is called an electric circuit.
—
Q2. Define the unit of current.
Answer:
1 Ampere is the current when 1 Coulomb of charge flows through a conductor in 1 second.
—
Q3. Calculate the number of electrons constituting one coulomb of charge.
Answer:
Charge of one electron =
n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} \text{ electrons}
—
Q4. Name a device that helps to maintain a potential difference across a conductor.
Answer: Cell or battery.
—
Q5. What is meant by saying that potential difference between two points is 1V?
Answer:
It means 1 Joule of work is done to move 1 Coulomb of charge between those two points.
—
Q6. How much energy is given to each coulomb of charge passing through a 6V battery?
Answer:
W = V \times Q = 6 \times 1 = 6 \text{ Joules}
—
Q7. On what factors does the resistance of a conductor depend?
Answer:
Length of the conductor
Area of cross-section
Nature of material
Temperature
—
Q8. Why are coils of electric toasters made of alloy rather than pure metal?
Answer:
Alloys have higher resistivity and do not oxidize easily at high temperatures.
—
Q9. Why is tungsten used for filament of electric lamps?
Answer:
Because it has high melting point and high resistivity.
—
**Q10. Calculate the resistance of 1 km long copper wire of cross-section .
Given: ρ = **
Answer:
R = \rho \frac{l}{A} = \frac{1.7 \times 10^{-8} \times 1000}{0.5 \times 10^{-6}} = 34 \, \Omega
—
Q11. Draw a schematic diagram of a circuit with 3 cells, 2 resistors in series, and a switch.
Answer:
—
Q12. What are the advantages of connecting electrical devices in parallel?
Answer:
Each device gets same voltage.
Devices can operate independently.
Total resistance decreases.
—
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