Electricity Class 10th Notes - Free NCERT Class 10 Science Chapter 12 Notes - Download PDF

Electricity Class 10th Notes - Free NCERT Class 10 Science Chapter 12 Notes - Download PDF

Edited By Vishal kumar | Updated on Mar 11, 2024 08:16 AM IST

Ncert Class 10 Physics Chapter 12 Notes- Download Free PDF

The chapter on Electricity in the CBSE NCERT Science Book for Class 10 provides a thorough understanding of electricity, including how it flows in circuits and the factors that influence its flow. This Electricity class 10 notes delves into important concepts like current, potential difference, resistance, and more.

This Story also Contains
  1. Ncert Class 10 Physics Chapter 12 Notes- Download Free PDF
  2. NCERT Class 10 Science Chapter 12 Notes
  3. Electric Current
  4. Electric Potential Difference
  5. Electric Circuit
  6. Ohm’s Law
  7. Resistances in Series
  8. Resistances in Parallel
  9. Heating Effect of Electric Current
  10. Practical Applications of the Heating Effects of Electric Current
  11. Electric Power
  12. NCERT Solutions of Cass 10 Subject Wise
  13. NCERT Class 10 Exemplar Solutions for Other Subjects:

The CBSE class 10 physics ch 12 notes provide a detailed overview of the Electricity chapter, including fundamental principles and mathematical equations. These notes, meticulously prepared by Careers360 subject experts, are invaluable resources not only for exams and class tests, but also for higher-level courses. These class 10 physics chapter 12 notes are available for download in PDF format.

Also, students can refer,

NCERT Class 10 Science Chapter 12 Notes

Electric Current

  • Electric current is the amount of charge that is flowing through a particular area in unit time.
  • Electric current is given as the rate of flow of electric charge.

Current I=Carge flowing (Q)/Time taken

  • S.I. unit = Ampere (A), and 1 A=1 C/s

  • 1mA=10-3A and 1μA=10-6A

  • The direction of the current is opposite to the flow of electrons.

Electric Potential Difference

  • The amount of work done to transport a unit charge from one point to another is known as the potential difference (V).

Potential difference V=Work done (W)Charge(Q)

  • S.I. unit= Volt (V), where 1V=1JoulCoulomb

Electric Circuit

  • A conducting path between the terminals of a source of electricity and other electronic devices in continuity is called an electric circuit.

  • A schematic diagram showing the way different electronic devices are connected in a circuit is known as a circuit diagram.

Ohm’s Law

  • The current (I) flowing through a conductor is directly proportional to the potential (V) across its ends at a constant temperature.

V∝I

VI=constant

V/I=R

V=IR

R is the constant of proportionality, which is called Resistance.

  • The property of conductors to resist the flow of current is known as Resistance.

  • S.I. unit of resistance = ohm (Ω).

  • R=V/I gives us 1 Ohm=1Vol/tAmpere

Resistivity

R∝l/A

  • R=ρl/A

Where, ρ = constant of proportionality which is known as the specific resistance or resistivity

l is the length of the conductor and A is the area of the cross-section.

  • S.I. unit of resistivity = Ohm meter (Ω m)

Resistances in Series

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∙ The current flowing through each resistance is the same.

∙ Whereas, the potential difference of the resistance varies according to the resistance value.

∙ The total resistance (Rs) of a series combination consisting of resistances R1, R2, R3... is

Rs=R1+R2+R3+…

Resistances in Parallel

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∙ The potential difference across each resistance in the circuit is the same and is equal to the potential difference across the whole combination or the circuit.

∙ The main current is divided among the different resistors.

. ∙ The equivalent resistance (Rp) of a parallel combination consisting of resistances R1, R2, R3... is

1/Rp=1/R1+1/R2+1/R3+…

∙ The equivalent resistance is lesser than that of the least resistance of all the resistances in the circuit.

Heating Effect of Electric Current

∙ The effect of current due to which heat is induced in a conductor when current passes through it is known as the heating effect of electric current.

∙ The total work (W) done by the electric current in a circuit is known as electric energy and is expressed as

W=VIt=I2Rt=V2t/R

This energy comes out as heat. Therefore, we have

H=VIt=I2Rt=V2t/R

The heat produced in a resistor is directly proportional to the resistance, according to Joule's Law of Heating.

  • Square of the current in the resistor
  • Resistance of the resistor
  • Time for which the current flows through the resistance

Practical Applications of the Heating Effects of Electric Current

∙ Electrical appliances like iron, oven, and heater are a few devices that are based on Joule’s Law of Heating.

∙ This phenomenon is also applied to produce light in a bulb.

∙ One of the most important applications of Joule’s Law of Heating is the fuse used in electric circuits.

Electric Power

  • The rate at which electrical energy is generated or consumed in an electric circuit is referred to as electric power.

  • P=VI=I2R=V2/R

  • S.I. unit of power = Watt (W).

  • 1 Watt of power is consumed when 1 A of current flows at a potential difference of 1 V.

  • The commercial unit of electric energy is in kilowatt-hour (kWh), which is generally known as a Unit.

  • 1kWh=3.6MJ

Significance of NCERT Electricity Class 10 Notes

  • Comprehensive Coverage: Physics Class 10 chapter 12 notes pdf provides a comprehensive understanding of the chapter, covering all major topics covered in the CBSE Physics Syllabus in Class 10.
  • Exam Preparation: These ch 12 physics class 10 notes are excellent study materials for exam preparation, allowing students to grasp key concepts more effectively.
  • Offline Study: The CBSE class 10 physics ch 12 notes download option allows students to study even when they are offline, ensuring continuous learning at any time and from any location.
  • Conceptual Clarity: Electricity Notes class 10 helps to achieve conceptual clarity by simplifying complex topics and providing clear explanations.
  • Supplementary Resources: These class 10 physics chapter 12 notes supplement students' textbooks and help them better understand the subject matter.

Class 10 Chapter Wise Notes

NCERT Solutions of Cass 10 Subject Wise

NCERT Class 10 Exemplar Solutions for Other Subjects:


Frequently Asked Questions (FAQs)

1. What are the important concepts covered in the Science chapter 12 notes for Class 10?

 The main topics covered in Chapter 10 of NCERT for Class 12 Science are

  • Current

  • Potential difference

  • Resistance

  • Resistivity

  • Combination of Resistances

  • Heating effect of electric current

2. How will CBSE Class 10 Science chapter 12 notes benefit students?

The notes for Chapter 12 of Science in Class 10 were created by subject experts and will give you further knowledge on the subject. You can solidify your foundation with these notes from class 10 Electricity notes. Here, key ideas are thoroughly discussed. The segment covers all aspects of electricity.

3. State Ohm’s Law.

As given in the Class 10 Science chapter 12 notes pdf download;

At constant temperature, the current (I) flowing through a conductor is directly proportional to the potential difference (V) across its ends.

V∝I

VI=constant

V/I=R

V=IR

 R is the constant of proportionality, which is called Resistance.

4. Sate Joule’s law of heating

The law states that the heat produced in a resistor is directly proportional to the

 o Square of the current in the resistor

 o Resistance of the resistor

 o Time for which the current flows through the resistance 

It’s mentioned vividly in the NCERT notes for class 10 Science chapter 12.

5. Define Potential Difference.

Given in, Class 10 Science chapter 12 notes that The amount of work done to transport a unit charge from one point to another is known as the potential difference (V).

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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