# CBSE Class 10 Science Electricity Notes Set D

Download CBSE Class 10 Science Electricity Notes Set D in PDF format. All Revision notes for Class 10 Physics have been designed as per the latest syllabus and updated chapters given in your textbook for Physics in Standard 10. Our teachers have designed these concept notes for the benefit of Grade 10 students. You should use these chapter wise notes for revision on daily basis. These study notes can also be used for learning each chapter and its important and difficult topics or revision just before your exams to help you get better scores in upcoming examinations, You can also use Printable notes for Class 10 Physics for faster revision of difficult topics and get higher rank. After reading these notes also refer to MCQ questions for Class 10 Physics given our website

## Electricity Class 10 Physics Revision Notes

Class 10 Physics students should refer to the following concepts and notes for Electricity in standard 10. These exam notes for Grade 10 Physics will be very useful for upcoming class tests and examinations and help you to score good marks

### Electricity Notes Class 10 Physics

Class -X

Chapter 3: Electricity

Chapter Notes

Key Learnings:

1. Electric current is the rate of flow of charge.

2. Battery provides the driving force required to move the charges along the wire from one terminal to another.

3. The constant voltage difference between the two terminals of the wire maintains the constant electric current through the wire.

4. Electric current is measured in terms of amperes where 1 ampere = 1 coulomb / second

5. Voltage is measured in terms of volt where 1 volt = 1 joule /coulomb

6. Resistance is a property that resists the flow of electrons in a conductor. It controls the magnitude of the current. The SI unit of resistance is ohm (Ω).

7. Resistivity is defined as the resistance offered by a cube of the material of side 1 m when the current flows perpendicular to the opposite faces of the cube.

8. Ohm’s law: The potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains the same.

9. The resistance of a conductor depends directly on its length, inversely on its area of cross – section, and also on the material of the conductor.

10. In Series combination of resistors: - The current flowing through each resistor is the same - The potential difference across the ends of the series combination is distributed across the resistors - The equivalent resistance is greater than the greatest resistance in the combination.

11. In Parallel combination of resistors: - The potential difference across each resistor is same and is equal to the potential difference across the combination.

- The main current divides itself and a different current flow through each resistor.

- The equivalent resistance is lesser than the least of all the resistances.

12. The effect of heating current due to which heat is produced in a wire when current is passed through it is called heating effect of current.

13. Electric power is the rate at which electrical energy is produced or consumed in an electric circuit.

14. The unit of power is watt (W). One watt of power is consumed when 1 A of current flows at a potential difference of 1 V.

15. The commercial unit of electric energy is kilowatt hour (kW h), commonly known a ‘unit’.

Top Formulae:

1. The current I through the cross – section of a conductor is

I = Q/t

Where Q is net charge flowing across the cross – section of a conductor in time t.

2. Potential difference (V) between two points = work done (W)/ Charge (Q)

V = W/Q

3. Ohm’s law: V = I R

4. The equivalent resistance in series circuit is the sum of the individual resistances - R = R1 + R2 + R3

5. The equivalent resistance of a parallel circuit containing resistances R1, R2, R3 is given as

1/Req = 1/R1 + 1/R2 + 1/R3

6. The electric power P is given by

P = VI

Or

P = I2R = V²/ R

7. The electrical energy dissipated in a resistor is given by W = V * I * t

8. Joule’s law of heating; H = I2Rt

9. 1 kW h = 3, 600, 000 J = 3.6 x 106 J

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