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NCERT Solutions for Class 10 Science Chapter 11 Electricity

NCERT Solutions for Class 10 Science Chapter 11 Electricity covers electric current, potential difference, Ohm's law, resistance, series and parallel connections, Joule's law of heating, and electric power with step-by-step numerical solutions.

Overview of Topics & Subtopics (Click to Jump)

1. Electric Current and Circuit

Short Note:

• Electric Current (I): The rate of flow of electric charges per unit time through a conductor cross-section (I = Q / t). Its SI unit is Ampere (A).

• Electric Circuit: A continuous and closed path along which electric current flows.

• Ammeter: An instrument used to measure electric current, always connected in series in a circuit.

2. Electric Potential and Potential Difference

Short Note:

• Potential Difference (V): The work done to move a unit positive charge from one point to another in an electric circuit (V = W / Q). Its SI unit is Volt (V).

• Voltmeter: Device used to measure potential difference, connected in parallel across points.

• 1 Volt is defined as 1 Joule of work done in moving a charge of 1 Coulomb from one point to another.

3. Ohm's Law and Resistance Factors

Short Note:

• Ohm's Law: At constant temperature, electric current (I) flowing through a metallic conductor is directly proportional to potential difference (V) across its ends (V = IR).

• Resistance (R): Property of a conductor to oppose the flow of electric current. SI unit: Ohm (Ω).

• Factors Affecting Resistance: Resistance depends directly on length (L), inversely on cross-sectional area (A), temperature, and nature of material (R = ρL / A, where ρ is resistivity).

4. Resistance in Series and Parallel Combinations

Short Note:

• Series Combination: Resistors connected end-to-end. Total resistance Rs = R1 + R2 + R3. Current remains constant across all resistors.

• Parallel Combination: Resistors connected across common nodes. Total resistance 1 / Rp = 1 / R1 + 1 / R2 + 1 / R3. Potential difference remains constant across all branches.

5. Heating Effect of Electric Current & Power

Short Note:

• Joule's Law of Heating: Heat produced in a resistor is proportional to the square of current, resistance, and time duration (H = I2Rt).

• Electric Power (P): Rate at which electrical energy is consumed (P = VI = I2R = V2 / R). SI unit: Watt (W).

• Commercial Unit: Kilowatt-hour (kWh). 1 kWh = 3.6 × 106 Joules.

Intext Questions - Page Number: 200

Q1: What does an electric circuit mean?
Answer: An electric circuit is a continuous and closed conductive loop containing a power source (cell/battery), conducting wires, switches, and electrical loads through which electric current can flow.
Q2: Define the unit of current.
Answer: The SI unit of current is Ampere (A). One Ampere is defined as the flow of one Coulomb of electric charge per second through any cross-section of a conductor (1 A = 1 C / 1 s).

Intext Questions - Page Number: 202

Q1: Name a device that helps to maintain a potential difference across a conductor.
Answer: An electric cell or a battery.
Q2: How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Given: Charge (Q) = 1 C, Potential Difference (V) = 6 V
Formula: Energy (W) = Q × V
Calculation: W = 1 × 6 = 6 Joules.

Textbook Chapter End Exercise Questions

Q1: A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is:

(a) 1/25
(b) 1/5
(c) 5
(d) 25

Correct Answer: (d) 25
Explanation: Each part has resistance R/5. When connected in parallel, 1/R' = 5/(R/5) = 25/R. Therefore, R/R' = 25.

Q2: Which of the following terms does not represent electrical power in a circuit?

(a) I2R
(b) IR2
(c) VI
(d) V2/R

Correct Answer: (b) IR2.

Q3: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:

(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

Correct Answer: (d) 25 W
Explanation: Resistance R = V2 / P = 2202 / 100 = 484 Ω. When operated at 110 V, new power P' = V'2 / R = 1102 / 484 = 25 W.

Q4: Why are copper and aluminium wires usually employed for electricity transmission?
Answer: Copper and aluminium have extremely low electrical resistivity and high conductivity. This minimizes power loss as heat during long-distance transmission. Additionally, they are ductile and cost-effective.
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Class 10 › Science › Electricity
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Topics in this chapter: Electric Current and CircuitElectric Potential and Potential DifferenceOhm's Law and Resistance FactorsResistance in Series and Parallel CombinationsHeating Effect of Electric Current & Power