Electricity – Class 10 Science Chapter 11 Notes & NCERT PDF

Class 10ScienceChapter 11NCERT book: Science - Class 10

Notes and a simple summary of Chapter 11 of the NCERT Class 10 Science book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 11: Electricity

Introduction

Electricity is a controllable and convenient form of energy used in homes, schools, hospitals and industries. This chapter answers three questions: what constitutes electricity, how it flows in a circuit, and what controls the current. It also explains the heating effect of current and its uses. It is one of the most numerical chapters in the Class 10 board paper.

Key Concepts

1. Electric current and circuit

  • A continuous and closed path of electric current is an electric circuit. A switch makes the conducting link; if the circuit is broken, current stops.
  • Electric current is the rate of flow of charge:
    • I = Q / t
  • SI unit of charge: coulomb (C) — the charge of nearly 6 × 10¹⁸ electrons. An electron carries 1.6 × 10⁻¹⁹ C of negative charge.
  • SI unit of current: ampere (A), named after André-Marie Ampère. 1 A = 1 C/1 s. Also 1 mA = 10⁻³ A, 1 µA = 10⁻⁶ A.
  • In metal wires, electrons carry the charge. By convention, current direction is opposite to electron flow (from + terminal to − terminal outside the cell).
  • Ammeter measures current and is always connected in series.

    • 2. Potential and potential difference

  • Charges flow only when there is a potential difference (like water flowing because of a pressure difference). A cell's chemical action maintains it.
    • V = W / Q — work done to move a unit charge between two points.
  • SI unit volt (V), after Alessandro Volta: 1 V = 1 J C⁻¹.
  • Voltmeter measures potential difference and is always connected in parallel.

    • 3. Circuit symbols

    The book lists symbols for a cell, battery, open and closed plug key, wire joint, wires crossing without joining, bulb, resistor, rheostat (variable resistance), ammeter and voltmeter. Learn to draw them neatly — diagrams carry marks.

    4. Ohm's law

    In Activity 11.1 a nichrome wire is tested with 1, 2, 3 and 4 cells; V/I stays nearly constant and the V–I graph is a straight line through the origin.

    Ohm's law (Georg Simon Ohm, 1827): the potential difference across a metallic wire is directly proportional to the current through it, provided its temperature remains the same. V ∝ I, so V = IR.
  • R is resistance; SI unit ohm (Ω). 1 Ω = 1 V / 1 A.
  • I = V/R: doubling resistance halves the current.
  • A rheostat changes resistance to regulate current without changing the source.

    • 5. Factors affecting resistance

    Activity 11.3 shows: doubling the length halves the current; a thicker wire increases current; a different material changes it.

    R = ρ l / A — resistance is directly proportional to length and inversely proportional to area of cross-section.
  • ρ (rho) is resistivity, SI unit Ω m; it depends on the material and temperature, not on length or area.
  • Metals and alloys have very low resistivity (10⁻⁸ to 10⁻⁶ Ω m) — good conductors. Insulators like rubber and glass have 10¹² to 10¹⁷ Ω m.
  • Alloys have higher resistivity than their metals and don't oxidise (burn) readily at high temperature, so they are used in heating elements (e.g., nichrome). Tungsten is used for bulb filaments; copper and aluminium for transmission wires (copper: 1.62 × 10⁻⁸ Ω m).

    • 6. Resistors in series

  • Same current through each resistor; potential differences add: V = V₁ + V₂ + V₃.
    • Rₛ = R₁ + R₂ + R₃ — the equivalent resistance is greater than any individual resistance.

    7. Resistors in parallel

  • Same potential difference across each; currents add: I = I₁ + I₂ + I₃.
    • 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ — the equivalent resistance is less than the smallest individual resistance.
  • Series is unsuitable for home appliances: if one fails the whole circuit breaks, and all share one current. Parallel lets each gadget get full voltage and its own current, and be switched separately.

    • 8. Heating effect of current (Joule's law)

  • Energy supplied by the source is partly dissipated as heat in the resistor.
    • H = I²Rt — heat is proportional to the square of current, to resistance, and to time.
  • Uses: electric iron, toaster, oven, kettle, heater; bulb filament (tungsten, melting point 3380 °C, bulbs filled with inert nitrogen/argon); fuse.
  • Fuse: a wire of suitable melting point placed in series; unduly high current melts it and breaks the circuit. Domestic ratings: 1 A, 2 A, 3 A, 5 A, 10 A. A 1 kW iron at 220 V draws 1000/220 ≈ 4.54 A, so a 5 A fuse is used.

    • 9. Electric power

    P = VI = I²R = V²/R
  • SI unit watt (W): 1 W = 1 V × 1 A. 1 kW = 1000 W.
  • Commercial unit of energy: kilowatt hour (kW h) = 1 'unit' = 3.6 × 10⁶ J.
  • We pay for the energy used — electrons are not consumed.

    • Worked Numericals

  • Charge (Ex. 11.1): I = 0.5 A for 10 min. t = 600 s, Q = It = 0.5 × 600 = 300 C.
  • Work (Ex. 11.2): Q = 2 C, V = 12 V. W = VQ = 24 J.
  • Resistivity (Ex. 11.5): R = 26 Ω, l = 1 m, d = 0.3 mm = 3 × 10⁻⁴ m. ρ = Rπd²/(4l) = 1.84 × 10⁻⁶ Ω m → manganese.
  • Changed wire (Ex. 11.6): R₁ = 4 Ω with l, A. New wire l/2, 2A: R₂ = ρ(l/2)/(2A) = R₁/4 = 1 Ω.
  • Series (Ex. 11.7): 20 Ω lamp + 4 Ω conductor on 6 V. Rₛ = 24 Ω, I = 6/24 = 0.25 A, V(lamp) = 5 V, V(conductor) = 1 V.
  • Parallel (Ex. 11.8): 5 Ω, 10 Ω, 30 Ω on 12 V. I₁ = 2.4 A, I₂ = 1.2 A, I₃ = 0.4 A; total 4 A; Rₚ = 3 Ω.
  • Mixed (Ex. 11.9): 10 Ω ∥ 40 Ω = 8 Ω; 30 Ω ∥ 20 Ω ∥ 60 Ω = 10 Ω; total 18 Ω; I = 12/18 ≈ 0.67 A.
  • Power & cost (Ex. 11.13): 400 W fridge, 8 h/day, 30 days = 96 000 W h = 96 kW h; at ₹3 per unit, cost = ₹288.

    • Board Exam Focus

  • Derivations: equivalent resistance in series and parallel; H = I²Rt from W = VQ.
  • Why alloys for heaters, tungsten for filaments, copper for wires.
  • V–I graph and Ohm's law statement (with the temperature condition).
  • Numericals on resistivity, combinations, power and kW h cost.

    • Formula Sheet

  • I = Q/t · V = W/Q · V = IR · R = ρl/A
  • Rₛ = R₁ + R₂ + … · 1/Rₚ = 1/R₁ + 1/R₂ + …
  • H = I²Rt · P = VI = I²R = V²/R · 1 kW h = 3.6 × 10⁶ J

    • Key Terms

  • Electric current (विद्युत धारा) — rate of flow of charge
  • Coulomb (कूलॉम) — SI unit of charge
  • Potential difference (विभवांतर) — work per unit charge
  • Volt (वोल्ट) — SI unit of potential difference
  • Ammeter (ऐमीटर) — measures current, in series
  • Voltmeter (वोल्टमीटर) — measures p.d., in parallel
  • Resistance (प्रतिरोध) — opposition to current, in ohm
  • Resistivity (प्रतिरोधकता) — material property, Ω m
  • Rheostat (धारा नियंत्रक) — variable resistance
  • Ohm's law (ओम का नियम) — V = IR at constant temperature
  • Joule's law of heating (जूल का तापन नियम) — H = I²Rt
  • Fuse (फ़्यूज़) — safety device in series
  • Electric power (विद्युत शक्ति) — rate of energy consumption
  • Kilowatt hour (किलोवाट घंटा) — commercial 'unit' of energy

    • Common Mistakes

  • Connecting the voltmeter in series or the ammeter in parallel.
  • Forgetting to convert minutes to seconds or mm to m before calculating.
  • Using diameter instead of radius in A = πr² (A = πd²/4).
  • Thinking resistivity changes with length — only resistance does.
  • Stating Ohm's law without "temperature remains the same".
  • Saying electricity bills pay for electrons — we pay for energy.

    • 💡 Key Learning Points

    • ✓Define current (I = Q/t) and potential difference (V = W/Q) with SI units coulomb, ampere and volt
    • ✓State Ohm's law (V = IR at constant temperature) and interpret the straight-line V–I graph
    • ✓Use R = ρl/A to explain how length, area and material affect resistance
    • ✓Find equivalent resistance of series (Rₛ = R₁ + R₂ + …) and parallel (1/Rₚ = Σ1/R) combinations
    • ✓Apply H = I²Rt and P = VI = I²R = V²/R, and compute energy cost in kW h

    👨‍🏫 Teaching Tips

    • →Use the book's water-tank analogy for potential difference before introducing V = W/Q
    • →Perform Activity 11.1 with 1–4 cells and let students plot the V–I graph themselves
    • →Show wires of different length/thickness (Activity 11.3) and ask students to predict ammeter readings first
    • →Draw circuit symbols on flashcards and run a quick 'draw the circuit' drill
    • →Solve every NCERT example (11.1–11.13) step by step, insisting on units at each line
    • →Bring an old electricity bill and let students verify the kW h and cost calculation
    • →Use a real fuse cartridge to explain why the fuse is placed in series and rated by current

    📋 Assessment Questions

    1. Can the student state Ohm's law with its condition and draw the V–I graph?
    2. Can the student derive the formulae for series and parallel equivalent resistance?
    3. Can the student solve resistivity problems converting mm to m correctly?
    4. Can the student explain why household appliances are connected in parallel?
    5. Can the student calculate power and the cost of energy in kW h?

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    2. 2. Acids, Bases and Salts
    3. 3. Metals and Non-metals
    4. 4. Carbon and its Compounds
    5. 5. Life Processes
    6. 6. Control and Coordination
    7. 7. How do Organisms Reproduce?
    8. 8. Heredity
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