CBSE Class 10 · Science · board exam 2027
Chapter 11: Electricity — notes
The chapter’s 9 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.
1. Electric current
The figure.
- • Electric circuit = a continuous and closed path of electric current
- • Electric current = rate of flow of charge
- • I = Q/t (Q = charge, t = time)
- • 1 A = (1 C)/(1 s)
- • 1 mA = 10⁻³ A, 1 µA = 10⁻⁶ A
- • Direction of current: from + to −
- • Electrons flow: from − to + (opposite)
- • Ammeter: measures current, connected in series
- • Given: I = 0.5 A, t = 10 min = 10 × 60 s = 600 s
- • Formula: Q = I × t
- • Q = 0.5 × 600 = 300 C
- • Answer: 300 coulomb of charge
2. Symbols used in circuit diagrams
We do not draw the real objects every time we draw a circuit.
- • Cell: longer thin line (+), shorter thick line (−)
- • Battery = two or more cells joined in series
- • Key (closed): dot in the middle, Key (open): no dot
- • Resistor: zigzag line
- • Rheostat: zigzag with a slanting arrow
- • Ⓐ ammeter, Ⓥ voltmeter
3. Electric potential and potential difference
Just like water in a perfectly flat tube does not flow.
- • Charge flows because of: potential difference
- • Potential difference is produced by: a cell or a battery
- • Potential difference V = W/Q (W = work done, Q = charge)
- • SI unit: volt (V)
- • 1 V = (1 J)/(1 C) = 1 J C⁻¹
- • Voltmeter: in parallel
- • Ammeter: in series
- • Given: V = 12 V, Q = 2 C
- • Formula: W = V × Q
- • W = 12 × 2 = 24 J
- • Answer: 24 joule
4. Ohm's law
This experiment in the figure.
- • Experiment: each time read V and I, and find V/I
- • Observation: V/I stays almost constant
- • Ohm's law: at constant temperature, V ∝ I
- • V/I = constant = R
- • V = IR
- • V–I graph: a straight line through the origin
- • Slope = ΔV/ΔI = R
- • SI unit of resistance: ohm (Ω)
- • 1 Ω = (1 V)/(1 A)
- • I = V/R
- • Given: R = 1200 Ω, V = 220 V
- • Formula: I = V/R
- • I = 220/1200 = 11/60
- • I ≈ 0.18 A
- • Answer: the bulb draws about 0.18 A
- • R = 100 Ω: I = 220/100 = 2.2 A
- • Less resistance ⇒ more current
5. Factors on which resistance depends
The figure, a wire drawn like a long cylinder.
- • l = length of the wire, A = area of cross-section
- • R ∝ l (length doubled ⇒ R doubled)
- • R ∝ 1/A (thicker wire ⇒ less R)
- • R ∝ l/A
- • R = ρl/A (ρ = electrical resistivity)
- • ρ = RA/l
- • Unit = (Ω × m²)/m = Ω m
- • ρ depends on the material and temperature, not on l or A
- • Conductors (metals): ρ ≈ 10⁻⁸ to 10⁻⁶ Ω m
- • Insulators (rubber, glass): ρ ≈ 10¹² to 10¹⁷ Ω m
- • Heating devices: alloys (nichrome)
- • Bulb filament: tungsten (melting point ≈ 3380 °C)
6. Resistors in series
The figure.
- • Series: one single path, the same current I in each
- • V = V₁ + V₂ + V₃ ... (1)
- • V₁ = IR1, V₂ = IR2, V₃ = IR3
- • Across the whole circuit: V = I Rs
- • Putting in (1): I Rₛ = I R₁ + I R₂ + I R₃
- • Dividing by I:
- • Rₛ = R₁ + R₂ + R₃
- • Rs = 20 Ω + 4 Ω = 24 Ω
- • I = V/Rs = 6/24 = 1/4 = 0.25
- • Answer: I = 0.25 A
- • Drawback: one device fails ⇒ the whole circuit stops
7. Resistors in parallel
This figure.
- • Parallel: all ends at X and Y, the same potential difference V across each
- • I = I₁ + I₂ + I₃ ... (1)
- • I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃
- • Across the combination: I = V/Rp
- • Putting in (1): V/Rₚ = V/R₁ + V/R₂ + V/R₃
- • Dividing by V:
- • 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃
- • 1/Rp = 1/10 + 1/40 = 4/40 + 1/40 = 5/40 = 1/8
- • Rp = 8 Ω
- • Rp is less than the smallest resistance (10 Ω)
- • Domestic circuits: parallel
- • Why: full V for each, separate currents, one fails and the rest work
8. Heating effect of electric current
When current flows through a resistor, the energy from the source becomes heat, and the conductor gets hot.
- • Heating effect: current makes a conductor hot
- • Work W = V × Q
- • Q = It ⇒ H = VIt
- • Putting V = IR: H = (IR) × I × t
- • H = I²Rt (Joule's law of heating)
- • H ∝ I², H ∝ R, H ∝ t
- • Current doubled ⇒ heat 4 times
- • Uses: iron, toaster, heater, bulb, fuse
- • Fuse: low melting point, in series
- • Too much current ⇒ wire melts ⇒ circuit breaks
- • Given: R = 4 Ω, H = 100 J, t = 1 s
- • I² = H/Rt = 100/(4 × 1) = 25
- • I = √25 = 5 A
- • V = IR = 5 × 4 = 20 V
- • Answer: I = 5 A, V = 20 V
9. Electric power
Power is the rate of doing work.
- • Electric power = rate of consumption of electric energy
- • P = VIt/t = VI
- • P = VI = I²R = V²/R
- • 1 W = 1 V × 1 A = 1 J/s
- • 1 kW = 1000 W
- • 1 kWh = 1 kW × 1 h
- • = 1000 W × 3600 s
- • = 3600000 J = 3.6 × 10⁶ J
- • Given: V = 220 V, I = 0.50 A
- • P = VI = 220 × 0.50 = 110 W
- • Answer: 110 watt
- • 220 V, 60 W: at 220 V it uses 60 J every second
Summary
Very good! You learnt a lot today, so let us revise it once together. Current is the rate of flow of charge, I equals Q upon t. Potential difference is the work done on a unit charge, V equals W upon Q. And Ohm's law, V equals I R. Resistance is directly proportional to length and inversely proportional to area, R equals ρ l upon A. In series the resistances add, and in parallel their reciprocals add. And heat H equals I² R t, power P equals V I. Write today's seven formulas once in your notebook without looking, each with its unit. Today's homework is on your screen. If anything is not clear, press the Ask button and ask me. Done, you worked very well today!
AskMedha · Medha Ma’am — CBSE Class 10 Science, Chapter 11. Written from the textbook, checked twice.
Questions parents and students ask
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