Electricity | Class 10 Science Notes

⚡ ELECTRICITY

Class 10 CBSE Science • Easy Language Notes with Diagrams, Tricks & 40+ PYQs

Charge Current Potential Difference Why Current Flows Resistance Resistivity Ohm’s Law Series & Parallel Circuit Symbols Power & Energy Heating Effect Definitions Numericals / PYQs

1. Charge (Q)

There are two types of charge: positive (+) and negative (–).
Like charges repel (+ + or – –) and unlike charges attract (+ –).
SI unit of charge: Coulomb (C).
Smallest independent charge = charge on one electron (fundamental charge).

Charge of 1 electron: e = –1.6 × 10⁻¹⁹ C

2. Electric Current (I)

Current is the rate of flow of (positive) charge.

I = Q / t and Q = I × t

I = current (Ampere, A) • Q = charge (Coulomb, C) • t = time (second, s)
Direction of current = opposite to the flow of electrons.
SI unit of current: Ampere (A). Other popular units: mA (milli-ampere) and µA (micro-ampere).
1 mA = 10⁻³ A • 1 µA = 10⁻⁶ A

Electrons go one way, conventional current goes the opposite way

3. Potential Difference (V)

Potential difference between two points = the amount of work done in moving a unit charge (1 C) from one point to the other.

V = W / Q and W = Q × V

V = potential difference (Volt, V) • W = work done (Joule, J) • Q = charge (Coulomb, C)
SI unit of P.D. is Volt (V).

Remember: Q = I t → (C, A, s) | W = Q V → (J, C, V)

Why Does Current Flow? (Potential Difference)

Electrons flow from lower potential → higher potential (LP → HP).
Current flows from higher potential → lower potential (HP → LP).
A cell keeps a potential difference across the ends of the conductor. Cell symbol: long line (+) and short thick line (–).
A combination of cells is a Battery. Example: four 1.5 V cells in series = 1.5 + 1.5 + 1.5 + 1.5 = 6 V.

A voltage source V driving a resistor R with current I (Wikimedia Commons)

Voltage source V → resistor R → current I (Wikimedia Commons)

4. Resistance (R)

Resistance is the property of a conductor to oppose the flow of electric current. (Inside a conductor, free electrons keep colliding with atoms.)

Factors on which Resistance of a Conductor Depends

Length (l): R ∝ l (longer wire → more resistance)
Area of cross-section (A): R ∝ 1/A (thicker wire → less resistance)
Material: R ∝ ρ (resistivity of the material)
Temperature: if temperature ↑ then:
- Metals: R increases ↑
- Alloys (nichrome): R increases ↑ (only a little)
- Semiconductors (C, Si): R decreases ↓

R = ρ l / A

ρ (rho) = resistivity (Ω m) • l = length (m) • A = area of cross-section (m²)

SI unit of R: ohm (Ω) • SI unit of resistivity (ρ): ohm-metre (Ω m)
Proof: ρ = RA / l = (Ω × m²) / m = Ω m

5. Resistivity (ρ)

Resistivity depends only on the material. It does not change with shape and size (length, area, diameter or radius).
Order: Metals < Alloys < Insulators.
Copper and aluminium have low ρ → used for transmission lines.
Insulators (rubber, glass) have very high ρ → poor conductors.
SI unit of resistivity = ohm metre (Ω m).

🔹 Potential difference is measured by a Voltmeter (connected in parallel).
🔹 Electric current is measured by an Ammeter (connected in series).

Resistivity Table (CBSE 2025)

Material	Resistivity (Ω m)
A – Insulator	10¹⁷
B – Alloy	44 × 10⁻⁸
C – Metal	1.62 × 10⁻⁸

Metal (C) has the lowest ρ → best conductor. Alloy (B) has higher ρ than metal → used in heating devices. Insulator (A) has the highest ρ → does not conduct.

Resistance, Radius and Diameter

Area of circle: A = πr² → R = ρ l / (π r²) and r = d / 2
Unit conversion: 1 mm = 10⁻³ m • 1 cm = 10⁻² m (always convert r and d to metres!)

If this changes…	Then R becomes…
Length doubled (2l)	2R
Length halved (l/2)	R/2
Area doubled (2A)	R/2
Area halved (A/2)	2R
Radius (or diameter) doubled	R/4
Radius (or diameter) halved	4R

Radius and diameter follow the same relation: R ∝ 1/r² and R ∝ 1/d². Double the diameter → R becomes one-fourth.

6. Ohm’s Law

Ohm’s Law: The potential difference (V) across the ends of a metallic conductor is directly proportional to the current (I) flowing through it, provided its temperature remains the same.

V ∝ I → V = I R → V/I = R

V = P.D. (volt) • I = current (ampere) • R = resistance (ohm, Ω). Example: 2 V → 1 A, 4 V → 2 A, 6 V → 3 A (V ∝ I).

V–I Graph

V–I graph: straight line through the origin. Slope = V/I = R. Slope ↑ → R ↑.
I–V graph: Slope = 1/R. Slope ↑ → R ↓.

CBSE 2024, 2023V–I graph for three resistors R₁, R₂, R₃ is given. Arrange in order.

On a V–I graph the line with the biggest slope has the biggest R. From the graph: R₃ > R₁ > R₂.

Ohm’s Law – Experimental Set-up

V (volt)	I (ampere)	V / I = R
1.5 V	1 A	1.5 Ω
3 V	2 A	1.5 Ω
4.5 V	3 A	1.5 Ω

Ammeter in series, voltmeter in parallel with resistor R (CBSE Term 2 2021-22 / 2021 / 2020)

7. Combination of Resistors

	Series	Parallel
Current (I)	Same	Different (divides)
Voltage (V)	Different (divides)	Same
Equivalent resistance	Rₑ = R₁ + R₂ + R₃	1/Rₑ = 1/R₁ + 1/R₂ + 1/R₃

Imp Note 1: Rₚ < Rₛ (parallel resistance is always less than series resistance).
Example: 6 Ω and 3 Ω → Series: Rₑ = 6 + 3 = 9 Ω. Parallel: 1/Rₑ = 1/6 + 1/3 = (1 + 2)/6 = 1/2 → Rₑ = 2 Ω.
Maximum resistance → series • Minimum resistance → parallel.

Imp Note 2 – Shortcut for TWO resistors in parallel:

Rₑ = (R₁ × R₂) / (R₁ + R₂) (product ÷ sum)

Example: (6 × 3)/(6 + 3) = 18/9 = 2 Ω.

Imp Note 3 – Identical resistors (each R): 3 in series → Rₑ = 3R • 3 in parallel → Rₑ = R/3.

Advantages of Parallel over Series Combination

Independence of appliances: if one appliance fails or there is a short circuit in one branch, the other appliances keep working normally because each is on a separate branch.
Devices of different types need different currents: for example a bulb and a heater need different currents and cannot be joined in series. This is possible with a parallel connection.
Reduced equivalent resistance: the equivalent resistance of a parallel circuit is less than the smallest resistance of any branch. So a higher total current flows at the same voltage, which can mean more efficient energy use.

8. Circuit Diagram & Symbols

Circuit = a continuous and closed path of electric current.

S.No.	Component	Symbol
1	An electric cell
2	A battery or a combination of cells
3	Plug key or switch (open) – OFF
4	Plug key or switch (closed) – ON
5	A wire joint
6	Wires crossing without joining
7	Electric bulb
8	A resistor of resistance R
9	Variable resistance or Rheostat
10	Ammeter
11	Voltmeter

9. Electric Power (P)

Electric power is the rate at which electric energy is consumed.

P = V I P = I² R P = V² / R

P = power (SI unit watt, W) • V = potential difference (volt) • I = current (ampere)
How the formulas come: P = VI, and V = IR → P = I²R; I = V/R → P = V²/R.

Which formula to use? Same current (series) → use P = I²R (P ∝ R). Same voltage (parallel / house wiring) → use P = V²/R (P ∝ 1/R). Use P = VI when V and I are known.

10. Electrical Energy (E)

Electrical energy is supplied by the cell. Energy = Power × Time.

E = P × t

SI unit of energy: Joule (J). Commercial unit: kilowatt-hour (kWh) (P in kW, t in hours).
1 kWh = 3.6 × 10⁶ J
kW = watt / 1000
Electric meter measures energy: 1 Unit = 1 kWh.

Bill = No. of units (kWh) × Price of 1 unit

11. Heating Effect of Electric Current

When an electric current passes through a conductor or an electric device, it becomes hot after some time and produces heat. This is called the heating effect of electric current.

Joule’s Law of Heating

Heat (H) produced in a pure resistor:

H ∝ I² (square of current)
H ∝ R (resistance)
H ∝ t (time)

H = I² R t

Heat (H) is measured in Joule (J). Other forms: H = VIt = (V²/R) t.

Practical Applications

Heating devices (iron, electric oven, kettle, heater) use alloys (e.g. nichrome) because alloys have higher resistivity than metals and a high melting point. CBSE 2025, 2023
Electric bulb: filament is made of tungsten (very high melting point). Filament gets heated and emits light. Most energy appears as heat; only a small part comes as light.
Electric fuse – safety device: based on the heating effect. It is used to prevent short circuit / overloading. The fuse wire has a low melting point, so when high current flows it melts (heat by Joule’s law H = I²Rt) and stops the flow of current.
- Fuse wire is connected in series with the appliance.
- Fuse wire is an alloy of Al, Cu, lead, iron. It should have high resistivity and low melting point.
- Rating of fuse = max current allowed (1 A, 2 A, 3 A, 4 A, 5 A, 10 A…). If more current than required flows, the wire heats up and melts.

12. Important SI Unit Definitions

1 Ampere: If 1 coulomb of charge flows through a conductor in 1 second, the current is 1 A. (1 A = 1 C / 1 s)

1 Volt: The potential difference between two points is 1 V if 1 joule of work is done in moving a positive charge of 1 coulomb from one point to the other. (1 V = 1 J / 1 C)

1 Ohm: The resistance of an object is 1 Ω when a current of 1 A flows through it on applying a potential difference of 1 V. (1 Ω = 1 V / 1 A)

1 Watt: The power consumed in a circuit when a current of 1 A flows on applying a potential difference of 1 V. (1 W = 1 V × 1 A)

13. Numericals & PYQs (40+)

Charge, Current, Potential Difference

Q1. CBSE 2024An electric source can supply 500 C of charge. If the current drawn by a device is 25 mA, find the time in which the source will be discharged. (A) 30,000 s (B) 48,000 s (C) 60,000 s (D) 20,000 s

Q = 500 C, I = 25 mA = 25 × 10⁻³ A. t = Q/I = 500 / (25 × 10⁻³) = 20 × 10³ = 20,000 s → (D).

Q2. CBSE 2025Define potential difference between two points. Name and define its SI unit. Express in terms of SI units of work and charge.

P.D. = work done to move a unit charge between two points (V = W/Q). SI unit = volt: 1 V is the P.D. when 1 J of work is done in moving 1 C of charge. 1 V = 1 J / 1 C.

Q3. Work done by a cell when 20 C of charge is moved through a P.D. of 3 V? (A) 6.66 J (B) 66 J (C) 60 J (D) 0.15 J

W = QV = 20 × 3 = 60 J → (C).

Q4. CBSE 2024Assertion (A): Electrons move from lower potential to higher potential in a conductor. Reason (R): A dry cell maintains electric potential difference across the ends of a conductor.

Both are true. The cell maintains the P.D., and because of that P.D. the electrons move from LP to HP. So R correctly explains A → (a).

Resistance & Resistivity

Q5. CBSE 2023The resistance of a wire does not depend on its: (A) length (B) area of cross-section (C) shape (D) material

R = ρl/A depends on length, area and material (ρ). It is not a separate function of the wire’s shape → (C) shape.

Q6. CBSE 2025, 2024, 2023Write the relation between R and electrical resistivity ρ for a cylindrical conductor of length l and area A. Derive the SI unit of ρ.

R = ρ l / A → ρ = R A / l. Unit = Ω × m² / m = Ω m.

Q7. CBSE 2025, 2023, 2022, 2021Resistance of a metal wire of length 3 m is 60 Ω. Area of cross-section = 4 × 10⁻⁷ m². Find resistivity. How will resistivity change if length and diameter are both doubled?

ρ = RA / l = (60 × 4 × 10⁻⁷) / 3 = 80 × 10⁻⁷ = 8 × 10⁻⁶ Ω m.
Resistivity does not change – it depends only on the material, not on length or diameter.

Q8. CBSE 2024, 2022, 2021Three cylindrical conductors of the same material: A (length L, area A), B (length L/2, area 2A), C (length L/2, area A/2). Find Rₐ/R_b and Rₐ/R_c.

Rₐ = ρL/A = x. R_b = ρ(L/2)/(2A) = x/4. R_c = ρ(L/2)/(A/2) = x.
Rₐ/R_b = 4/1 and Rₐ/R_c = 1/1.

Q9. CBSE 2021A wire of length l and area A has resistance 4 Ω. Find the resistance of another wire of the same material with length l/2 and area 2A.

R′ = ρ(l/2)/(2A) = (1/4) × (ρl/A) = 4/4 = 1 Ω.

Q10. CBSE 2024How much does resistance change if the diameter of the wire is doubled?

Area ∝ d² → area becomes 4 times → R becomes R/4.

Q11. CBSE Term 2 2021-22Resistance of a wire of 0.01 cm radius is 10 Ω. If resistivity is 50 × 10⁻⁸ Ω m, find its length.

r = 0.01 cm = 10⁻⁴ m, A = πr² = 3.14 × 10⁻⁸ m².
l = RA/ρ = (10 × 3.14 × 10⁻⁸) / (50 × 10⁻⁸) = 31.4/50 ≈ 0.62 m.

Q12. CBSE 2024Copper wire: diameter 0.2 mm, resistivity 1.6 × 10⁻⁸ Ω m. Length needed for resistance 14 Ω? How does the resistance change if diameter is doubled?

d = 0.2 mm = 0.2 × 10⁻³ m → r = 10⁻⁴ m, A = πr² = 3.14 × 10⁻⁸ m².
l = RA/ρ = (14 × 3.14 × 10⁻⁸)/(1.6 × 10⁻⁸) ≈ 27.5 m.
Diameter doubled → ρ is unchanged, area ×4 → R becomes R/4.

Ohm’s Law

Q13. CBSE 2024, 2023, 2020, 2019, 2016P.D. across a circuit component is decreased to one-third, its resistance remains constant. What change in current? Name and state the law.

V → V/3, so by I = V/R, current also becomes one-third (I/3).
Ohm’s Law: P.D. across the ends of a metallic conductor is directly proportional to the current through it, provided temperature remains the same.

Equivalent Resistance

Q14. Find the equivalent resistance between X and Y (various circuits).

(a) 3 Ω ∥ 6 Ω: 1/Rₑ = 1/3 + 1/6 = 3/6 = 1/2 → Rₑ = 2 Ω.
(b) 8 Ω + 8 Ω in series = 16 Ω; this is parallel with 16 Ω → 1/Rₑ = 1/16 + 1/16 = 1/8 → Rₑ = 8 Ω.
(c) 12 Ω ∥ 4 Ω = (12 × 4)/16 = 3 Ω; then in series with 3 Ω → Rₑ = 6 Ω.
(d) CBSE Term II 2021-22 5 Ω, then R₂ (10 Ω) ∥ R₃ (10 Ω) = 5 Ω, then 5 Ω in series: 5 + 5 + 5 = 15 Ω.
(e) 4 Ω + 2 Ω = 6 Ω (series); 2 Ω + 10 Ω = 12 Ω (series); then 6 Ω ∥ 12 Ω → 1/Rₑ = 1/12 + 1/6 = 3/12 = 1/4 → Rₑ = 4 Ω.

Q15. CBSE 2025Equivalent resistance between A and B: four 2 Ω resistors, three of them in series and parallel with the fourth.

Series part: 2 + 2 + 2 = 6 Ω. Parallel with 2 Ω: Rₑ = (6 × 2)/(6 + 2) = 12/8 = 1.5 Ω.

Q16. CBSE 2025, 2023, 2020Maximum and minimum resistance using four resistors of 4 Ω each?

Maximum (all in series) = 4 + 4 + 4 + 4 = 16 Ω. Minimum (all in parallel) = 4/4 = 1 Ω.

Q17. CBSE 2023Four identical 8 Ω resistors: in series gives Rₛ, in parallel gives Rₚ. Ratio Rₛ/Rₚ? (A) 32 (B) 2 (C) 0.5 (D) 16

Rₛ = 4 × 8 = 32 Ω. Rₚ = 8/4 = 2 Ω. Ratio = 32/2 = 16 → (D).

Q18. CBSE 2025A wire of resistance R is cut lengthwise into three identical parts, then joined in parallel. If equivalent resistance is R′, R/R′ = ? (A) 1/9 (B) 1/3 (C) 3 (D) 9

Each part has resistance R/3. Parallel: R′ = (R/3)/3 = R/9. So R/R′ = 9 → (D).

Q19. How would you connect three 4 Ω resistors to get 6 Ω?

Connect two 4 Ω in parallel (= 2 Ω) and then join the third 4 Ω in series → 2 + 4 = 6 Ω. (2 parallel + 1 series.)
Check: all series = 12 Ω ✗, all parallel = 1.33 Ω ✗.

Q20. CBSE 2024, 2020, 2016Connect three 6 Ω resistors to get 9 Ω. Justify.

Connect two 6 Ω in parallel (= 3 Ω) and put the third 6 Ω in series → 3 + 6 = 9 Ω. (All series = 18 Ω, all parallel = 2 Ω – neither works.)

Q21. CBSE 2020Three 2 Ω resistors A, B, C connected so that total resistance is 3 Ω. Show arrangement. (Homework)

Join two 2 Ω in parallel (= 1 Ω) and connect the third 2 Ω in series → 1 + 2 = 3 Ω.

Q22. CBSE 2024, 2020Series or parallel – which circuit for operating different electrical gadgets at home? Give three reasons.

Parallel circuit. (1) If one appliance fails, the others keep working. (2) Different appliances need different currents, and each gets the same (full) voltage. (3) The total resistance is reduced, so more current can be drawn efficiently.

Q23. CBSE 2025What do the symbols (ammeter) and (rheostat) represent? Write one function of each.

Ammeter – measures electric current; connected in series. Rheostat (variable resistance) – changes/regulates the current in the circuit by changing resistance.

Q24. Circuit: 4 Ω in series with (6 Ω ∥ 3 Ω), 12 V battery. Find Rₑ, ammeter reading, P.D. across each resistor and current through each.

6 Ω ∥ 3 Ω: 1/R = 1/6 + 1/3 = 1/2 → 2 Ω. Rₑ = 4 + 2 = 6 Ω.
Ammeter: I = V/Rₑ = 12/6 = 2 A (this flows through the 4 Ω).
P.D. across 4 Ω = 2 × 4 = 8 V. P.D. across the parallel part = 12 – 8 = 4 V (same across 6 Ω and 3 Ω).
Current in 6 Ω = 4/6 = 2/3 A. Current in 3 Ω = 4/3 A. Check: 2/3 + 4/3 = 2 A ✓.

Q25. CBSE 202515 V battery: 10 Ω ∥ 15 Ω, in series with (60 Ω ∥ 40 Ω). Find (a) total resistance (b) total current (c) P.D. across the parallel combination of 10 Ω and 15 Ω.

10 ∥ 15: (10 × 15)/25 = 6 Ω. 60 ∥ 40: (60 × 40)/100 = 24 Ω.
(a) Rₑ = 6 + 24 = 30 Ω (b) I = 15/30 = 0.5 A (c) V = I × R = 0.5 × 6 = 3 V.

Q26. CBSE 2024Draw a circuit with a battery of four 1.5 V cells, 5 Ω, 10 Ω and 15 Ω resistors and a plug key, all in series. Find (i) the current and (ii) the P.D. across the 10 Ω resistor.

Battery V = 4 × 1.5 = 6 V. Rₑ = 5 + 10 + 15 = 30 Ω.
(i) I = V/R = 6/30 = 0.2 A (ii) V₁₀ = IR = 0.2 × 10 = 2 V.

Q28. CBSE 2025, 2024Define electric power. Express it in terms of potential difference (V) and resistance (R).

Electric power = rate at which electric energy is consumed (P = W/t = VI). Since I = V/R: P = V²/R.

Q29. CBSE 20246 V battery with 1 Ω in series with (2 Ω ∥ 2 Ω)… find power consumed in the 2 Ω resistor.

Total R = 1 + 2 = 3 Ω (as in the circuit). I = V/R = 6/3 = 2 A through the 2 Ω resistor.
P = I²R = 2² × 2 = 8 W (also V = IR = 4 V, P = VI = 4 × 2 = 8 W).

Q30. CBSE Term II 2021-22An electric heater rated 1100 W operates at 220 V. Find (i) its resistance (ii) current drawn.

P = VI → I = 1100/220 = 5 A. R = V/I = 220/5 = 44 Ω.

Q31. CBSE 2025Lamps A (50 W, 220 V) and B (25 W, 220 V). Find Rₐ : R_b.

R = V²/P. Rₐ = 220²/50, R_b = 220²/25. Rₐ/R_b = 25/50 = 1/2 → Rₐ : R_b = 1 : 2.

Q32. CBSE 2023An electric iron uses 880 W at maximum heating and 330 W at minimum. Source 220 V. Find current and resistance in each case. (Homework)

Maximum: I = P/V = 880/220 = 4 A; R = V/I = 220/4 = 55 Ω.
Minimum: I = 330/220 = 1.5 A; R = 220/1.5 ≈ 146.7 Ω.

Q33. CBSE 2023, 2020Two LED bulbs of 10 W and 5 W are connected in series. If the current through the 5 W bulb is 0.005 A, current through the 10 W bulb is: (A) 0.02 A (B) 0.01 A (C) 0.005 A (D) 0.0025 A

In series, the current is the same everywhere → 0.005 A → (C).

Energy & Bill

Q34. CBSE Term II 2021-22Heater rated 4 kW, 220 V. (a) Energy used in 2 hours (b) cost at ₹4.50 per kWh.

(a) E = P × t = 4 kW × 2 h = 8 kWh. (b) Cost = 8 × 4.50 = ₹36.

Q35. CBSE 2023Two 50 W bulbs used 6 h daily and a 1 kW geyser used 1 h daily. Total energy in 30 days and cost at ₹8.00 per kWh.

Bulbs: (100 W = 0.1 kW) × 6 h × 30 = 18 kWh. Geyser: 1 kW × 1 h × 30 = 30 kWh.
Total energy = 48 kWh. Cost = 48 × 8 = ₹384.

Q36. CBSE 2025, 2024Air-conditioner rated at 220 V uses 22 units (kWh) in 10 hours. Find (i) power (ii) current (iii) resistance.

(i) P = E/t = 22/10 = 2.2 kW (2200 W). (ii) I = P/V = 2200/220 = 10 A. (iii) R = V/I = 220/10 = 22 Ω.

Heating Effect

Q37. CBSE 2023, 2024What is the heating effect of electric current? Find an expression for heat produced when current passes through a resistor for some time.

When current flows through a conductor, it gets hot and produces heat – this is the heating effect.
Work done W = QV = (It)(V). With V = IR: H = I² R t (Joule’s law).

Q38. CBSE 2024, 2022, 2021A 5 Ω resistor is connected across a 6 V battery. Energy dissipated as heat in 10 s?

I = V/R = 6/5 A. H = I²Rt = (6/5)² × 5 × 10 = (36/25) × 50 = 72 J.

Q39. CBSE 2024State Joule’s law of heating. How is this effect useful in circuits where a fuse is used as a safety device? Also: appliance 220 V, 1100 W – current rating of fuse wire? (A) 4.5 A (B) 6 A (C) 5.2 A (D) 6.5 A

Joule’s law: heat produced H = I²Rt (∝ I², R and t). In a fuse, if too much current flows, the low-melting-point fuse wire heats up and melts, breaking the circuit and protecting the appliance.
Rating: I = P/V = 1100/220 = 5 A. The fuse must be rated just above this, so the closest suitable choice is (C) 5.2 A (4.5 A would melt in normal use; 6 A or more gives less protection).

Quick Revision
Q = It • V = W/Q • V = IR • R = ρl/A • Rₛ = R₁ + R₂ + R₃ • 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃
P = VI = I²R = V²/R • E = Pt • 1 kWh = 3.6 × 10⁶ J • H = I²Rt • Bill = units × rate

class 10 physics : Electricity