CBSE Class 10 Science Chapter 11 PDF Notes – Electricity 2026-27

These Class 10 Science Chapter 11 PDF Notes explain Electricity step by step: electric current and charge, potential difference, circuit diagrams, Ohm’s law, resistance and resistivity, series and parallel combinations, the heating effect of current, and electric power, with solved numericals.

Electricity is a numerical-heavy chapter with high weightage. Expect at least one calculation on resistance, current or power, and a question on series versus parallel circuits.

Use the table of contents below to jump to the part you need, or read top to bottom for a complete revision of the chapter.

Table of Contents

• Chapter Overview
• Important Topics and Concepts
• Key Equations and Formulas
• Important Definitions and Terms
• Short Answer Questions
• Long Answer Questions
• MCQs with Answers
• Quick Revision Points
• Chapter Summary
• Exam Tips and PYQ Pattern
• Frequently Asked Questions
• Conclusion

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Chapter Overview

Electricity powers our homes, schools and industries. A steady flow of electric charge through a conductor is called electric current, and it flows only when there is a potential difference across the conductor.

The chapter introduces circuit symbols, explains how current, voltage and resistance are related by Ohm’s law, shows what resistance depends on, compares series and parallel connections, and studies the heating effect of current and electric power, including how electricity is billed.

Important Topics and Concepts

Electric current and charge

Electric current I = Q/t, the rate of flow of charge. SI unit: ampere (A); 1 A = 1 C/s. In metals, current is a flow of electrons, but conventional current is taken in the opposite direction (positive to negative). Charge of an electron = 1.6 × 10⁻¹⁹ C; 1 coulomb ≈ 6 × 10¹⁸ electrons. An ammeter measures current and is connected in series.

Potential difference

The work done to move a unit charge from one point to another: V = W/Q. SI unit: volt (V); 1 V = 1 J/C. A cell or battery maintains the potential difference by chemical action. A voltmeter measures it and is connected in parallel across the points.

Circuit diagram symbols

Cell, battery (combination of cells), plug key or switch (open/closed), wire joint, wires crossing without joining, electric bulb, resistor, variable resistor (rheostat), ammeter and voltmeter. Learn to draw a simple circuit with a cell, key, resistor, ammeter and voltmeter.

Ohm’s law

At constant temperature, the current through a conductor is directly proportional to the potential difference across its ends: V ∝ I, so V = IR. R is the resistance, measured in ohm (Ω). A V–I graph for an ohmic conductor is a straight line through the origin. A rheostat changes resistance in a circuit to control current.

Factors affecting resistance

Resistance depends on the length (R ∝ l), the area of cross-section (R ∝ 1/A), and the nature of the material: R = ρl/A. ρ (rho) is resistivity, measured in Ω m. Metals and alloys have very low resistivity (10⁻⁸ to 10⁻⁶ Ω m); insulators like glass and rubber have very high resistivity. Alloys such as nichrome have high resistivity and do not oxidise easily at high temperature, so they are used in heaters, irons and toasters. Copper and aluminium are used for transmission wires. Tungsten is used in bulb filaments.

Resistors in series

Same current flows through each resistor; total voltage divides. Equivalent resistance Rs = R₁ + R₂ + R₃. The total resistance is greater than any individual resistance.

Resistors in parallel

Same potential difference across each resistor; total current divides. 1/Rp = 1/R₁ + 1/R₂ + 1/R₃. The total resistance is less than the smallest individual resistance. Household appliances are connected in parallel so each gets the full voltage, works independently and draws the current it needs.

Heating effect of electric current (Joule’s law)

When current flows through a resistor, electrical energy turns into heat: H = I²Rt. Heat produced is proportional to the square of current, the resistance and the time. Applications: electric iron, toaster, heater, kettle, bulb filament (tungsten, melting point 3380°C; bulbs filled with nitrogen and argon), and fuse (a wire of low melting point that melts when excess current flows, protecting the circuit).

Electric power and energy

Power is the rate of using electrical energy: P = VI = I²R = V²/R. SI unit: watt (W); 1 W = 1 V × 1 A. The commercial unit of energy is the kilowatt hour (kWh), commonly called a “unit”: 1 kWh = 3.6 × 10⁶ J.

Key Equations and Formulas

Important Definitions and Terms

Short Answer Questions

Q1. What does an electric circuit mean?

An electric circuit is a continuous, closed path made of conducting wires and electrical components through which current flows.

Q2. How much charge flows in 10 minutes if a current of 0.5 A flows through a bulb?

Q = I × t = 0.5 × 600 = 300 C.

Q3. How much work is done in moving 2 C of charge across a potential difference of 12 V?

W = V × Q = 12 × 2 = 24 J.

Q4. On what factors does the resistance of a conductor depend?

On its length (directly), area of cross-section (inversely), the material (resistivity) and temperature.

Q5. Why are coils of electric toasters and irons made of an alloy rather than a pure metal?

Alloys have higher resistivity and do not oxidise (burn) easily at high temperatures, so they produce more heat and last longer.

Q6. Why is the series arrangement not used for domestic circuits?

In series, the same current flows through all appliances and the voltage divides. If one appliance fails, the whole circuit breaks, and appliances with different power needs cannot work properly.

Q7. What is the commercial unit of electrical energy? Convert it into joules.

Kilowatt hour (kWh). 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Q8. Why does the cord of an electric heater not glow while the heating element does?

The heating element has a high resistance, so much more heat is produced in it (H = I²Rt). The cord is made of copper with very low resistance, so it produces little heat.

Long Answer Questions

Q1. Three resistors of 5 Ω, 10 Ω and 30 Ω are connected in parallel to a 12 V battery. Find the current through each resistor, the total current and the total resistance.

Same voltage (12 V) across each resistor.

I₁ = 12/5 = 2.4 A; I₂ = 12/10 = 1.2 A; I₃ = 12/30 = 0.4 A.

Total current I = 2.4 + 1.2 + 0.4 = 4 A.

1/Rp = 1/5 + 1/10 + 1/30 = (6 + 3 + 1)/30 = 10/30, so Rp = 3 Ω.

Check: I = V/Rp = 12/3 = 4 A. ✓

Q2. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, what is the power consumed?

Resistance of bulb: R = V²/P = (220 × 220)/100 = 484 Ω.

At 110 V: P = V²/R = (110 × 110)/484 = 12100/484 = 25 W.

Answer: The bulb uses 25 W, one-fourth of its rated power, because power varies as the square of voltage.

Q3. State Ohm’s law. Describe an experiment to verify it.

Ohm’s law: at constant temperature, the current through a conductor is directly proportional to the potential difference across it (V = IR).

Set up a circuit with a nichrome wire (resistor), an ammeter in series, a voltmeter in parallel across the wire, a plug key and a battery of cells.

Start with one cell and note the ammeter (I) and voltmeter (V) readings. Repeat with two, three and four cells.

Calculate V/I for each reading; it comes out nearly constant. Plot V against I: the graph is a straight line through the origin, verifying Ohm’s law.

MCQs with Answers

1. The SI unit of electric current is
(a) Volt (b) Coulomb (c) Ampere (d) Ohm
Answer: (c) Ampere

2. A voltmeter is connected in
(a) Series (b) Parallel (c) Either way (d) Neither
Answer: (b) Parallel

3. If the length of a wire is doubled, its resistance becomes
(a) Half (b) Same (c) Double (d) Four times
Answer: (c) Double

4. The resistivity of a material depends on
(a) Length (b) Area (c) Nature of material and temperature (d) Current
Answer: (c) Nature of material and temperature

5. Two 4 Ω resistors in parallel give an equivalent resistance of
(a) 8 Ω (b) 4 Ω (c) 2 Ω (d) 1 Ω
Answer: (c) 2 Ω

6. Heat produced in a resistor is given by
(a) H = IRt (b) H = I²Rt (c) H = V²t (d) H = IR²t
Answer: (b) H = I²Rt

7. 1 kWh is equal to
(a) 3.6 × 10³ J (b) 3.6 × 10⁶ J (c) 3.6 × 10⁵ J (d) 1000 J
Answer: (b) 3.6 × 10⁶ J

8. The filament of an electric bulb is made of
(a) Copper (b) Nichrome (c) Tungsten (d) Aluminium
Answer: (c) Tungsten

9. A fuse wire should have
(a) High melting point (b) Low melting point (c) Low resistance only (d) No resistance
Answer: (b) Low melting point

10. Which is not an expression for electric power?
(a) VI (b) I²R (c) V²/R (d) IR²
Answer: (d) IR²

Quick Revision Points

• I = Q/t; V = W/Q; V = IR.
• Ammeter in series; voltmeter in parallel.
• R = ρl/A; R doubles if length doubles, halves if area doubles.
• Series: Rs = sum; parallel: 1/Rp = sum of reciprocals.
• Household wiring is parallel.
• H = I²Rt; P = VI = I²R = V²/R.
• 1 kWh = 3.6 × 10⁶ J; nichrome for heaters, tungsten for filaments.

Chapter Summary

Electric current is the flow of charge, driven by a potential difference maintained by a cell or battery. Ohm’s law links voltage, current and resistance, and resistance depends on length, area of cross-section and the material’s resistivity. Resistors in series add up, while in parallel the equivalent resistance is less than the smallest one, which is why homes use parallel wiring.

Current produces heat according to Joule’s law, which is used in heaters, irons, bulbs and fuses. Electric power is the rate of energy use, measured in watts, and household energy is billed in kilowatt hours.

Exam Tips and PYQ Pattern

• A numerical on series-parallel combinations or power is almost always asked.
• Write the formula, substitute values with units and box the final answer.
• Draw circuit diagrams with correct symbols; ammeter in series and voltmeter in parallel.
• Know why alloys are used in heating devices and why tungsten is used in bulbs.

Frequently Asked Questions

Why is the resistance of a parallel combination less than each resistor?

Each new parallel path gives the current another route, so the total opposition to current decreases.

What happens to resistance if the area of the wire is doubled?

It becomes half, because R is inversely proportional to the area of cross-section.

Why are bulbs filled with nitrogen and argon?

These unreactive gases prolong the life of the filament by preventing it from oxidising.

Conclusion

Electricity is about a few formulas used well. Learn the units and formulas table, practise series-parallel and power numericals daily, and draw clean circuit diagrams. With that preparation, this chapter becomes one of the most dependable sources of marks.

More Notes: Class 10 Science Chapter 10: The Human Eye and the Colourful World notes | Class 10 Science Chapter 12: Magnetic Effects of Electric Current notes | All Class 10 Science notes

Author Bio: Written and reviewed by the Nextoper Editorial Team, a group of CBSE science educators who build chapter notes directly from the NCERT textbook and past board paper patterns.

Freshness Note: These notes follow the NCERT Class 10 Science textbook (Reprint 2026-27) and the CBSE syllabus for the 2026-27 session. Last checked on 7 November 2026.