By the end of this chapter you'll be able to…

  • 1Define current, voltage, resistance
  • 2Apply Ohm's Law
  • 3Calculate series and parallel resistance
  • 4Compute electric power and energy
  • 5Understand heating effect (Joule's law)
💡
Why this chapter matters
Highest-marked physics chapter. Foundation for all electrical engineering. Critical for everyday devices.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Electricity — Class 10 Science

"Electricity: the energy that powers your phone, lights your home, runs every modern device."

1. About the Chapter

This is a HEAVY-MARK chapter on electrical circuits. Covers:

  • Electric current and charge
  • Potential difference (voltage)
  • Ohm's Law
  • Resistance and resistivity
  • Series and parallel circuits
  • Electric power and energy
  • Heating effect

Why Important

  • Foundation for all electrical engineering
  • Used in every device (phone to factories)
  • High-mark chapter (~10-12 marks)

2. Electric Current

Definition

Flow of ELECTRIC CHARGE (specifically, electrons) through a conductor.

Direction

  • ELECTRONS flow from negative (-) to positive (+) terminal
  • CONVENTIONAL CURRENT direction: positive (+) to negative (-) — OPPOSITE to electron flow
  • (Historical convention from before electron was discovered.)

SI Unit

AMPERE (A) — named after André-Marie Ampère

1 Ampere = 1 Coulomb / 1 Second

Formula

I = Q / t

where:

  • I = current (in amperes)
  • Q = charge (in coulombs)
  • t = time (in seconds)

Electron Charge

1 electron = 1.6 × 10⁻¹⁹ Coulomb 1 Coulomb ≈ 6.25 × 10¹⁸ electrons


3. Potential Difference (Voltage)

Definition

Work done per unit charge in moving a charge from one point to another.

SI Unit

VOLT (V) — named after Alessandro Volta

1 Volt = 1 Joule / 1 Coulomb

Formula

V = W / Q

where:

  • V = potential difference (volts)
  • W = work done (joules)
  • Q = charge (coulombs)

Source

A BATTERY (or cell) maintains potential difference, driving current.


4. Ohm's Law

Statement

For a conductor at constant temperature, CURRENT is DIRECTLY PROPORTIONAL to VOLTAGE.

V ∝ I

Or: V = IR

where R = RESISTANCE (constant for a given conductor).

SI Unit of Resistance

OHM (Ω) — named after Georg Simon Ohm

1 Ohm = 1 Volt / 1 Ampere

Graphical Form

A V-I graph is a STRAIGHT LINE passing through origin.

  • Slope = R (resistance)
  • Steeper slope = higher resistance

What is Resistance?

The property of a conductor that OPPOSES the flow of electric current.

Factors Affecting Resistance

R = ρL/A

where:

  • ρ (rho) = resistivity (depends on material)
  • L = length of conductor
  • A = cross-sectional area

Resistance:

  • Increases with length (R ∝ L)
  • Decreases with cross-section (R ∝ 1/A)
  • Depends on material (ρ varies)
  • Temperature affects (most: R increases with T)

Resistivity (ρ)

  • Property of the material
  • Unit: Ω·m
  • Lower for good conductors (copper, silver)
  • Higher for poor conductors (nichrome)
  • Very high for insulators

Examples (Common Materials at 20°C)

  • Silver: 1.6 × 10⁻⁸ Ω·m (best conductor)
  • Copper: 1.7 × 10⁻⁸ Ω·m
  • Aluminium: 2.7 × 10⁻⁸ Ω·m
  • Nichrome: 1.1 × 10⁻⁶ Ω·m (high — used in heaters)
  • Rubber: 10¹³-10¹⁶ Ω·m (insulator)

5. Series and Parallel Circuits

Series Connection

Components connected ONE AFTER ANOTHER. Single path for current.

Current: SAME through all components Voltage: DIVIDES across components

Total Resistance: R_total = R₁ + R₂ + R₃ + ...

Total ALWAYS GREATER than any individual resistance.

If one fails, ENTIRE circuit breaks.

Parallel Connection

Components on SEPARATE branches. Multiple paths for current.

Voltage: SAME across all components (= battery voltage) Current: DIVIDES through branches

Total Resistance: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ...

Total ALWAYS LESS than smallest individual resistance.

If one fails, others continue (independent paths).

Examples

Two resistors in series:

  • R₁ = 4Ω, R₂ = 6Ω
  • R_total = 4 + 6 = 10Ω

Two resistors in parallel:

  • R₁ = 4Ω, R₂ = 6Ω
  • 1/R_total = 1/4 + 1/6 = 3/12 + 2/12 = 5/12
  • R_total = 12/5 = 2.4Ω

Why Parallel for Homes?

  • Each appliance gets full voltage (220V)
  • If one fails, others work
  • Used in all home wiring

Why Series Sometimes?

  • Old-style Christmas lights
  • Voltage dividers
  • Fuses

6. Heating Effect of Current

Joule's Law

H = I²Rt

(Energy in JOULES when current I flows through resistance R for time t.)

Formula Forms

Energy = Power × Time E = Pt = VIt = I²Rt = V²t/R

Examples of Heating Effect

  • Electric iron: nichrome wire heats clothes
  • Electric heater/geyser: nichrome coil heats water
  • Electric bulb (incandescent): tungsten filament heats and glows
  • Fuse: thin wire melts when current too high (safety)
  • Toaster

Why Tungsten in Bulbs?

  • VERY HIGH MELTING POINT (~3400°C)
  • High resistivity
  • Glows white-hot

Filament Bulb vs Modern Alternatives

  • Filament: ~95% energy as HEAT, only 5% as LIGHT
  • LED: 80-90% energy as light, very efficient
  • LED bulbs save energy in India significantly

7. Electric Power

Definition

Rate of doing electrical work, or rate of energy transfer.

SI Unit

WATT (W)

1 W = 1 Joule / 1 Second

Formulas

P = VI = I²R = V²/R

(All equivalent; use whichever fits given variables.)

Unit Conversion

  • 1 W = 1 V × 1 A
  • 1 kW (kilowatt) = 1000 W
  • 1 MW (megawatt) = 10⁶ W

Electrical Energy (Billing)

Electricity consumed measured in kWh (kilowatt-hour) — also called UNIT.

1 kWh = 1 kW × 1 hour = 3.6 × 10⁶ Joules

This is the unit used in electricity bills.

Example

A 100 W bulb on for 10 hours:

  • Energy = 100 × 10 = 1000 Wh = 1 kWh = 1 unit
  • Cost (if ₹5/unit): ₹5

8. Worked Examples

Example 1: Current

A current of 0.5 A flows through a circuit for 2 minutes. Find charge.

  • I = Q/t
  • Q = It = 0.5 × 120 = 60 Coulombs

Example 2: Voltage

A 100 J of work is done in moving 5 C of charge between two points. Find potential difference.

  • V = W/Q = 100/5 = 20 Volts

Example 3: Ohm's Law

A bulb has resistance 100Ω and 220V supply. Find current.

  • I = V/R = 220/100 = 2.2 A

Example 4: Series

Two resistors 5Ω and 10Ω in series, connected to 30V battery. Find current.

  • R_total = 5 + 10 = 15Ω
  • I = V/R = 30/15 = 2 A
  • (Same current through both)

Example 5: Parallel

Two resistors 6Ω and 12Ω in parallel, connected to 6V battery. Find total current.

  • 1/R_total = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4
  • R_total = 4Ω
  • I = V/R = 6/4 = 1.5 A

Example 6: Power

A heater has resistance 50Ω, voltage 220V. Find power.

  • P = V²/R = (220)²/50 = 48400/50 = 968 W

Example 7: Energy Cost

A 1000 W heater used for 5 hours daily. Monthly energy and cost (₹5/unit)?

  • Daily energy = 1000 × 5 = 5000 Wh = 5 kWh
  • Monthly (30 days): 5 × 30 = 150 kWh = 150 units
  • Cost: 150 × 5 = ₹750

9. Common Mistakes

  1. Confusing series and parallel formulas

    • Series: R = R₁ + R₂ + ... (ADD)
    • Parallel: 1/R = 1/R₁ + 1/R₂ + ... (RECIPROCALS)
  2. Wrong power formula

    • P = VI = I²R = V²/R — three forms, use what fits.
  3. Units confusion

    • Watt = Joule/second. kWh ≠ kW.
  4. Current direction

    • Conventional current: + to -. Electrons: - to +.
  5. Resistance vs resistivity

    • RESISTANCE depends on shape (L, A) AND material
    • RESISTIVITY (ρ) is just material property

10. Real-World Applications

Indian Electricity Sector

  • 95% of India electrified (2026)
  • Major sources: coal (50%), solar (15%), hydro (12%), wind, nuclear
  • Target: 500 GW renewable by 2030

Indian Power Companies

  • NTPC (National Thermal Power)
  • Tata Power
  • Adani Power
  • State utilities

Modern Uses

  • Smartphones (battery + charging circuit)
  • Electric vehicles (Tata, Mahindra EVs)
  • Solar panels (DC to AC inverters)
  • LED lights (saved India many GW)

11. Conclusion

Electricity is the FOUNDATION of modern life:

  • Current: flow of charge
  • Voltage: drives current
  • Ohm's Law: V = IR
  • Series and parallel: different rules
  • Power: P = VI
  • Energy: measured in kWh (units)

Master:

  • Ohm's law
  • Series/parallel formulas
  • Power and energy
  • Heating effect (Joule's law)
  • Unit calculations (for bills!)

Practice 20+ problems. This is HIGH-MARK chapter (~10-12 marks).

Electricity: invisible but indispensable.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Current
I = Q/t (Coulomb/sec = Ampere)
Voltage
V = W/Q (Joule/Coulomb = Volt)
Ohm's Law
V = IR
Resistance R in Ohms
Resistivity
R = ρL/A
ρ = resistivity, L = length, A = area
Series resistance
R_total = R₁ + R₂ + ...
Parallel resistance
1/R_total = 1/R₁ + 1/R₂ + ...
Power
P = VI = I²R = V²/R
Watts
Energy (Joule's law)
H = I²Rt
Heat in joules
Electrical energy
E = Pt (kWh)
1 kWh = 3.6×10⁶ J
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Series and parallel formulas confused
SERIES: add Rs directly. PARALLEL: add RECIPROCALS.
WATCH OUT
Conventional current = electron flow
OPPOSITE! Conventional current: + to -. Electrons: - to +.
WATCH OUT
Watt vs kWh
WATT = power (rate). kWh = ENERGY (= power × time). Bills measure kWh.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Ohm's
A bulb of resistance 110 Ω is connected to 220V supply. Find current.
Show solution
✦ Answer: I = V/R = 220/110 = 2 A.
Q2EASY· Series
Find total resistance of 4Ω, 6Ω, 8Ω in series.
Show solution
✦ Answer: R = 4 + 6 + 8 = 18Ω.
Q3MEDIUM· Parallel
Find equivalent resistance of 2Ω, 4Ω, 6Ω in parallel.
Show solution
Step 1 — Use formula. 1/R = 1/2 + 1/4 + 1/6 Step 2 — Find common denominator (12). 1/R = 6/12 + 3/12 + 2/12 = 11/12 Step 3 — Take reciprocal. R = 12/11 ≈ 1.09Ω Step 4 — Verify. Parallel total is ALWAYS less than smallest individual resistance. Smallest = 2Ω; total 1.09Ω < 2Ω ✓ ✦ Answer: R = 12/11 ≈ 1.09 Ω.
Q4HARD· Power
A heater of 1000W operates on 220V for 4 hours daily. Find: (a) resistance, (b) current, (c) monthly energy in kWh, (d) cost at ₹6/unit.
Show solution
Part (a) — Resistance. P = V²/R 1000 = (220)²/R R = 48400/1000 = 48.4Ω Part (b) — Current. P = VI 1000 = 220 × I I = 4.55 A Part (c) — Monthly energy. Daily: 1000 W × 4 hours = 4000 Wh = 4 kWh Monthly (30 days): 4 × 30 = 120 kWh = 120 units Part (d) — Cost. At ₹6/unit: 120 × 6 = ₹720 Step — Verify. V × I × t = 220 × 4.55 × (4 × 30 × 3600) seconds = compute... ≈ 4.32 × 10⁸ J = 120 kWh ✓ ✦ Answer: (a) R = 48.4Ω; (b) I = 4.55 A; (c) Monthly energy = 120 kWh; (d) Cost = ₹720.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Current I = Q/t (Coulomb/sec = Ampere)
  • Voltage V = W/Q (Joule/Coulomb = Volt)
  • Ohm's Law: V = IR
  • Resistance R depends on material, length, area
  • R = ρL/A
  • Series: R = R₁ + R₂ + ...
  • Parallel: 1/R = 1/R₁ + 1/R₂ + ...
  • Power P = VI = I²R = V²/R (Watts)
  • Energy: Joule's law H = I²Rt
  • Electricity bill: kWh (kilowatt-hour)
  • 1 kWh = 3.6 × 10⁶ J
  • Indian household voltage: 220V AC, 50 Hz

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 10-12 marks

Question typeMarks eachTypical countWhat it tests
MCQ13Definitions, units
Short2-32Ohm's law, circuits
Long51Combined circuits, power
Prep strategy
  • Memorise all formulas COLD
  • Practice 30+ circuit problems
  • Master series and parallel
  • Know unit conversions

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Indian electricity grid

95% of India electrified. NTPC, Tata Power, Adani are major players.

Electricity bills

Measured in kWh (units). Knowledge of P = VI helps you understand consumption.

Indian EVs

Tata, Mahindra, Ola Electric — all use electricity principles for batteries, motors.

LED bulbs

India saved many GW by switching from incandescent to LEDs. Government UJALA programme.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Memorise formulas
  2. Identify series vs parallel
  3. Be careful with units
  4. Practice numerical problems

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Kirchhoff's laws
  • Wheatstone bridge
  • Capacitors and inductors
  • AC vs DC analysis

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 10 BoardVery High
Science OlympiadVery High
JEE/NEETVery High

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Three main reasons: (1) Each appliance gets FULL VOLTAGE (220V) - works at intended power. (2) Independent operation - if one bulb fuses, others continue. (3) Different appliances can have different currents based on their needs. Series would mean: voltage divides (insufficient), one fail = all stop, hard to add new devices.
Verified by the tuition.in editorial team
Last reviewed on 20 May 2026. Written and reviewed by subject-matter experts — read about our process.
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