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Chapter 11: Electricity

Grade 10 Science | Chapter 11

Electricity

From the flow of electrons in a wire to Joule heating in your electric iron — master every concept in Grade 10 Electricity. Explore electric current, potential difference, Ohm’s Law, resistance, series and parallel circuits, heating effect, and electric power.

⚡ Current

🔋 Voltage

🔌 Resistance

💡 Ohm’s Law

🔥 Joule Heating

📈 Power

🔎	Introduction

Electricity powers modern civilisation — homes, hospitals, industries, transport, and communication all depend on it. But what exactly is electricity? How does it flow through a wire, and what rules govern it? In this chapter we build a complete understanding of electric circuits, starting from the basic nature of electric charge and ending with the practical calculation of electricity bills.

You will discover that moving electrons obey simple, predictable laws. Ohm’s Law, Joule’s heating law, and the rules for combining resistors give us all the tools needed to design and analyse any electric circuit.

Topics in This Chapter

► Electric Current & Circuit

► Electric Potential & Potential Difference

► Circuit Diagrams & Standard Symbols

► Ohm’s Law

► Factors Affecting Resistance & Resistivity

► Series & Parallel Resistor Combinations

► Heating Effect of Current (Joule’s Law)

► Electric Power & Commercial Energy Units

⚡	Section 1: Electric Current & Circuit

Definition: Electric Current

Electric current is the rate of flow of electric charge through a cross-section of a conductor. In metallic wires, current is due to the drift of electrons. By convention, the direction of current is taken as opposite to electron flow.

I = Q / t

I = current (Ampere, A) | Q = charge (Coulomb, C) | t = time (s)

Quantity	Symbol	SI Unit	Key Fact
Electric Charge	Q	Coulomb (C)	1 C ≈ 6×10¹⁸ electrons
Electric Current	I	Ampere (A)	1 mA = 10⁻³ A 1 μA = 10⁻⁶ A
Potential Difference	V	Volt (V)	1 V = 1 J C⁻¹
Resistance	R	Ohm (Ω)	1 Ω = 1 V A⁻¹
Electric Power	P	Watt (W)	1 W = 1 V×1 A

📌 Key Facts: Current & Circuit

► A continuous closed path of electric current is called an electric circuit.

► If the circuit is broken anywhere, current stops — the bulb goes out.

► Ammeter measures current — always connected in series.

► Voltmeter measures potential difference — always connected in parallel.

Diagram: Simple Electric Circuit

Cell

○

Bulb

Ammeter

Switch

Figure 1 — Current flows from + terminal through bulb → ammeter → switch → back to − terminal

🔋	Section 2: Electric Potential & Potential Difference

Just as water flows from a high level to a low level due to a pressure difference, electrons flow through a conductor only when there is a difference in electric potential between two points. This potential difference is created by a cell or battery through a chemical reaction.

Definition & Formula: Potential Difference

V = W / Q

V = potential difference (V) | W = work done (J) | Q = charge moved (C)

1 Volt = 1 joule of work done to move 1 coulomb of charge between two points. 1 V = 1 J/C

Water Flow Analogy

Water flows from high level to low level because of a pressure difference. The pump maintains that difference.

Electric Current Analogy

Electrons flow from low potential to high potential. The cell/battery acts as the “pump” maintaining potential difference.

📄	Section 3: Circuit Symbols

No.	Component	Symbol Description	Connection Rule
1	Electric Cell	Long thin line (+) beside short thick line (−)	Source of EMF
2	Battery (combination of cells)	Repeated cell symbols in series	Source of higher EMF
3	Switch / Plug Key (open)	—( )— open circle	Breaks the circuit
4	Switch / Plug Key (closed)	—(•)— dot in circle	Completes the circuit
5	Resistor of resistance R	—/\/\/\— zigzag line	Opposes current flow
6	Variable Resistor / Rheostat	Zigzag with arrow overlay	Adjustable resistance
7	Ammeter	Circle labelled A	Always in SERIES
8	Voltmeter	Circle labelled V	Always in PARALLEL
9	Electric Bulb	Circle with cross or filament inside	Converts electrical energy to light & heat
10	Wire Joint	Solid dot at junction	Wires connected at that point

💡	Section 4: Ohm’s Law

In 1827, German physicist Georg Simon Ohm (1787–1854) discovered a fundamental relationship between current and voltage in metallic conductors. When he plotted V against I for a nichrome wire, he obtained a straight line through the origin — demonstrating that V/I is a constant.

Ohm’s Law

V = I × R

V = Voltage (V)

I = Current (A)

R = Resistance (Ω)

Valid only when temperature remains constant

V = IR

To find Voltage

I = V/R

To find Current

R = V/I

To find Resistance

Diagram: V–I Graph for a Metallic Conductor

V ↑

I (A) →

Straight line through origin → V/I = R = constant

Figure 2 — V–I graph confirms Ohm’s Law: steeper slope = higher resistance

🔌	Section 5: Factors Affecting Resistance & Resistivity

Resistance is the property of a conductor that opposes the flow of electric current. Experimental investigation shows that resistance depends on three factors: length, cross-sectional area, and the material of the conductor.

Resistance Formula (Resistivity Equation)

R = ρ l / A

ρ (rho) = electrical resistivity (Ω m) — property of material
l = length of conductor (m)
A = cross-sectional area (m²)

R ∝ l

Longer wire → higher resistance. Double length → double R.

R ∝ 1/A

Thicker wire → lower resistance. Double area → halve R.

Depends on material

Different materials have different ρ values. Alloys generally have higher ρ than their component metals.

Resistivity of Selected Materials at 20°C

Category	Material	Resistivity (Ω m)
Conductors	Silver	1.60 × 10⁻⁸ (best conductor)
	Copper	1.62 × 10⁻⁸
	Aluminium	2.63 × 10⁻⁸
	Tungsten	5.20 × 10⁻⁸
Alloys	Nichrome (Ni, Cr, Mn, Fe)	100 × 10⁻⁶ (heaters, irons)
	Constantan (Cu, Ni)	49 × 10⁻⁶
	Manganin (Cu, Mn, Ni)	44 × 10⁻⁶
Insulators	Glass	10¹⁰ – 10¹⁴
Insulators	Hard Rubber	10¹³ – 10¹⁶

Why Alloys Are Used in Heating Devices

Alloys like Nichrome have higher resistivity than pure metals, so they generate more heat for a given current. Crucially, they also do not oxidise (burn) readily at high temperatures — making them durable for electric irons, toasters, and room heaters. Tungsten is used exclusively for bulb filaments due to its extremely high melting point of 3380°C.

⏻	Section 6: Series & Parallel Resistors

Property	Series	Parallel
Current	Same through all components	Splits: I = I₁ + I₂ + I₃
Voltage	Splits: V = V₁ + V₂ + V₃	Same across all branches
Equivalent Resistance	Rₛ = R₁ + R₂ + R₃ Always greater than any single R	1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ Always less than the smallest R
If one component fails	Entire circuit breaks	Other branches still work
Use in homes	Not used — impractical	Always used — every appliance gets full supply voltage

Diagram: Resistors in Series

—∖∖∖∖—
R₁

—∖∖∖∖—
R₂

—∖∖∖∖—
R₃

Rₛ = R₁ + R₂ + R₃ | Same current I through every resistor

Figure 3 — In series, all resistors carry the same current; voltages add up to supply voltage

Diagram: Resistors in Parallel

	R₁

	R₂

	R₃

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ | Same voltage V across every branch

Figure 4 — In parallel, every branch has the same voltage; currents split and add up to total I

🔥	Section 7: Heating Effect of Electric Current

When current flows through a resistor, electrical energy is continuously converted into heat. This is the heating effect of electric current. In a purely resistive circuit, all source energy appears as heat — this is known as Joule’s Law of Heating.

Joule’s Law of Heating

H = I² R t

H = heat produced (J) | I = current (A) | R = resistance (Ω) | t = time (s)

Also written as: H = VIt | H = (V²/R) t

H ∝ I²

Heat ∝ square of current (fixed R, t). Doubling I → 4× heat.

H ∝ R

Heat ∝ resistance (fixed I, t). Higher R element heats more.

H ∝ t

Heat ∝ time (fixed I, R). Longer on → more heat.

Device	How Joule Heating Is Applied
Electric Iron / Toaster / Oven	Nichrome coil heats up because of high resistivity; transfers heat for pressing clothes or toasting bread
Electric Bulb	Tungsten filament heats to ~3000°C and glows; bulb filled with inert N₂/Ar gas to prevent oxidation
Electric Fuse	Low-melting-point alloy wire melts when current exceeds rated value, breaking the circuit and protecting appliances

📈	Section 8: Electric Power

Electric power is the rate at which electrical energy is consumed or dissipated in a circuit. It tells us how fast a device uses energy.

Electric Power Formulas

P = VI

P = I²R

P = V²/R

1 Watt = 1 Volt × 1 Ampere = 1 joule per second. SI unit of power is the watt (W).

📌 Energy Units

► Commercial unit of energy = kilowatt hour (kWh), shown as “units” on electricity bills

► 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J

► We pay for energy consumed (kWh) — electrons are never used up, they just transfer energy!

✏	Worked Examples

Example 1

Finding charge from current and time

Problem: A current of 0.5 A flows through a bulb filament for 10 minutes. Find the charge that flows through the circuit.

Show Solution ▼

Given: I = 0.5 A, t = 10 min = 600 s

Q = I × t = 0.5 × 600 = 300 C

✓ 300 coulombs of charge flows through the circuit in 10 minutes.

Example 2

Work done in moving charge across potential difference

Problem: Find the work done in moving 2 C of charge across two points having a potential difference of 12 V.

Show Solution ▼

V = W/Q → W = V × Q = 12 × 2 = 24 J

✓ 24 joules of work is done to move the charge.

Example 3

Current through a bulb and a heater — Ohm’s Law

Problem: (a) A bulb filament of 1200 Ω is connected to 220 V. (b) A heater coil of 100 Ω is connected to 220 V. Find the current in each case.

Show Solution ▼

(a) I = V/R = 220/1200 = 0.18 A

(b) I = V/R = 220/100 = 2.2 A

✓ The heater draws 12× more current from the same 220 V source — its resistance is much lower.

Example 4

Finding new current when voltage is doubled

Problem: A heater draws 4 A at 60 V. What current will it draw if voltage is increased to 120 V?

Show Solution ▼

R = V/I = 60/4 = 15 Ω

New I = 120/15 = 8 A

✓ Doubling voltage doubles current (resistance stays constant).

Example 5

Calculating resistivity and identifying material

Problem: A metal wire 1 m long has resistance 26 Ω and diameter 0.3 mm. Find its resistivity and identify the material.

Show Solution ▼

r = 0.15 mm = 1.5 × 10⁻⁴ m → A = πr² ≈ 7.07 × 10⁻⁸ m²

ρ = RA/l = (26 × 7.07 × 10⁻⁸) / 1 ≈ 1.84 × 10⁻⁶ Ω m

✓ From the resistivity table, this matches Manganese.

Example 6

Effect of halving length and doubling area on resistance

Problem: A 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.

Show Solution ▼

R₂ = ρ(l/2)/(2A) = (ρl/A) × (1/4) = R₁/4 = 4/4 = 1 Ω

✓ Halving length and doubling area reduces resistance to one-quarter of its original value.

Example 7

Series circuit — lamp and conductor with 6 V battery

Problem: A lamp (20 Ω) and a conductor (4 Ω) are in series across a 6 V battery. Find: (a) total resistance, (b) current, (c) voltage across each component.

Show Solution ▼

(a) Rₛ = 20 + 4 = 24 Ω

(b) I = V/Rₛ = 6/24 = 0.25 A

✓ Check: 5 + 1 = 6 V ✓ Voltages add up to the battery voltage.

Example 8

Parallel circuit — three resistors, 12 V battery

Problem: R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 30 Ω in parallel with a 12 V battery. Find current through each resistor and total resistance.

Show Solution ▼

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

Total I = 2.4 + 1.2 + 0.4 = 4 A

1/Rₚ = 1/5 + 1/10 + 1/30 = 6/30 + 3/30 + 1/30 = 10/30 → Rₚ = 3 Ω

✓ Total resistance (3 Ω) is less than the smallest individual resistor (5 Ω).

Example 9

Joule heating — electric iron

Problem: An electric iron (resistance 20 Ω) carries a current of 5 A. Calculate the heat developed in 30 seconds.

Show Solution ▼

H = I²Rt = 5² × 20 × 30 = 25 × 20 × 30 = 15,000 J = 15 kJ

✓ 15,000 joules of heat is produced in 30 seconds.

Example 10

Find potential difference from heat produced per second

Problem: 100 J of heat is produced each second in a 4 Ω resistor. Find the potential difference across it.

Show Solution ▼

H = I²Rt → 100 = I² × 4 × 1 → I² = 25 → I = 5 A

V = IR = 5 × 4 = 20 V

✓ The potential difference across the resistor is 20 V.

Example 11

Power of an electric bulb

Problem: An electric bulb connected to a 220 V generator carries a current of 0.5 A. Find the power of the bulb.

Show Solution ▼

P = VI = 220 × 0.5 = 110 W

✓ The bulb consumes 110 watts of electrical power.

Example 12

Cost of running a refrigerator for 30 days

Problem: A 400 W refrigerator runs 8 hours per day for 30 days. Find the cost of energy at ₹3.00 per kWh.

Show Solution ▼

Energy = 400 W × 8 h/day × 30 days = 96,000 Wh = 96 kWh

Cost = 96 kWh × ₹3 = ₹288

✓ It costs ₹288 to run the refrigerator for 30 days.

📝	Practice Sets

Practice Set A — Multiple Choice

1. The SI unit of electrical resistance is:
(a) Volt (b) Ampere (c) Ohm (d) Watt

2. Which term does NOT represent electrical power?
(a) I²R (b) IR² (c) VI (d) V²/R

3. In a parallel circuit, which quantity is identical for all branches?
(a) Current (b) Resistance (c) Voltage (d) Power

4. 1 kilowatt hour equals:
(a) 1000 J (b) 3600 J (c) 3.6 × 10⁶ J (d) 3.6 × 10⁵ J

5. A wire of resistance R is cut into 5 equal parts and all reconnected in parallel. The ratio R/R′ is:
(a) 1/25 (b) 1/5 (c) 5 (d) 25

Show Answers ▼

1 — (c) Ohm | 2 — (b) IR² | 3 — (c) Voltage | 4 — (c) 3.6×10⁶ J | 5 — (d) 25

Practice Set B — Short Answer

1. Define potential difference. What instrument measures it and how is it connected?

2. State Ohm’s Law and write its formula. Under what condition does it apply?

3. Why is tungsten used almost exclusively for the filaments of electric bulbs?

4. Why are heating coils of electric toasters and irons made from alloys rather than pure metals?

5. State two advantages of connecting electrical appliances in parallel rather than in series.

Show Answers ▼

1. P.D. is work done per unit charge to move charge between two points (V = W/Q). Measured by a voltmeter connected in parallel.

2. Potential difference V across a metallic conductor is directly proportional to current I through it at constant temperature: V = IR.

3. Tungsten has an extremely high melting point (3380°C) so it can withstand the temperature needed to emit white light without melting.

4. Alloys have higher resistivity (more heat per unit current) and do not oxidise/burn easily at high temperatures — making them safe and durable.

5. (i) Each appliance receives the full supply voltage independently. (ii) If one appliance fails, the rest continue to work normally.

Practice Set C — Numerical Problems

1. A 9 V battery is connected in series with resistors 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. Find the current through the 12 Ω resistor.

2. A copper wire has diameter 0.5 mm and resistivity 1.6 × 10⁻⁸ Ω m. What length of this wire will give a resistance of 10 Ω?

3. An electric heater (44 Ω) draws 5 A for 2 hours. Find the rate at which heat is developed.

4. Two lamps — one rated 100 W at 220 V, the other 60 W at 220 V — are in parallel. What current is drawn from a 220 V supply?

5. Which uses more energy: a 250 W TV set used for 1 hour, or a 1200 W toaster used for 10 minutes?

Show Answers ▼

1. Rₛ = 0.2+0.3+0.4+0.5+12 = 13.4 Ω I = 9/13.4 ≈ 0.67 A

2. A = π(0.25×10⁻³)² ≈ 1.96×10⁻⁷ m² l = RA/ρ = 10×1.96×10⁻⁷/1.6×10⁻⁸ ≈ 122.7 m

3. P = I²R = 25 × 44 = 1100 W = 1.1 kW

4. I₁ = 100/220 ≈ 0.455 A I₂ = 60/220 ≈ 0.273 A Total = 0.727 A ≈ 0.73 A

5. TV: 250×1 = 250 Wh Toaster: 1200×(1/6) = 200 Wh TV uses more energy.

Practice Set D — Extended Response

1. Compare series and parallel resistor circuits across: current distribution, voltage distribution, equivalent resistance formula, and practical use in homes. Include diagrams.

2. State Joule’s Law of Heating. Name three everyday devices that rely on this effect and explain how each one works.

3. A hot plate has two coils A and B, each of 24 Ω resistance, connected to a 220 V supply. Find the current drawn when: (i) only coil A, (ii) A and B in series, (iii) A and B in parallel.

4. Explain how an electric fuse protects household circuits. What rating fuse would you choose for an iron that consumes 1 kW at 220 V?

Show Model Answers ▼

1. Series: same current throughout; voltage splits (V=V₁+V₂+V₃); Rₛ=sum (always larger); one failure breaks all. Parallel: same voltage across each branch; current splits (I=I₁+I₂+I₃); 1/Rₚ=sum of reciprocals (always smaller); failure of one branch doesn’t affect others — hence parallel is used in homes.

2. H = I²Rt. Three devices: (i) Electric Iron — Nichrome coil heats up for pressing. (ii) Electric Bulb — Tungsten filament heats to ~3000°C and emits light. (iii) Fuse — Low-melting-point wire melts at excess current, breaking the circuit.

3. (i) Only A: I = 220/24 ≈ 9.17 A (ii) Series (R=48Ω): I = 220/48 ≈ 4.58 A (iii) Parallel (R=12Ω): I = 220/12 ≈ 18.33 A

4. Fuse is placed in series. Excess current melts the wire, breaking circuit before damage occurs. For a 1 kW iron at 220 V: I = 1000/220 ≈ 4.54 A → use a 5 A fuse (next standard rating above 4.54 A).

📚	Chapter Summary

Electric Current

I = Q/t (A). Flows + to − conventionally; electrons flow opposite. Ammeter in series measures it.

Potential Difference

V = W/Q (Volt). Created by cell/battery. Voltmeter in parallel measures it. 1 V = 1 J/C.

Resistance & Resistivity

R = ρl/A (Ω). Increases with length; decreases with area. Alloys: high ρ, no oxidation at heat.

Ohm’s Law

V = IR at constant temperature. V-I graph is a straight line through origin. R = slope.

Series & Parallel

Series: Rₛ=ΣR (same I). Parallel: 1/Rₚ=Σ1/R (same V). Homes always use parallel.

Power & Joule Heating

H = I²Rt P = VI = I²R = V²/R. Commercial unit: 1 kWh = 3.6×10⁶ J.

⚡	8-Point Exam Quick-Check

① Ohm’s Law

V = IR. Applies to metallic conductors at constant temperature only.

② Series Resistance

Rₛ = R₁ + R₂ + R₃. Always greater than any individual resistance.

③ Parallel Resistance

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃. Always less than the smallest R.

④ Joule’s Law of Heating

H = I²Rt. Heat increases with square of current. Iron, toaster, fuse all use this.

⑤ Electric Power

P = VI = I²R = V²/R. Unit is Watt (W). 1 W = 1 J/s.

⑥ Commercial Energy Unit

1 kWh = 3.6 × 10⁶ J. This is what you pay for on your electricity bill.

⑦ Meter Connections

Ammeter → SERIES. Voltmeter → PARALLEL. Swapping these will damage the meters!

⑧ Resistivity Formula

R = ρl/A. Unit of ρ = Ω m. Characteristic property of material — changes with temperature.

This comprehensive Grade 10 Science study guide covers Electricity — Chapter 11 of the CBSE Class 10 Science syllabus. Topics include electric current and charge (I = Q/t), potential difference (V = W/Q), Ohm’s Law (V = IR), resistance and resistivity (R = ρl/A), series circuits (Rₛ = R₁+R₂+R₃), parallel circuits (1/Rₚ = 1/R₁+1/R₂+1/R₃), Joule’s Law of heating (H = I²Rt), electric power (P = VI = I²R = V²/R), and energy units (1 kWh = 3.6×10⁶ J). Features circuit diagrams, resistivity data table, 12 fully solved worked examples, Practice Sets A–D with answers, chapter summary card, and 8-point exam quick-check. Perfect for School Revise CBSE Board exam revision, class tests, and competitive entrance preparation.

Curriculum

Science Grade 10

Grade 10 Science

Interactive Worksheets

Chapter 11: Electricity

Electricity

Introduction

Section 1: Electric Current & Circuit

Section 2: Electric Potential & Potential Difference

Section 3: Circuit Symbols

Section 4: Ohm’s Law

Section 5: Factors Affecting Resistance & Resistivity

Section 6: Series & Parallel Resistors

Section 7: Heating Effect of Electric Current

Section 8: Electric Power

Worked Examples

Practice Sets

Chapter Summary

8-Point Exam Quick-Check