Electricity

INTRODUCTION:

Electricity is a fundamental Physics chapter that explains how electric current flows through conductors, governed by Ohm's Law (V=IR) and influenced by factors like resistance (R=ρ.l/A), material resistivity, and temperature. The chapter explores series and parallel circuits, where resistors combine differently (series: R1+R2+R3; parallel: 1/Req=1/R1+1/R2), and demonstrates practical applications through Joule's heating effect (H=I²)—seen in electric heaters, bulbs, and fuses—and electric power calculations (P=VI=I²=V²/). Understanding these concepts unlocks numerical problem-solving (6-8 marks), circuit analysis (3-5 marks), and theory questions (2-3 marks) that appear predictably in every CBSE Board exam. This chapter rewards formula mastery + practice over rote memorization, making it one of the most reliable scoring areas (12-15 marks potential) when you know how to apply the relationships systematically.

What is Electric Current?

Electric current (I) is the rate of flow of electric charge through a conductor.

· Q = Electric charge (Coulombs, C)

· t = Time (seconds, s)

· Unit of current: Ampere (A) = 1 Coulomb per second

Conventional current: Flows from positive (+) to negative (−) terminal of battery (opposite to actual electron flow).

Ammeter: Measures electric current — always connected in series in a circuit.

Voltmeter: Measures potential difference — always connected in parallel across a component.

History: HAND LABOUR & STEAM POWER Chapter 4: The Age of Industrialisation

Electric Potential and Potential Difference

Electric potential: Work done per unit charge to move a positive test charge from infinity to that point.

Potential difference (V): Work done per unit charge to move charge from one point to another.

· Unit: Volt (V) = 1 Joule per Coulomb (J/C)

Why current flows? Current flows from higher potential to lower potential (conventional). Battery maintains potential difference continuously.

Class 10 Science – Chapter: Human Eye and the Colourful World complete notes.

Ohm's Law (Most Important!)

Statement: At constant temperature, the electric current flowing through a conductor is directly proportional to the potential difference across its ends.

V∝I

Where R = Resistance (constant of proportionality)

· Unit of Resistance: Ohm (Ω) = 1 Volt per Ampere

V-I Graph: Straight line passing through origin → slope = Resistance (R = V/I).

Ohmic vs Non-Ohmic Conductors

Property	Ohmic conductors	Non-ohmic devices
Definition	Follow Ohm's law strictly	Don't follow Ohm's law
V-I graph	Straight line through origin	Curved/irregular
R value	Constant at all V	Changes with V
Examples	Copper, aluminium wires	Diode, filament bulb, LED

History: BEFORE THE INDUSTRIAL REVOLUTIONChapter 4: The Age of Industrialisation

Resistance and Resistivity

Resistance (R): Opposition offered by a conductor to the flow of electric current.

Factors affecting resistance:

Factor	Effect on Resistance
Length (l)	R ∝ l (double length = double resistance)
Cross-section area (A)	R ∝ 1/A (thicker wire = less resistance)
Material (ρ)	Depends on resistivity of material
Temperature	Increases for metals; decreases for semiconductors

Resistivity (ρ): Resistance of a conductor of unit length and unit cross-sectional area. Unit: Ohm-metre (Ω-m). It is a material property, independent of dimensions.

Resistivity Table:

Material	Resistivity (Ω-m)	Type
Silver	1.60 × 10⁻⁸	Best conductor
Copper	1.70 × 10⁻⁸	Conductor
Aluminium	2.60 × 10⁻⁸	Conductor
Tungsten	5.60 × 10⁻⁸	Metal (bulb filament)
Iron	10.0 × 10⁻⁸	Metal
Nichrome	100 × 10⁻⁸	Alloy (heater element)
Rubber	10¹³ − 10¹⁶	Insulator
Glass	10¹⁰ − 10¹⁴	Insulator

Why copper used for electric wires? Low resistivity → less heat loss → efficient current transmission.

Why Nichrome used in heaters? High resistivity + high melting point → produces maximum heat safely.

Class 10 Science – Chapter 10: Light - Reflection and Refraction complete notes

Resistors in Series Circuit

Series connection: All resistors connected end-to-end in a single path.

Key formulas:

Voltage across each resistor: V₁ = IR₁, V₂ = IR₂, V₃ = IR₃

Important: Total series resistance is always greater than the largest individual resistance.

Application: Not used in domestic wiring (single break stops all; unequal voltages; overall resistance too high).

History: THE INTER-WAR ECONOMY Chapter 3: The Making of a Global World

Resistors in Parallel Circuit

Parallel connection: All resistors connected between same two points (same voltage).

Key formulas:

Current through each resistor: I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃

Important: Total parallel resistance is always less than the smallest individual resistance.

Application: Used in domestic wiring — each appliance works independently at same voltage.

Class 10 Science – Chapter 9: Heredity and Evolution complete notes.

Series vs Parallel Complete Comparison

Property	Series Connection	Parallel Connection
Equivalent resistance	R₁ + R₂ + R₃	Less than smallest R
Current	Same through all	Divides I = I₁ + I₂ + I₃
Voltage	Divides V = V₁+V₂+V₃	Same across all
If one fails	Whole circuit breaks	Others continue normally
Bulb brightness	Dimmer (shares voltage)	Brighter (full voltage each)
Domestic use	Not used	Used (house wiring)
Application	Decorative bulb chains	All household circuits

History: THE NINETEENTH CENTURY (1815-1914) Chapter 3: The Making of a Global World

Heating Effect of Electric Current (Joule's Law)

Joule's Law: When electric current flows through a conductor, heat is produced. The heat produced is:

I²

Alternate forms:

V²

Unit: Joule (J) → 1 Calorie = 4.18 J

Applications of Heating Effect

Appliance	Why heating effect used	Material used	Why that material
Electric heater/toaster	Converts electricity to heat	Nichrome wire	High ρ, high melting point
Electric iron	Converts electricity to heat	Nichrome wire	Same as above
Electric bulb	Produces light (via heat)	Tungsten filament	Very high melting point (3380°C)
Electric fuse	Safety device (melts at excess current)	Tin-lead alloy	Low melting point
Welding machines	High heat for welding	Carbon electrodes	High resistance

Why tungsten for bulb filament?

· Melting point 3380°C (highest of all metals)

· Stays solid even at 2500°C operating temperature

· Gives white incandescent light

Why fuse has low melting point?

· Melts quickly when current exceeds rated value

· Breaks circuit before appliances get damaged

Electric Power

Electric power (P): Rate at which electric energy is consumed or work is done.

I² V²

Unit: Watt (W) = 1 Joule per second (J/s)

Which formula to use?

· P = VI (when both V and I known)

· P = I²R (when I and R known)

· P = V²/R (when V and R known)

Commercial Unit of Electric Energy

10⁶

1 unit of electricity = 1 kWh = 3.6 × 10⁶ J

Cost calculation:
Energy consumed (kWh) = Power (kW) × Time (hours)
Cost = Energy (kWh) × Rate per unit (₹/kWh)

Class 10 Science – Chapter: How Do Organisms Reproduce? complete notes

MCQs PYQ

1. SI unit of electric charge is:
(a) Ampere
(b) Volt
(c) Coulomb
(d) Ohm
Ans: (c) Coulomb (CBSE 2020)

2. Ohm's law is applicable when:
(a) Temperature varies
(b) Temperature constant
(c) Resistance zero
(d) Current zero
Ans: (b) Temperature constant (CBSE 2021)

3. In series combination, quantity that remains same:
(a) Voltage
(b) Power
(c) Current
(d) Resistance
Ans: (c) Current (CBSE 2023)

4. 1 kWh equals:
(a) 1000 J
(b) 3600 J
(c) 3.6 × 10⁶ J
(d) 10⁶ J
Ans: (c) 3.6 × 10⁶ J (CBSE 2020)

5. Nichrome is used in heaters because of:
(a) Low resistivity
(b) High resistivity + high MP
(c) Low melting point
(d) High conductivity
Ans: (b) High resistivity + high melting point (CBSE 2024)

6. Power formula P =:
(a) VI only
(b) I²R only
(c) V²/R only
(d) All three
Ans: (d) All three (VI = I²R = V²/R) (CBSE 2023)

7. Fuse wire has low melting point because:
(a) Good conductor
(b) Melts quickly at excess current
(c) Low cost
(d) High resistance
Ans: (b) Melts quickly at excess current (CBSE 2020)

8. Resistivity of a material depends on:
(a) Length only
(b) Area only
(c) Temperature and material
(d) Shape
Ans: (c) Temperature and material (CBSE 2022)

9. In parallel combination, which remains same?
(a) Current
(b) Resistance
(c) Voltage
(d) Power
Ans: (c) Voltage (CBSE 2023)

10. Heat produced by current given by:
(a) H = IRt
(b) H = I²Rt
(c) H = VRt
(d) H = I²R²t
Ans: (b) H = I²Rt (CBSE 2020)

Class 10 Science – Chapter 6: Control and Coordination

Short Answer Questions (PYQ)

Q1. State Ohm's law. Write its mathematical form.
Ans: At constant temperature, current through conductor is directly proportional to potential difference across ends → V ∝ I → V = IR. Slope of V-I graph = Resistance R.

Q2. Why series arrangement not used in domestic circuits?
Ans: (1) Single break stops all appliances. (2) Voltage divides, appliances get less than rated voltage. (3) High total resistance → inefficient. (4) Cannot use appliances independently.

Q3. Distinguish between ohmic and non-ohmic conductors.
Ans: Ohmic: Follow V = IR at all values (straight V-I graph); e.g., copper, aluminium. Non-ohmic: Don't follow Ohm's law (curved V-I graph); resistance changes with voltage; e.g., diode, LED, filament bulb.

Q4. What is a fuse? Why is a low melting point material used?
Ans: Fuse is a safety device with a thin wire of low melting alloy (tin-lead) in series with circuit. When excess current flows, wire melts quickly → circuit breaks → appliances protected.

Q5. Two bulbs rated 60W and 100W used at 220V. Which has higher resistance?
Ans: R = V²/P. Higher wattage = lower resistance. R₆₀ = 220²/60 = 806.7 Ω, R₁₀₀ = 220²/100 = 484 Ω. 60W bulb has higher resistance.

Chapter 6 Life Processes complete notes

Long Answer Questions (PYQ)

Q1. Three resistors 3Ω, 4Ω, 5Ω connected in series to 24V. Find: (a) Total R (b) Total I (c) Voltage across each (d) Power dissipated in each.

Solution:
(a) R_total = 3+4+5 = 12 Ω
(b) I = V/R = 24/12 = 2 A
(c) V₁ = 2×3 = 6 V, V₂ = 2×4 = 8 V, V₃ = 2×5 = 10 V (Check: 6+8+10 = 24 V ✓)
(d) P₁ = I²R₁ = 4×3 = 12 W, P₂ = 4×4 = 16 W, P₃ = 4×5 = 20 W

Q2. State Joule's law. An electric iron rated 1000W, 220V used 1.5h daily for 25 days. Find: (a) Resistance (b) Current (c) Total energy in kWh (d) Cost at ₹6/kWh.

Solution:
Joule's law: H = I²Rt (heat ∝ square of current × resistance × time)
(a) R = V²/P = 220²/1000 = 48.4 Ω
(b) I = P/V = 1000/220 = 4.55 A
(c) Daily energy = 1 kW × 1.5h = 1.5 kWh; Monthly = 1.5 × 25 = 37.5 kWh
(d) Cost = 37.5 × 6 = ₹225

Q3. Three resistors 2Ω, 3Ω, 6Ω connected in parallel to 12V. Find: (a) R_eq (b) Total I (c) I through each (d) Power of each.

Solution:
(a) 1/R_eq = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1 → R_eq = 1 Ω
(b) I_total = V/R_eq = 12/1 = 12 A
(c) I₁ = 12/2 = 6 A, I₂ = 12/3 = 4 A, I₃ = 12/6 = 2 A (Check: 6+4+2 = 12 A ✓)
(d) P₁ = VI₁ = 12×6 = 72 W, P₂ = 12×4 = 48 W, P₃ = 12×2 = 24 W

Q4. Wire of resistance R is stretched to double its length. Find new resistance. How does resistivity change?

Solution:
Original: R = ρl/A (volume = lA = constant)
When stretched: new l' = 2l → volume constant → new A' = A/2
New R' = ρ(2l)/(A/2) = ρ × 2l × 2/A = 4(ρl/A) = 4R
New resistance = 4 times original resistance
Resistivity does not change — it's a material property independent of dimensions.

Q5. A house has: 5 LED bulbs (10W each), 2 fans (75W each), 1 refrigerator (150W, 24h), 1 iron (1000W, 1h). All other appliances used 8h/day for 30 days. Find: (a) Total energy consumed (b) Monthly bill at ₹8/kWh.

Solution:
Daily energy:
Bulbs: 5 × 10W × 8h = 400 Wh
Fans: 2 × 75W × 8h = 1200 Wh
Refrigerator: 150W × 24h = 3600 Wh
Iron: 1000W × 1h = 1000 Wh
Daily total = 400+1200+3600+1000 = 6200 Wh = 6.2 kWh

Monthly energy = 6.2 × 30 = 186 kWh
Monthly bill = 186 × 8 = ₹1488

Conclusion

Electricity becomes completely manageable when divided into four skill areas: (1) Ohm's law and resistance (V=IR, R=ρl/A — master the formula triangle, know V/I/R relationships), (2) Series/parallel circuits (R_total and current/voltage behavior — practice 10-15 circuit problems), (3) Heating effect (H=I²Rt — three formula forms, applications: heater/bulb/fuse materials and reasons), (4) Electric power and energy (P=VI=I²R=V²/R, 1 kWh = 3.6×10⁶ J, cost calculations). NCERT emphasizes numerical problems above everything else — series/parallel combination numericals (3-5 marks), power + cost calculations (3-5 marks), and heating effect numericals (3 marks) appear in every board paper. The key strategy: write given data clearly, choose correct formula, apply correct units, and check answer plausibility. Most common mistakes are wrong formula selection (P=VI vs P=I²R) and unit errors (watts vs kilowatts, seconds vs hours).

Chapter 4 Carbon and its Compounds

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Electricity Class 10 Science Notes | Formulas, Numericals & MCQs [PDF Download].