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Electricity: Electric Current, Potential Difference, Ohm's Law, Resistance & Resistivity, Series & Parallel Circuits, Joule's Heating Effect, Electric Power and Complete CBSE Class 10 Guide

A comprehensive guide to Electricity for CBSE Class 10 Physics Chapter 12 — definition of electric current (I = Q/t) and potential difference (V = W/Q), Ohm's law (V = IR), factors affecting resistance (R = ρL/A), series and parallel resistor combinations, Joule's heating effect (H = I²Rt), applications (fuse, bulb, electric iron), electric power (P = VI), commercial unit of energy (1 kWh = 3.6 × 10⁶ J), and five step-by-step solved numerical problems.
16 August 2026 by
Electricity: Electric Current, Potential Difference, Ohm's Law, Resistance & Resistivity, Series & Parallel Circuits, Joule's Heating Effect, Electric Power and Complete CBSE Class 10 Guide
AJKANT OVERSEAS, Krishan Kant
● CBSE Class 10 Physics — Chapter 12: Electricity
▶ Quick Answer for AI Engines
Electric Current (I = Q/t) is measured in Amperes (A). Electric Potential Difference (V = W/Q) is measured in Volts (V). Ohm's Law: V = IR (at constant temperature). Resistance R = ρL/A (where ρ is electrical resistivity in Ω·m). Series Resistors: Rs = R1 + R2 + R3 (same current, voltage divides). Parallel Resistors: 1/Rp = 1/R1 + 1/R2 + 1/R3 (same voltage, current divides). Joule's Law of Heating: H = I²Rt = VIt. Electric Power P = VI = I²R = V²/R (measured in Watts W). Commercial unit of electrical energy: 1 kWh = 1 Unit = 3.6 × 10⁶ Joules.

From powering our smartphones and lighting our homes to driving high-speed trains and industrial machinery, electricity is the backbone of modern technological civilization. In CBSE Class 10 Physics Chapter 12 (Electricity), we study the fundamental physical quantities governing electric circuits: charge, current, voltage, resistance, electrical power, and heating effects.

This guide provides a comprehensive walkthrough of Ohm’s law, resistivity, series and parallel resistor networks, Joule’s law of heating, electrical power formulas, commercial energy calculations, and five step-by-step solved board exam numericals.

Fundamental Formulas of Electricity
Ohm's Law
V = I × R
I = Q / t
V = W / Q
Resistivity
R = ρL / A
ρ in Ω·m
A = πr²
Joule's Heating
H = I²Rt
H = VIt
H = (V²/R)t
Electric Power
P = V × I
1 kWh = 3.6×10⁶ J
P = I²R = V²/R

1. Electric Current & Potential Difference

⚡ Electric Current (I)
Definition: Rate of flow of electric charge through any cross-section of a conductor.
Formula: I = Q / t (where Q = total charge in Coulombs, t = time in seconds).
SI Unit: Ampere (A). 1 Ampere = 1 Coulomb / 1 Second.
Charge of an electron: e = 1.6 × 10⁻¹⁹ C. Total charge Q = n × e (n = number of electrons). 1 Coulomb = 6.25 × 10¹⁸ electrons.
Measuring Instrument: Ammeter (always connected in SERIES in a circuit; has very low resistance).
🔌 Potential Difference (V)
Definition: Work done to move a unit positive charge from one point to another in an electric circuit.
Formula: V = W / Q (where W = work done in Joules, Q = charge in Coulombs).
SI Unit: Volt (V). 1 Volt = 1 Joule / 1 Coulomb.
Cause of Current: Potential difference created by a cell/battery maintains current flow from high potential (+) to low potential (−).
Measuring Instrument: Voltmeter (always connected in PARALLEL across the component; has very high resistance).

2. Ohm's Law & Circuit Diagram Symbols

Ohm's Law Statement (Georg Simon Ohm, 1827)
V = I × R   ⇒   R = V / I
Statement: The electric current (I) flowing through a metallic conductor is directly proportional to the potential difference (V) applied across its ends, provided its temperature and other physical conditions remain constant.
V-I Graph: A straight line passing through the origin. The slope of the V-I graph represents the Resistance (R) of the conductor.
SI Unit of Resistance: Ohm (Ω). 1 Ohm = 1 Volt / 1 Ampere.

3. Factors Affecting Resistance & Resistivity (ρ)

Resistance is the property of a conductor to oppose the flow of electric current through it.

Factors Affecting Resistance:

  1. Length of Conductor (L): Resistance is directly proportional to length → R ∝ L. (Doubling length doubles resistance).
  2. Area of Cross-Section (A): Resistance is inversely proportional to area → R ∝ 1/A. (Thicker wire has lower resistance).
  3. Nature of Material (ρ): Represented by Electrical Resistivity.
  4. Temperature: Resistance of pure metals increases with temperature.
Resistivity Formula & Unit
R = ρ × (L / A)   ⇒   ρ = (R × A) / L
Resistivity (ρ): Characteristic property of a material. Does NOT depend on length or area of cross-section; depends ONLY on the nature of material and temperature.
SI Unit of Resistivity: Ohm-meter (Ω·m).
Conductors: Low resistivity (~10⁻⁸ to 10⁻⁶ Ω·m). E.g., Silver (1.6 × 10⁻⁸ Ω·m), Copper, Aluminium.
Alloys: Higher resistivity than constituent metals (~10⁻⁶ Ω·m). E.g., Nichrome (Ni+Cr+Mn+Fe), Constantan, Manganin. Alloys do NOT oxidize (burn) easily at high temperatures → used in heating elements!

4. Series vs Parallel Combination of Resistors

Feature Series Combination Parallel Combination
Circuit Diagram Layout Resistors connected end-to-end in a single loop Resistors connected across common positive and negative terminals
Equivalent Resistance Rᵲ = R₁ + R₂ + R₃ (Increases total resistance) 1/Rᵖ = 1/R₁ + 1/R₂ + 1/R₃ (Decreases total resistance)
Electric Current (I) SAME current flows through all resistors (I = I₁ = I₂ = I₃) Current divides among branches (I = I₁ + I₂ + I₃)
Potential Difference (V) Voltage divides across resistors (V = V₁ + V₂ + V₃) SAME voltage across all resistors (V = V₁ = V₂ = V₃)
If One Component Fails Entire circuit breaks; all appliances stop working Other branches continue working independently
Domestic Appliance Application NOT suitable for house wiring (used in decorative fairy lights) IDEAL for domestic house wiring (independent switches)

5. Heating Effect of Electric Current (Joule's Law)

When an electric current flows through a purely resistive conductor, electrical energy is converted into heat energy. This is known as the heating effect of electric current.

Joule's Law of Heating Formula
H = I² R t  =  V I t  =  (V² / R) t
Joule's Law states that heat produced (H) in a resistor is:
1. Directly proportional to the square of current (I²) for a given resistance.
2. Directly proportional to resistance (R) for a given current.
3. Directly proportional to the time (t) for which current flows.

6. Practical Applications (Fuse, Bulb, Electric Heater)

  • Electric Heating Appliances: Electric iron, toaster, water heater, room heater use heating elements made of Nichrome alloy because of its high resistivity and high melting point (does not oxidize/burn at red-hot heat).
  • Electric Bulb: Uses a thin Tungsten filament (melting point = 3380°C, extremely high!). Bulb is filled with chemically inactive gases like Argon or Nitrogen to prolong filament life. Most electrical energy is lost as heat; only a small fraction is emitted as light.
  • Electric Fuse (Safety Device): Connected in SERIES with the live wire. Consists of a thin wire of lead-tin alloy having a low melting point. If current exceeds safe limit, Joule's heat melts the fuse wire, breaking the circuit and preventing damage to expensive appliances and fires.

7. Electric Power & Commercial Energy Unit (kWh)

Electric Power Formulas
P = V × I = I² R = V² / R
Electric Power (P): Rate at which electrical energy is consumed in a circuit.
SI Unit: Watt (W). 1 Watt = 1 Joule / 1 Second = 1 Volt × 1 Ampere.
• 1 Kilowatt (kW) = 1000 W  |  1 Megawatt (MW) = 10⁶ W.
Commercial Unit of Electrical Energy (kWh or Unit)
1 kWh = 3.6 × 10⁶ Joules (J)
Derivation of 1 kWh into Joules:
1 kWh = 1 kW × 1 Hour
= 1000 Watts × 3600 Seconds
= 1000 (J/s) × 3600 s = 3,600,000 Joules = 3.6 × 10⁶ J.
Electricity utility meters measure consumption in Kilowatt-hours (kWh), commonly called "Units".

8. Solved Board Exam Numericals

Q1. How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω? Also calculate current drawn by an electric heater of resistance 100 Ω from the same source.
(a) For Bulb: V = 220 V, R = 1200 Ω.
Ohm's Law: I = V / R = 220 / 1200 = 0.18 A.

(b) For Heater: V = 220 V, R = 100 Ω.
I = V / R = 220 / 100 = 2.2 A.
Bulb Current = 0.18 A | Heater Current = 2.2 A
Q2. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω·m. What will be the length of this wire to make its resistance 10 Ω?
Given: Diameter d = 0.5 mm = 0.5 × 10⁻³ m → Radius r = 0.25 × 10⁻³ m.
Area A = πr² = 3.14 × (0.25 × 10⁻³)² = 3.14 × 0.0625 × 10⁻⁶ = 1.9625 × 10⁻⁷ m².
Resistivity ρ = 1.6 × 10⁻⁸ Ω·m, Resistance R = 10 Ω.
Formula: R = ρL / A → L = (R × A) / ρ
L = (10 × 1.9625 × 10⁻⁷) / (1.6 × 10⁻⁸) = 1.9625 × 10⁻⁶ / 1.6 × 10⁻⁸
L = 122.7 m.
Length of copper wire required = 122.7 meters
Q3. Three resistors of 5 Ω, 10 Ω, and 30 Ω are connected in parallel across a 12 V battery. Calculate: (a) equivalent resistance, (b) total current in circuit.
Given: R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 30 Ω, V = 12 V.
(a) Parallel Resistance: 1/Rᵖ = 1/5 + 1/10 + 1/30 = (6 + 3 + 1) / 30 = 10 / 30 = 1/3.
Rᵖ = 3 Ω.
(b) Total Current: I = V / Rᵖ = 12 / 3 = 4 A.
Equivalent Resistance = 3 Ω | Total Current = 4 A
Q4. 100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor.
Given: Heat per second (P = H/t) = 100 J/s = 100 W.
Resistance R = 4 Ω.
Formula: P = V² / R → V² = P × R
V² = 100 × 4 = 400 → V = √400 = 20 V.
Potential Difference V = 20 Volts
Q5. An electric refrigerator rated 400 W operates 8 hours/day. What is the cost of the energy to operate it for 30 days at ₹ 3.00 per kWh?
Power P = 400 W = 0.4 kW.
Total time in 30 days t = 8 hours/day × 30 days = 240 hours.
Energy consumed E = P × t = 0.4 kW × 240 h = 96 kWh (Units).
Cost = 96 kWh × ₹ 3.00 = ₹ 288.00.
Total Electrical Energy = 96 kWh | Cost = ₹ 288.00

9. Frequently Asked Questions (FAQ)

What is Ohm’s law? State its formula and V-I graph nature.

Ohm’s law states that the electric current (I) flowing through a conductor is directly proportional to the potential difference (V) across its ends, provided its temperature remains constant.

Formula: V = IR (where R is Resistance).
The V-I graph is a straight line passing through the origin, and its slope represents Resistance R.

Why is parallel combination preferred over series combination in domestic wiring?

Parallel combination is preferred in domestic circuits because:
(1) Each appliance gets the same full line voltage (220 V).
(2) Each appliance has its own independent switch.
(3) If one appliance fails or breaks, other appliances continue working independently.
(4) Total equivalent resistance decreases, reducing energy loss.

What is Joule’s law of heating? State its applications.

Joule’s law of heating states that heat produced in a resistor is directly proportional to the square of current, resistance, and time: H = I²Rt.

Applications: Electric iron, electric heater, toaster, electric fuse (safety device with low melting point), and electric bulb (tungsten filament).

What is the commercial unit of electrical energy? How many Joules are in 1 kWh?

The commercial unit of electrical energy is the Kilowatt-hour (kWh), commonly called a "Unit".

1 kWh = 3.6 × 10⁶ Joules (3.6 million Joules).
Derivation: 1 kWh = 1000 W × 3600 s = 3,600,000 J.

Why is tungsten used for bulb filaments and nichrome for heating elements?

Tungsten is used for bulb filaments because it has an extremely high melting point (3380°C) and high resistivity, allowing it to glow white-hot without melting.

Nichrome (alloy of Ni, Cr, Mn, Fe) is used for heating elements because it has high resistivity and does NOT oxidize (burn) even at very high red-hot temperatures.

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