NCERT Solutions for Class 10 Science Chapter 11 – Electricity
Electricity is one of the most scoring and conceptually rich chapters in CBSE Class 10 Science. NCERT Solutions for Class 10 Science Chapter 11 takes students through the fundamentals of electric charge, current, potential difference, resistance, and the laws that govern the flow of current in circuits. What makes this chapter truly fascinating is how abstract concepts — invisible electrons rushing through a wire — are expressed through measurable quantities and precise mathematical relationships. Students who build a clear foundation here will find it much easier to handle advanced physics in Class 11 and 12. The chapter also has direct links to everyday life: from the resistance value printed on a bulb to the way your home's parallel circuit ensures that switching off one light doesn't kill all the others. At Myclass24, the solutions are written step-by-step so students understand not just the answer but why each step is taken — something that makes a real difference in board exams where method marks matter just as much as the final answer.
Download NCERT Solutions for Class 10 Science Chapter 11 Electricity – Free PDF
Chapter 11 Electricity – Key Concepts, Formulas & NCERT Exercise Answers
Chapter 11 is divided into several logical sections — from understanding what electric current actually means at the atomic level, to applying Kirchhoff-style rules in series and parallel circuits. Every NCERT in-text question and exercise question has a specific purpose: together they test conceptual understanding, formula application, and real-world reasoning. Below is a structured breakdown to help you navigate the chapter and its solutions efficiently.
Core Formulas You Must Know
| Formula | What It Represents | SI Unit |
|---|---|---|
| I = Q / t | Electric current = charge / time | Ampere (A) |
| V = W / Q | Potential difference = work done / charge | Volt (V) |
| V = IR (Ohm's Law) | Voltage across a conductor | Volt (V) |
| R = ρL / A | Resistance depends on material, length, area | Ohm (Ω) |
| P = VI = I²R = V²/R | Electric power | Watt (W) |
| H = I²Rt | Heat produced (Joule's Law) | Joule (J) |
Series vs Parallel Circuits – Quick Comparison
| Property | Series Circuit | Parallel Circuit |
|---|---|---|
| Current | Same through all resistors | Splits across branches |
| Voltage | Splits across resistors | Same across all branches |
| Total Resistance | R = R₁ + R₂ + R₃ | 1/R = 1/R₁ + 1/R₂ + 1/R₃ |
| Effect of one break | Entire circuit stops | Other branches work |
| Used in | String lights, simple circuits | Home wiring, appliances |
NCERT Exercise – Important Q&A Highlights
| Question Focus | Key Concept Tested | Marks Weightage |
|---|---|---|
| Q1 – Current & Charge | I = Q/t calculation | 2 marks |
| Q3 – Ohm's Law verification | V-I graph interpretation | 3 marks |
| Q6 – Resistance in series | R_total = R₁+R₂+R₃ | 3 marks |
| Q8 – Resistance in parallel | 1/R formula, effective R | 3 marks |
| Q13 – Joule's heating | H = I²Rt; energy cost | 3 marks |
| Q16 – Power dissipation | P = V²/R comparison | 2 marks |
Resistivity of Common Materials (Fact Table)
| Material | Resistivity (Ω·m) at 20°C | Type |
|---|---|---|
| Silver | 1.60 × 10⁻⁸ | Best conductor |
| Copper | 1.69 × 10⁻⁸ | Conductor (used in wires) |
| Tungsten | 5.60 × 10⁻⁸ | Used in bulb filaments |
| Nichrome | 1.00 × 10⁻⁶ | Used in heating elements |
| Glass | 10¹⁰ – 10¹⁴ | Insulator |
| Rubber | 10¹³ – 10¹⁶ | Insulator |
The resistivity table isn't just a data dump — NCERT uses it to explain why copper is preferred in electrical wiring (low resistivity, cost-effective) while tungsten is used in bulb filaments (high melting point, high resistivity that generates heat and light). These application-based answers regularly appear in CBSE board exams as short-answer questions.
For complete, step-by-step solved answers to all NCERT in-text and exercise questions — including all numericals — visit Myclass24 and download the free PDF for Chapter 11: Electricity.
FAQs for NCERT Solutions for Class 10 Science Chapter 11 Electricity
Ohm's Law states that the electric current flowing through a conductor is directly proportional to the potential difference across it, provided the temperature and other physical conditions remain constant. Mathematically, V = IR, where V is voltage in volts, I is current in amperes, and R is resistance in ohms. This law is fundamental to solving electric circuit problems. By rearranging: I = V/R (to find current) and R = V/I (to find resistance). For a series circuit, total resistance R = R₁ + R₂ + R₃. For a parallel circuit, 1/R = 1/R₁ + 1/R₂ + 1/R₃. Using Ohm's Law along with Kirchhoff's laws allows calculation of current, voltage, and resistance anywhere in a circuit. It also helps understand why thinner wires have higher resistance and how resistors control current in electronic devices.
In a series circuit, all components are connected end-to-end in a single path. The same current flows through all components, but the voltage is divided across each. If one component fails, the circuit breaks and all components stop working. The total resistance is the sum of individual resistances: R_total = R₁ + R₂ + R₃. Series circuits are used in decorative lights (old-style) and some battery connections. In a parallel circuit, components are connected across the same two points, providing multiple paths for current flow. Each component gets the same voltage but different currents flow through each branch. If one branch fails, others continue to work. Total resistance decreases: 1/R_total = 1/R₁ + 1/R₂. Household wiring uses parallel circuits so all appliances receive the same voltage and can be operated independently.
Electric power is the rate at which electric energy is consumed or transferred. It is measured in watts (W). The formula is P = VI, where P is power in watts, V is voltage in volts, and I is current in amperes. Using Ohm's Law (V = IR), power can also be expressed as P = I²R or P = V²/R. These formulas are very useful in solving numericals. For example, a 100W bulb connected to 220V draws a current of I = P/V = 100/220 ≈ 0.45 A. Commercial electricity is measured in kilowatt-hours (kWh) — the unit of electrical energy. One kilowatt-hour = 1000 W × 3600 s = 3.6 × 10⁶ J. Energy consumed = Power × Time. Electric bill calculations are based on units consumed, where 1 unit = 1 kWh. Power ratings on appliances help users understand energy consumption.
The resistance of a conductor depends on four main factors. Length: resistance is directly proportional to the length of the conductor — a longer wire has more resistance because electrons must travel further and collide more often. Cross-sectional area: resistance is inversely proportional to area — a thicker wire has less resistance as more electrons can flow simultaneously. Material: different materials have different resistivities (ρ). Copper has low resistivity (good conductor) while nichrome has high resistivity (used in heaters). Temperature: for most metals, resistance increases with temperature because increased thermal energy causes more atomic vibrations, obstructing electron flow. However, semiconductors and electrolytes show the opposite trend. The formula combining these factors is R = ρL/A. This explains why power transmission lines use thick copper wires and heating elements use high-resistivity alloys.
When electric current flows through a conductor, electrons collide with atoms in the material, losing kinetic energy which is converted into heat. This heating effect of current is the basis of Joule's Law, which states that the heat produced (H) in a conductor is directly proportional to the square of the current (I²), resistance (R), and time (t): H = I²Rt. This means doubling the current produces four times the heat. Higher resistance and longer time also produce more heat. This principle is used in electric heaters, geysers, toasters, electric irons, and incandescent bulbs (where the tungsten filament heats to 2000°C to produce light). Fuses are also based on this — excessive current generates enough heat to melt the fuse wire, breaking the circuit and protecting appliances from damage due to overloading.