BharatNotes Why this chapter matters for UPSC: This is Chapter 11, "Electricity", of NCERT's Class X Science (Reprint 2026-27). It builds the whole of school circuit theory from four relations: current as the rate of flow of charge, potential difference as work per unit charge, Ohm's law, and Joule's law of heating. From them come the rules for series and parallel circuits, the reason heaters use alloys and bulbs use tungsten, the fuse, and the kilowatt hour on the electricity bill. Prelims asks these relations directly and through small numericals; GS3 meets the same physics in power transmission and energy efficiency.
Contemporary hook: On 31 August 2026 India's installed power capacity stood at 5,54,544 MW, of which 3,04,334 MW (54.88%) was non-fossil, according to the Central Electricity Authority. Every megawatt of it reaches consumers through wires that obey this chapter: the power a line carries is P = VI, and the power it wastes as heat is I²R.
🧠 First Principles — Read This First
- Current is charge on the move. I = Q/t. By convention current flows opposite to the electrons, because the convention was fixed before electrons were known (p. 171).
- Potential difference does the pushing. V = W/Q, the work done to move a unit charge between two points. A cell's chemical action maintains it.
- Ohm's law needs a fixed temperature. For a metallic wire, V = IR holds provided its temperature remains the same. Resistance depends on the wire's length, its area of cross-section and its material; resistivity depends on the material alone (and on temperature).
- Series adds resistances; parallel adds currents. In series the current is the same everywhere and Rs = R1 + R2 + R3. In parallel the voltage is the same across each branch and 1/Rp = 1/R1 + 1/R2 + 1/R3. Homes are wired in parallel.
- Every current heats its conductor. H = I²Rt. Heaters, irons, filament bulbs and fuses all use this; in other circuits it is a loss.
- The bill counts energy, not electrons. Power is the rate of using energy, P = VI = I²R = V²/R. Energy = power × time, sold in kilowatt hours. Electrons are not used up.
PART 1 — Quick Reference
Table 1: Charge, current, potential difference and the instruments (11.1-11.4)
| Quantity | Relation | SI unit | What NCERT adds |
|---|---|---|---|
| Electric current | I = Q/t (Eq. 11.1) | ampere (A), after the French scientist Andre-Marie Ampere (1775–1836); 1 A = 1 C/1 s | The rate of flow of electric charges; small currents in milliampere (10⁻³ A) or microampere (10⁻⁶ A) |
| Electric charge | Q | coulomb (C) | One coulomb is the charge of nearly 6 × 10¹⁸ electrons; an electron carries 1.6 × 10⁻¹⁹ C of negative charge |
| Direction of current | — | — | Taken as the direction of flow of positive charges, so opposite to the flow of electrons |
| Potential difference | V = W/Q (Eq. 11.2) | volt (V), after the Italian physicist Alessandro Volta (1745–1827); 1 V = 1 J C⁻¹ (Eq. 11.3) | The work done to move a unit charge from one point to the other; NCERT likens it to a pressure difference that makes water flow |
| Ammeter | — | — | Measures current; always connected in series |
| Voltmeter | — | — | Measures potential difference; always connected in parallel across the points |
| Resistor; rheostat | — | — | "A conductor having some appreciable resistance is called a resistor"; a rheostat (variable resistance) regulates the current without changing the voltage source. Table 11.1 gives the symbols for these and the other components |
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27, sections 11.1-11.4 and Table 11.1 (pp. 171-177).
Table 2: Ohm's law and what resistance depends on (11.4-11.5)
| Idea | Relation | What NCERT says |
|---|---|---|
| Ohm's law | V ∝ I (Eq. 11.4); V = IR (Eq. 11.5) | Found in 1827 by the German physicist Georg Simon Ohm (1787–1854). Holds for a metallic wire provided its temperature remains the same; the V–I graph is a straight line through the origin (Activity 11.1) |
| Resistance | R = V/I (Eq. 11.6); unit ohm (Ω); 1 Ω = 1 V/1 A | The property of a conductor to resist the flow of charges. Electrons are restrained by the attraction of the atoms among which they move |
| Current from voltage | I = V/R (Eq. 11.7) | If the resistance is doubled, the current is halved |
| Length | R ∝ l (Eq. 11.8) | Twice the length, half the current (Activity 11.3) |
| Area of cross-section | R ∝ 1/A (Eq. 11.9) | A thicker wire passes more current |
| Resistivity | R = ρ l/A (Eq. 11.10); unit Ω m | A characteristic property of the material. Both resistance and resistivity vary with temperature |
| Good and poor conductors | — | Of components of the same size, a good conductor has low resistance, a poor conductor higher, an insulator higher still |
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27, sections 11.4-11.5 and Activities 11.2-11.3 (pp. 175-179).
Table 3: NCERT's Table 11.2: resistivity at 20 °C
| Group | Material | Resistivity (Ω m) |
|---|---|---|
| Conductors | Silver | 1.60 × 10⁻⁸ |
| Copper | 1.62 × 10⁻⁸ | |
| Aluminium | 2.63 × 10⁻⁸ | |
| Tungsten | 5.20 × 10⁻⁸ | |
| Nickel | 6.84 × 10⁻⁸ | |
| Iron | 10.0 × 10⁻⁸ | |
| Chromium | 12.9 × 10⁻⁸ | |
| Mercury | 94.0 × 10⁻⁸ | |
| Manganese | 1.84 × 10⁻⁶ | |
| Alloys | Constantan (Cu, Ni) | 49 × 10⁻⁶ |
| Manganin (Cu, Mn, Ni) | 44 × 10⁻⁶ | |
| Nichrome (Ni, Cr, Mn, Fe) | 100 × 10⁻⁶ | |
| Insulators | Glass | 10¹⁰ – 10¹⁴ |
| Hard rubber | 10¹³ – 10¹⁶ | |
| Ebonite | 10¹⁵ – 10¹⁷ | |
| Diamond | 10¹² – 10¹³ | |
| Paper (dry) | 10¹² |
NCERT notes that the values need not be memorised. What matters is the order: silver, then copper, then aluminium are the best conductors; tungsten is several times more resistive than copper; an alloy is generally far more resistive than the metals it is made of.
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27, Table 11.2 (p. 179).
Table 4: Series and parallel (11.6)
| Feature | Series | Parallel |
|---|---|---|
| Current | The same in every part of the circuit (Activity 11.4) | Divides among the branches: I = I1 + I2 + I3 (Eq. 11.15) |
| Potential difference | Shared: V = V1 + V2 + V3 (Eq. 11.11) | The same across each branch |
| Equivalent resistance | Rs = R1 + R2 + R3 (Eq. 11.14), greater than any single resistance | 1/Rp = 1/R1 + 1/R2 + 1/R3 (Eq. 11.18); the total is decreased |
| If one component fails | The circuit is broken and none of the components works | The other branches stay connected to the source |
| Practical use | Impracticable for a bulb and a heater together, which need widely different currents | Each gadget can draw the different current it needs |
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27, sections 11.6.1-11.6.2 and Activities 11.4-11.6 (pp. 181-187).
Table 5: NCERT's worked examples (Examples 11.1-11.13)
| Example | Data | Result |
|---|---|---|
| 11.1 | 0.5 A for 10 minutes | Q = It = 0.5 × 600 = 300 C |
| 11.2 | 2 C moved across 12 V | W = VQ = 24 J |
| 11.3 | 220 V across a 1200 Ω bulb filament; across a 100 Ω heater coil | 0.18 A; 2.2 A |
| 11.4 | A heater draws 4 A at 60 V; current at 120 V? | R = 15 Ω; 8 A |
| 11.5 | Wire 1 m long, 26 Ω at 20 °C, diameter 0.3 mm | ρ = 1.84 × 10⁻⁶ Ω m: manganese, from Table 11.2 |
| 11.6 | Wire of 4 Ω; another of the same material with length l/2 and area 2A | 1 Ω |
| 11.7 | 20 Ω lamp and 4 Ω conductor in series on 6 V | 24 Ω; 0.25 A; 5 V across the lamp, 1 V across the conductor |
| 11.8 | 5 Ω, 10 Ω and 30 Ω in parallel on 12 V | 2.4 A, 1.2 A, 0.4 A; total 4 A; Rp = 3 Ω |
| 11.9 | 10 Ω and 40 Ω in parallel, in series with 30 Ω, 20 Ω and 60 Ω in parallel, on 12 V | R′ = 8 Ω, R″ = 10 Ω; total 18 Ω; 0.67 A |
| 11.10 | Electric iron at 840 W (maximum) and 360 W (minimum) on 220 V | 3.82 A and 57.60 Ω; 1.64 A and 134.15 Ω |
| 11.11 | 100 J of heat each second in a 4 Ω resistor | 5 A; 20 V |
| 11.12 | Bulb on 220 V drawing 0.50 A | 110 W |
| 11.13 | 400 W refrigerator, 8 hours a day for 30 days, at Rs 3.00 per kW h | 96 kW h; Rs 288 |
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27, Examples 11.1-11.13 (pp. 172-192).
Table 6: The heating effect and its uses (11.7)
| Point | What NCERT says |
|---|---|
| Where the source's energy goes | Partly into useful work (turning a fan's blades), the rest into heat. In a purely resistive circuit it is dissipated entirely as heat: the heating effect of electric current |
| Power and heat | P = VI (Eq. 11.19); H = VIt (Eq. 11.20); H = I²Rt (Eq. 11.21) |
| Joule's law of heating | Heat is directly proportional to (i) the square of the current, (ii) the resistance, (iii) the time |
| A loss as well as a use | Heating is "an inevitable consequence of electric current"; it can raise the temperature of components and alter their properties |
| Appliances | Electric laundry iron, toaster, oven, kettle and heater |
| Filament bulb | Tungsten (melting point 3380 °C), thermally isolated by insulating supports; bulbs filled with nitrogen and argon to prolong the filament's life; most of the power appears as heat and a small part as light |
| Fuse | Placed in series with the device; a wire of a metal or alloy of appropriate melting point (for example aluminium, copper, iron, lead) that melts and breaks the circuit when the current exceeds the specified value; usually in a porcelain cartridge with metal ends |
| Fuse ratings | Domestic fuses are rated 1 A, 2 A, 3 A, 5 A, 10 A, etc. A 1 kW iron on 220 V draws 4.54 A, so a 5 A fuse must be used |
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27, sections 11.7-11.7.1 (pp. 188-190).
Table 7: Electric power and energy (11.8)
| Quantity | Relation or value |
|---|---|
| Electric power | The rate of consumption of energy: P = VI = I²R = V²/R (Eq. 11.22) |
| Watt | The power used by a device carrying 1 A at 1 V: 1 W = 1 V A (Eq. 11.23); 1 kilowatt = 1000 W |
| Energy | Power × time; unit watt hour (W h) |
| Commercial unit | Kilowatt hour (kW h), commonly called a unit: 1 kW h = 3.6 × 10⁶ J |
| What we pay for | Energy. Electrons are not consumed in a circuit |
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27, section 11.8 and its More to Know box (p. 191).
Table 8: The chapter exercises and NCERT's answers
| Exercise | Asked | Answer |
|---|---|---|
| 1 | Wire of resistance R cut into five equal parts joined in parallel; R/R′ | (d) 25 |
| 2 | Term that is not electrical power | (b) IR² |
| 3 | 220 V, 100 W bulb run on 110 V | (d) 25 W |
| 4 | Heat in series against parallel, two identical wires, same voltage | (c) 1:4 |
| 5 | How a voltmeter is connected | In parallel |
| 6 | Copper wire, diameter 0.5 mm, ρ = 1.6 × 10⁻⁸ Ω m: length for 10 Ω; effect of doubling the diameter | 122.7 m; resistance becomes ¼ |
| 7 | Resistance from a V–I table | 3.33 Ω |
| 8 | 12 V gives 2.5 mA | 4.8 kΩ |
| 9 | 9 V across 0.2, 0.3, 0.4, 0.5 and 12 Ω in series | 0.67 A |
| 10 | 176 Ω resistors in parallel to carry 5 A on 220 V | 4 |
| 11 | Three 6 Ω resistors to give 9 Ω and 4 Ω | Not in the key. Two in parallel (3 Ω) in series with the third gives 9 Ω; two in series (12 Ω) in parallel with the third gives 4 Ω |
| 12 | 10 W lamps on a 220 V line with a 5 A limit | 110 |
| 13 | Two 24 Ω coils on 220 V: separately, in series, in parallel | 9.2 A, 4.6 A, 18.3 A |
| 14 | Power in a 2 Ω resistor in two circuits | 8 W in each |
| 15 | 100 W and 60 W lamps in parallel on 220 V | 0.73 A |
| 16 | 250 W TV for 1 hour or 1200 W toaster for 10 minutes | The TV (250 W h against 200 W h) |
| 17 | 44 Ω heater drawing 5 A for 2 hours | 1100 W |
| 18 | Why tungsten, why alloys, why not series at home, area of cross-section, why copper and aluminium | The key prints only (b) high resistivity of alloys and (d) inversely; see the explainer below |
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27, Exercises 1-18 (pp. 193-194); NCERT Answers (p. 219). The working for Exercises 11 and 16 is arithmetic.
Table 9: NCERT lines to read with care
| Where | What the book prints | How to read it |
|---|---|---|
| Section 11.5 (p. 179) | Metals and alloys have resistivity in the range 10⁻⁸ to 10⁻⁶ Ω m | Table 11.2 on the same page lists the alloys at 44 × 10⁻⁶ to 100 × 10⁻⁶ Ω m, above that range. The range fits the pure metals; for alloys, take the table's point that they are far more resistive than their constituent metals |
| Section 11.5 (p. 179) | Rubber and glass have resistivity of the order of 10¹² to 10¹⁷ Ω m | Table 11.2 lists glass from 10¹⁰ Ω m. Read both as orders of magnitude |
| Section 11.4 (p. 176) | "In Eq. (11.4), R is a constant" | R first appears in Eq. (11.5), V = IR; Eq. (11.4) is V ∝ I |
| Example 11.3 (p. 179) | Part (a) works "From Eq. (12.6)" | A leftover of the old chapter number. The relation is R = V/I, Eq. (11.6), used as I = V/R (Eq. 11.7) |
| Example 11.10 (p. 189) | 57.60 Ω and 134.15 Ω | The book divides 220 V by currents already rounded to 3.82 A and 1.64 A. R = V²/P gives about 57.6 Ω and 134.4 Ω (arithmetic) |
| Section 11.8 (p. 191) | Equation 11.21 "gives the rate at which electric energy is dissipated" | Eq. (11.21), H = I²Rt, is the heat produced in time t. The rate is H/t = I²R, part of Eq. (11.22) |
| Section 11.2 (p. 172) | One coulomb is the charge of nearly 6 × 10¹⁸ electrons | With the book's own 1.6 × 10⁻¹⁹ C per electron, 1/(1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ (arithmetic) |
Source: NCERT, Science Class X, ch. 11, Reprint 2026-27 (pp. 172-191).
PART 2 — Concepts & Narrative
Current and potential difference (11.1-11.3)
A circuit is a continuous, closed path; break it anywhere and the current stops. In metal wires the moving charges are electrons, but current was defined before electrons were known, so it is still taken to flow the other way. What drives it is a difference of electric pressure: NCERT compares it with water, which flows through a tube only when there is a pressure difference between its ends. A cell's chemical action sets up this potential difference, and to maintain the current the cell has to expend its stored chemical energy. One volt means one joule of work for each coulomb moved.
Ohm's law (11.4)
In Activity 11.1 the potential difference across a nichrome wire is raised in steps, from one cell to four, and V/I comes out nearly the same each time. The V–I graph is a straight line through the origin. That constant ratio is the resistance.
Ohm's law has a condition. NCERT states it for a metallic wire "provided its temperature remains the same". Resistance changes with temperature, so a conductor that heats up as the current rises, such as a bulb filament, does not keep a constant V/I. In the chapter's numericals the resistance is held fixed, which is why a bulb's rating can be used to find its resistance (Exercise 3).
What resistance depends on (11.5)
Electrons in a conductor are not completely free: the attraction of the atoms among which they move holds them back, and this is resistance. Activity 11.3 isolates three factors. Doubling the length halves the current; a thicker wire of the same length passes more; a wire of another material with the same length and area gives a different reading. Hence R = ρl/A, where the resistivity ρ belongs to the material, not to the shape of the wire. Example 11.5 runs the formula backwards: from a wire's resistance, length and diameter it finds ρ = 1.84 × 10⁻⁶ Ω m and identifies the metal as manganese from Table 11.2. Example 11.6 shows the two geometric factors together: halving the length and doubling the area cuts a 4 Ω wire to 1 Ω.
Exercise 18 in one place: choosing a material. Heating elements use alloys because an alloy is generally more resistive than its constituent metals and does not oxidise (burn) readily at high temperatures. Bulb filaments use tungsten because the filament must get hot enough to glow without melting, and tungsten is strong with a melting point of 3380 °C. Transmission lines use copper and aluminium because, after silver, they have the lowest resistivities in Table 11.2. Homes are not wired in series because one failure breaks the whole circuit and appliances need different currents. Resistance falls as the area of cross-section rises (inversely proportional).
Series and parallel (11.6)
In series, one current flows through everything, so the potential differences add up to the source voltage and the resistances simply add. Example 11.7 shows it: a 20 Ω lamp and a 4 Ω conductor on 6 V give 24 Ω and 0.25 A, and the 6 V splits as 5 V and 1 V. In parallel, every branch has the full source voltage, so the currents add, and the combined resistance is less than before: in Example 11.8, 5 Ω, 10 Ω and 30 Ω together make only 3 Ω.
NCERT's case against series wiring at home is practical. A bulb and a heater need widely different currents (Example 11.3 gives 0.18 A against 2.2 A on the same 220 V), which one series current cannot supply. And one failed component breaks the whole circuit, which is why an electrician spends so long finding the dead bulb in a string of fairy lights.
Solving a combination circuit. Replace each parallel group by its equivalent, then add the series parts. In Example 11.9, 10 Ω and 40 Ω in parallel give 1/R′ = 1/10 + 1/40 = 5/40, so R′ = 8 Ω; 30 Ω, 20 Ω and 60 Ω in parallel give R″ = 10 Ω; in series these make 18 Ω, and 12 V drives 0.67 A. Two checks catch most slips: a series total must exceed the largest resistor, and a parallel total must be smaller than the smallest (both follow from Eqs. 11.14 and 11.18).
The heating effect of current (11.7)
The source keeps spending energy to maintain a current. In a fan, part of it turns the blades and the rest warms the motor; in a purely resistive circuit all of it becomes heat. Since the source does work VQ in time t, the power is VI and the heat is H = VIt; with Ohm's law this becomes H = I²Rt. Example 11.11 uses it in reverse: 100 J of heat each second in a 4 Ω resistor means a current of 5 A and a potential difference of 20 V.
Joule's law: three proportionalities. Heat produced in a resistor is directly proportional to the square of the current (for a given resistance), to the resistance (for a given current) and to the time. Doubling the current quadruples the heat. The square is why a fuse wire, which carries the full current, heats up sharply when the current rises above its rating.
The same heating is a nuisance elsewhere: it converts useful electrical energy into heat and can alter the properties of components. Exercise 4 shows how the arrangement matters. Two identical wires across the same voltage produce four times as much heat in parallel as in series, since the parallel pair has a quarter of the resistance and P = V²/R.
Electric power (11.8)
Power is the rate at which energy is used. It can be written three ways, P = VI = I²R = V²/R, and choosing the right one is most of a numerical: I²R when the current is known (a series resistor), V²/R when the voltage is fixed (an appliance on the mains). Energy is power multiplied by time, so a 400 W refrigerator running 8 hours a day for 30 days uses 96 kW h, and at Rs 3.00 per unit costs Rs 288 (Example 11.13).
The 110 V trap (Exercise 3). A 220 V, 100 W bulb has a resistance R = V²/P = 220²/100 = 484 Ω. On 110 V, with the resistance taken as fixed, P = 110²/484 = 25 W. Halving the voltage halves the current too, so the power falls to a quarter, not a half. The same reasoning, with P = V²/R, answers Exercise 15 and the domestic-load questions.
NCERT's More to Know box closes the chapter on a misconception: "Many people think that electrons are consumed in an electric circuit. This is wrong!" The electricity company is paid for the energy that moves electrons through bulbs, fans and engines.
Did the 2020-21 edition differ?
Only slightly. The chapter was Chapter 12 (pp. 199-222) in the 2020-21 edition and is Chapter 11 (pp. 171-194) now. One More to Know box has gone: "'Flow' of charges inside a wire" (p. 201), kept below. Exercise 17 has changed. The old heater had a resistance of 8 Ω and drew 15 A, and the old Answers page printed 120 W, which does not match the data (15² × 8 = 1,800 W). The current book uses 44 Ω and 5 A, and its key of 1100 W does match. The rest is kept, including Table 11.2 with the same values, the worked examples and the four MCQs.
Source: NCERT, Science Class X, 2020-21 edition (whole-book zip, Wayback Machine capture of 9 October 2021), compared with Reprint 2026-27.
Retained content: 'Flow' of charges inside a wire (2020-21 edition, ch. 12, p. 201; not in Reprint 2026-27)
The box asked how a metal conducts at all. Inside a solid the atoms are packed with very little space between them, yet electrons are able to travel through a perfect solid crystal "smoothly and easily, almost as if they were in a vacuum". In a steady current they move with an average drift speed, and for a typical copper wire carrying a small current this speed is very small, of the order of 1 mm per second. Why, then, does a bulb light the moment the switch is pressed? Not because an electron has to travel from the supply to the bulb: the physical drift is very slow, while the current flow "takes place with a speed close to the speed of light". The box left the mechanism "beyond the scope of this book".
Source: NCERT, Science Class X, 2020-21 edition, ch. 12, More to Know box "'Flow' of charges inside a wire" (p. 201).
Beyond the textbook: why power travels at high voltage, and India's capacity mix
- The arithmetic of transmission. A line delivering power P at voltage V carries a current I = P/V, and its own resistance R wastes I²R as heat. Raising the voltage ten times cuts the current to a tenth and the loss to a hundredth, for the same power over the same line. This is arithmetic on NCERT's Eqs. (11.21) and (11.22), and it is the reason long-distance lines run at high voltage.
- Installed capacity, 31 August 2026 (CEA). Total 5,54,544 MW. Non-fossil 3,04,334 MW (54.88%): solar 1,68,040 MW (30.30%), wind 58,520 MW, hydro including pumped storage 52,065 MW, nuclear 8,780 MW, with smaller bio-power and small-hydro shares. Fossil 2,50,210 MW (45.12%), of which coal 2,24,408 MW (40.47%). These are capacities, not shares of electricity generated.
- The non-fossil target. India's updated Nationally Determined Contribution, approved by the Union Cabinet and announced on 3 August 2022, aims for about 50 percent cumulative electric power installed capacity from non-fossil fuel-based energy resources by 2030. On 14 July 2025 the Ministry of New and Renewable Energy said India had reached 50% of installed capacity from non-fossil sources, five years ahead of that target, and restated the goal of 500 GW of non-fossil capacity by 2030.
- The next NDC (2031-35). On 25 March 2026 the Union Cabinet approved India's NDC for 2031-35. It raises the capacity aim to "60 percent cumulative electric power installed capacity from non-fossil fuel-based energy resources by 2035", alongside a 47 percent cut in the emissions intensity of GDP by 2035 from the 2005 level. The same release put the non-fossil share at 52.57% in February 2026 (PIB).
- A high-voltage example. On 18 October 2023 the Cabinet Committee on Economic Affairs approved Green Energy Corridor Phase-II transmission for a 13 GW renewable energy project in Ladakh: 713 km of lines including a 480 km HVDC line, with 5 GW HVDC terminals at Pang (Ladakh) and Kaithal (Haryana), at a total estimated cost of Rs 20,773.70 crore with central assistance of 40% (Rs 8,309.48 crore). It is targeted to be set up by FY 2029-30; it was an approval, not a completed line.
Source: Central Electricity Authority, "Installed Capacity (in MW) of the country as on 31.08.2026"; PIB releases of 3 August 2022 (Ministry of Environment, Forest and Climate Change), 18 October 2023 (Cabinet Committee on Economic Affairs), 14 July 2025 (Ministry of New and Renewable Energy) and 25 March 2026 (Cabinet).
PART 3 — UPSC Integration
Cross-paper relevance
- Prelims (general science) — Ammeter in series and voltmeter in parallel; Ohm's law and its temperature condition; what resistance and resistivity depend on; series and parallel arithmetic; why alloys, tungsten, copper and aluminium; the fuse; P = V²/R traps; 1 kW h = 3.6 × 10⁶ J.
- GS3 (energy and infrastructure) — Why transmission runs at high voltage; line losses as I²R; installed capacity mix and the non-fossil target; energy efficiency of lighting and appliances.
- GS3 (science and technology) — Joule heating in appliances and safety devices; materials chosen for their resistivity.
Past questions on these themes: No past question in the bank is set directly on this chapter.
Frames for Mains Answers
1. Why electricity is transmitted at high voltage. Start from the physics: for a given power P = VI, a higher voltage means a smaller current, and the heat lost in the line is I²R, so ten times the voltage gives a hundredth of the loss over the same line. Then the application: long-distance corridors that carry renewable power from where it is generated to where it is used, such as the Ladakh HVDC project approved in October 2023 with a target of FY 2029-30.
2. Efficiency of lighting. NCERT's own description of the filament bulb is the argument: "Most of the power consumed by the filament appears as heat, but a small part of it is in the form of light radiated." Any lamp that turns more of its power into light gives the same light for less energy, and the saving is counted in kilowatt hours.
3. India's clean-capacity share. Non-fossil sources made up 54.88% of installed capacity on 31 August 2026 (CEA), against the updated NDC's aim of about 50% by 2030; the NDC for 2031-35, approved on 25 March 2026, raises the aim to 60% by 2035. Add the caveat that installed capacity is not generation: a share of capacity is not a share of the electricity actually produced.
Exam Strategy
Prelims fact-traps:
- Ammeter in series, voltmeter in parallel; a fuse in series with the device.
- Ohm's law holds only while the temperature stays the same.
- Resistance depends on length, area and material; resistivity on the material (and temperature) only.
- Series: same current, voltages add. Parallel: same voltage, currents add. Homes are wired in parallel.
- Silver is the best conductor in Table 11.2, then copper, then aluminium; alloys are far more resistive than their metals.
- A bulb on half its rated voltage gives a quarter of its rated power.
- The kilowatt hour is a unit of energy, not of power: 1 kW h = 3.6 × 10⁶ J.
Mains: Use the chapter for the science line of a GS3 energy answer: one relation (P = VI, loss I²R), one dated capacity figure, then the policy point.
Practice Questions
Questions 1-4 are the NCERT exercise MCQs. Practice (UPSC-pattern, not past papers): questions 5-10.
1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Answer: (d) (NCERT Answers, p. 219). Each part has R/5, and five such parts in parallel give R/25.
2. Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer: (b) (NCERT Answers, p. 219).
3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer: (d) (NCERT Answers, p. 219).
4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer: (c) (NCERT Answers, p. 219).
5. Consider the following statements:
1. Doubling the length of a wire doubles its resistivity.
2. The resistivity of a material varies with temperature.
3. In NCERT's Table 11.2, nichrome is more resistive than nickel and chromium.
Which of the statements given above is/are correct?
(a) 1 only
(b) 1 and 2 only
(c) 2 and 3 only
(d) 1, 2 and 3
Answer: (c). Doubling the length doubles the resistance; resistivity is a property of the material (pp. 178-179).
6. Consider the following statements about a domestic fuse:
1. It is connected in parallel with the appliance it protects.
2. It works through the heating effect of current.
3. For a 1 kW electric iron on 220 V, a 5 A fuse is appropriate.
Which of the statements given above is/are correct?
(a) 1 and 2 only
(b) 2 and 3 only
(c) 3 only
(d) 1, 2 and 3
Answer: (b). The fuse is placed in series with the device; the iron draws 4.54 A (p. 190).
7. Household appliances are connected in parallel mainly because
(a) all of them then carry the same current
(b) each gets the full supply voltage and can draw the current it needs
(c) the total resistance of the circuit increases
(d) the total power consumed falls
Answer: (b). In parallel the total resistance decreases, and one failed appliance does not break the others' circuit (p. 187).
8. Consider the following statements about an electric filament bulb:
1. Tungsten is used for the filament because of its high melting point.
2. The bulb is filled with nitrogen and argon to prolong the life of the filament.
3. Most of the power it consumes is emitted as light.
Which of the statements given above is/are correct?
(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3
Answer: (a). Most of the power appears as heat and only a small part as light (p. 190).
9. A 1.5 kW heater is used for 2 hours a day for 30 days. The electrical energy it consumes is
(a) 9 kW h
(b) 90 kW h
(c) 900 kW h
(d) 3,000 kW h
Answer: (b). Energy = 1.5 kW × 2 h × 30 = 90 kW h (arithmetic, on the method of Example 11.13).
10. A 4 Ω and a 12 Ω resistor are connected in parallel, and the pair is connected in series with a 3 Ω resistor across a 12 V battery. The current drawn from the battery is
(a) 0.6 A
(b) 1 A
(c) 2 A
(d) 4 A
Answer: (c). The parallel pair gives 1/Rp = 1/4 + 1/12 = 4/12, so Rp = 3 Ω; the total is 6 Ω and I = 12/6 = 2 A (arithmetic, on the method of Example 11.9).
📦 Revision Capsule
Hard Facts
- I = Q/t; V = W/Q; V = IR; R = ρl/A; Rs = R1 + R2 + R3; 1/Rp = 1/R1 + 1/R2 + 1/R3.
- H = I²Rt (Joule's law); P = VI = I²R = V²/R; 1 W = 1 V A; 1 kW h = 3.6 × 10⁶ J.
- Electron charge 1.6 × 10⁻¹⁹ C; one coulomb is nearly 6 × 10¹⁸ electrons (6.25 × 10¹⁸ by arithmetic).
- Ampere (French, 1775–1836); Volta (Italian, 1745–1827); Ohm (German, 1787–1854), law found in 1827.
- Tungsten melts at 3380 °C; bulbs filled with nitrogen and argon.
- Fuse ratings 1, 2, 3, 5, 10 A; a 1 kW iron on 220 V draws 4.54 A, so a 5 A fuse.
Core Concepts
- Current direction is opposite to electron flow by convention.
- Ohm's law needs a constant temperature.
- Resistivity is the material's; resistance is the wire's.
- Series: one current. Parallel: one voltage.
- The square in I²R drives both the fuse and the case for high-voltage transmission.
Confused Pairs
- Resistance (Ω; depends on shape) vs resistivity (Ω m; material).
- Ammeter (series) vs voltmeter (parallel).
- Power (W) vs energy (W h, kW h).
- Series (resistance adds, one failure breaks all) vs parallel (resistance falls, branches independent).
- Alloys in heaters (high resistivity, no ready oxidation) vs copper and aluminium in lines (low resistivity).
Data Points
- Resistivity at 20 °C: silver 1.60, copper 1.62, aluminium 2.63, tungsten 5.20 (all × 10⁻⁸ Ω m); nichrome 100 × 10⁻⁶ Ω m.
- CEA, 31 August 2026: 5,54,544 MW installed; non-fossil 54.88%; solar 1,68,040 MW; coal 40.47%.
- NDC (announced 3 August 2022): about 50% non-fossil installed capacity by 2030; reached in July 2025 (MNRE, 14 July 2025).
- NDC 2031-35 (Cabinet, 25 March 2026): 60% non-fossil installed capacity and a 47% cut in emissions intensity of GDP (from 2005), both by 2035.
- Ladakh GEC Phase-II: approved 18 October 2023; 480 km HVDC line; 5 GW terminals at Pang and Kaithal; target FY 2029-30.
PYQ Pattern
- No past question in the bank matches this chapter directly.
Sources
- NCERT, Science, Textbook for Class X, ch. 11 "Electricity", Reprint 2026-27 — ncert.nic.in PDF.
- NCERT, Science, Class X, Answers, Reprint 2026-27 — ncert.nic.in PDF.
- NCERT, Science, Class X, 2020-21 edition (whole-book zip), as archived on 9 October 2021, ch. 12 "Electricity" — Wayback Machine.
- Central Electricity Authority, "Installed Capacity (in MW) of the country as on 31.08.2026" — cea.nic.in PDF.
- PIB (Ministry of Environment, Forest and Climate Change), release on India's updated Nationally Determined Contribution, 3 August 2022 — pib.gov.in.
- PIB (Cabinet), Cabinet approves India's Nationally Determined Contribution (2031-2035), 25 March 2026 — pib.gov.in.
- PIB (Cabinet Committee on Economic Affairs), "Cabinet approves Green Energy Corridor (GEC) Phase-II – Inter-State Transmission System (ISTS) for 13 GW Renewable Energy Project in Ladakh", 18 October 2023 — pib.gov.in.
- PIB (Ministry of New and Renewable Energy), release on reaching 50% non-fossil installed capacity, 14 July 2025 — pib.gov.in.
