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Every NCERT book that matters for UPSC — open a book, then a chapter. Learn each chapter's concepts, then how UPSC tests them — with diagrams, flowcharts and the past questions to attempt.
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← Science — all chaptersElectricity
Start here: why this chapter matters
In simple wordsThink of a circuit as a loop of pipe full of water with a pump in it. The pump (the cell) gives each drop a push (voltage); the water moving past any point each second is the current; narrow or rough stretches of pipe (resistors) hold the flow back and get warm. This chapter gives you the rules for that loop — how much flows, how it splits, how much heat it makes and what it costs on your electricity bill.
This is Chapter 11 of NCERT's Class 10 Science textbook (pages 171–194 of the 2026-27 reprint; the contents page and the PDF agree). It has eight parts — electric current and circuit (11.1), potential difference (11.2), circuit diagrams (11.3), Ohm's law (11.4), factors affecting resistance (11.5), series and parallel resistors (11.6), the heating effect (11.7) and electric power (11.8) — with 6 activities, Figures 11.1–11.13, Tables 11.1–11.2, 13 worked examples, a ‘More to Know!’ box, seven ‘Questions’ boxes and 18 exercise questions. It is the most numerical chapter in the book: every number on this page is computed and asserted in the page's source, and every circuit is drawn with the book's standard symbols.
- Neighbouring chapters: Magnetic Effects of Electric Current (Chapter 12 of this book) continues from here — what a current does to a compass and how motors work. Metals and Non-metals (Chapter 3) explains why metals conduct; Light – Reflection and Refraction (Chapter 9) set the physics conventions used on this page.
- Prelims — science: UPSC has asked about this chapter's everyday applications again and again — the nichrome heating element (1980), the job of a fuse (1981, 1998), kilowatt as a unit of power (1982), comparing the resistance of a 100 W and a 60 W bulb (1987), why power travels at high voltage (1987), domestic wiring in parallel (1996) and why a current heats a wire (1998).
- GS3 — science, technology and energy: electricity units and billing, energy efficiency (filament lamps waste most of their power as heat), safety devices (fuses, circuit breakers), and transmission losses in the power sector.
| Word | What it means here |
|---|---|
| Electric current (I) | Rate of flow of charge, Q/t; unit ampere (A) |
| Potential difference (V) | Work done per unit charge between two points, W/Q; unit volt (V) |
| Resistance (R) | A conductor's opposition to current, V/I; unit ohm (Ω) |
| Resistivity (ρ) | Resistance property of the material itself, from R = ρl/A; unit Ω m |
| Series / parallel | End to end, one path / side by side between the same two points |
| Equivalent resistance | The single resistor that would draw the same current from the same source |
| Joule heating | Heat produced by a current in a resistor, H = I²Rt |
| Power / kW h | Rate of using energy (watt) / the commercial unit of energy (‘unit’) |
Reading NCERT's chapter opening (page 171)The opener shows a small circuit sketch — a battery, an ammeter, two resistors in parallel and three more in parallel, the network that later becomes Figure 11.12 — and a QR code labelled 1064CH12 (from when this was Chapter 12). The text asks what electricity is, how it flows and what controls it, and promises the heating effect. The point: the chapter builds from what a current is to what it does and costs.
Common confusions
- Current vs voltage: current is the flow (amperes); voltage is the push per coulomb (volts). A cell supplies a potential difference, not ‘current’ stored inside it.
- Resistance vs resistivity: resistance belongs to a particular wire (Ω); resistivity belongs to the material (Ω m).
- Power vs energy: watt is a rate; kilowatt hour is an amount. The bill counts kW h.
- Magnetic effects, motors, generators and domestic wiring details are in the next chapter, not this one.
At a glance — the six ideas to carry away
- Current I = Q/t; 1 A = 1 C/s; 1 C ≈ 6.25 × 10¹⁸ electrons (book: ‘nearly 6 × 10¹⁸’). Conventional current flows from + to − outside the cell — opposite to electron flow. Ammeter in series.
- Potential difference V = W/Q; 1 V = 1 J/C; kept up by a cell's chemical action. Voltmeter in parallel. Table 11.1 gives the standard circuit symbols.
- Ohm's law: V ∝ I at constant temperature → V = IR; 1 Ω = 1 V/1 A. R = ρl/A: doubles with length, halves when area doubles; resistivity ρ (Ω m) is a property of the material. Rheostat = variable resistance.
- Series: Rₛ = R₁ + R₂ + R₃; same current, voltages add; one break stops all. Parallel: 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃; same voltage, currents add; Rₚ is below the smallest resistor — domestic wiring is parallel.
- Heating: H = VIt = I²Rt (Joule's law). Alloys (nichrome) for heating elements — high resistivity, don't oxidise; tungsten (m.p. 3380 °C, book) for filaments, bulbs filled with nitrogen and argon; fuse in series melts on overload (1, 2, 3, 5, 10 A).
- Power P = VI = I²R = V²/R, in watts; 1 W = 1 V × 1 A. Energy = power × time; commercial unit kW h = ‘unit’ = 3.6 × 10⁶ J. We pay for energy, not for electrons.
1. Electric current and the circuit
In simple wordsStand by a road and count the cars passing each minute — that is traffic ‘current’. Electric current counts charge passing a point each second. In a metal wire the moving charges are electrons. The flow needs an unbroken loop: a switch simply closes or opens a gap in that loop.
When charge flows through a conductor such as a metal wire, there is an electric current in it. In a torch, the cells drive charge through the bulb, but only when the switch closes the gap: a continuous, closed path for current is an electric circuit, and if it is broken anywhere the current stops. Current is the rate of flow of charge: if a net charge Q crosses a cross-section in time t, I = Q/t (eq. 11.1). The SI unit of charge is the coulomb (C); an electron carries 1.6 × 10−19 C. The unit of current is the ampere (A), after André-Marie Ampère (1775–1836): 1 A = 1 C per second; small currents are in milliamperes (10−3 A) and microamperes (10−6 A). In metal wires the charges that move are electrons, but current was defined before electrons were known, as a flow of positive charge — so the conventional direction of current is opposite to the electrons' motion, from the cell's positive terminal round the circuit to its negative terminal. An ammeter measures current and is always connected in series.
⤵ What a torch switch does
Switch offa gap in the loop: no closed path, no current, bulb dark
Switch onthe gap is bridged: the circuit is closed
Cell's potential difference actsit sets the electrons in the whole loop drifting
Current through the bulbthe same current everywhere in the single loop
Filament heats and glowselectrical energy becomes heat and light
Reading NCERT's Figure 11.1 (page 172)A rectangular circuit with a cell at the bottom left, a plug key at the bottom, a bulb on the top wire and an ammeter on the right-hand side, with arrows and the letter I showing current going round from the cell's + terminal. The point: the ammeter sits in the loop (series), and current flows from + to − through the external circuit.
Reading NCERT's Example 11.1 and ‘Questions’ box (page 172)Example 11.1: a bulb filament draws 0.5 A for 10 minutes; Q = It = 0.5 A × 600 s = 300 C (the time must be in seconds). The box then asks what a circuit is, to define the ampere, and how many electrons make one coulomb. The point: Q = It with SI units.
Common confusions
- Conventional current is opposite to electron flow — electrons leave the cell's negative terminal.
- Ammeter in series, voltmeter in parallel — never the other way round.
- Convert minutes to seconds before Q = It (Example 11.1 uses 600 s).
- 1 C is about 6.25 × 1018 electrons; the book's ‘nearly 6 × 1018’ is a rounding — either is accepted, but 1 ÷ (1.6 × 10−19) = 6.25 × 1018 exactly.
- Current is the same at every point of a single loop; it is not ‘used up’ by the bulb (see the ‘More to Know!’ box).
2. Potential difference and the volt
In simple wordsWater in a perfectly level pipe does not flow; lift one end to a tank and it does. Charges in a wire are the same: they move only if there is a difference of ‘electric pressure’ between the ends. A cell is like a pump that keeps one end ‘higher’. Voltage tells you how much energy each coulomb gets for the trip.
Charges do not flow in a copper wire by themselves. They move only if there is a difference of electric pressure — a potential difference — along the conductor. A battery of one or more cells provides it: the chemical action inside a cell creates a potential difference across its terminals even when no current is drawn, and the cell spends its stored chemical energy to keep a current going. The potential difference between two points is the work done to move a unit charge from one to the other: V = W/Q (eq. 11.2). Its unit is the volt (V), after Alessandro Volta (1745–1827): 1 V = 1 J C−1, the potential difference when 1 J of work moves 1 C. It is measured with a voltmeter connected in parallel across the two points.
| Water picture | Electric circuit | Measured by |
|---|---|---|
| Height difference / pressure difference | Potential difference (V = W/Q) | Voltmeter, in parallel |
| Pump lifting water | Cell or battery (chemical energy) | — |
| Water flowing past a point per second | Current (I = Q/t) | Ammeter, in series |
| Narrow pipe resisting the flow | Resistor (R = V/I) | V and I together |
Reading NCERT's Example 11.2 and ‘Questions’ box (pages 173–174)Example 11.2: moving 2 C across 12 V needs W = VQ = 24 J. The box asks for a device that maintains a potential difference (a cell or battery), the meaning of 1 V, and the energy given to each coulomb by a 6 V battery (6 J). The point: volts are joules per coulomb.
Common confusions
- Potential difference is work per unit charge (J/C), not force and not current.
- A cell has a potential difference across its terminals even with no current drawn (book).
- The voltmeter goes across (in parallel); connected in series it would block the circuit.
- A 6 V battery gives 6 J to each coulomb, whatever the current.
3. Circuit diagrams and their symbols
In simple wordsA circuit diagram is like a metro map: it does not look like the real wires, but it shows exactly what is connected to what. Every part has a small standard picture so that anyone, anywhere, can read the drawing.
A circuit has a cell or battery, a plug key, components and connecting wires. Drawing them realistically is slow, so we use schematic diagrams with conventional symbols (Table 11.1). The cell's longer line is the positive terminal; a battery is several cells joined + to −; a plug key is open without its plug and closed with it; a dot marks a joint, while wires that cross without a dot (or with a hop) are not connected; a zigzag is a resistor and a zigzag with an arrow a variable resistance (rheostat); circles with A and V are the ammeter and voltmeter, each with its + terminal towards the cell's + side.
Reading NCERT's Table 11.1 (pages 174–175)A green table of eleven components with their symbols: cell, battery, plug key open and closed, wire joint, wires crossing without joining, electric bulb (a filament loop, or a small bulb picture), resistor, variable resistance or rheostat (two alternative drawings), ammeter and voltmeter with + and − terminals. The point: these are the drawing vocabulary for every circuit question in the chapter and the exercises (in-text Q1 on page 185 asks you to draw one).
Common confusions
- In a cell symbol the long thin line is +, the short thick line is −.
- Crossing wires are joined only where a dot is drawn.
- A rheostat changes resistance (and so current) without changing the voltage source.
- A plug key is closed when the plug is in (dot drawn) — that completes the circuit.
4. Ohm's law and resistance
In simple wordsPush twice as hard and twice as much flows — as long as nothing else changes. For a metal wire kept at the same temperature, doubling the voltage doubles the current. The fixed ratio between them is the wire's resistance: how stubbornly it holds the current back.
In Activity 11.1 a nichrome wire XY (about 0.5 m) is connected with an ammeter, a plug key and one, two, three and then four 1.5 V cells, with a voltmeter across XY. Each time V/I comes out about the same, and the V–I graph is a straight line through the origin (Fig. 11.3). In 1827 Georg Simon Ohm (1787–1854) found this rule: the potential difference across a metallic wire is directly proportional to the current, provided its temperature stays the same — Ohm's law: V ∝ I, so V/I = R, or V = IR. R is the wire's resistance, the property of a conductor to resist the flow of charge, measured in ohms (Ω): 1 Ω = 1 V/1 A. Rearranged, I = V/R: double the resistance and the current halves. A component that changes the current without changing the source is a variable resistance; a rheostat does this. In Activity 11.2 a nichrome wire, a torch bulb and a 10 W bulb are placed in turn in a gap XY in a circuit with four cells and an ammeter: the readings differ, because electrons in a conductor are held back by the attraction of its atoms. A low-resistance component is a good conductor; one with appreciable resistance is a resistor; a poor conductor has more, and an insulator of the same size still more.
⤵ Activity 11.1: from readings to Ohm's law
One cellnote I (ammeter) and V across the nichrome wire (voltmeter)
Two, three, four cellsnote I and V again each time
Work out V/Iit comes out about the same every time
Plot V against Ia straight line through the origin
ConclusionV ∝ I at constant temperature: V = IR, with R the wire's resistance
Reading NCERT's Activity 11.1, Figure 11.2 and Figure 11.3 (pages 175–176)Figure 11.2 shows four cells (with dashed taps to use 1–4 of them), the nichrome wire XY marked R, an ammeter and a key in series, and a voltmeter across XY. The activity's blank table has columns for the number of cells, I, V and V/I. Figure 11.3 is a V–I graph — potential difference (0–2 V) against current (0–0.6 A) — with plotted points on a straight line through the origin. The point: a straight V–I line through 0 is Ohm's law. (A printing slip: in the table's header the words ‘(ampere)’ and ‘wire, V (volt)’ have drifted into the wrong columns; the units belong to I and V.)
Reading NCERT's Activity 11.2 and Figure 11.4 (pages 176–177)A circuit of four cells, an ammeter and a key with a gap XY, beside small drawings of the three things to put in the gap: a nichrome wire on a board, a 10 W bulb in a holder and a torch bulb. The point: different components let different currents through from the same source — they have different resistances.
Reading NCERT's Examples 11.3 and 11.4 (pages 179–180)Example 11.3: from 220 V, a 1200 Ω bulb filament draws about 0.18 A but a 100 Ω heater coil draws 2.2 A — note the big difference. Example 11.4: a heater taking 4 A at 60 V has R = 15 Ω, so at 120 V it takes 8 A. The point: I = V/R; with R fixed, current is proportional to voltage. (A leftover from the old numbering: part (a) cites ‘Eq. (12.6)’ — read 11.6, though the current is really worked out with I = V/R, eq. 11.7.)
Common confusions
- Ohm's law holds only at constant temperature — a hot filament's resistance is higher than a cold one's.
- V–I graph: slope = R when V is on the y-axis. (If I is on the y-axis, the slope is 1/R.)
- Resistance does not change when you change the voltage: the current changes (in-text Q3 on page 181: half the voltage → half the current).
- Ohm = volt per ampere; the symbol Ω is the Greek capital omega.
5. What resistance depends on: length, area, material
In simple wordsA long, narrow corridor slows a crowd more than a short, wide one. Wires are the same: a longer wire resists more, a thicker one less. And the ‘floor’ matters too — some materials let electrons through easily (copper, silver), others hardly at all (rubber, glass). That built-in property of the material is resistivity.
In Activity 11.3 a cell, an ammeter and a key are used with four wires in turn: (1) nichrome of length l; (2) nichrome of the same thickness but length 2l; (3) a thicker nichrome wire of length l; (4) copper of the same length and thickness as (1). Doubling the length halves the current; a thicker wire increases it; a different material changes it. So the resistance of a uniform metallic conductor is directly proportional to its length and inversely proportional to its area of cross-section: R = ρl/A (eq. 11.10), where ρ (rho) is the electrical resistivity — a characteristic of the material, in Ω m. Metals and alloys have very low resistivity (the book's range: 10−8 to 10−6 Ω m) and are good conductors; insulators such as rubber and glass are of the order of 1012 to 1017 Ω m. Both resistance and resistivity change with temperature. An alloy's resistivity is generally higher than its constituent metals', and alloys do not oxidise (burn) readily when hot — so they are used in heating devices (electric irons, toasters). Tungsten is used almost exclusively for bulb filaments; copper and aluminium for transmission lines.
| Group | Material (Table 11.2) | Resistivity at 20 °C (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 (alloy of Cu, Ni) | 49 × 10⁻⁶ |
| Manganin (alloy of Cu, Mn, Ni) | 44 × 10⁻⁶ | |
| Nichrome (alloy of Ni, Cr, Mn, Fe) | 100 × 10⁻⁶ | |
| Insulators | Glass | 1010 – 1014 |
| Hard rubber | 1013 – 1016 | |
| Ebonite | 1015 – 1017 | |
| Diamond | 1012 – 1013 | |
| Paper (dry) | 1012 |
Reading NCERT's Activity 11.3 and Figure 11.5 (pages 177–178)Figure 11.5 shows a board holding four wires marked (1) to (4) — a plain nichrome wire, a long folded one, a thick one and a copper one — that can be put in turn into a circuit with a cell, an ammeter and a key. The point: change one thing at a time (length, thickness, material) and watch the current.
Reading NCERT's Table 11.2 (page 179)Resistivity at 20 °C of nine conductors (silver lowest at 1.60 × 10−8 Ω m, up to manganese at 1.84 × 10−6), three alloys (constantan, manganin, nichrome — nichrome highest at 100 × 10−6) and five insulators (glass, hard rubber, ebonite, diamond, dry paper, 1010 to 1017), with a note that you need not memorise the values. The point: silver is the best conductor, alloys resist far more than their metals, insulators are in another universe. (Two inconsistencies with the text: the prose puts metals and alloys at 10−8 to 10−6 Ω m, but the table's alloys are 44–100 × 10−6 Ω m; and it puts insulators at 1012 to 1017, but glass starts at 1010. Go by the table.)
Reading NCERT's Examples 11.5 and 11.6 and ‘Questions’ box (pages 180–181)Example 11.5: a 1 m wire of diameter 0.3 mm with resistance 26 Ω has ρ = Rπd²/4l = 1.84 × 10−6 Ω m — manganese, from the table (recomputed: 1.838 × 10−6). Example 11.6: a wire of 4 Ω made half as long and twice as thick becomes 1 Ω. The box asks what resistance depends on, whether thick or thin wire conducts more easily, what halving V does to I, why toaster and iron coils are alloys, and which of iron and mercury conducts better and which material is best. The point: R = ρl/A in both directions.
Common confusions
- Resistance (Ω) depends on size and material; resistivity (Ω m) depends only on the material (and temperature). Cutting a wire in half halves R but leaves ρ unchanged.
- Area goes as diameter squared: double the diameter → four times the area → a quarter of the resistance (exercise 6).
- Alloys have higher resistivity than their constituent metals — that is why nichrome, not copper, is the heating element (UPSC 1980).
- Silver is the best conductor in the table; copper is used for wiring because it is nearly as good and far cheaper (Beyond NCERT on cost).
- Tungsten is for filaments (high melting point); nichrome for heating elements (high resistivity, no oxidation) — do not swap them.
6. Resistors in series
In simple wordsSeries is a single-file queue through several turnstiles: everyone passes every turnstile, so the same number of people per minute pass each one, and the delays add up. In a series circuit the same current passes through every resistor, their resistances add, and the battery's voltage is shared out among them.
Resistors joined end to end are in series (Fig. 11.6). In Activity 11.4 (three resistors such as 1, 2 and 3 Ω with a 6 V battery) the ammeter reads the same wherever it is placed: in series, the current is the same through every resistor. In Activity 11.5 a voltmeter across the whole combination reads the same as across the battery, and across each resistor in turn reads V₁, V₂, V₃ with V = V₁ + V₂ + V₃ (eq. 11.11). Applying V = IR to each resistor and to the whole, IR = IR₁ + IR₂ + IR₃, so Rs = R₁ + R₂ + R₃ (eq. 11.14): the equivalent resistance is the sum, and so greater than any single resistance. In Example 11.7 a 20 Ω lamp and a 4 Ω conductor on a 6 V battery give Rs = 24 Ω, I = 0.25 A, and 5 V across the lamp, 1 V across the conductor — together 6 V.
Reading NCERT's Figure 11.6, Activity 11.4 and Activity 11.5 with Figure 11.8 (pages 181–183)Figure 11.6: R₁, R₂, R₃ end to end between X and Y, a voltmeter across XY, an ammeter, a battery and a key, with I marked. Figure 11.8 puts a separate voltmeter (V₁, V₂, V₃) across each resistor (XP, QS, TY). The point: one current, shared voltage.
Reading NCERT's Example 11.7, Figure 11.9 and ‘Questions’ box (pages 184–185)Figure 11.9: a 6 V battery, a 4 Ω resistor and a bulb in series with an ammeter and a key. The solution finds 24 Ω, 0.25 A, 5 V and 1 V, and checks that a single 24 Ω resistor would draw the same 0.25 A — the idea of equivalent resistance. The box asks you to draw three 2 V cells with 5, 8 and 12 Ω in series, then add an ammeter and a voltmeter across 12 Ω (answers: 0.24 A and 2.88 V). (A figure slip: the arrow on the bottom wire of Figure 11.9 points towards the ammeter, against the direction shown everywhere else in the loop — current returns to the cell's − terminal along that wire.)
Common confusions
- In series the current is common; the voltage is shared. In parallel it is the other way round.
- Rs is more than the largest resistor; Rp is less than the smallest.
- The biggest resistor in series gets the biggest share of the voltage (V = IR).
- Series is unsuited to homes: one fused bulb in a string of ‘fairy lights’ stops them all, and a bulb and a heater need very different currents (exercise 18c).
7. Resistors in parallel and mixed networks
In simple wordsParallel is a toll plaza with several booths side by side: traffic splits, each booth handles its own share, and opening more booths lets more traffic through overall. Resistors in parallel all feel the same voltage, the current divides among them, and adding another branch lowers the total resistance.
Resistors connected between the same two points X and Y are in parallel (Fig. 11.7). In Activity 11.6 the voltmeter reads the same V across each resistor as across the whole combination, and ammeters show that the total current equals the sum of the branch currents: I = I₁ + I₂ + I₃ (eq. 11.15). With I = V/Rp and I₁ = V/R₁ etc., 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ (eq. 11.18): the reciprocal of the equivalent resistance is the sum of the reciprocals. In Example 11.8, 5, 10 and 30 Ω on 12 V take 2.4, 1.2 and 0.4 A — 4 A in all — and Rp = 3 Ω. Example 11.9 combines both rules: 10 Ω ∥ 40 Ω = 8 Ω, 30 ∥ 20 ∥ 60 Ω = 10 Ω, in series 18 Ω, so 12 V drives about 0.67 A. Series is impractical for a bulb and a heater together (they need very different currents) and one failure breaks the whole circuit; a parallel circuit divides the current, lowers the total resistance, and lets each gadget draw the current it needs.
| Series | Parallel | |
|---|---|---|
| How joined | End to end, one path | Between the same two points, several paths |
| Current | Same through every resistor | Divides: I = I₁ + I₂ + I₃ |
| Voltage | Shared: V = V₁ + V₂ + V₃ | Same across every branch |
| Equivalent resistance | Rs = R₁ + R₂ + R₃ (more than the largest) | 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ (less than the smallest) |
| One component fails | Everything stops | The others keep working |
| Book's examples | Ex. 11.7: 24 Ω, 0.25 A | Ex. 11.8: 3 Ω, 4 A |
| Used for | Fairy-light strings, a fuse with its appliance | Domestic wiring (UPSC 1996) |
⤵ Reducing a mixed network (Example 11.9)
Find groups in parallelR1 with R2; R3 with R4 and R5
Replace each group1/R' = 1/10 + 1/40 gives R' = 8 ohm; 1/R'' = 1/30 + 1/20 + 1/60 gives R'' = 10 ohm
Now they are in seriesR = 8 + 10 = 18 ohm
Apply Ohm's law to the wholeI = 12 V / 18 ohm = 0.67 A
Work back if askedvoltage across each group = I x its equivalent resistance
Reading NCERT's Figure 11.7, Activity 11.6, Figures 11.10 and 11.11 (pages 182–186)Figure 11.7: three resistors fanning out from X and rejoining at Y, with a voltmeter across XY and the battery, key and ammeter in the main line. Figure 11.10 labels the branch ends L–M, P–Q and S–T. Figure 11.11 moves the ammeter into one branch (with I₁, I₂, I₃ marked on the three branches) and the voltmeter across R₁ alone. The point: same V on every branch, currents that add.
Reading NCERT's Examples 11.8 and 11.9 and Figure 11.12 (pages 186–187)Example 11.8 works out 2.4, 1.2 and 0.4 A, a total of 4 A and Rp = 3 Ω. Figure 11.12 shows R₁ and R₂ in parallel at the top, R₃, R₄ and R₅ in parallel at the bottom, a vertical battery, an ammeter and a key — the same network as the chapter's opening sketch. The text that follows uses ‘fairy lights’ to show why series circuits are impractical. The point: reduce parallel groups first, then add in series.
Reading NCERT's ‘Questions’ box (page 188)Five questions: judge parallel combinations of 1 Ω with 106 Ω (and also 103 Ω); find the single iron that draws as much as a 100 Ω lamp, 50 Ω toaster and 500 Ω water filter in parallel on 220 V (31.25 Ω, 7.04 A); the advantages of parallel; how to make 4 Ω and 1 Ω from 2, 3 and 6 Ω; and the highest and lowest totals from 4, 8, 12 and 24 Ω (48 Ω and 2 Ω). The point: parallel totals always fall below the smallest resistor.
Common confusions
- Adding a resistor in parallel lowers the total resistance and raises the total current — that is why plugging many appliances into one point overloads it (UPSC 1981). That question's keyed answer, ‘voltage will decrease whereas current will increase’, also needs a fact the book does not teach: the extra current drops a little voltage in the supply wiring (Beyond NCERT). Hold on to the current rising.
- 1 Ω in parallel with 106 Ω is just under 1 Ω, not about 106 Ω — the small resistor dominates.
- n equal resistors R in parallel give R/n; in series nR. Cutting a wire into 5 and paralleling the pieces gives R/25 (exercise 1).
- Don't add parallel resistances directly; add their reciprocals, then take the reciprocal of the sum.
- Homes are wired in parallel, so every appliance gets the full supply voltage and has its own switch.
Beyond NCERT
- For two resistors only, Rp = R₁R₂/(R₁ + R₂) (‘product over sum’) — a quick check: 10 × 40/50 = 8 Ω, Example 11.9.
- In real house wiring, the many parallel loads draw a large total current through the shared supply wires, whose small resistance then wastes power and drops a little voltage — another reason for separate circuits and protective devices.
8. The heating effect: Joule's law, filaments and fuses
In simple wordsRub your hands and they warm up. Electrons pushed through a wire keep bumping into its atoms, and the wire warms up the same way. Sometimes that heat is the whole point (heater, iron, toaster, a glowing bulb filament, a melting fuse); sometimes it is waste (a warm fan, hot wires).
A cell keeps spending energy to maintain a current. Part may do useful work (a fan's blades); the rest heats the gadget — a fan grows warm with long use. In a purely resistive circuit (only resistors on a battery) all the source's energy becomes heat: the heating effect of electric current. Moving charge Q through V needs work VQ, so the power input is P = VQ/t = VI (eq. 11.19), and the heat in time t is H = VIt (eq. 11.20); with V = IR, H = I²Rt (eq. 11.21) — Joule's law of heating: heat is proportional to the square of the current, to the resistance and to the time. For an appliance on a known voltage, find I = V/R first. Uses: laundry iron, toaster, oven, kettle and heater; electric bulbs, whose filament must get very hot without melting — tungsten (melting point 3380 °C, book), thermally isolated on insulating supports, in bulbs filled with chemically inactive nitrogen and argon to prolong its life; most of the power still appears as heat, only a little as light. The fuse protects circuits and appliances by stopping any unduly high current: a wire of a metal or alloy of suitable melting point (aluminium, copper, iron, lead…) placed in series with the device, usually in a porcelain cartridge with metal ends, melts when the current exceeds its rating and breaks the circuit. Domestic fuses are rated 1, 2, 3, 5, 10 A; a 1 kW iron on 220 V draws 1000/220 ≈ 4.5 A, so it needs a 5 A fuse.
⤵ How a fuse protects a circuit
Normal usethe current is below the fuse's rating; the fuse wire stays warm, not hot
Fault or overloadtoo many appliances, or a short circuit: current rises sharply
Joule heating in the fuse wireH = I squared R t grows with the square of the current
Fuse wire meltsit is in series, so the whole circuit breaks
Appliances and wiring savedreplace the fuse only after removing the cause
Reading NCERT's Figure 11.13 (page 189)A resistor R between X and Y with a voltmeter across it, an ammeter, a cell and a key in one loop, current I marked. The point: in a purely resistive circuit all the energy VIt supplied by the source becomes heat in R.
Reading NCERT's Examples 11.10 and 11.11 and ‘Questions’ box (pages 189–190)Example 11.10: an iron using 840 W and 360 W at 220 V draws 3.82 A and 1.64 A, with resistances printed as 57.60 Ω and 134.15 Ω. Example 11.11: 100 J each second in 4 Ω means I = √(H/Rt) = 5 A and V = 20 V. The box asks why a heater's cord does not glow while its element does, the heat from 96 000 C through 50 V (4.8 × 106 J), and the heat from 5 A in 20 Ω for 30 s (15 000 J). The point: H = VIt = I²Rt. (Rounding note: the book divides 220 V by the rounded currents; exactly, R = V²/P = 57.62 Ω and 134.44 Ω. Either is accepted; the rule ‘less power → more resistance at the same voltage’ is the point.)
Common confusions
- Heat goes as the square of the current: double I and four times the heat (for the same R and t).
- At a fixed voltage (mains), the lower resistance gets more power (P = V²/R) — the heater (100 Ω) heats more than the bulb (1200 Ω).
- A fuse goes in series, and needs a low (suitable) melting point — not high resistance, not in parallel with the switch (UPSC 1998).
- The heater's cord does not glow because copper's low resistance makes little heat; the nichrome element has high resistance (in-text page 190 Q1).
- The heating in a wire comes from electrons colliding with the metal's atoms, not from atoms colliding with each other (UPSC 1998 Assertion–Reason).
- The book's tungsten melting point is 3380 °C; some references give a slightly higher figure — quote NCERT's in a school-level answer.
Beyond NCERT
- Miniature circuit breakers (MCBs) now replace fuse wires in most new homes: they trip on overload and can be reset instead of replaced.
- Efficiency: since a filament lamp turns most of its power into heat (book), lamps that make light without a hot filament — fluorescent tubes and especially LEDs — give far more light per watt. India's mass LED-distribution programme (UJALA) was built on this.
- Transmission at high voltage: the same power at ten times the voltage needs a tenth of the current, so the I²R loss in the line falls a hundredfold (asserted for an illustrative line) — the reason power is sent at very high voltages (UPSC 1987).
9. Electric power and the kilowatt hour
In simple wordsPower is how fast an appliance uses energy; energy is how much it uses in total. A tap's flow rate versus the water in the bucket. The electricity company charges for the bucket — kilowatt hours, the ‘units’ on your bill — not for the flow rate and not for electrons, which are never used up.
Power is the rate of doing work, or of using energy. In a circuit the rate at which electrical energy is used is the electric power: P = VI, or with Ohm's law P = I²R = V²/R (eq. 11.22). Its SI unit is the watt (W): the power of a device carrying 1 A at 1 V, so 1 W = 1 V A (eq. 11.23). The watt is small, so the kilowatt (1000 W) is used. Energy is power × time, so its unit is the watt hour; the commercial unit is the kilowatt hour (kW h), commonly called a ‘unit’: 1 kW h = 1000 W × 3600 s = 3.6 × 106 J. In Example 11.12 a bulb on 220 V taking 0.50 A uses 110 W; in Example 11.13 a 400 W refrigerator running 8 hours a day for 30 days uses 96 kW h, costing Rs 288 at Rs 3.00 per kW h.
⤵ From an appliance's rating to the bill
Power ratingwatts on the label, e.g. a 400 W refrigerator = 0.4 kW
Hours of use8 h a day for 30 days = 240 h
Energy = power x time0.4 kW x 240 h = 96 kW h, i.e. 96 units
TariffRs 3.00 per unit (the book's rate)
Cost96 x Rs 3.00 = Rs 288
Reading NCERT's ‘More to Know!’ box (page 191)An orange spiral-bound note: many people think electrons are consumed in a circuit — wrong. We pay the electricity company for the energy used to move electrons through bulbs, fans and engines. The point: current is not used up; energy is.
Reading NCERT's Examples 11.12 and 11.13 and ‘Questions’ box (pages 191–192)Example 11.12: 220 V × 0.50 A = 110 W. Example 11.13: 400 W × 8 h × 30 days = 96 000 W h = 96 kW h; × Rs 3.00 = Rs 288. The box asks what decides the rate at which a current delivers energy (power, P = VI) and the power and 2-hour energy of a motor taking 5 A from 220 V (1100 W; 2.2 kW h = 7.92 × 106 J). The point: convert W to kW and hours before multiplying. (A slip: the section opens by saying eq. 11.21 gives the rate of energy use — eq. 11.21 is the heat H = I²Rt; the rate is eq. 11.19, P = VI.)
Common confusions
- Kilowatt = unit of power (UPSC 1982); kilowatt hour = unit of energy. ‘kW h’ is not ‘kW per hour’.
- 1 kW h = 3.6 × 106 J (not 3.6 × 103 J): 1000 W × 3600 s.
- A 100 W bulb has less resistance than a 60 W bulb on the same supply (R = V²/P) — UPSC 1987.
- A bulb rated 100 W at 220 V uses only 25 W at 110 V — power goes as V² for a fixed R (exercise 3).
- Higher power does not mean more energy: it depends on how long (exercise 16).
- Of I²R, IR², VI and V²/R, only IR² is not a power (exercise 2).
Beyond NCERT
- India's household supply is nominally 230 V AC at 50 Hz; the book's 220 V is the older round figure used for its numericals. (AC and its frequency belong to the next chapter's territory.)
- Appliance star labels from the Bureau of Energy Efficiency (BEE) compare yearly kW h consumption — the same power × time arithmetic as Example 11.13.
NCERT's in-text questions and activities — short answers
The chapter has seven ‘Questions’ boxes (pages 172, 174, 181, 185, 188, 190 and 192) and questions inside its activities. Outlines follow; every number is computed and asserted in the page's source.
Page 172, Q1: What does an electric circuit mean?
A continuous, closed path along which an electric current flows — for example a cell, a switch, a bulb and connecting wires. If it is broken anywhere, the current stops.
Page 172, Q2: Define the unit of current.
The ampere: 1 A is the current when 1 coulomb of charge flows through a cross-section in 1 second (1 A = 1 C/1 s).
Page 172, Q3: Calculate the number of electrons constituting one coulomb of charge.
n = Q/e = 1 C ÷ 1.6 × 10−19 C = 6.25 × 1018 electrons (the book's ‘nearly 6 × 1018’).
Page 174, Q1: Name a device that helps to maintain a potential difference across a conductor.
A cell or battery (its chemical action keeps the potential difference).
Page 174, Q2: What is meant by saying that the potential difference between two points is 1 V?
1 joule of work is done in moving 1 coulomb of charge from one point to the other.
Page 174, Q3: How much energy is given to each coulomb of charge passing through a 6 V battery?
W = VQ = 6 V × 1 C = 6 J.
Page 181, Q1: On what factors does the resistance of a conductor depend?
Its length (R ∝ l), its area of cross-section (R ∝ 1/A), the material (resistivity ρ) — and its temperature. R = ρl/A.
Page 181, Q2: Will current flow more easily through a thick wire or a thin wire of the same material, connected to the same source? Why?
Through the thick wire: its larger area of cross-section gives it lower resistance (R ∝ 1/A), so more current flows.
Page 181, Q3: If the resistance stays constant while the potential difference is halved, what happens to the current?
By I = V/R, the current is also halved.
Page 181, Q4: Why are coils of electric toasters and irons made of an alloy rather than a pure metal?
Alloys have higher resistivity (more heat for a given current) and do not oxidise (burn) readily at high temperatures.
Page 181, Q5: From Table 11.2, (a) which of iron and mercury is the better conductor? (b) which material is the best conductor?
(a) Iron (10.0 × 10−8 Ω m against mercury's 94.0 × 10−8). (b) Silver (1.60 × 10−8 Ω m, the lowest).
Page 185, Q1: Draw a circuit of a battery of three 2 V cells, a 5 Ω, an 8 Ω and a 12 Ω resistor and a plug key, all in series.
One loop: battery (three cells + to −) → 5 Ω → 8 Ω → 12 Ω → plug key → back to the battery, drawn with Table 11.1's symbols (see the series diagram above).
Page 185, Q2: Redraw it with an ammeter for the current and a voltmeter across the 12 Ω resistor. What are the readings?
Ammeter in series anywhere in the loop; voltmeter in parallel across 12 Ω. Rs = 25 Ω, V = 6 V ⇒ I = 6/25 = 0.24 A; voltmeter: 0.24 × 12 = 2.88 V.
Page 188, Q1: Judge the equivalent resistance in parallel of (a) 1 Ω and 10⁶ Ω; (b) 1 Ω, 10³ Ω and 10⁶ Ω.
Both are just under 1 Ω — a parallel total is below the smallest resistor, and the large ones add almost nothing: (a) ≈ 0.999999 Ω; (b) ≈ 0.9990 Ω.
Page 188, Q2: A 100 Ω lamp, a 50 Ω toaster and a 500 Ω water filter are in parallel on 220 V. What is the resistance of an iron that takes as much current as all three, and what current?
1/R = 1/100 + 1/50 + 1/500 = 16/500 ⇒ R = 31.25 Ω; I = 220/31.25 = 7.04 A.
Page 188, Q3: What are the advantages of connecting devices in parallel with the battery rather than in series?
Each device gets the full voltage; each draws the current it needs; each can be switched independently; if one fails the others keep working; the total resistance is lower.
Page 188, Q4: How can 2 Ω, 3 Ω and 6 Ω be connected to give (a) 4 Ω, (b) 1 Ω?
(a) 3 Ω ∥ 6 Ω = 2 Ω, in series with 2 Ω → 4 Ω. (b) All three in parallel: 1/2 + 1/3 + 1/6 = 1 → 1 Ω.
Page 188, Q5: What are the highest and lowest totals from coils of 4, 8, 12 and 24 Ω?
Highest: all in series = 48 Ω. Lowest: all in parallel: 1/4 + 1/8 + 1/12 + 1/24 = 1/2 → 2 Ω.
Page 190, Q1: Why does the cord of an electric heater not glow while the heating element does?
The same current flows in both, but the cord is copper of very low resistance, so I²Rt in it is small; the element (nichrome) has high resistance, so it gets red hot.
Page 190, Q2: Compute the heat generated in transferring 96 000 C in one hour through 50 V.
H = VQ = 50 × 96 000 = 4.8 × 106 J (the hour is not needed once the charge is given).
Page 190, Q3: An iron of resistance 20 Ω takes 5 A. Heat developed in 30 s?
H = I²Rt = 25 × 20 × 30 = 15 000 J (1.5 × 104 J).
Page 192, Q1: What determines the rate at which energy is delivered by a current?
The electric power, P = VI (= I²R = V²/R): the product of the potential difference and the current.
Page 192, Q2: An electric motor takes 5 A from a 220 V line. Find its power and the energy used in 2 h.
P = VI = 220 × 5 = 1100 W; energy = 1.1 kW × 2 h = 2.2 kW h = 7.92 × 106 J.
Activity 11.2 (pages 176–177): Are the ammeter readings different for the nichrome wire, the torch bulb and the 10 W bulb? What does that show?
Yes — with the same cells, each component lets a different current through, so each has a different resistance.
Activity 11.3 (pages 177–178): Does the current depend on the length and the area of cross-section of the wire?
Yes: doubling the length halves the current; a thicker wire passes more current; copper of the same size passes more than nichrome. Hence R = ρl/A.
Activities 11.4 and 11.5 (pages 182–183): Does the current change with the ammeter's position in a series circuit? How are V, V₁, V₂, V₃ related?
The current is the same everywhere in a series circuit; V = V₁ + V₂ + V₃, and the voltage across the combination equals the battery's.
Activity 11.6 (pages 185–186): How are I, I₁, I₂ and I₃ related in a parallel combination?
I = I₁ + I₂ + I₃; the voltage across each branch equals that across the whole combination.
NCERT exercise — answer outlines
Reading NCERT's ‘What you have learnt’ and Exercises (pages 192–194)Eleven summary points: current as a stream of electrons, opposite to the conventional direction; the ampere; the cell's potential difference in volts; resistance in ohms; Ohm's law at constant temperature; R depending on length, area and material; the series and parallel formulae; energy dissipated W = V × I × t; the watt; and 1 kW h = 3.6 × 106 J. Eighteen exercises follow — four multiple-choice, ten numericals and four explanations; answers to most are on pages 218–219. The point: the exercises drill Ohm's law, series/parallel and P = VI.
Try each question first, then open the outline. Every number below is computed with exact fractions and checked against the book's answer key.
1. A wire of resistance R is cut into five equal parts, which are joined in parallel. If the equivalent resistance is R′, R/R′ is (a) 1/25 (b) 1/5 (c) 5 (d) 25
(d) 25. Each piece is R/5; five equal resistors in parallel give (R/5)/5 = R/25, so R/R′ = 25.
2. Which does not represent electrical power? (a) I²R (b) IR² (c) VI (d) V²/R
(b) IR². The others follow from P = VI with V = IR.
3. A bulb rated 220 V, 100 W runs on 110 V. The power consumed is (a) 100 W (b) 75 W (c) 50 W (d) 25 W
(d) 25 W. R = V²/P is fixed, so P ∝ V²: 100 × (110/220)² = 25 W.
4. Two identical wires are joined first in series and then in parallel across the same potential difference. The ratio of heat produced (series : parallel) is (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
(c) 1:4. Series: 2r; parallel: r/2. At fixed V, H = V²t/R ∝ 1/R, so Hs : Hp = (1/2r) : (2/r) = 1 : 4.
5. How is a voltmeter connected to measure the potential difference between two points?
In parallel, across the two points (+ terminal towards the cell's + side).
6. A copper wire of diameter 0.5 mm and resistivity 1.6 × 10⁻⁸ Ω m is to have a resistance of 10 Ω. What length is needed? How does the resistance change if the diameter is doubled?
l = RA/ρ = 10 × π(0.25 × 10−3)² ÷ 1.6 × 10−8 ≈ 122.7 m. Doubling the diameter makes the area four times, so R becomes one quarter: 2.5 Ω (book: 122.7 m; ¼ times).
7. For I = 0.5, 1.0, 2.0, 3.0, 4.0 A the voltages are 1.6, 3.4, 6.7, 10.2, 13.2 V. Plot V against I and find the resistance.
The points lie close to a straight line through the origin (graph above). V/I for each pair: 3.2, 3.4, 3.35, 3.4, 3.3; their mean is 3.33 Ω (book: 3.33 Ω). A best-fit line through the origin gives 3.34 Ω — the same to the accuracy of the data.
8. A 12 V battery across an unknown resistor gives 2.5 mA. Find the resistance.
R = V/I = 12 ÷ 2.5 × 10−3 = 4800 Ω = 4.8 kΩ.
9. A 9 V battery is in series with 0.2, 0.3, 0.4, 0.5 and 12 Ω. How much current flows through the 12 Ω resistor?
In series the current is common: Rs = 13.4 Ω, I = 9/13.4 ≈ 0.67 A (book: 0.67 A).
10. How many 176 Ω resistors in parallel are needed to carry 5 A on a 220 V line?
Needed R = 220/5 = 44 Ω; n identical resistors in parallel give 176/n = 44 ⇒ n = 4.
11. Show how to connect three 6 Ω resistors to get (i) 9 Ω, (ii) 4 Ω.
(i) Two in parallel (3 Ω) in series with the third: 3 + 6 = 9 Ω. (ii) Two in series (12 Ω) in parallel with the third: 12 × 6/18 = 4 Ω. Diagrams above.
12. Bulbs rated 10 W for 220 V: how many can be connected in parallel across the line if the maximum allowable current is 5 A?
Maximum power = 220 × 5 = 1100 W; 1100/10 = 110 bulbs.
13. A hot plate on 220 V has two 24 Ω coils A and B, used separately, in series or in parallel. Find the currents.
Separately: 220/24 ≈ 9.17 A; in series (48 Ω): ≈ 4.58 A; in parallel (12 Ω): ≈ 18.33 A (book: 9.2 A, 4.6 A, 18.3 A).
14. Compare the power used in the 2 Ω resistor in (i) a 6 V battery in series with 1 Ω and 2 Ω, (ii) a 4 V battery in parallel with 12 Ω and 2 Ω.
(i) I = 6/3 = 2 A, P = I²R = 4 × 2 = 8 W. (ii) The 2 Ω has the full 4 V: P = V²/R = 16/2 = 8 W. Equal.
15. Lamps of 100 W and 60 W (both 220 V) are in parallel on 220 V mains. What current is drawn from the line?
Total power 160 W; I = P/V = 160/220 ≈ 0.73 A.
16. Which uses more energy: a 250 W TV in 1 hour, or a 1200 W toaster in 10 minutes?
TV: 250 × 1 = 250 W h; toaster: 1200 × 1/6 = 200 W h. The TV uses more (diagram above).
17. A heater of resistance 44 Ω draws 5 A for 2 hours. At what rate is heat developed?
Rate = P = I²R = 25 × 44 = 1100 W (the 2 hours would give the total energy, 2.2 kW h, which is not asked).
18. Explain: (a) why tungsten is used for lamp filaments; (b) why heating elements of toasters and irons are alloys; (c) why series is not used for domestic circuits; (d) how a wire's resistance varies with its area; (e) why copper and aluminium are used for transmission.
(a) Tungsten is strong with a very high melting point (3380 °C, book), so it can glow white-hot without melting. (b) Alloys have high resistivity and do not oxidise readily when hot. (c) In series every appliance gets the same current though they need different currents, the voltage is shared, and one failure stops all. (d) Inversely: R ∝ 1/A. (e) They have very low resistivity, so little energy is lost as heat.
What to remember
Numbers
Charge on an electron1.6 × 10⁻¹⁹ C
Electrons in 1 C6.25 × 10¹⁸ (book: ‘nearly 6 × 10¹⁸’)
1 A / 1 V / 1 Ω / 1 W1 C/s / 1 J/C / 1 V/A / 1 V A
1 mA / 1 µA10⁻³ A / 10⁻⁶ A
1 kW h (one ‘unit’)3.6 × 10⁶ J
Ohm's law discovered1827, Georg Simon Ohm (1787–1854)
Ampère / Volta1775–1836 / 1745–1827
Best conductor (Table 11.2)silver, 1.60 × 10⁻⁸ Ω m; copper 1.62 × 10⁻⁸
Nichrome100 × 10⁻⁶ Ω m — alloy of Ni, Cr, Mn, Fe
Insulators (Table 11.2)10¹⁰ to 10¹⁷ Ω m
Tungsten melting point (book)3380 °C
Domestic fuse ratings (book)1, 2, 3, 5, 10 A; 1 kW iron at 220 V → 5 A fuse
Example 11.85, 10, 30 Ω ∥ on 12 V → 2.4 + 1.2 + 0.4 = 4 A; Rₚ = 3 Ω
Example 11.13400 W × 8 h × 30 days = 96 kW h → Rs 288 at Rs 3
Chapter size6 activities, Figs 11.1–11.13, Tables 11.1–11.2, 13 examples, 18 exercises
One-liners
- Current is the rate of flow of charge, I = Q/t; conventional current is opposite to electron flow.
- Potential difference is work per unit charge, V = W/Q; a cell maintains it by chemical action.
- Ammeter in series; voltmeter in parallel.
- Ohm's law: V = IR, valid at constant temperature.
- R = ρl/A: resistance doubles with length and halves when the area doubles; resistivity is a property of the material.
- Alloys have higher resistivity than their metals and do not oxidise readily — so heating elements are nichrome.
- Tungsten for filaments; copper and aluminium for transmission lines.
- Series: same current, voltages add, Rₛ = R₁ + R₂ + R₃.
- Parallel: same voltage, currents add, 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃.
- Domestic wiring is parallel: full voltage to each appliance, independent switching.
- Joule's law: H = I²Rt; a fuse in series melts on overload.
- Power P = VI = I²R = V²/R in watts; energy is billed in kilowatt hours.
- Electrons are not consumed in a circuit; we pay for energy.
Traps to avoid
- Conventional current flows OPPOSITE to electrons.
- Ammeter in SERIES, voltmeter in PARALLEL.
- Resistance (Ω) depends on size and material; resistivity (Ω m) only on material and temperature.
- Double the diameter → four times the area → a QUARTER of the resistance.
- Series total exceeds the largest resistor; parallel total is below the smallest.
- Adding appliances in parallel LOWERS total resistance and RAISES total current.
- Heat goes as the SQUARE of the current.
- At fixed mains voltage, the higher-wattage bulb has the LOWER resistance.
- kW is power; kW h is energy (3.6 × 10⁶ J), not 'kW per hour'.
- A fuse needs a suitably LOW melting point and goes in SERIES.
- Nichrome for heating elements; tungsten for filaments — don't swap them.
How to strategise
Prelims — what UPSC asks
- Heating effect and safety is UPSC's favourite corner of this chapter: the heater element of an iron is nichrome (1980); a fuse wire prevents an unduly high current (1981); a fuse must have a low melting point — not be in parallel with the switch or have very high resistance (1998); a wire warms because of the current, but not because metal atoms collide with each other (1998, Assertion–Reason).
- Circuits and power: domestic wiring is a parallel connection (1996); drawing current for many appliances from one point is unwise because the current rises (1981); kilowatt is the unit of power (1982); a 100 W bulb has less resistance than a 60 W bulb (1987, R = V²/P); power is transmitted at high voltage to cut losses (1987, I²R).
- Method: for ratings questions use P = V²/R at fixed voltage; for heating use H = I²Rt; for any ‘many appliances’ question think parallel → lower R → higher current. Learn Table 11.1 symbols and the series/parallel table cold.
- Likely next: kW vs kW h and bill arithmetic; resistance vs resistivity statements; why alloys are heating elements; why copper/aluminium for transmission; MCBs vs fuses and LED efficiency (Beyond NCERT).
Mains — where it appears and how to answer
- No GS Mains question in the bank so far is built on this chapter. Its use in GS3 is as the science line in energy answers: transmission and distribution losses (I²R, why high-voltage lines), energy efficiency (filament lamps waste most power as heat; LEDs and star-labelled appliances), and electrical safety (fuses, circuit breakers).
- Answer frame: one-line physics (with the formula — H = I²Rt or P = V²/R) → the technology or programme that uses it → benefit in kW h or rupees → limits → way forward.
Self-check
Tap an option. NCERT's own exercise questions come first; the rest test the figures and common confusions.
1. In a metallic wire, the conventional current flows:
2. A voltmeter is connected:
3. A wire's diameter is doubled, keeping its length and material the same. Its resistance becomes:
4. Resistors of 5 Ω, 10 Ω and 30 Ω are connected in parallel across 12 V. The total current is:
5. Heating elements of irons and toasters are made of nichrome because it has:
6. Two bulbs rated 100 W and 60 W for 220 V. Compared with the 60 W bulb, the 100 W bulb has:
7. One ‘unit’ of electricity on a bill is:
8. A fuse protects a circuit because it:
UPSC questions from this chapter
Open any question to attempt it in the Question bank — your attempt is saved to your progress.