Where Mindful Learning Begins
Where Mindful Learning Begins
Class 10 Science NCERT Solutions Chapter 11 (CBSE 2026-27) — every in-text question, every embedded Activity question, and all 18 back-of-chapter Exercises on this one page, solved with full working.
Electricity is the most numerically intensive chapter in Class 10 Science, and it builds up in careful steps: electric current and charge, potential difference, circuit diagrams and symbols, Ohm's law and resistance, the factors (length, area, material) that determine a conductor's resistance, resistors combined in series and in parallel, the heating effect of current (Joule's law), and finally electric power and the commercial unit of energy — the kilowatt-hour on every electricity bill. These NCERT solutions work through every Activity-based conceptual question and the full 18-question Exercise set with complete step-by-step working for every calculation, using consistent units throughout.
Electricity is where Class 10 Science gets seriously numerical, and every concept builds directly on the one before it. It starts with the basic quantities — current as the rate of flow of charge, and potential difference as the "electric pressure" that drives that flow — then introduces Ohm's law, the single relationship (V = IR) that governs almost every calculation in the chapter. From there it explains why different conductors resist current differently (their length, cross-sectional area, and material), and builds up the two ways resistors combine in real circuits — in series, where resistances simply add, and in parallel, where the combination is always smaller than the smallest resistor — which explains why household wiring uses parallel circuits. The chapter closes with what happens to the electrical energy that isn't used for useful work: it becomes heat (Joule's law), and the rate at which this energy is used or dissipated is electric power, measured in the same kilowatt-hours that show up on every electricity bill. Every question woven through the chapter is solved here, section by section, in the order the textbook presents it: all seven "QUESTIONS" boxes spread across the chapter, the questions embedded directly inside the Activities, and the full 18-question Exercise set at the end, with complete numerical working for every calculation.
Charge, current, potential difference, circuit diagrams, and the V = IR relationship that anchors almost every calculation.
What resistance depends on (length, area, material), and how resistors add up differently in series versus parallel.
Joule's law of heating, electric power (P = VI), and the kilowatt-hour — the commercial unit every electricity bill is based on.
An electric circuit is a continuous and closed path along which an electric current flows.
The SI unit of current is the ampere (A). One ampere is the current constituted by the flow of one coulomb of charge per second: 1 A = 1 C/1 s.
A cell or a battery.
It means that 1 joule of work is done in moving a charge of 1 coulomb from one point to the other.
(i) Length of the conductor (directly proportional), (ii) Area of cross-section (inversely proportional), and (iii) the nature (material) of the conductor. It also depends on temperature.
Current flows more easily through a thick wire. A thicker wire has a larger cross-sectional area, and since resistance is inversely proportional to area (R ∝ 1/A), it has lower resistance, allowing current to flow more easily.
Since I = V/R and R is constant, if V is halved, the current I is also halved.
Alloys have a higher resistivity than their constituent pure metals, and they do not oxidise (burn) readily even at high temperatures — making them more durable and suitable for repeated use in heating devices.
(a) Iron is a better conductor than mercury (iron: 10.0×10−8 Ω m; mercury: 94.0×10−8 Ω m — lower resistivity means better conductivity).
(b) Silver is the best conductor, with the lowest resistivity (1.60×10−8 Ω m) in the table.
The circuit consists of three 2 V cells connected in series (total EMF = 6 V), connected in one continuous loop with a plug key (switch) and the three resistors — 5 Ω, 8 Ω, and 12 Ω — one after another, using the standard circuit symbols for a battery, resistors, and a switch.
Total resistance = 5+8+12 = 25 Ω. Total EMF = 3×2V = 6V.
(a) 1/Rp = 1/1 + 1/106 ≈ 1 (the second term is negligible). So Rp ≈ 1 Ω.
(b) 1/Rp = 1/1 + 1/1000 + 1/1000000 ≈ 1 (again dominated by the smallest resistance). So Rp ≈ 1 Ω.
In both cases, the equivalent resistance is always less than the smallest individual resistance, and here it is very close to it.
The iron's resistance should be 31.25 Ω, drawing a current of 7.04 A.
(a) 4Ω: Connect the 3Ω and 6Ω resistors in parallel (giving 2Ω), then connect this in series with the 2Ω resistor.
(b) 1Ω: Connect all three resistors (2Ω, 3Ω, 6Ω) in parallel.
(a) Highest: connect all four in series.
(b) Lowest: connect all four in parallel.
The heating element has a much higher resistance than the connecting cord (usually copper, a good conductor with low resistance). Since heat produced (H = I²Rt) is directly proportional to resistance for the same current, much more heat is generated in the high-resistance element (enough to make it glow), while very little heat is generated in the low-resistance cord.
The power (P = VI) — the potential difference across, and the current through, the device.
Yes, the readings differ, even with the same voltage source. This indicates that different components offer different amounts of resistance to current flow — components with lower resistance allow more current, while those with higher resistance allow less.
Yes, on both. The ammeter reading halves when the wire's length is doubled (same thickness) — resistance increases with length, so current decreases. The reading increases with a thicker wire of the same length and material — resistance decreases as cross-sectional area increases, so current increases.
No — the ammeter reading stays the same regardless of where it's placed in the series circuit, showing that the same current flows through every part of a series circuit.
The total potential difference across a series combination equals the sum of the potential differences across the individual resistors.
The total current entering a parallel combination equals the sum of the currents through each individual branch.
Each part has resistance R/5. In parallel: 1/R′ = 5/(R/5) = 25/R, so R′ = R/25.
Answer: (d) 25
Answer: (b) IR² — the valid formulas for power are P = I²R, P = VI, and P = V²/R.
Answer: (d) 25 W
Let each wire have resistance R. Since H = V²t/R for the whole combination:
Answer: (c) 1:4
A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured.
Radius r = 0.25 mm = 2.5×10⁻⁴ m; Area A = πr² ≈ 1.963×10⁻⁷ m²
If the diameter is doubled, area increases 4× (A' = 4A). Since R = ρl/A, resistance becomes 1/4 of the original:
Plotting V against I gives a straight line through the origin, confirming Ohm's law. Taking the average slope (V/I) across the data points:
In series, total resistance = 0.2+0.3+0.4+0.5+12 = 13.4 Ω, and the same current flows through every resistor.
(i) 9Ω: two resistors in parallel (giving 3Ω) in series with the third (6Ω): 3+6 = 9Ω ✓
(ii) 4Ω: two resistors in series (giving 12Ω) in parallel with the third (6Ω): 1/R = 1/12+1/6 = 1/4 ⟹ R = 4Ω ✓
(a) Separately: I = V/R = 220/24 ≈ 9.17 A
(b) In series: R = 24+24 = 48Ω; I = 220/48 ≈ 4.58 A
(c) In parallel: 1/Rp = 1/24+1/24 = 1/12 ⟹ Rp = 12Ω; I = 220/12 ≈ 18.33 A
(i) Series: Total R = 1+2 = 3Ω; I = 6/3 = 2A. Power in 2Ω = I²R = (2)² × 2 = 8 W
(ii) Parallel: the 2Ω resistor sees the full 4V directly. Power in 2Ω = V²/R = (4)²/2 = 8 W
The power dissipated in the 2Ω resistor is the same (8 W) in both cases — ratio 1:1.
The TV (250 Wh) uses more energy than the toaster (200 Wh).
(a) Tungsten has a very high melting point (3380°C), letting it be heated to incandescence (to emit bright light) without melting.
(b) Alloys have higher resistivity than pure metals and don't oxidise readily at high temperatures, making them more durable for repeated heating.
(c) In series, the same current flows through all appliances, but different appliances need different currents to work properly; also, if one appliance fails, the whole circuit breaks. Parallel arrangements avoid both problems.
(d) Resistance is inversely proportional to the area of cross-section (R ∝ 1/A) — larger area means lower resistance.
(e) Copper and aluminium have very low resistivity, allowing current to flow with minimal energy loss as heat — making them efficient and cost-effective for long-distance transmission.
Next up is Chapter 12 — Magnetic Effects of Electric Current. Or explore the full chapter list, browse the Class 10 Science hub, or book a free demo class for personalised coaching.
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