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Class 10 Electricity numericals 2026-27, worked step by step
Here are 14 numericals on NCERT Class 10 Science Chapter 11, Electricity, in rising order of difficulty. Each one shows what is given, the formula and the textbook page it comes from, the working with units, the answer, and the mistake students most often make on that kind of sum. They cover Ohm's law, charge and current, resistivity, resistors in series and in parallel, the heating effect, power and a month's electricity bill in kilowatt hours.
These numericals are our own, drafted with AI help. Every answer was re-computed by a script and every formula was checked against the chapter, but no practising teacher has reviewed them yet. None of them is a textbook example, in-text question or exercise. If you find a mistake, please tell us. These are paper calculations only: never open, rewire or test mains wiring or an appliance yourself.
The formulas you will use
From Chapter 11 of the NCERT textbook (Science, Textbook for Class X, Reprint 2026-27):
- Charge and current: Q = I t (page 172).
- Ohm's law: V = I R, so I = V ÷ R and R = V ÷ I (page 176).
- Resistance of a wire: R = ρ l ÷ A, with ρ the resistivity of the material in Ω m (page 178); the values for common metals and alloys are in Table 11.2 (page 179).
- Series: Rs = R₁ + R₂ + R₃; parallel: 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ (pages 184 and 186).
- Heat: H = I² R t (page 189).
- Power: P = V I = I² R = V² ÷ R, in watts; energy in kilowatt hours, with 1 kW h = 3.6 × 10⁶ J (page 191).
Level 1: one formula
Numerical 1. A torch bulb draws a current of 0.3 A from a 4.5 V battery. What is the resistance of the bulb?
- Given: V = 4.5 V, I = 0.3 A.
- Formula: R = V ÷ I (Ohm's law, page 176).
- Working: R = 4.5 ÷ 0.3 = 15.
- Answer: 15 Ω.
- Common mistake: dividing the wrong way round (0.3 ÷ 4.5). Check the size: a bigger voltage for the same current means a bigger resistance.
Numerical 2. A 470 Ω resistor in a toy circuit carries a current of 20 mA. What is the potential difference across it?
- Given: R = 470 Ω, I = 20 mA = 0.020 A.
- Formula: V = I R (page 176).
- Working: V = 0.020 × 470 = 9.4.
- Answer: 9.4 V.
- Common mistake: using 20 instead of 0.020. That gives 9,400 V, far more than any toy battery. Always turn milliamperes into amperes (divide by 1,000) first.
Numerical 3. A doorbell is pressed for 2 minutes in all during a day, and while it rings it carries a current of 0.25 A. How much charge flows through it?
- Given: I = 0.25 A, t = 2 min = 120 s.
- Formula: Q = I t (page 172).
- Working: Q = 0.25 × 120 = 30.
- Answer: 30 C.
- Common mistake: leaving the time in minutes (0.25 × 2 = 0.5). The ampere is a coulomb per second, so time must be in seconds.
Level 2: resistance of a wire
Numerical 4. A heater coil is made of nichrome wire 2 m long with a cross-section of 0.5 mm². Using the resistivity of nichrome from Table 11.2, find its resistance.
- Given: l = 2 m, A = 0.5 mm² = 0.5 × 10⁻⁶ m², ρ = 100 × 10⁻⁶ Ω m (Table 11.2, page 179).
- Formula: R = ρ l ÷ A (page 178).
- Working: R = (100 × 10⁻⁶ × 2) ÷ (0.5 × 10⁻⁶) = 200 ÷ 0.5 = 400.
- Answer: 400 Ω.
- Common mistake: turning mm² into m² by dividing by 1,000. A square millimetre is a millimetre times a millimetre, so divide by 1,000 × 1,000 = 10⁶.
Numerical 5. A copper wire has a resistance of 10 Ω. It is pulled through a die so that its length doubles; the amount of copper stays the same, so its cross-section halves. What is its new resistance?
- Given: R = 10 Ω; new length 2l; new area A ÷ 2.
- Formula: R = ρ l ÷ A (page 178); ρ does not change, because the material is the same.
- Working: new R = ρ (2l) ÷ (A ÷ 2) = 4 × ρ l ÷ A = 4 × 10 = 40.
- Answer: 40 Ω.
- Common mistake: answering 20 Ω by doubling only for the length. Stretching changes the length and the area together, and both changes raise the resistance.
Level 3: series and parallel
Numerical 6. A string of decorative lights has 20 identical bulbs joined in series across a 220 V supply, and a current of 0.5 A flows. (a) What is the total resistance? (b) What is the resistance of each bulb? (c) What is the potential difference across each bulb?
- Given: V = 220 V, I = 0.5 A, 20 equal bulbs in series.
- Formulas: R = V ÷ I (page 176); in series the resistances add and the same current flows through each (page 184).
- Working: (a) R = 220 ÷ 0.5 = 440 Ω. (b) 440 ÷ 20 = 22 Ω. (c) V = I R = 0.5 × 22 = 11 V; check: 20 × 11 = 220 V.
- Answer: (a) 440 Ω; (b) 22 Ω; (c) 11 V.
- Common mistake: thinking each bulb gets the full 220 V. In series the supply voltage is shared out, so the bulbs' voltages add up to 220 V.
Numerical 7. You have a 12 Ω resistor and want a combination of 4 Ω. What resistor must you join in parallel with the 12 Ω one?
- Given: R₁ = 12 Ω, Rp = 4 Ω.
- Formula: 1/Rp = 1/R₁ + 1/R₂ (page 186).
- Working: 1/R₂ = 1/4 − 1/12 = 3/12 − 1/12 = 2/12 = 1/6, so R₂ = 6 Ω. Check: 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4.
- Answer: 6 Ω.
- Common mistake: subtracting the resistances (12 − 4 = 8 Ω). In parallel it is the reciprocals that add, not the resistances.
Numerical 8. A 10 Ω resistor is joined in series with a parallel pair of 20 Ω and 30 Ω, and the whole circuit is connected to a 22 V battery. Find (a) the total resistance, (b) the current from the battery, (c) the current in each resistor of the parallel pair.
- Given: 10 Ω in series with (20 Ω parallel with 30 Ω); V = 22 V.
- Formulas: parallel (page 186), series (page 184), Ohm's law (page 176).
- Working: (a) 1/Rp = 1/20 + 1/30 = 3/60 + 2/60 = 5/60, so Rp = 12 Ω; total = 10 + 12 = 22 Ω. (b) I = 22 ÷ 22 = 1 A. (c) The pair has V = 1 × 12 = 12 V across it, so 12 ÷ 20 = 0.6 A and 12 ÷ 30 = 0.4 A; check: 0.6 + 0.4 = 1 A.
- Answer: (a) 22 Ω; (b) 1 A; (c) 0.6 A in the 20 Ω and 0.4 A in the 30 Ω resistor.
- Common mistake: using the battery's 22 V for each branch. Only the 12 V left after the 10 Ω resistor is across the pair.
Level 4: heating effect
Numerical 9. An immersion rod of resistance 50 Ω is connected to a 220 V supply for 5 minutes. How much heat does it produce?
- Given: R = 50 Ω, V = 220 V, t = 5 min = 300 s.
- Formulas: I = V ÷ R (page 176), H = I² R t (page 189).
- Working: I = 220 ÷ 50 = 4.4 A. H = 4.4² × 50 × 300 = 19.36 × 50 × 300 = 290,400.
- Answer: 290,400 J, about 2.9 × 10⁵ J.
- Common mistake: putting V instead of I into H = I² R t. Find the current first, or use H = V² t ÷ R, which gives the same 290,400 J.
Numerical 10. A resistor gives out 600 J of heat in one minute when the current through it is 2 A. How much heat does it give out in one minute when the current is 3 A?
- Given: H = 600 J at I = 2 A; same R and t.
- Formula: H = I² R t, so with R and t fixed, H is proportional to I² (page 189).
- Working: new H = 600 × (3 ÷ 2)² = 600 × 9/4 = 1,350.
- Answer: 1,350 J.
- Common mistake: scaling by 3/2 (900 J). The heat goes with the square of the current.
Level 5: power and energy
Numerical 11. An LED bulb is rated 9 W for a 220 V supply. What current does it draw, in milliamperes?
- Given: P = 9 W, V = 220 V.
- Formula: P = V I, so I = P ÷ V (page 191).
- Working: I = 9 ÷ 220 ≈ 0.0409 A = 40.9 mA.
- Answer: about 41 mA.
- Common mistake: writing 0.04 mA. To turn amperes into milliamperes, multiply by 1,000.
Numerical 12. A geyser rated 2,000 W uses 2.5 units of electrical energy. (a) How many joules is that? (b) For how long was the geyser on?
- Given: P = 2,000 W = 2 kW, E = 2.5 kW h.
- Formulas: 1 kW h = 3.6 × 10⁶ J; energy = power × time (page 191).
- Working: (a) 2.5 × 3.6 × 10⁶ = 9 × 10⁶ J. (b) t = 2.5 ÷ 2 = 1.25 h = 1 hour 15 minutes.
- Answer: (a) 9 × 10⁶ J; (b) 1 hour 15 minutes.
- Common mistake: dividing 2.5 by 2,000 and getting 0.00125 h. Keep kW with kW h, or W with W h.
Numerical 13. A family uses, every day: 4 ceiling fans of 75 W each for 10 hours, 6 LED bulbs of 9 W each for 6 hours, and an air conditioner of 1,500 W for 4 hours. Electricity costs Rs 6 a unit. What is the bill for 30 days?
- Given: the three loads, 30 days, Rs 6 per kW h.
- Formula: energy = power × time, in kW h ("units") (page 191).
- Working: fans 4 × 75 × 10 = 3,000 W h = 3 kW h; bulbs 6 × 9 × 6 = 324 W h = 0.324 kW h; air conditioner 1,500 × 4 = 6,000 W h = 6 kW h. Per day: 3 + 0.324 + 6 = 9.324 kW h. In 30 days: 9.324 × 30 = 279.72 kW h. Cost: 279.72 × 6 = Rs 1,678.32.
- Answer: about 280 units, Rs 1,678.32.
- Common mistake: forgetting to divide watt hours by 1,000 to get units. 3,000 W h is 3 units, not 3,000.
Level 6: putting it together
Numerical 14. A 100 W bulb and a 40 W bulb, both rated for 220 V, are joined in series across a 220 V supply. (a) Find the resistance of each bulb. (b) Find the current. (c) Find the power used by each bulb now. Which one glows brighter?
- Given: P₁ = 100 W, P₂ = 40 W, both at 220 V; in series across 220 V.
- Formulas: R = V² ÷ P (from P = V² ÷ R, page 191); series (page 184); P = I² R (page 191).
- Working: (a) R₁ = 220² ÷ 100 = 48,400 ÷ 100 = 484 Ω; R₂ = 48,400 ÷ 40 = 1,210 Ω. (b) Total 484 + 1,210 = 1,694 Ω; I = 220 ÷ 1,694 ≈ 0.130 A. (c) P₁ = I² × 484 ≈ 8.2 W; P₂ = I² × 1,210 ≈ 20.4 W.
- Answer: (a) 484 Ω and 1,210 Ω; (b) about 0.13 A; (c) about 8.2 W and 20.4 W, so the 40 W bulb glows brighter.
- Common mistake: assuming the 100 W bulb must be brighter. Its rating holds only when it gets the full 220 V on its own. In series the same current flows through both, and the bulb with the larger resistance, the 40 W one, turns more energy into heat and light (P = I² R).
Practise more
- Eight more questions on the chapter, with an assertion-reason question: Class 10 Electricity questions with answers.
- The ideas behind each formula, step by step: NCERT Class 10 Science Chapter 11, Electricity, explained.
- For the maths side of your revision: the Class 10 Maths formula sheet 2026-27.
- How the chapter fits the board paper: our Class 10 Science syllabus 2026-27 guide.
Where CMDLearn fits
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Sources
All opened and read on 3 November 2026. Page numbers are the printed page numbers of the textbook.
- NCERT, Science, Textbook for Class X, Chapter 11 "Electricity" (Reprint 2026-27), printed pages 171-194 (Q = I t, page 172; Ohm's law, page 176; R = ρ l ÷ A, page 178; Table 11.2 of resistivities, page 179; resistors in series, page 184; resistors in parallel, page 186; the heating effect and H = I² R t, pages 188-189; electric power, P = V I = I² R = V² ÷ R, and the kilowatt hour, page 191): ncert.nic.in/textbook/pdf/jesc111.pdf
The formulas and the resistivity of nichrome are from the textbook pages above; Table 11.2 was read from the rendered page. The numericals, the made-up figures and the worked answers are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.
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