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CBSE · Class 10 · Science

Electricity

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
A charge of 88 C flows through a wire in 22 s. The current is:
  1. 0.25 A
  2. 1,936 A
  3. 5 A
  4. 4 A
2.
A current of 10 A flows through a 37 Ω resistor. The potential difference across it is:
  1. 370 V
  2. 47 V
  3. 3.7 V
  4. 3700 V
3.
A potential difference of 8 V drives a current of 4 A through a conductor. Its resistance is:
  1. 4 Ω
  2. 32 Ω
  3. 6 Ω
  4. 2 Ω
4.
Resistors of 27 Ω, 4 Ω and 15 Ω are connected in series. The equivalent resistance is:
  1. 1620 Ω
  2. 2.83 Ω
  3. 27 Ω
  4. 46 Ω
5.
The SI unit of electric current is:
  1. coulomb (C)
  2. volt (V)
  3. ampere (A)
  4. ohm (Ω)
6.
Resistors of 9 Ω and 18 Ω are connected in parallel. The equivalent resistance is:
  1. 6 Ω
  2. 13.5 Ω
  3. 27 Ω
  4. 9 Ω
7.
An appliance has a resistance of 6 Ω and works at 54 V. Its power is:
  1. 54 W
  2. 972 W
  3. 486 W
  4. 63 W
8.
Assertion (A): A 11 Ω and a 16 Ω resistor in series have an equivalent resistance of 27 Ω.
Reason (R): Electric current is the rate of flow of charge.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
How many electrons make up a charge of 19 C? (Charge on an electron = 1.6 × 10−19 C)
10.
Find the resistance of a wire 1 m long with a cross-section of 2 mm2, made of a material of resistivity 10 × 10−8 Ω m. (Give the answer to 2 decimal places where needed.)
11.
A 104 V battery is connected to 13 Ω and 13 Ω resistors in series. Find the current and the potential difference across each resistor.
12.
Resistors of 9 Ω and 18 Ω are connected in parallel across a 162 V supply. Find the current in each and the total current.
13.
A current of 7 A flows through a 39 Ω resistor for 293 s. Find the heat produced.
14.
A 750 W appliance runs 2 hours a day for 30 days. Find the energy used and its cost at Rs 8 per kWh.
15.
A 17 Ω resistor is connected in series with a parallel pair of 15 Ω and 30 Ω. Find the total resistance.
16.
An appliance is rated 1,100 W at 220 V. Find the current it draws.
Section C — Word problems4 × 4 = 16 marks
17.
An electric kettle is rated 1,320 W at 220 V. Find the current it draws and its resistance. (Give the answer to 2 decimal places where needed.)
18.
9 bulbs of 100 W each are connected in parallel to a 220 V supply. Find the total current drawn. (Give the answer to 2 decimal places where needed.)
19.
A heater of resistance 22 Ω is connected to a 220 V supply for 41 minutes. Find the heat produced, in joules and in kWh. (Give the answer to 2 decimal places where needed.)
20.
Two bulbs are rated 40 W and 200 W, both at 220 V. Find their resistances. Which has the higher resistance? (Give the answer to 2 decimal places where needed.)
AmbifulWorksheets
Practice that never repeats
CBSE · Class 10 · Science

Answer key

Electricity · AW-10S-ELE-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 4 AI = Q ÷ t = 88 ÷ 22
2(a) 370 VV = IR = 10 × 37
3(d) 2 ΩR = V ÷ I = 8 ÷ 4
4(d) 46 Ωadd them: 27 + 4 + 15
5(c) ampere (A)—
6(a) 6 Ω1/R = 1/9 + 1/18, so R = 9 × 18 ÷ 27 = 6
7(c) 486 WP = V2/R = 542 ÷ 6
8(b) Both A and R are true, but R is not the correct explanation of A.A is true: in series, R = 11 + 16 = 27. R is true but does not explain A.
91.1875 × 102019 ÷ (1.6 × 10−19) = 1.1875 × 1020
100.05 ΩR = ρL/A = (10 × 10−8 × 1) ÷ (2 × 10−6)
114 A; 52 V and 52 VI = 104 ÷ (13 + 13); V1 = 4 × 13, V2 = 4 × 13
1218 A and 9 A; total 27 AI = V ÷ R for each branch: 162 ÷ 9, 162 ÷ 18
13559,923 JH = I2Rt = 72 × 39 × 293
1445 kWh; Rs 360750 × 2 × 30 ÷ 1000 kWh
1527 Ωparallel pair = 15 × 30 ÷ 45 = 10; add 17
165 AI = P ÷ V = 1100 ÷ 220
176 A; 36.67 ΩI = P ÷ V = 1320 ÷ 220; R = V2 ÷ P = 2202 ÷ 1320
184.09 Atotal power 900 W; I = 900 ÷ 220
195,412,000 J; 1.50 kWhP = V2/R = 2,200 W; H = P × 2460 s; 1 kWh = 3.6 × 106 J
201,210 Ω and 242 Ω; the 40 W bulbR = V2/P: 2202 ÷ 40 and 2202 ÷ 200; the lower-power bulb has more resistance