Electricity: Circuits and Their Components – Numerical Problems with Stepwise Solutions
CBSE Class 7 Science – Chapter 3
Electricity: Circuits and Their Components – Numerical Problems with Stepwise Solutions
Class: 7 (Middle School)
Subject: Science – Physics (Basic Electricity)
Board: CBSE – Based on NCERT
Chapter No.: 3
CBSE Board Examinations – Practice Sequence:
1. NCERT Textbook Concepts
2. Topic-wise Numerical Practice
3. School Unit & Term Exams
4. Annual / Pre-Board Exams
Content Bank: Important Points & Simple Formulas
Use these simple Class 7 level formulas and facts while solving numerical problems from the chapter “Electricity: Circuits and Their Components”.
- 1. Battery (cells in series): Number of cells in battery = (Number of cells in one group) × (Number of such groups)
- 2. Total cells used in devices: Total cells needed = (Cells used in one device) × (Number of devices)
- 3. Series connection of bulbs: If one bulb in series fuses → entire circuit becomes open → all bulbs go OFF.
- 4. Parallel idea (basic): If bulbs are connected separately to the battery (like branches) → if one fuses, others can still glow.
- 5. Time calculation (simple): Total time = Time taken for one event × Number of events (example: minutes of glow per cell × number of cells used one after another)
- 6. Cost calculation: Total cost = Cost of one item × Number of items
- 7. Fraction / percentage questions: Number of items with a property = Given fraction × Total number of items
- 8. Open & closed circuit (key fact): Closed circuit → current flows → bulb glows. Open circuit → current does not flow → bulb does not glow.
Numerical Problems with Stepwise Solutions
Instructions: The following 20 numerical problems are strictly based on the NCERT Class 7 Science Chapter 3 – Electricity: Circuits and Their Components. The questions are kept simple and concept-based, matching CBSE Class 7 standard.
- Problems are arranged topic-wise to help in gradual learning.
- Each question has a “Given” section followed by stepwise solution.
- Focus is on understanding concepts like cells, batteries, bulbs, circuits, conductors, insulators and their simple numerical applications.
Topic 1: Electric Cell and Battery – Counting Cells & Simple Calculations
Q1.
Cells in a Torch
A torch needs 2 cells to work. How many cells are needed to fill 7 such torches completely?
Given: 2 cells per torch, 7 torches.
Stepwise Solution:
- Each torch uses 2 cells.
- Total number of torches = 7.
- Use the simple multiplication idea: Total cells needed = cells per torch × number of torches.
- So, total cells = 2 × 7 = 14.
Answer: 14 cells are needed to fill 7 torches.
Q2.
Battery in Toys
A toy car uses 3 cells placed in series to form a battery. How many cells are required to prepare 10 such toy cars for a school exhibition?
Given: 3 cells per car, 10 cars.
Stepwise Solution:
- Cells needed for 1 toy car = 3.
- Number of toy cars = 10.
- Total cells required = cells per car × number of cars = 3 × 10.
- 3 × 10 = 30.
Answer: 30 cells are required for 10 toy cars.
Q3.
Batteries in Radios
A small radio uses a battery of 4 cells. If a family has 3 such radios and wants to replace all the cells at once, how many cells will they need?
Given: 4 cells per radio, 3 radios.
Stepwise Solution:
- Cells used in 1 radio = 4.
- Number of radios = 3.
- Total cells needed = 4 × 3 = 12.
Answer: The family needs 12 cells in total.
Q4.
Cells in Series Groups
A battery is made by joining 3 cells in series. A science teacher prepares 5 such identical batteries for classroom experiments. How many cells does the teacher use altogether?
Given: 3 cells per battery, 5 batteries.
Stepwise Solution:
- Cells in one battery = 3.
- Number of batteries = 5.
- Total number of cells = cells in one battery × number of batteries = 3 × 5 = 15.
Answer: The teacher uses 15 cells in total.
Q5.
Cost of Cells
The cost of one cell is ₹12. A student buys 6 cells to make different circuits at home. How much money does the student pay?
Given: Cost of 1 cell = ₹12, Number of cells = 6.
Stepwise Solution:
- Cost of one cell = ₹12.
- Number of cells bought = 6.
- Total cost = cost of 1 cell × number of cells = 12 × 6.
- 12 × 6 = 72.
Answer: The student pays ₹72 for 6 cells.
Topic 2: Electric Bulb and Circuits – Counting Bulbs & Simple Reasoning
Q6.
Torch Bulbs
A shopkeeper has 18 torch bulbs. He wants to pack them equally in 3 boxes for selling. How many bulbs will each box contain?
Given: 18 bulbs, 3 boxes, equal packing.
Stepwise Solution:
- Total bulbs = 18.
- Total boxes = 3.
- Bulbs in each box = total bulbs ÷ number of boxes = 18 ÷ 3.
- 18 ÷ 3 = 6.
Answer: Each box will contain 6 bulbs.
Q7.
Series Bulbs
In a decorative series circuit, 12 small bulbs are connected one after another. If 3 bulbs fuse, how many bulbs are left in working condition (even though the circuit will not glow)?
Given: Total bulbs = 12, fused bulbs = 3.
Stepwise Solution:
- Total bulbs = 12.
- Number of fused bulbs = 3.
- Working bulbs = total bulbs – fused bulbs = 12 – 3 = 9.
- Note: The circuit will still not glow because it is a series circuit and any one fused bulb breaks the circuit.
Answer: 9 bulbs are still in working condition.
Q8.
Bulbs Per Circuit
A teacher has 24 bulbs and wants to make 4 identical series circuits with equal number of bulbs in each. How many bulbs will each circuit have?
Given: 24 bulbs, 4 identical circuits.
Stepwise Solution:
- Total bulbs = 24.
- Number of circuits = 4.
- Bulbs in each circuit = total bulbs ÷ number of circuits = 24 ÷ 4.
- 24 ÷ 4 = 6.
Answer: Each circuit will have 6 bulbs.
Q9.
Bulbs & Families
In an apartment building, each of the 8 flats keeps 5 spare bulbs for emergencies. How many spare bulbs are there in the whole building?
Given: 8 flats, 5 bulbs per flat.
Stepwise Solution:
- Number of flats = 8.
- Spare bulbs per flat = 5.
- Total spare bulbs = bulbs per flat × number of flats = 5 × 8.
- 5 × 8 = 40.
Answer: There are 40 spare bulbs in the building.
Q10.
Fused Bulb Count
Out of 30 bulbs in a shop, 4 are found to be fused and cannot glow. What fraction of bulbs are fused? Also, how many good bulbs are left?
Given: Total bulbs = 30, fused bulbs = 4.
Stepwise Solution:
- Total bulbs = 30.
- Number of fused bulbs = 4.
- Fraction of fused bulbs = fused ÷ total = 4 ÷ 30 = 4/30.
- Simplify 4/30 by dividing numerator and denominator by 2: 4 ÷ 2 = 2, 30 ÷ 2 = 15 → fraction = 2/15.
- Good bulbs = total bulbs – fused bulbs = 30 – 4 = 26.
Answer: Fused bulbs = 2/15 of total; good bulbs left = 26.
Topic 3: Simple Time & Usage Calculations with Cells and Torches
Q11.
Torch Usage
A cell can light a torch continuously for 50 minutes. If 3 cells are used one after another (not together) in the same torch, for how many minutes in total can the torch glow?
Given: 50 minutes per cell, 3 cells used one after another.
Stepwise Solution:
- Time for which 1 cell can be used = 50 minutes.
- Number of such cells used one after another = 3.
- Total time = time for 1 cell × number of cells = 50 × 3.
- 50 × 3 = 150 minutes.
Answer: The torch can glow for 150 minutes in total.
Q12.
Daily Torch Use
A student uses a torch for 10 minutes every night during power cuts. If one cell can run the torch for 60 minutes, for how many days can one cell last?
Given: 10 minutes usage per day, cell life = 60 minutes.
Stepwise Solution:
- Total working time of one cell = 60 minutes.
- Daily usage = 10 minutes.
- Number of days = total minutes ÷ minutes used per day = 60 ÷ 10.
- 60 ÷ 10 = 6.
Answer: One cell can last for 6 days.
Q13.
Lamp Usage
A small battery-operated study lamp can run for 40 minutes on one fully charged battery. How many minutes can it run if 4 such batteries are used one after another?
Given: 40 minutes per battery, 4 batteries one after another.
Stepwise Solution:
- Time for which 1 battery can run the lamp = 40 minutes.
- Number of batteries used one after another = 4.
- Total time = 40 × 4 = 160 minutes.
Answer: The lamp can run for 160 minutes in total.
Q14.
Time per Day
A torch has new cells which can provide light for 90 minutes. If it is used for 15 minutes daily, after how many days will the cells be exhausted?
Given: Total available time = 90 minutes, 15 minutes per day.
Stepwise Solution:
- Total time cells can provide light = 90 minutes.
- Daily usage = 15 minutes.
- Number of days = total time ÷ daily usage = 90 ÷ 15.
- 90 ÷ 15 = 6.
Answer: The cells will be exhausted after 6 days.
Q15.
Project Work
For a science project, a group of students needs their battery-operated model to run for 120 minutes. If one battery can run it for 30 minutes, how many batteries are required (used one after another)?
Given: Required time = 120 min, 30 min per battery.
Stepwise Solution:
- Time provided by 1 battery = 30 minutes.
- Total time needed = 120 minutes.
- Number of batteries = total time ÷ time per battery = 120 ÷ 30.
- 120 ÷ 30 = 4.
Answer: The students need 4 batteries.
Topic 4: Conductors, Insulators and Simple Counting / Fraction Problems
Q16.
Testing Materials
A student tests 20 different objects in a simple circuit to check whether they are conductors or insulators. He finds that 7 of them make the bulb glow. How many objects act as insulators?
Given: Total objects = 20, conductors (bulb glows) = 7.
Stepwise Solution:
- Total objects tested = 20.
- Number of conductors (bulb glowed) = 7.
- Number of insulators = total objects – conductors = 20 – 7.
- 20 – 7 = 13.
Answer: 13 objects act as insulators.
Q17.
Fraction of Conductors
In a classroom activity, out of 15 objects, 9 are found to be conductors. What fraction of the objects are conductors and what fraction are insulators?
Given: Total objects = 15, conductors = 9.
Stepwise Solution:
- Total objects = 15.
- Conductors = 9.
- Fraction of conductors = conductors ÷ total = 9 ÷ 15 = 9/15.
- Simplify 9/15 by dividing by 3: 9 ÷ 3 = 3, 15 ÷ 3 = 5 → fraction = 3/5.
- Insulators = total – conductors = 15 – 9 = 6.
- Fraction of insulators = 6 ÷ 15 = 6/15 = (divide by 3) = 2/5.
Answer: Conductors = 3/5 of objects; Insulators = 2/5 of objects.
Q18.
Sets of Wires
A shop sells wires in small packs. Each pack contains 8 copper wires (conductors) and 2 plastic-coated sticks (insulators for demonstration). If a teacher buys 5 such packs, how many conductors and insulators does she get in total?
Given: 8 conductors + 2 insulators per pack; number of packs = 5.
Stepwise Solution:
- Conductors per pack = 8; Insulators per pack = 2.
- Number of packs bought = 5.
- Total conductors = 8 × 5 = 40.
- Total insulators = 2 × 5 = 10.
Answer: The teacher gets 40 conductors and 10 insulators.
Q19.
Percentage Idea
In a science fair, a team displays 25 items related to electricity. If 20 of them are conductors, what percentage of the items are conductors?
Given: Total items = 25, conductors = 20.
Stepwise Solution:
- Total items = 25.
- Conductors = 20.
- Fraction of conductors = 20 ÷ 25 = 20/25.
- To convert to percentage: (fraction × 100) % = (20/25) × 100.
- 20 ÷ 25 = 0.8 → 0.8 × 100 = 80%.
Answer: 80% of the items are conductors.
Q20.
Safe Materials
A student has 12 objects. Out of these, 5 are safe insulators to hold in an electric experiment. How many objects are not safe to hold directly (i.e., likely conductors)?
Given: Total objects = 12, safe insulators = 5.
Stepwise Solution:
- Total objects = 12.
- Safe insulators = 5.
- Objects that are not safe to hold directly = conductors = total – insulators = 12 – 5.
- 12 – 5 = 7.
Answer: 7 objects are likely conductors and not safe to hold directly.