Electricity is one of the most important chapters in CBSE Class 10 Science. Many students understand the theory but find it difficult to solve numerical problems involving electric current, potential difference, resistance, electric power and electrical energy.
The good news is that most Class 10 Electricity numericals can be solved easily if you understand the formulas, identify the given values and follow a systematic approach.
In this guide, you will learn how to solve Class 10 Electricity numericals step by step, with clear explanations, solved examples, important formulas and practice questions.
Whether you are preparing for your school examination, pre-board examination or CBSE Board Exam 2027, this article will help you improve your numerical-solving skills.
Table of Contents
- Important Concepts of Electricity
- Important Electricity Formulas
- How to Solve Electricity Numericals
- Numericals Based on Electric Current
- Numericals Based on Ohm's Law
- Resistance and Resistivity Numericals
- Series Combination of Resistors
- Parallel Combination of Resistors
- Electric Power Numericals
- Electrical Energy and Heating Effect
- Competency-Based Questions
- Common Mistakes
- Practice Questions with Answers
- Preparation Strategy
- Frequently Asked Questions
1. Important Concepts of Electricity
Before solving numerical problems, students should understand the following basic concepts.
Electric Current
Electric current is the rate of flow of electric charge through a conductor.
Where:
- I = Electric current
- Q = Electric charge
- t = Time
SI unit of electric current is ampere (A). One ampere means one coulomb of charge flows through a conductor in one second.
Potential Difference
Potential difference between two points is the work done per unit charge to move a charge from one point to another.
Where V is potential difference, W is work done and Q is electric charge. Its SI unit is volt (V).
Resistance
Resistance is the property of a conductor that opposes the flow of electric current.
Its SI unit is ohm (Ω).
Ohm's Law
According to Ohm's law, the current flowing through a conductor is directly proportional to the potential difference across its ends, provided its temperature and other physical conditions remain constant.
This is one of the most important formulas for solving Class 10 Electricity numericals.
2. Important Electricity Formulas for Class 10
Students should memorise the following formulas before attempting numerical problems.
| Physical Quantity | Formula | SI Unit |
|---|---|---|
| Electric current | I = Q/t | Ampere |
| Electric charge | Q = It | Coulomb |
| Potential difference | V = W/Q | Volt |
| Ohm's law | V = IR | Volt |
| Resistance | R = V/I | Ohm |
| Resistance of conductor | R = ρl/A | Ohm |
| Series resistance | Rs = R1 + R2 + R3 | Ohm |
| Parallel resistance | 1/Rp = 1/R1 + 1/R2 + 1/R3 | Ohm |
| Electric power | P = VI | Watt |
| Power | P = I²R | Watt |
| Power | P = V²/R | Watt |
| Electrical energy | E = Pt | Joule |
| Heating effect | H = I²Rt | Joule |
Important Unit Conversions
- 1 kW = 1000 W
- 1 hour = 3600 seconds
- 1 kWh = 3.6 × 10⁶ J
- 1 mA = 0.001 A
- 1 kΩ = 1000 Ω
3. How to Solve Class 10 Electricity Numericals
Follow this five-step method to solve numerical problems correctly.
- Read the question carefully: Identify what the question is asking.
- Write the given values: Note down all the numerical values with units.
- Identify the formula: Select the correct formula.
- Substitute the values: Put the given values into the formula.
- Write the final answer: Include the correct SI unit.
Potential difference = 12 V
Resistance = 4 Ω
Required current:
I = V/R
I = 12/4
I = 3 A
4. Numericals Based on Electric Current
Example 1: Calculate Electric Current
Question: A charge of 600 C flows through a conductor in 5 minutes. Calculate the electric current.
Q = 600 C
t = 5 minutes = 300 seconds
Formula:I = Q/t
Substitution:I = 600/300
I = 2 A
Answer: The electric current is 2 A.Example 2: Calculate Electric Charge
Question: A current of 2.5 A flows through a wire for 4 minutes. Calculate the total charge flowing through the wire.
I = 2.5 A
t = 4 × 60 = 240 seconds
Formula:Q = It
Q = 2.5 × 240
Q = 600 C
Answer: The total electric charge is 600 coulombs.5. Numericals Based on Ohm's Law
Example 3: Calculate Current Using Ohm's Law
Question: A potential difference of 12 V is applied across a resistor of 6 Ω. Find the current flowing through it.
V = 12 V, R = 6 Ω
Formula:I = V/R
I = 12/6
I = 2 A
Answer: The current is 2 A.Example 4: Calculate Resistance
Question: A potential difference of 24 V produces a current of 3 A through a conductor. Calculate its resistance.
V = 24 V, I = 3 A
Formula:R = V/I
R = 24/3
R = 8 Ω
Answer: The resistance is 8 Ω.Example 5: Calculate Potential Difference
Question: A current of 0.5 A flows through a resistor of 20 Ω. Find the potential difference.
I = 0.5 A, R = 20 Ω
Formula:V = IR
V = 0.5 × 20
V = 10 V
Answer: The potential difference is 10 V.6. Numericals Based on Resistance and Resistivity
Resistance depends on the length, area of cross-section, material and temperature of a conductor.
Where R is resistance, ρ is resistivity, l is length and A is area of cross-section.
Example 6: Calculate Resistance Using Resistivity
Question: A wire has a length of 2 m and a cross-sectional area of 1 × 10⁻⁶ m². If its resistivity is 1.7 × 10⁻⁸ Ωm, calculate its resistance.
l = 2 m
A = 1 × 10⁻⁶ m²
ρ = 1.7 × 10⁻⁸ Ωm
Formula:R = ρl/A
R = (1.7 × 10⁻⁸ × 2)/(1 × 10⁻⁶)
R = 0.034 Ω
Answer: The resistance is 0.034 Ω.Example 7: Effect of Changing the Length of a Wire
Question: The resistance of a wire is 5 Ω. What will be its resistance if its length is doubled without changing its area of cross-section or material?
We know that resistance is directly proportional to length.
R ∝ l
New resistance = 2 × 5
New resistance = 10 Ω
Answer: The new resistance is 10 Ω.7. Numericals Based on Series Combination of Resistors
When resistors are connected one after another in a single path, they are said to be connected in series.
- Current remains the same through all resistors.
- Potential difference is divided among the resistors.
- Equivalent resistance is greater than the individual resistances.
Example 8: Calculate Equivalent Resistance in Series
Question: Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series. Calculate their equivalent resistance.
R1 = 2 Ω, R2 = 3 Ω, R3 = 5 Ω
Formula:Rs = R1 + R2 + R3
Rs = 2 + 3 + 5
Rs = 10 Ω
Answer: The equivalent resistance is 10 Ω.Example 9: Calculate Current in a Series Circuit
Question: Two resistors of 4 Ω and 6 Ω are connected in series with a 20 V battery. Calculate the current.
Rs = 4 + 6 = 10 Ω
Step 2: Apply Ohm's law.I = V/R
I = 20/10
I = 2 A
Answer: The current is 2 A.8. Numericals Based on Parallel Combination of Resistors
When two or more resistors are connected between the same two points, they are said to be connected in parallel.
For two resistors:
- Potential difference is the same across each resistor.
- Current is divided among the branches.
- Equivalent resistance is less than the smallest individual resistance.
Example 10: Calculate Equivalent Resistance in Parallel
Question: Two resistors of 6 Ω and 3 Ω are connected in parallel. Calculate their equivalent resistance.
R1 = 6 Ω, R2 = 3 Ω
Formula:Rp = (R1 × R2)/(R1 + R2)
Rp = (6 × 3)/(6 + 3)
Rp = 18/9
Rp = 2 Ω
Answer: The equivalent resistance is 2 Ω.Example 11: Calculate Current in a Parallel Circuit
Question: Two resistors of 4 Ω and 12 Ω are connected in parallel across a 12 V battery. Calculate the total current supplied by the battery.
Rp = (4 × 12)/(4 + 12)
Rp = 48/16 = 3 Ω
Step 2: Calculate total current.I = V/Rp
I = 12/3
I = 4 A
Answer: The total current is 4 A.9. Numericals Based on Electric Power
Electric power is the rate at which electrical energy is consumed or converted into another form of energy.
The SI unit of power is watt (W).
Example 12: Calculate Electric Power
Question: An electric appliance operates at 220 V and draws a current of 2 A. Calculate its power.
V = 220 V, I = 2 A
Formula:P = VI
P = 220 × 2
P = 440 W
Answer: The power is 440 W.Example 13: Calculate Resistance from Power and Voltage
Question: An electric bulb is rated 100 W, 200 V. Calculate its resistance.
P = 100 W, V = 200 V
Formula:R = V²/P
R = (200 × 200)/100
R = 40000/100
R = 400 Ω
Answer: The resistance is 400 Ω.10. Numericals Based on Electrical Energy and Heating Effect
When electric current passes through a resistor, electrical energy is converted into heat.
Joule's Law of Heating
Electrical energy is calculated using:
The commercial unit of electrical energy is kilowatt-hour (kWh).
Example 14: Calculate Electrical Energy Consumed
Question: An electric heater of 1000 W is used for 2 hours. Calculate the electrical energy consumed in kWh.
Power = 1000 W = 1 kW
Time = 2 hours
Formula:E = P × t
E = 1 × 2
E = 2 kWh
Answer: The electrical energy consumed is 2 kWh or 2 units.Example 15: Calculate Heat Produced
Question: A current of 2 A flows through a resistance of 5 Ω for 10 seconds. Calculate the heat produced.
I = 2 A, R = 5 Ω, t = 10 seconds
Formula:H = I²Rt
H = 2² × 5 × 10
H = 4 × 5 × 10
H = 200 J
Answer: The heat produced is 200 joules.Example 16: Calculate the Cost of Electricity
Question: A 2 kW electric heater is used for 3 hours daily for 10 days. If the cost of electricity is ₹6 per unit, calculate the total electricity bill.
Power = 2 kW
Daily usage = 3 hours
Number of days = 10
Cost per unit = ₹6
Step 1: Calculate energy consumed.Energy = 2 × 3 × 10 = 60 kWh
Step 2: Calculate cost.Cost = 60 × 6
Total cost = ₹360
Answer: The total electricity cost is ₹360.11. Mixed Competency-Based Electricity Questions
Competency-based questions test whether students can apply scientific concepts to practical situations.
Question 1: Electric Heater
An electric heater is connected to a 220 V supply and draws a current of 5 A. Find (a) resistance and (b) power.
R = V/I = 220/5
R = 44 Ω
(b) Power:P = VI = 220 × 5
P = 1100 W
Question 2: Two Bulbs
Two bulbs rated 60 W and 100 W are connected to the same voltage supply. Which bulb has greater resistance?
We know that R = V²/P.
For the same voltage, resistance is inversely proportional to power.
The 60 W bulb has greater resistance.
Question 3: Household Electrical Consumption
A family uses a 100 W television for 5 hours daily. Calculate the energy consumed in 30 days.
Power = 100 W = 0.1 kW
Time = 5 × 30 = 150 hours
Energy = P × t
Energy = 0.1 × 150
Energy = 15 kWh
Answer: The television consumes 15 units of electrical energy.12. Common Mistakes Students Should Avoid
Even students who understand the formulas sometimes lose marks due to simple calculation and presentation mistakes.
- Forgetting unit conversion: Convert minutes into seconds when using SI units.
- Using the wrong formula: Identify the required physical quantity before selecting a formula.
- Confusing series and parallel: In series, resistances are added directly. In parallel, reciprocals are added.
- Ignoring square terms: Remember that P = I²R contains the square of current.
- Writing incomplete answers: Always mention the correct SI unit.
- Making calculation errors: Check multiplication, division and decimal placement.
- Confusing kW and kWh: kW measures power, whereas kWh measures electrical energy.
13. Practice Questions for CBSE Class 10 Students
Try solving these questions independently before checking the answers.
Section A: Basic Numericals
- A charge of 120 C flows through a conductor in 2 minutes. Calculate the current.
- A resistor of 10 Ω is connected to a 20 V battery. Find the current.
- Calculate the potential difference across a resistor of 15 Ω carrying a current of 2 A.
Section B: Resistance Combination
- Three resistors of 2 Ω, 4 Ω and 6 Ω are connected in series. Find the equivalent resistance.
- Two resistors of 8 Ω and 8 Ω are connected in parallel. Calculate their equivalent resistance.
- Two resistors of 5 Ω and 10 Ω are connected in parallel. Find the equivalent resistance.
Section C: Power and Energy
- An electric appliance operates at 220 V and draws 1 A current. Calculate its power.
- A 500 W electric appliance is used for 4 hours. Calculate the energy consumed in kWh.
- Calculate the resistance of a 100 W bulb operating at 200 V.
- A current of 3 A flows through a resistance of 4 Ω for 5 seconds. Calculate the heat produced.
Answer Key
| Question | Correct Answer |
|---|---|
| Q1 | 1 A |
| Q2 | 2 A |
| Q3 | 30 V |
| Q4 | 12 Ω |
| Q5 | 4 Ω |
| Q6 | 3.33 Ω (approximately) |
| Q7 | 220 W |
| Q8 | 2 kWh |
| Q9 | 400 Ω |
| Q10 | 180 J |
14. How to Prepare Electricity Numericals for CBSE Board Exam 2027
You do not need to solve hundreds of questions every day to become good at Electricity numericals. A systematic practice routine is more effective.
| Day | What to Practise |
|---|---|
| Day 1 | Electric current, charge and potential difference |
| Day 2 | Ohm's law and resistance |
| Day 3 | Resistivity and factors affecting resistance |
| Day 4 | Series and parallel combinations |
| Day 5 | Electric power and electrical energy |
| Day 6 | Heating effect and mixed numericals |
| Day 7 | Full chapter revision and practice test |
Five Effective Tips for Better Numerical Solving
1. Memorise formulas through regular practice
Do not simply read formulas. Write them in a notebook and practise using them.
2. Understand the meaning of each formula
For example, P = VI tells you that electric power depends on both voltage and current.
3. Practise different types of questions
Solve basic, intermediate and application-based questions.
4. Maintain a separate formula notebook
Write all important formulas and unit conversions in one place.
5. Analyse your mistakes
Whenever you solve a question incorrectly, identify whether the mistake was related to the formula, unit conversion or calculation.
15. Frequently Asked Questions (FAQs)
Q1. Which formulas are most important in Class 10 Electricity?
The most important formulas are I = Q/t, V = IR, R = ρl/A, the series and parallel resistance formulas, P = VI, P = I²R, P = V²/R and E = Pt.
Q2. How can I solve Electricity numericals easily?
Read the question carefully, write the given values, identify the correct formula, substitute the values and calculate the answer with the appropriate unit.
Q3. What is the difference between series and parallel combinations?
In a series combination, the same current flows through every resistor. In a parallel combination, the potential difference across each resistor is the same.
Q4. What is the commercial unit of electrical energy?
The commercial unit of electrical energy is kilowatt-hour (kWh). One kWh is equal to 3.6 × 10⁶ joules.
Q5. How many numerical problems should I practise daily?
Try solving 5–10 numerical problems daily. Focus on understanding the method rather than simply memorising answers.
Q6. What is the relationship between electric power and resistance?
For a fixed potential difference, power is inversely proportional to resistance (P = V²/R). For a fixed current, power is directly proportional to resistance (P = I²R).
Q7. Are Electricity numericals important for CBSE Class 10 Board Exams?
Yes. Numerical problems are an important way of testing students' understanding of electric current, resistance, Ohm's law, combinations of resistors, power and electrical energy.
Conclusion
Electricity numericals become much easier when you understand the basic concepts and apply the correct formulas.
Start with simple questions based on electric current and Ohm's law. Then gradually move towards series and parallel combinations, electric power, electrical energy and heating effect.
Remember, regular practice is more important than memorising solutions.
Understand the formula, practise the steps and check your answers. With consistent effort, you can develop confidence in solving Class 10 Electricity numericals.
Keep learning and keep improving with Choice Better – Helping Students Study Smarter!
Official Reference
For the latest prescribed topics, refer to the CBSE Class 10 Science Curriculum 2026–27 and your prescribed NCERT Science textbook.
Disclaimer: This article is an independent educational resource. The practice questions and worked examples are original learning exercises, not official or predicted CBSE examination questions.
