Electricity explains how charges flow through a circuit and how current, voltage and resistance control that flow. In Important Questions Class 10 Science Chapter 11, students practise electric current, potential difference, Ohm’s law, resistance, resistors, heating effect and electric power.
Electricity is a scoring chapter, but only when students practise formulas with units. A small error in ampere, volt, ohm, second or kilowatt-hour can change the answer. These Important Questions Class 10 Science Chapter 11 help students revise Electricity through short answers, numericals, reasoning questions and 5-mark practice. The answers focus on formulas, circuit logic, series and parallel combinations, heating effect and electric power for CBSE 2026 exam preparation.
Key Takeaways from Class 10 Science Chapter 11 Electricity
| Area |
What Students Should Revise |
| Chapter Name |
Electricity |
| Chapter Number |
Chapter 11 |
| Subject |
Class 10 Science |
| Main Concepts |
Current, circuit, voltage, resistance and power |
| Important Laws |
Ohm’s law and Joule’s law of heating |
| Scoring Areas |
Numericals and circuit-based questions |
| Common Formulas |
I = Q/t, V = W/Q, V = IR, H = I²Rt, P = VI |
| Important Devices |
Ammeter, voltmeter, rheostat, fuse |
| Best Practice |
Write formula, substitute values and add units |
Important Questions Class 10 Science Chapter 11: Chapter Overview
Important Questions Class 10 Science Chapter 11 focus on how electric charge flows and how circuits behave.
The chapter begins with electric current and potential difference. It then explains Ohm’s law, resistance, resistivity, combinations of resistors, heating effect of current and electric power.
The chapter covers these core areas:
- Electric current and circuit
- Electric potential and potential difference
- Ohm’s law
- Resistance and resistivity
- Factors affecting resistance
- Series combination of resistors
- Parallel combination of resistors
- Heating effect of electric current
- Electric power
- Commercial unit of electrical energy
Class 10 Science Chapter 11 Electricity Formula Sheet
Class 10 Science Chapter 11 Electricity numericals become easier when students revise formulas first.
Most questions use direct formula substitution, but units must be converted correctly.
| Formula |
Use |
| I = Q/t |
Current from charge and time |
| Q = It |
Charge flowing in a circuit |
| V = W/Q |
Potential difference |
| W = VQ |
Work done in moving charge |
| V = IR |
Ohm’s law |
| R = V/I |
Resistance |
| R = ρl/A |
Resistance using resistivity |
| Rs = R₁ + R₂ + R₃ |
Series combination |
| 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ |
Parallel combination |
| H = I²Rt |
Joule’s law of heating |
| P = VI |
Electric power |
| P = I²R |
Power using current and resistance |
| P = V²/R |
Power using voltage and resistance |
| 1 kWh = 3.6 × 10⁶ J |
Commercial unit of electrical energy |
Electricity Class 10 Important Questions with Answers on Current and Circuit
Electricity class 10 important questions with answers often begin with current, charge and simple circuits. Electric current is the rate of flow of electric charge. A circuit must remain closed for current to flow.
Important Questions Class 10 Science Chapter 11 on Electric Current
Q1. What is an electric circuit?
An electric circuit is a continuous and closed path for electric current.
If the circuit breaks, current stops flowing.
Q2. Define electric current.
Electric current is the rate of flow of electric charge through a conductor.
Its SI unit is ampere.
Q3. Write the relation between current, charge and time.
The relation is I = Q/t.
Here, I is current, Q is charge and t is time.
Q4. A current of 0.5 A flows through a bulb for 10 minutes. Find the charge.
| Given |
Value |
| Current |
0.5 A |
| Time |
10 minutes = 600 s |
| Formula |
Q = It |
| Charge |
0.5 × 600 = 300 C |
Answer: The charge flowing through the circuit is 300 C.
Q5. Why is an ammeter connected in series?
An ammeter measures current through a circuit.
It is connected in series so the same current passes through it.
Class 10 Electricity Question Answer on Potential Difference
Class 10 electricity question answer practice should include potential difference because it explains why charge moves in a circuit.
Potential difference pushes charges through a conductor. A cell or battery maintains this difference in a circuit.
Q6. What is potential difference?
Potential difference is the work done to move a unit charge from one point to another.
Its SI unit is volt.
Q7. What does 1 volt mean?
One volt means 1 joule of work moves 1 coulomb of charge between two points.
Q8. Write the formula for potential difference.
The formula is V = W/Q.
Here, V is potential difference, W is work done and Q is charge.
Q9. How much work is done in moving 2 C charge across 12 V?
| Given |
Value |
| Charge |
2 C |
| Potential difference |
12 V |
| Formula |
W = VQ |
| Work done |
12 × 2 = 24 J |
Answer: The work done is 24 J.
Q10. Why is a voltmeter connected in parallel?
A voltmeter measures potential difference between two points.
It is connected in parallel across those points.
Ohm’s Law Class 10 Questions with Answers
Ohm’s law class 10 questions are among the most important parts of Electricity.
Ohm’s law connects voltage, current and resistance. It is also used in many class 10 science chapter 11 question answer numericals.
Q11. State Ohm’s law.
Ohm’s law states that current through a conductor is directly proportional to the potential difference across its ends, provided temperature remains constant.
Q12. Write the mathematical form of Ohm’s law.
The mathematical form is V = IR.
Here, V is potential difference, I is current and R is resistance.
Q13. What does the slope of a V-I graph represent?
The slope of a V-I graph represents resistance.
A steeper slope means higher resistance.
Q14. A bulb has resistance 1200 Ω and is connected to 220 V. Find the current.
| Given |
Value |
| Voltage |
220 V |
| Resistance |
1200 Ω |
| Formula |
I = V/R |
| Current |
220/1200 = 0.18 A |
Answer: The current is 0.18 A.
Q15. What happens to current if resistance doubles at constant voltage?
The current becomes half.
According to Ohm’s law, current is inversely proportional to resistance.
Resistance Class 10 Questions from Electricity
Resistance class 10 questions test how materials oppose current flow.
Resistance opposes the flow of current. It depends on length, area of cross-section, material and temperature.
Q16. What is resistance?
Resistance is the property of a conductor that opposes the flow of electric current.
Its SI unit is ohm.
Q17. On what factors does resistance depend?
Resistance depends on length of the conductor, area of cross-section, nature of material and temperature.
Q18. How does resistance change when wire length increases?
Resistance increases when wire length increases.
Resistance is directly proportional to length.
Q19. How does resistance change when wire thickness increases?
Resistance decreases when wire thickness increases.
A thicker wire has a larger area of cross-section.
Q20. Why does current flow more easily through a thick wire?
Current flows more easily through a thick wire because it has lower resistance.
A larger cross-sectional area allows charges to move more easily.
Electricity Class 10 Numericals on Resistivity
Electricity class 10 numericals often test the formula R = ρl/A.
Students should keep length in metre and area in square metre before solving.
Q21. A wire has resistance 26 Ω, length 1 m and diameter 0.3 mm. Find its resistivity.
Use ρ = RA/l and A = πd²/4.
| Step |
Value |
| Diameter |
0.3 mm = 3 × 10⁻⁴ m |
| Area |
πd²/4 |
| Resistance |
26 Ω |
| Length |
1 m |
| Resistivity |
1.84 × 10⁻⁶ Ω m |
Answer: The resistivity is 1.84 × 10⁻⁶ Ω m.
Q22. A wire of resistance 4 Ω has length l and area A. Find resistance of another wire of same material with length l/2 and area 2A.
| Wire |
Resistance Relation |
| First wire |
R = ρl/A = 4 Ω |
| Second wire |
R = ρ(l/2)/2A = ρl/4A |
| New resistance |
4/4 = 1 Ω |
Answer: The resistance of the second wire is 1 Ω.
Series and Parallel Combination Class 10 Questions
Series and parallel combination class 10 questions are common in numericals and reasoning questions.
Students must first identify whether current is same or voltage is same.
| Basis |
Series Combination |
Parallel Combination |
| Current |
Same through each resistor |
Divides in branches |
| Voltage |
Divides across resistors |
Same across each branch |
| Equivalent Resistance |
Rs = R₁ + R₂ + R₃ |
1/Rp = 1/R₁ + 1/R₂ + 1/R₃ |
| Total Resistance |
Greater than each resistor |
Less than the smallest resistor |
| Domestic Use |
Not preferred |
Preferred |
Q23. Find total resistance of 5 Ω, 8 Ω and 12 Ω connected in series.
Total resistance = 5 + 8 + 12
= 25 Ω
Answer: Total resistance is 25 Ω.
Q24. Find equivalent resistance of 5 Ω, 10 Ω and 30 Ω in parallel.
| Step |
Calculation |
| 1/Rp |
1/5 + 1/10 + 1/30 |
| 1/Rp |
6/30 + 3/30 + 1/30 = 10/30 |
| Rp |
3 Ω |
Answer: Equivalent resistance is 3 Ω.
Q25. Why are household appliances connected in parallel?
Household appliances are connected in parallel because each appliance gets the same voltage.
If one appliance fails, the others still work.
Heating Effect of Electric Current Class 10 Questions
Heating effect of electric current class 10 questions appear in reasoning and numerical form.
This concept explains electric heaters, irons, bulbs, geysers and fuses.
Q26. What is the heating effect of electric current?
The heating effect of electric current is the production of heat when current flows through a resistor.
Q27. State Joule’s law of heating.
Joule’s law states that heat produced is directly proportional to the square of current, resistance and time.
Q28. Write the formula for heat produced in a resistor.
The formula is H = I²Rt.
Here, H is heat, I is current, R is resistance and t is time.
Q29. An electric iron of resistance 20 Ω takes current 5 A. Calculate heat produced in 30 s.
| Given |
Value |
| Current |
5 A |
| Resistance |
20 Ω |
| Time |
30 s |
| Formula |
H = I²Rt |
| Heat |
25 × 20 × 30 = 15000 J |
Answer: Heat produced is 15000 J.
Q30. Why does the heating element glow but the cord does not?
The heating element has high resistance, so it produces more heat.
The cord has low resistance, so it does not become hot enough to glow.
Electric Power Class 10 Questions with Answers
Electric power class 10 questions test watt, kilowatt, kilowatt-hour and energy use.
Electric power tells how fast electrical energy gets used or converted.
Q31. What is electric power?
Electric power is the rate at which electrical energy is consumed or converted.
Its SI unit is watt.
Q32. Write three formulas for electric power.
| Formula |
Use |
| P = VI |
When voltage and current are given |
| P = I²R |
When current and resistance are given |
| P = V²/R |
When voltage and resistance are given |
Q33. A bulb is connected to 220 V and draws 0.50 A current. Find its power.
Power = VI
= 220 × 0.50
= 110 W
Answer: Power of the bulb is 110 W.
Q34. What is the commercial unit of electrical energy?
The commercial unit of electrical energy is kilowatt-hour.
It is commonly called one unit.
Q35. Convert 1 kWh into joules.
1 kWh = 3.6 × 10⁶ J.
Electricity Class 10 5 Mark Questions
Electricity class 10 5 mark questions usually combine theory, formulas and circuit logic.
Students should write formulas clearly and explain each step.
Q36. Explain series combination of resistors and derive the formula for equivalent resistance.
In a series combination, resistors connect end to end.
The same current flows through each resistor. Total potential difference equals the sum of potential differences across individual resistors.
| Step |
Relation |
| Total voltage |
V = V₁ + V₂ + V₃ |
| Ohm’s law |
V = IR |
| Individual voltages |
V₁ = IR₁, V₂ = IR₂, V₃ = IR₃ |
| Substitute |
IRs = IR₁ + IR₂ + IR₃ |
| Result |
Rs = R₁ + R₂ + R₃ |
Answer: Equivalent resistance in series equals the sum of individual resistances.
Q37. Explain parallel combination of resistors and derive the formula for equivalent resistance.
In a parallel combination, resistors connect across the same two points.
The same potential difference acts across each resistor. Total current equals the sum of currents through each branch.
| Step |
Relation |
| Total current |
I = I₁ + I₂ + I₃ |
| Ohm’s law |
I = V/Rp |
| Branch currents |
I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃ |
| Substitute |
V/Rp = V/R₁ + V/R₂ + V/R₃ |
| Result |
1/Rp = 1/R₁ + 1/R₂ + 1/R₃ |
Answer: Reciprocal of equivalent resistance equals the sum of reciprocals of individual resistances.
Q38. Why are alloys used in electric heating devices?
Alloys have higher resistivity than their constituent metals.
They also do not oxidise easily at high temperatures. So, electric irons, toasters and heaters use alloy coils.
Q39. Why is tungsten used in electric bulb filaments?
Tungsten has a very high melting point.
It can become white hot and emit light without melting quickly.
Electricity Class 10 Extra Questions for Practice
Electricity class 10 extra questions help students practise formula selection.
Use these after completing the solved questions above.
| Question |
Practice Focus |
| A current of 2 A flows for 5 minutes. Find charge. |
Current and charge |
| A charge of 5 C moves through 10 V. Find work done. |
Potential difference |
| A 12 V battery gives 3 A current. Find resistance. |
Ohm’s law |
| Three 6 Ω resistors are connected in series. Find total resistance. |
Series combination |
| Three 6 Ω resistors are connected in parallel. Find total resistance. |
Parallel combination |
| A 100 W bulb works for 10 hours. Find energy in kWh. |
Electric energy |
| A 5 A current flows through 10 Ω for 20 s. Find heat. |
Joule’s law |
| A wire length doubles. What happens to resistance? |
Resistance factors |
Electricity Class 10 Assertion Reason Questions
Electricity class 10 assertion reason questions test concept clarity.
Read both statements carefully before selecting the relation.
Q40. Assertion: An ammeter is connected in series. Reason: An ammeter must carry the same current as the circuit element.
Answer: Both Assertion and Reason are true, and Reason explains Assertion.
Q41. Assertion: Household appliances are connected in parallel. Reason: In parallel combination, each appliance gets the same potential difference.
Answer: Both Assertion and Reason are true, and Reason explains Assertion.
Q42. Assertion: The heating effect of current is proportional to I². Reason: Joule’s law of heating is H = I²Rt.
Answer: Both Assertion and Reason are true, and Reason explains Assertion.
Q43. Assertion: A thicker wire has less resistance. Reason: Resistance is directly proportional to area of cross-section.
Answer: Assertion is true, but Reason is false.
Resistance is inversely proportional to area of cross-section.
Class 10 Science Chapter 11 Question Answer: Common Mistakes
Class 10 Science Chapter 11 question answer practice needs careful unit handling.
Students often lose marks in Electricity because they skip conversions or choose the wrong circuit formula.
| Mistake |
Correct Approach |
| Writing time in minutes |
Convert time into seconds |
| Mixing kW and W |
Convert 1 kW = 1000 W |
| Forgetting kWh conversion |
1 kWh = 3.6 × 10⁶ J |
| Using series formula for parallel circuit |
Check connection before solving |
| Connecting voltmeter in series |
Voltmeter connects in parallel |
| Connecting ammeter in parallel |
Ammeter connects in series |
| Ignoring square in H = I²Rt |
Square the current first |
Quick Revision Notes for Electricity Class 10 Important Questions
Electric current is the rate of flow of charge. A closed circuit allows current to flow.
Potential difference moves charges through a conductor. A voltmeter measures potential difference in parallel.
Ohm’s law gives the relation V = IR. Resistance opposes current and depends on length, area, material and temperature.
In series, equivalent resistance increases. In parallel, equivalent resistance decreases.
Joule’s law of heating gives H = I²Rt. Electric power can be calculated using P = VI, P = I²R or P = V²/R.