Q1. What is an electric current?
Answer:
Electric current is the rate of flow of electric charges through a conductor.
It is represented by I.
THE BUSY BRAINS • BY MANSI SARASWAT
Electricity
Class 10 Science | Important Questions with Answers
Pre-Board & Board Exam Preparation
Class 10 Science Electricity Important Questions with Answers – Prepare the chapter Electricity for your Pre-Board and Board Exams with this comprehensive revision resource. This page covers electric current and circuit, electric potential difference, circuit diagrams, Ohm’s law, resistance, resistivity, series and parallel combination of resistors, heating effect of electric current, Joule’s law, electric power and electrical energy.
The questions are organised for quick exam revision and include important definitions, formulas, conceptual questions, numerical problems, MCQs, case-based questions and NCERT exercise-based questions.
01. Electric Current & Circuit
Q2. What is an electric circuit?
Answer:
A continuous and closed path through which electric current flows is called
an electric circuit.
Q3. Write the formula for electric current.
I = Q / t
I = Current, Q = Charge, t = Time
Here, current is the charge flowing through a particular cross-section
per unit time.
Q4. What is the SI unit of electric current?
Answer:
The SI unit of electric current is ampere (A).
One ampere is the current produced when one coulomb of charge flows
through a conductor in one second.
Q5. What is the conventional direction of electric current?
Answer:
Conventionally, the direction of electric current is taken to be
opposite to the direction of flow of electrons.
Q6. A current of 0.5 A flows through a bulb for 10 minutes. Find the charge flowing through the circuit.
Given:
I = 0.5 A
t = 10 min = 600 s
I = 0.5 A
t = 10 min = 600 s
Q = It
Q = 0.5 × 600
Q = 300 C
Q = 0.5 × 600
Q = 300 C
Q7. How many electrons constitute one coulomb of charge?
Answer:
Approximately 6 × 1018 electrons constitute
one coulomb of charge.
Q8. What is the role of a switch in an electric circuit?
Answer:
A switch makes or breaks the conducting path of a circuit.
When the switch is closed, current can flow; when it is open,
the circuit is broken and current stops flowing.
02. Electric Potential & Potential Difference
Q9. What is electric potential difference?
Answer:
Electric potential difference between two points is the work done
to move a unit charge from one point to the other.
Q10. Write the formula for potential difference.
V = W / Q
V = Potential difference, W = Work done, Q = Charge
Q11. What is the SI unit of potential difference?
Answer:
The SI unit of potential difference is volt (V).
1 V = 1 J/C
1 V = 1 J/C
Q12. Which instrument measures potential difference?
Answer:
A voltmeter measures potential difference.
It is connected in parallel across the two points
between which the potential difference is to be measured.
Q13. How much work is done in moving a charge of 2 C across a potential difference of 12 V?
W = VQ
W = 12 × 2
W = 24 J
W = 12 × 2
W = 24 J
Q14. What makes electric charge flow through a conductor?
Answer:
A potential difference across the conductor causes
electric charges to move and produces an electric current.
03. Circuit Diagram & Measuring Instruments
Q15. Why are schematic circuit diagrams used?
Answer:
Schematic diagrams provide a convenient way of representing the
components of an electric circuit using standard symbols.
Q16. How is an ammeter connected in a circuit?
Answer:
An ammeter is connected in series with the component
through which the current is to be measured.
Q17. How is a voltmeter connected in a circuit?
Answer:
A voltmeter is connected in parallel across the
component or points between which potential difference is to be measured.
Q18. Name the common electrical components represented in circuit diagrams.
Answer:
Common components include an electric cell, battery, switch,
wire joint, electric bulb, resistor, variable resistance or rheostat,
ammeter and voltmeter.
04. Ohm’s Law
Q19. State Ohm’s law.
Answer:
The potential difference across the ends of a given metallic wire is
directly proportional to the current flowing through it, provided its
temperature remains constant.
V ∝ I
V = IR
V = IR
Q20. What is resistance?
Answer:
Resistance is the property of a conductor that resists the flow of
electric charges through it.
Q21. What is the SI unit of resistance?
Answer:
The SI unit of resistance is ohm (Ω).
Q22. Write the different forms of Ohm’s law.
V = IR
R = V/I
I = V/R
R = V/I
I = V/R
Q23. What happens to current if resistance is doubled while voltage remains constant?
From I = V/R, current is inversely proportional to resistance.
Therefore, if resistance is doubled, the current becomes half.
Q24. A bulb has a resistance of 1200 Ω and is connected to a 220 V source. Find the current.
I = V/R
I = 220/1200
I ≈ 0.18 A
I = 220/1200
I ≈ 0.18 A
05. Resistance & Resistivity
Q25. On what factors does the resistance of a conductor depend?
Resistance depends on:
- Length of the conductor
- Area of cross-section
- Nature of the material
Q26. How does resistance depend on the length of a conductor?
Resistance is directly proportional to length.
R ∝ l
Therefore, increasing the length increases resistance.
Q27. How does resistance depend on the area of cross-section?
Resistance is inversely proportional to the area of cross-section.
R ∝ 1/A
Hence, a thicker wire generally offers lower resistance than a thinner
wire of the same material and length.
Q28. Write the relation between resistance and resistivity.
R = ρl/A
Here, ρ is the resistivity of the material.
Q29. What is resistivity?
Answer:
Resistivity is the characteristic property of a material represented
by ρ. Its SI unit is Ω m.
Q30. Why are alloys commonly used in electrical heating devices?
Answer:
The chapter explains that alloys generally have higher resistivity than
their constituent metals and do not oxidise readily at high temperatures.
Therefore, they are commonly used in devices such as electric irons
and toasters.
Q31. Why is tungsten used for electric bulb filaments?
Answer:
Tungsten is used almost exclusively for bulb filaments because it has
a very high melting point and can withstand the high temperature required
for producing light.
Q32. Why are copper and aluminium used for electrical transmission lines?
Answer:
Copper and aluminium have relatively low resistivity and are therefore
good conductors of electricity.
06. Series & Parallel Combination of Resistors
Q33. What happens to current in a series combination of resistors?
Answer:
The same current flows through every resistor in a series combination.
Q34. Write the equivalent resistance of resistors connected in series.
Rs = R1 + R2 + R3
Q35. What happens to potential difference in a series combination?
The total potential difference is equal to the sum of the potential
differences across the individual resistors.
V = V1 + V2 + V3
Q36. Write the formula for equivalent resistance of resistors connected in parallel.
1/Rp =
1/R1 +
1/R2 +
1/R3
Q37. What happens to potential difference across resistors in parallel?
Answer:
The potential difference across each resistor in a parallel combination
is the same.
Q38. What happens to current in a parallel combination?
The total current is the sum of the currents through the individual branches.
I = I1 + I2 + I3
Q39. Three resistors of 5 Ω, 10 Ω and 30 Ω are connected in parallel to a 12 V battery. Find the current through each resistor and the total current.
Since the potential difference across each resistor is 12 V:
I1 = 12/5 = 2.4 A
I2 = 12/10 = 1.2 A
I3 = 12/30 = 0.4 A
I = 2.4 + 1.2 + 0.4
I = 4 A
I = 4 A
Q40. Why is a parallel arrangement preferred over a series arrangement for electrical appliances?
- Each appliance gets the required potential difference.
- Different appliances can draw different currents according to their resistance.
- If one appliance fails, the others can continue to work.
Q41. A 20 Ω lamp and a 4 Ω conductor are connected in series to a 6 V battery. Find the total resistance and current.
R = 20 + 4 = 24 Ω
I = V/R = 6/24
I = 0.25 A
I = V/R = 6/24
I = 0.25 A
07. Heating Effect & Joule’s Law
Q42. What is the heating effect of electric current?
Answer:
When electric current flows through a resistive conductor, electrical
energy can be dissipated in the form of heat. This is called the
heating effect of electric current.
Q43. Write the formula for heat produced in a resistor.
H = VIt
H = I²Rt
H = I²Rt
Q44. State Joule’s law of heating.
Joule’s law states that the heat produced in a resistor is:
- Directly proportional to the square of current for a given resistance.
- Directly proportional to resistance for a given current.
- Directly proportional to the time for which current flows.
Q45. Give some applications of the heating effect of electric current.
Answer:
Electric iron, toaster, oven, kettle, heater and electric bulb are
examples of applications of the heating effect of electric current.
Q46. How does an electric fuse protect an electric circuit?
Answer:
A fuse is connected in series with the device. If a current larger
than the specified value flows, the fuse wire heats up, melts and
breaks the circuit, thereby protecting the circuit and appliances.
Q47. An electric iron of resistance 20 Ω carries a current of 5 A. Calculate the heat developed in 30 s.
H = I²Rt
H = 5² × 20 × 30
H = 15,000 J
H = 5² × 20 × 30
H = 15,000 J
Q48. Why does the heating element of an electric heater glow while its connecting cord generally does not?
Answer:
The heating element has comparatively higher resistance and therefore
produces significant heat when current flows through it. The connecting
cord is designed to have much lower resistance.
08. Electric Power & Electrical Energy
Q49. What is electric power?
Answer:
Electric power is the rate at which electrical energy is consumed
or dissipated in an electric circuit.
Q50. Write the formulas for electric power.
P = VI
P = I²R
P = V²/R
P = I²R
P = V²/R
Q51. What is the SI unit of electric power?
Answer:
The SI unit of electric power is watt (W).
1 W = 1 V × 1 A
1 W = 1 V × 1 A
Q52. What is the commercial unit of electrical energy?
Answer:
The commercial unit of electrical energy is
kilowatt-hour (kWh), commonly called one
unit of electricity.
Q53. Convert 1 kWh into joules.
1 kWh = 3.6 × 106 J
Q54. A bulb is connected to a 220 V generator and draws a current of 0.50 A. Find its power.
P = VI
P = 220 × 0.50
P = 110 W
P = 220 × 0.50
P = 110 W
Q55. A refrigerator rated 400 W operates for 8 hours per day for 30 days. Find the energy consumed.
Energy = 400 × 8 × 30 Wh
= 96,000 Wh
= 96 kWh
= 96,000 Wh
= 96 kWh
If the cost is ₹3 per kWh, the cost would be:
96 × ₹3
= ₹288
= ₹288
09. Important MCQs for Board Exam
Q56. The SI unit of electric current is:
(a) Volt
(b) Ohm
(c) Ampere
(d) Watt
Answer: (c) Ampere
Answer: (c) Ampere
Q57. An ammeter is connected:
(a) In parallel
(b) In series
(c) Across the battery only
(d) Outside the circuit
Answer: (b) In series
(b) In series
(c) Across the battery only
(d) Outside the circuit
Answer: (b) In series
Q58. A voltmeter is connected:
(a) In series
(b) In parallel
(c) In the battery
(d) At the switch
Answer: (b) In parallel
(b) In parallel
(c) In the battery
(d) At the switch
Answer: (b) In parallel
Q59. According to Ohm’s law:
(a) V = IR
(b) V = I/R
(c) R = VI
(d) I = VR
Answer: (a) V = IR
(b) V = I/R
(c) R = VI
(d) I = VR
Answer: (a) V = IR
Q60. The SI unit of resistance is:
(a) Volt
(b) Ampere
(c) Ohm
(d) Watt
Answer: (c) Ohm
(b) Ampere
(c) Ohm
(d) Watt
Answer: (c) Ohm
Q61. The equivalent resistance of resistors connected in series is:
(a) Less than every resistor
(b) Equal to their sum
(c) Zero
(d) Always equal to the smallest resistor
Answer: (b) Equal to their sum
(b) Equal to their sum
(c) Zero
(d) Always equal to the smallest resistor
Answer: (b) Equal to their sum
Q62. In a parallel combination, the potential difference across each resistor is:
(a) Different
(b) Zero
(c) The same
(d) Infinite
Answer: (c) The same
(b) Zero
(c) The same
(d) Infinite
Answer: (c) The same
Q63. Which formula represents electric power?
(a) VI
(b) I²R
(c) V²/R
(d) All of these
Answer: (d) All of these
(b) I²R
(c) V²/R
(d) All of these
Answer: (d) All of these
Q64. The commercial unit of electrical energy is:
(a) Watt
(b) Joule
(c) Kilowatt-hour
(d) Volt
Answer: (c) Kilowatt-hour
(b) Joule
(c) Kilowatt-hour
(d) Volt
Answer: (c) Kilowatt-hour
Q65. Joule’s law of heating is represented by:
(a) H = IRt
(b) H = I²Rt
(c) H = V/R
(d) H = IR
Answer: (b) H = I²Rt
(b) H = I²Rt
(c) H = V/R
(d) H = IR
Answer: (b) H = I²Rt
10. Case-Based Questions
Case Study 1 – Ohm’s Law
A student connects a nichrome wire with a battery, ammeter and voltmeter. The potential difference across the wire is increased by using additional cells. The current also increases. The V-I graph is found to be a straight line for the given conditions.
(a) Which law is being verified?
Answer: Ohm’s law.
(b) Write the mathematical relation.
Answer: V = IR.
(c) What does R represent?
Answer: Resistance of the conductor.
Case Study 2 – Series and Parallel Combination
Three resistors are connected in different combinations. In one arrangement the same current flows through each resistor. In another arrangement the potential difference across each resistor is the same.
(a) Identify the first arrangement.
Answer: Series combination.
(b) Identify the second arrangement.
Answer: Parallel combination.
(c) Write the equivalent resistance for series connection.
Answer: Rs = R1 + R2 + R3.
Case Study 3 – Heating Effect
An electric iron uses the heating effect of current. When current flows through its heating element, electrical energy is converted into heat.
(a) Name the law used to calculate heat produced.
Answer: Joule’s law of heating.
(b) Write the formula.
Answer: H = I²Rt.
(c) Name two other appliances based on the heating effect.
Answer: Electric toaster and electric heater.
11. NCERT Exercise-Based Board Questions
Q66. A piece of wire of resistance R is cut into five equal parts and the parts are connected in parallel. If the equivalent resistance is R′, find R/R′.
Each part has resistance R/5.
For five equal resistors in parallel:
For five equal resistors in parallel:
R′ = R/25
R/R′ = 25
Answer: 25
R/R′ = 25
Q67. Which of the following does NOT represent electrical power?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer: (b) IR²
(b) IR²
(c) VI
(d) V²/R
Answer: (b) IR²
Q68. A 220 V, 100 W bulb is operated at 110 V. What power will it consume?
For the same bulb:
R = V²/P
R = 220²/100 = 484 Ω
P = 110²/484
P = 25 W
R = 220²/100 = 484 Ω
P = 110²/484
P = 25 W
Q69. A 12 V battery is connected across an unknown resistor and a current of 2.5 mA flows. Find the resistance.
R = V/I
= 12/(2.5 × 10-3)
R = 4800 Ω
= 12/(2.5 × 10-3)
R = 4800 Ω
Q70. A 9 V battery is connected in series with resistors 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. Find the current through the 12 Ω resistor.
Total resistance:
R = 0.2 + 0.3 + 0.4 + 0.5 + 12
R = 13.4 Ω
I = 9/13.4
I ≈ 0.67 A
In series, the same current flows through every resistor.
R = 13.4 Ω
I = 9/13.4
I ≈ 0.67 A
Q71. How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?
Total resistance required:
R = V/I = 220/5 = 44 Ω
For n identical resistors in parallel:
Rp = 176/n
44 = 176/n
n = 4
44 = 176/n
n = 4
Q72. Three resistors of 6 Ω each are connected to obtain 9 Ω and 4 Ω. How?
For 9 Ω:
Connect two 6 Ω resistors in parallel:
Connect two 6 Ω resistors in series:
Connect two 6 Ω resistors in parallel:
R = 6/2 = 3 Ω
Then connect this 3 Ω combination in series with 6 Ω:
R = 3 + 6 = 9 Ω
For 4 Ω:
Then connect this 3 Ω combination in series with 6 Ω:
R = 3 + 6 = 9 Ω
Connect two 6 Ω resistors in series:
R = 6 + 6 = 12 Ω
Then connect this 12 Ω combination in parallel with 6 Ω:
R = 4 Ω
Then connect this 12 Ω combination in parallel with 6 Ω:
R = 4 Ω
Q73. Why is series arrangement not used for domestic electrical circuits?
- The same current would flow through all appliances.
- Different appliances require different currents.
- If one appliance fails, the complete circuit would be broken.
- Appliances would not operate independently.
Q74. Which uses more energy: a 250 W TV set operating for 1 hour or a 1200 W toaster operating for 10 minutes?
TV:
E = 250 × 1 = 250 Wh
Toaster:
E = 1200 × 1/6 = 200 Wh
Answer: The 250 W TV set uses more energy.
Q75. An electric heater of resistance 44 Ω draws 5 A. Calculate the rate at which heat is developed.
The rate of heat production is power:
P = I²R
= 5² × 44
P = 1100 W
= 5² × 44
P = 1100 W
12. Electricity – Quick Revision Sheet
| Concept | Formula / Key Point |
|---|---|
| Electric Current | I = Q/t |
| Potential Difference | V = W/Q |
| Ohm’s Law | V = IR |
| Resistance | R = V/I |
| Resistance of a conductor | R = ρl/A |
| Series Combination | Rs = R1 + R2 + R3 |
| Parallel Combination | 1/Rp = 1/R1 + 1/R2 + 1/R3 |
| Heating Effect | H = VIt = I²Rt |
| Electric Power | P = VI = I²R = V²/R |
| Electrical Energy | E = Pt |
| Commercial Unit | 1 kWh = 3.6 × 106 J |
13. Board Exam Final Checklist
- ✔ Learn the definition and formula of electric current.
- ✔ Remember the difference between ammeter and voltmeter connections.
- ✔ Learn Ohm’s law and all three forms of the equation.
- ✔ Practise numericals based on V = IR.
- ✔ Learn all three factors affecting resistance.
- ✔ Remember R = ρl/A.
- ✔ Practise series and parallel resistor problems.
- ✔ Learn H = I²Rt and its applications.
- ✔ Learn all formulas of electric power.
- ✔ Remember 1 kWh = 3.6 × 106 J.
- ✔ Practise NCERT exercise numericals.
- ✔ Practise circuit-based and case-based questions.
14. Exam Strategy for Electricity
⭐ How to score better in Electricity
- Write the formula first in numerical questions.
- Clearly mention the given values.
- Convert units before substitution.
- Show the calculation step-by-step.
- Always write the correct SI unit in the final answer.
- Practise questions involving series and parallel combinations.
- Revise the relationships between V, I and R.
- Do not skip NCERT exercise questions.
📚 Ready for Your Class 10 Board Exam?
Revise Electricity, practise numericals and strengthen your concepts before your Pre-Board and Board Exams.
Keep learning, keep practising and keep improving with The Busy Brains.