Class 10th · Science · Chapter 11

Electricity – Notes, MCQs, Quiz & Worksheet

Overview

What is Electricity?

Electricity is the flow of electric charge that powers devices and makes things work.

Important Parts of Electricity:

  1. Electric Current: The flow of electric charge, measured in Amps (A).
  2. Voltage: The force that pushes the electric charge, measured in Volts (V).
  3. Resistance: The opposition to the flow of current, measured in Ohms (Ω).
  4. Ohm's Law:  V = I × R – Voltage = Current × Resistance.
  5. Power: The amount of energy used, calculated by P = V × I (measured in Watts).

Types of Current:

  1. DC (Direct Current): Flows in one direction (like from a battery).
  2. AC (Alternating Current): Reverses direction (like in home electricity).

Exam relevance

Electricity carries steady weightage in Class 10th exams. Practising its MCQs and important questions is one of the fastest ways to secure marks from this chapter.

Need formulas?

Get all the important Science formulas in one quick-revision sheet.

Open Formula Sheets
Key Topics

Topics Covered in Electricity

Study these topics one by one to fully master the chapter.

MCQ Practice

Practice MCQs – Electricity

Attempt these multiple-choice questions, then reveal the answer to check yourself.

Q1.Calculate the resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance 20 ohms.

To calculate the resistivity (ρ) of the material, we use the formula:

R = ρ × (L / A)

Where:

  • R = 20 ohms (resistance)
  • L = 1 meter (length)
  • A = πr² (cross-sectional area of the wire)
  • r = 0.01 cm = 1 × 10⁻⁴ meters (radius)

Step-by-step calculation:

Calculate the area (A) of the wire's cross section:
A = π × (1 × 10⁻⁴)² = π × 10⁻⁸ m² ≈ 3.1416 × 10⁻⁸ m²

Rearrange the formula to solve for ρ:
ρ = R × (A / L)

Substitute the values:
ρ = 20 × (3.1416 × 10⁻⁸ / 1)

Calculate the resistivity:
ρ ≈ 20 × 3.1416 × 10⁻⁸ = 6.28 × 10⁻⁷ Ω·m

Q2.List the factors on which the resistance of a conductor in the shape of a wire depends.

The resistance of a conductor depends on:

  1. Length (L):Directly proportional to resistance.
  2. Cross-sectional area (A): Inversely proportional to resistance.
  3. Resistivity (ρ):Material property; higher resistivity means higher resistance.
  4. Temperature (T): Generally, resistance increases with temperature.
Q3.Why are alloys commonly used in electrical heating devices ?

Alloys are commonly used in electrical heating devices (like electric heaters, toasters, and irons) because of the following reasons:

  1. High resistivity: Alloys have higher electrical resistivity than pure metals, which means they produce more heat when current flows through them.
  2. Do not oxidize easily: Alloys like nichrome do not burn or oxidize easily at high temperatures, making them more durable than pure metals.
  3. High melting point: They can withstand high temperatures without melting or deforming.
  4. Stable resistance: The resistance of alloys does not change much with temperature, making heating performance more consistent and reliable.
Q4.Define the unit of current.

Unit of Current:The SI unit of electric current is the ampere (A).

Definition :-  One ampere is the current when one coulomb of charge flows through a conductor in one second.
So,  1 ampere = 1 coulomb / 1 second

Q5.Calculate the resistance of a metal wire of length 2 meters and cross-sectional area 1.55 × 10⁻⁶ square meters, if the resistivity of the metal is 2.8 × 10⁻⁸ ohm meter.

Resistance (R) = Resistivity (ρ) × (Length (L) / Area (A))

Given:-
Resistivity (ρ) = 2.8 × 10⁻⁸ ohm meter
Length (L) = 2 meters
Area (A) = 1.55 × 10⁻⁶ square meters

Substitute the values:
R = (2.8 × 10⁻⁸) × (2 / 1.55 × 10⁻⁶)
R = (5.6 × 10⁻⁸) / (1.55 × 10⁻⁶)
R ≈ 0.0361 ohms

Hence answer is  0.0361 ohms

Q6.How is the resistivity of alloys compared with those of pure metals from which they may have been formed?

The resistivity of alloys is generally higher than that of the pure metals from which they are made.

Why?

  • In alloys, different types of atoms are mixed.
  • These atoms disturb the regular structure of the metal lattice.
  • This scatters electrons more, making it harder for electric current to flow.
  • As a result, resistance increases, and hence resistivity is higher.

Example:

  • Pure copper has low resistivity (good conductor).
  • An alloy like nichrome (nickel + chromium) has much higher resistivity.
Q7.What does an electric circuit mean?

An electric circuit is a closed path through which electric current can flow. It usually includes a power source like a battery, wires to carry the current, and a device like a bulb or fan that uses the electricity.

If the path is complete (closed), electricity flows and the device works. If the path is broken (open), electricity cannot flow and the device does not work.

Q8.Calculate the number of electrons constituting one coulomb of charge.

To find the number of electrons in one coulomb of charge:

Charge of one electron = 1.6 × 10⁻¹⁹ coulombs
Number of electrons = 1 ÷ (1.6 × 10⁻¹⁹)
= 6.25 × 10¹⁸ electrons

Hence  6.25 × 10¹⁸ electrons make up one coulomb of charge.

Q9.On what factors does the resistance of a conductor depend?

The resistance of a conductor depends on the following factors:

  1. Length of the conductor (L): Resistance increases with length.
    (More length → More resistance)
  2. Area of cross-section (A): Resistance decreases with a larger area.
    (More thickness → Less resistance)
  3. Material of the conductor:  Different materials have different resistivities.
    (Some materials oppose current more than others)
  4. Temperature: Resistance increases with temperature (for most conductors).
Q10.Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason.

Metals are good conductors of electricity because they have a large number of free electrons (also called conduction electrons). These electrons are not tightly bound to atoms and can move freely through the metal. When a potential difference is applied, these free electrons flow easily, allowing electric current to pass.

Glass is a bad conductor (insulator) because its electrons are tightly bound to their atoms. It does not have free electrons available to move and carry electric current. As a result, electricity cannot flow through it easily.

In short:

  • Metals have free-moving electrons  --->  Good conductors.
  • Glass lacks free-moving electrons  ---->  Bad conductor.
Q11.What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Advantages of connecting electrical devices in parallel instead of in series:

  1. Each device gets the full voltage of the battery:
    In parallel, every device gets the same voltage as the battery voltage, so all devices work properly.
  2. Devices work independently:
    If one device stops working or is switched off, the others continue to work because they have their own separate paths.
  3. Current is divided among devices:
    Total current is split according to resistance, preventing any one device from getting too much current.
  4. Adding more devices does not reduce voltage:
    You can add more devices without reducing the voltage across each device.
Q12.Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Current will flow more easily through a thick wire than a thin wire of the same material when connected to the same source. 
Because a thick wire has a larger area of cross-section, so it offers less resistance to the flow of current. Lower resistance means more current can flow.

Q13.Name a device that you can use to maintain a potential difference between the ends of a conductor. Explain the process by which this device does so.

A battery maintains a potential difference by using chemical reactions to move electrons from one terminal to another. This creates a positive terminal and a negative terminal, allowing current to flow through a connected conductor.

Process:-

  • The battery has two terminals: positive and negative.
  • Chemical reactions inside the battery cause a buildup of electrons at the negative terminal and a deficit at the positive terminal.
  • This creates an electric field between the terminals — a potential difference (voltage).
  • When a conductor is connected, electrons flow from the negative to the positive terminal through the conductor, driven by this potential difference.
Q14.A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly.

Given:
Current (I) = 1 A (Ampere)
Time (t) = 16 seconds

Charge of one electron (e) = 1.6 × 10⁻¹⁹ C (Coulombs)

Step 1: Calculate the total charge (Q) passing through the filament:

Q = I × t
Q = 1 A × 16 seconds
Q = 16 C (Coulombs)

Step 2: Calculate the number of electrons:

Number of electrons = Q / e
Number of electrons = 16 C / (1.6 × 10⁻¹⁹ C/electron)
Number of electrons = 1 × 10²⁰ electrons

Hence the number of electrons passing through the filament in 16 seconds is approximately 1 × 10²⁰ electrons.

Q15.Name a device that helps to maintain a potential difference across a conductor.

Cell or Battery, These devices supply electrical energy and create the necessary potential difference to allow current to flow through a circuit.

Q16.Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

If the resistance of an electrical component remains constant and the potential difference across it decreases to half of its original value, then the current flowing through it will also change.

According to Ohm’s Law:
Current (I) = Potential difference (V) ÷ Resistance (R)

Since resistance (R) is constant, if the potential difference (V) becomes half, then the new current will be:
I new = (1/2 × V) ÷ R = (1/2) × (V ÷ R) = (1/2) × I original

This means the current will become half of its original value.

Q17.Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Coils of electric toasters and electric irons are made of an alloy rather than a pure metal due to following reasons:

  1. Alloys have higher resistance than pure metals, so they produce more heat when electric current passes through them, which is useful for heating devices.
  2. Alloys do not melt easily,  they have higher melting points, so the coils can withstand high temperatures without melting.
  3. Alloys are more durable and strong,so the coils last longer under repeated heating and cooling.
Q18.Why does the cord of an electric heater not glow while the heating element does?

The cord of an electric heater does not glow because it is made of a material with low resistance, so it does not heat up much.

The heating element glows because it is made of a material with high resistance, which causes it to get very hot and produce light when current passes through it.

Q19.What determines the rate at which energy is delivered by a current?

The rate at which energy is delivered by a current is called power.

Power (P) = Voltage (V) × Current (I)

So, the rate of energy delivery depends on both the voltage and the current.

Q20.How much energy is given to each coulomb of charge passing through a 6 V battery?

The energy given to each coulomb of charge passing through a 6 V battery is calculated using the formula:
Energy = Voltage × Charge

Here, the voltage is 6 V and the charge is 1 coulomb.
Energy = 6 V × 1 C = 6 Joules

So, the energy given to each coulomb of charge is 6 joules.

Q21.Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V

Given:
Charge (Q) = 96000 coulombs
Potential difference (V) = 50 volts

Formula:
Heat generated (H) = Voltage (V) × Charge (Q)
H = 50 × 96000 = 4800000 joules

Hence the heat generated is 4800000 joules (4.8 million joules).

Q22.An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Given:
Resistance (R) = 20 ohms
Current (I) = 5 amperes
Time (t) = 30 seconds

Formula:
Heat developed (H) = I² × R × t
H = 5² × 20 × 30 = 25 × 20 × 30 = 15000 joules

Hence heat developed in 30 seconds = 15000 joules

 

Q23.What is the maximum resistance which can be made using five resistors each of 1/5 Ω ?

To find the maximum resistance that can be made using five resistors, each of 1/5 Ω, you should connect all the resistors in series, because resistances add up in series.

The formula for total resistance in a series connection is:

R_total = R1 + R2 + R3 + R4 + R5

Since each resistor has a value of 1/5 Ω, we can calculate:

R_total = 1/5 + 1/5 + 1/5 + 1/5 + 1/5

R_total = 5 × 1/5 = 1 Ω

Hence the maximum resistance that can be made using five resistors, each of 1/5 Ω, is 1 Ω.

Q24.What is meant by saying that the potential difference between two points is 1 V?

When we say that the potential difference between two points is 1 volt (1 V), it means that 1 joule of work is done to move 1 coulomb of charge from one point to the other.

In simple words:
1 volt = 1 joule / 1 coulomb

Q25.The SI unit of potential difference is:
A.Watt
B.Joule
C.Volt
D.Ampere
Answer: Volt
Q26.A device used to measure potential difference across two points in an electric circuit is called:
A.Ammeter
B.Voltmeter
C.Galvanometer
D.Ohmmeter
Answer: Voltmeter
Q27.Which of the following materials has the lowest electrical resistivity?
A.Glass
B.Copper
C.Wood
D.Rubber
Answer: Copper
Q28.Which of the following is a good conductor of electricity?
A.Pure water
B.Saltwater
C.Glass
D.Plastic
Answer: Saltwater
Q29.A 2 Ohm resistor carries a current of 3 A. The heat produced in 10 seconds is:
A.60 J
B.180 J
C.30 J
D.90 J
Answer: 180 J
Q30.If the current through a resistor is doubled, the heat produced in a given time interval will:
A.Be halved
B.Be doubled
C.Be quadrupled
D.Remain the same
Answer: Be quadrupled
Quiz

Take a Electricity Quiz

A short, timed quiz with instant scoring — perfect for checking how well you know the chapter.

Electricity Quiz

Attempt a 10–20 question quiz on Electricity. Try to finish within 15 minutes, get instant scoring, and see which topics need more revision.

Start Quiz Now

Quiz Test

Take a timed quiz with instant scoring to test your speed and accuracy.

Start Quiz
Quick Revision

Electricity – Quick Revision Notes

A one-page recap to revise the whole chapter in minutes.

  • Electricity is part of the Class 10th Science syllabus and carries steady exam weightage.
  • Revise Electricity Class 10 questions and answers: recall its definition, key points and one example.
  • Revise Ncert solutions for class 10th science electricity chapter 12: recall its definition, key points and one example.
  • Revise Ncert solutions for class 10th science electricity pdf: recall its definition, key points and one example.
  • Revise Electricity Class 10 Notes: recall its definition, key points and one example.
  • Revise Ncert solutions for class 10th science electricity chapter: recall its definition, key points and one example.
  • Revise Electricity Class 10 NCERT Solutions PDF: recall its definition, key points and one example.
  • Revise Ncert solutions for class 10th science electricity pdf download: recall its definition, key points and one example.
  • Revise NCERT Class 10 Science Chapter 12 exercise solutions: recall its definition, key points and one example.
  • Re-read all formulas, laws and definitions from this chapter.
  • Re-attempt the MCQs you got wrong and solve one worksheet.
  • Finish with a short quiz to confirm you remember everything.
Explore More

Explore More Science Resources

Educational intent

Created to help Class 10th students learn and revise Electricity from Science using notes, practice questions and free study tools.

Accuracy & learning-first

Our content is prepared and reviewed by experienced educators and kept aligned with the latest NCERT / CBSE syllabus and exam pattern.

Student-focused note

These resources support your school learning and self-study. Always cross-check with your prescribed textbook and your teacher's guidance for board exams.