⚡ Electricity – Class 10 Notes (Simplified for Quick Revision)

Electricity Class 10 Notes

Electricity is a vital chapter in Class 10 Science, and mastering its concepts is key to scoring well in exams. These simplified notes provide a quick and easy revision of all major topics—electric current, circuits, Ohm’s Law, series and parallel connections, heating effect, and electric power—making your exam preparation smooth and effective.

Electric Current and Circuit

If the electric charge flows through a conductor, we say that there is an electric current in the conductor.
Electric current is expressed by the amount of charge flowing through a particular area in unit time. In other words, it is the rate of flow of electric charges.
Conventionally, in an electric circuit, the direction of electric current is taken as opposite to the direction of the flow of electrons, which are negative charges.
A continuous and closed path of an electric current is called an electric circuit.

If, Charge = Q, time = t, current = I, then

I = Q/t

SI unit of Electric Charge: coulomb (C)

1 C ≈ 6 x 1018 electrons
1 electron = 1.6 x 10-19 C charge

SI unit of Electric Current: ampere (A)

One ampere is constituted by the flow of one coulomb of charge per second.
Smaller quantities of current are expressed in milliamperes (mA) or in microamperes (μA)
- 1mA = 10-3 A
- 1μA = 10-6 A

Ammeter:

An instrument called an ammeter measures electric current in a circuit.
It is always connected in series in a circuit through which the current is to be measured.

Schematic Diagram of an Electric Circuit:

electricity class 10 notes — *A schematic diagram of an electric circuit comprising - a cell, electric bulb, ammeter, and plug key*

Note that the electric current flows in the circuit from the positive terminal of the cell to the negative terminal of the cell through the bulb and ammeter.

Electric Potential and Potential Difference

The work done to move a unit charge from one point to the other is called the potential difference between two points.

V = W/Q

SI unit of Electric Potential Difference: volt (V)

One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

Voltmeter:

The potential difference is measured by means of an instrument called the voltmeter.
The voltmeter is always connected in parallel across the points between which the potential difference is to be measured.

Voltmeter	Ammeter
i. It is used to measure the potential difference across two points in an electric circuit.	i. It is used to measure electric current in an electric circuit.
ii. Its resistance is very high.	ii. Its resistance is very low.
iii. It is connected in parallel in an electric circuit.	iii. It is connected in series in an electric circuit.

Circuit Diagram

Ohm's Law

Ohm’s Law: Potential difference across the two points of a metallic conductor is directly proportional to the current passing through the circuit provided that temperature remains constant.

Mathematical expression of Ohm’s Law:

V ∝ I

⇒ V = IR

R is a constant called resistance for a given metal.

V-I graph for Ohm’s Law:

Resistance: It is the property of a conductor to resist the flow of charges through it.

Its SI unit is ohm (Ω).

1 ohm: If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 Ω.

Good Conductor	A component of a given size that offers a low resistance is a good conductor.
Resistor	A conductor having some appreciable resistance is called a resistor.
Poor Conductor	A component of identical size that offers a higher resistance is a poor conductor.
Insulator	An insulator of the same size offers even higher resistance.

Factors on which the Resistance of a Conductor Depends

Resistance of a uniform metallic conductor is:

Directly proportional to the length of the conductor (l).
Inversely proportional to the area of cross-section (A).
Directly proportional to the temperature.
Depend on the nature of the material (ρ).

R = ρl/A

Resistivity:

ρ (rho) is called the electrical resistivity of the material.
Its SI unit is Ω m.
It is a characteristic property of the material.
The metals and alloys have very low resistivity in the range of 10–8 Ω m to 10–6 Ω m. They are good conductors of electricity.
Insulators like rubber and glass have resistivity of the order of 1012 to 1017 Ω m.
Both the resistance and resistivity of a material vary with temperature.
The resistivity of an alloy is generally higher than that of its constituent metals.
Alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc.
Tungsten is used almost exclusively for filaments of electric bulbs.
Copper and aluminium are generally used for electrical transmission lines.

Resistance of a System of Resistors

1. Resistors in Series

In a series combination of resistors the current is the same in every part of the circuit or the same current through each resistor.
The total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors.
V = V1 + V2 + V3
Equivalent resistance (Rs) of three resistors in series (R1, R2, and R3) = Sum of R1, R2, and R3
Rs = R1 + R2 + R3

2. Resistors in Parallel

The total current I, is equal to the sum of the separate currents through each branch of the combination.
I = I1 + I2 + I3
Potential difference is same across each resistor.
Equivalent resistance of three resistors in parallel (R1, R2, and R3) = Rp
1/Rp = 1/R1 + 1/R2 + 1/R3

Advantages of parallel combination over series combination:

If one appliance stops working or goes out of order, then all other appliances keep on working.
All appliances can be operated at the same voltage as the electric supply.
Different appliances have different requirements of current. This cannot be satisfied in series as the current remains the same in series.
The total resistance in a parallel circuit is decreased.
All devices can be operated independently with separate switches.

Heating Effect of Electric Current

W (work done/hear/electrical energy) = QV = VIt = I2Rt = V2t/R

Joule's law of heating: The law implies that 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 the current flows through the resistor.

Practical Applications of Heating Effect of Electric Current

The electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some of the familiar devices based on Joule’s heating.

Electric Bulb:

The electric heating is also used to produce light, as in an electric bulb.
A strong metal with high melting point such as tungsten (melting point 3380°C) is used for making bulb filaments.
The bulbs are usually filled with chemically inactive nitrogen and argon gases to prolong the life of filament.

Fuse:

It protects circuits and appliances by stopping the flow of any unduly high electric current.
The fuse is placed in series with the device.
If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.
For an electric iron that consumes 1 kW electric power when operated at 220 V, a current of (1000/220) A, that is, 4.54 A will flow in the circuit. In this case, a 5 A fuse must be used.

Electric Power

The rate at which electric energy is dissipated or consumed in an electric circuit is called electric power.
P = VI = I2R (use this formula in series connection) = V2/R (use this formula in parallel connection)
The SI unit of electric power is watt (W).
1 W is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V.
The commercial unit of electric energy is kilowatt hour (kW h), commonly known as ‘unit’.
1kWh = 1000 Wh = 1000 x 60 x 60 Ws = 3600000 ws = 3.6 x 106 joule

Electricity

Electricity holds an important place in modern society.

It is a controllable and convenient form of energy used in homes, schools, hospitals, industries, and more.

This chapter aims to explain what constitutes electricity, how it flows, factors that control it, and its heating effect and applications.

Electric Current and Circuit

Electric Current Defined:

Similar to air current or water current, electric current is the flow of electric charge through a conductor, such as a metallic wire.

It is expressed by the amount of charge flowing through a particular area in unit time.

In other words, it is the rate of flow of electric charges.

In metallic wire circuits, electrons constitute the flow of charges.

Direction of Electric Current:

When electricity was first observed, electrons were not known, so electric current was considered the flow of positive charges.

Conventionally, the direction of electric current is taken as opposite to the direction of the flow of electrons (which are negative charges).

Electric Circuit Defined:

A continuous and closed path of an electric current is called an electric circuit.

A switch creates a conducting link between a cell and a bulb. If the circuit is broken (e.g., switch off), current stops, and the bulb does not glow.

Units and Measurement:

If a net charge Q flows across any cross-section of a conductor in time t, the current I is given by the formula: I = Q/t (Equation 11.1).

The SI unit of electric charge is coulomb (C).

One coulomb is equivalent to the charge contained in nearly 6 × 10^18 electrons.

An electron possesses a negative charge of 1.6 × 10^-19 C.

The electric current is expressed by a unit called ampere (A), named after Andre-Marie Ampere.

One ampere is constituted by the flow of one coulomb of charge per second (1 A = 1 C/1 s).

Smaller quantities of current are expressed in:

milliampere (1 mA = 10^-3 A).

microampere (1 µA = 10^-6 A).

An ammeter measures electric current in a circuit.

An ammeter is always connected in series in a circuit to measure the current.

In a schematic diagram, electric current flows from the positive terminal of the cell to the negative terminal through the components.

Electric Potential and Potential Difference

Necessity for Charge Flow:

Charges do not flow in a conductor by themselves, similar to how water doesn't flow in a perfectly horizontal tube.

For charges to flow, there must be a difference of electric pressure, called the potential difference, along the conductor.

This potential difference can be created by a battery (one or more electric cells).

Chemical action within a cell generates potential difference across its terminals.

When connected to a circuit, this potential difference sets charges in motion and produces electric current.

The cell expends its stored chemical energy to maintain the current.

Potential Difference Defined:

Electric potential difference (V) between two points in a current-carrying circuit is defined as the work done to move a unit charge from one point to the other.

Potential difference (V) = Work done (W) / Charge (Q).

V = W/Q (Equation 11.2).

Unit and Measurement:

The SI unit of electric potential difference is volt (V), named after Alessandro Volta.

One volt is the potential difference between two points where 1 joule of work is done to move a charge of 1 coulomb from one point to the other (1 V = 1 J / 1 C or 1 V = 1 J C^-1).

The potential difference is measured by an instrument called the voltmeter.

A voltmeter is always connected in parallel across the points where the potential difference is to be measured.

Circuit Diagram

An electric circuit typically includes a cell (or battery), a plug key, electrical components, and connecting wires.

It is convenient to draw a schematic diagram using conventional symbols to represent different components.

Symbols of commonly used components:

Electric cell: (symbol provided)

Battery or combination of cells: (symbol provided)

Plug key or switch (open): (symbol provided)

Plug key or switch (closed): (symbol provided)

A wire joint: (symbol provided)

Wires crossing without joining: (symbol provided)

Electric bulb: (symbol provided)

A resistor of resistance R: (symbol provided)

Variable resistance or rheostat: (symbol provided)

Ammeter: (symbol provided)

Voltmeter: (symbol provided)

Ohm's Law

Relationship between Potential Difference and Current:

In 1827, German physicist Georg Simon Ohm established a relationship between current (I) flowing in a metallic wire and the potential difference (V) across its terminals.

Ohm's law states that the potential difference (V) across the ends of a given metallic wire in an electric circuit is directly proportional to the current (I) flowing through it, provided its temperature remains the same.

Formulation of Ohm's Law:

V ∝ I (Equation 11.4).

V/I = constant = R.

V = IR (Equation 11.5).

Resistance (R):

R is a constant for a given metallic wire at a given temperature and is called its resistance.

It is the property of a conductor to resist the flow of charges through it.

Its SI unit is ohm (Ω), represented by the Greek letter omega.

According to Ohm's law, R = V/I (Equation 11.6).

One ohm (1 Ω) is defined as the resistance of a conductor when a potential difference of 1 V across its ends produces a current of 1 A through it (1 ohm = 1 volt / 1 ampere).

Current and Resistance Relationship:

From Ohm's law, I = V/R (Equation 11.7).

This shows that the current through a resistor is inversely proportional to its resistance. If resistance is doubled, current is halved.

Regulating Current:

A component used to regulate current without changing the voltage source is called a variable resistance.

A rheostat is a device often used in electric circuits to change the resistance.

Microscopic View of Resistance:

The motion of electrons within a conductor is retarded by its resistance because electrons are restrained by the attraction of atoms they move among.

Conductor Classification based on Resistance:

Good conductor: Offers a low resistance for a given size.

Resistor: A conductor having appreciable resistance.

Poor conductor: Offers a higher resistance for an identical size.

Insulator: Offers even higher resistance for the same size.

Factors on which the Resistance of a Conductor Depends

The resistance of a uniform metallic conductor depends on three factors:

Its length (l): Resistance is directly proportional to its length (R ∝ l) (Equation 11.8).

Doubling the length of a wire decreases the ammeter reading to one-half, indicating increased resistance.

Its area of cross-section (A): Resistance is inversely proportional to the area of cross-section (R ∝ 1/A) (Equation 11.9).

Using a thicker wire (larger cross-sectional area) of the same material and length increases the ammeter reading, indicating decreased resistance.

The nature of its material: Resistance varies with the material.

A change in ammeter reading is observed when a wire of different material (same length, same area) is used.

Resistivity (ρ):

Combining the proportionality relationships, resistance R = ρ (l/A) (Equation 11.10).

ρ (rho) is a constant of proportionality called the electrical resistivity of the material of the conductor.

The SI unit of resistivity is ohm-metre (Ω m).

Resistivity is a characteristic property of the material.

Resistivity Values and Material Properties (at 20°C):

Metals and Alloys have very low resistivity (10^-8 Ω m to 10^-6 Ω m), making them good conductors.

Examples: Silver (1.60 × 10^-8 Ω m), Copper (1.62 × 10^-8 Ω m), Aluminium (2.63 × 10^-8 Ω m).

Insulators like rubber and glass have very high resistivity (10^12 to 10^17 Ω m).

Examples: Glass (10^10 – 10^14 Ω m), Hard rubber (10^13 – 10^16 Ω m).

Both resistance and resistivity of a material vary with temperature.

Alloys generally have higher resistivity than their constituent metals.

Alloys do not oxidize (burn) readily at high temperatures, making them suitable for electrical heating devices (e.g., electric iron, toasters).

Tungsten is used for filaments of electric bulbs due to its high melting point.

Copper and aluminium are generally used for electrical transmission lines due to their low resistivity.

Resistance of a System of Resistors

There are two primary methods of joining resistors: series and parallel.

Resistors in Series

Connection: Resistors are joined end-to-end.

Current: In a series combination, the current is the same in every part of the circuit and through each resistor, regardless of the ammeter's position.

Potential Difference: The total potential difference (V) across a series combination of resistors is equal to the sum of the potential differences across the individual resistors (V = V1 + V2 + V3) (Equation 11.11).

Equivalent Resistance (R_s):

When several resistors are joined in series, the resistance of the combination (R_s) equals the sum of their individual resistances (R_s = R1 + R2 + R3) (Equation 11.14).

The equivalent resistance in series is greater than any individual resistance.

Disadvantages of Series Circuits:

Impracticable for diverse devices: Devices needing widely different currents (e.g., bulb and heater) cannot operate properly when connected in series because the current is constant throughout.

Single point of failure: If one component fails or breaks, the entire circuit is broken, and none of the components will work (e.g., "fairy lights" where one fused bulb stops all others).

Resistors in Parallel

Connection: Resistors are connected together between two common points.

Potential Difference: The potential difference (V) across each resistor in a parallel combination is the same and equal to the potential difference across the battery.

Current: The total current (I) entering the combination is equal to the sum of the separate currents through each branch of the combination (I = I1 + I2 + I3) (Equation 11.15).

Equivalent Resistance (R_p):

The reciprocal of the equivalent resistance (1/R_p) of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances (1/R_p = 1/R1 + 1/R2 + 1/R3) (Equation 11.18).

In a parallel circuit, the total resistance is decreased.

Advantages of Parallel Circuits:

Current division: Divides the current through electrical gadgets, allowing devices with different resistance and current requirements to operate properly.

Redundancy: If one component fails, the other components in parallel can continue to operate.

Lower total resistance: Helpful for drawing more current from the source, as the total resistance decreases.

This is why parallel arrangements are preferred for domestic circuits.

Heating Effect of Electric Current (Joule's Law of Heating)

Energy Dissipation:

A battery or cell provides electrical energy.

To maintain current in a purely resistive circuit, the source energy is continually dissipated entirely in the form of heat.

This is known as the heating effect of electric current.

This effect is utilized in devices like electric heaters and irons.

Formula for Heat Produced:

The work done in moving charge Q through potential difference V in time t is VQ.

The energy supplied by the source in time t is P × t, or VIt.

This energy is dissipated as heat (H) in the resistor: H = VIt (Equation 11.20).

By applying Ohm's law (V=IR), the heat produced can also be expressed as: H = I²Rt (Equation 11.21).

Joule's Law of Heating:

The equation H = I²Rt is known as Joule's law of heating.

This law implies that the heat produced in a resistor is:

Directly proportional to the square of the current (I²) for a given resistance.

Directly proportional to the resistance (R) for a given current.

Directly proportional to the time (t) for which the current flows through the resistor.

Practical Applications of Heating Effect:

Undesirable heating: In many cases, heating is undesirable as it converts useful electrical energy into heat and can increase component temperature, altering their properties.

Useful applications:

Electric laundry iron, electric toaster, electric oven, electric kettle, electric heater are common devices based on Joule's heating. The heating element of an electric heater glows, but the cord does not, because the element has a very high resistance, generating significant heat, while the cord has very low resistance.

Electric bulb: Used to produce light through heating. The filament (usually tungsten with a high melting point of 3380°C) gets very hot and emits light without melting. Bulbs are often filled with chemically inactive gases (nitrogen and argon) to prolong filament life. Most power consumed becomes heat, with a small part as light.

Electric fuse: Protects circuits and appliances by stopping unduly high electric current.

It is placed in series with the device.

Consists of a wire made of a metal or alloy (e.g., aluminum, copper, iron, lead) with an appropriate melting point.

If current exceeds a specified value, the fuse wire's temperature increases, it melts, and breaks the circuit.

Fuses for domestic purposes are rated (e.g., 1 A, 2 A, 3 A, 5 A, 10 A).

Electric Power

Definition: Power is the rate of doing work or the rate of consumption of energy.

Formulas: Electric power (P) is given by:

P = VI (Equation 11.19)

P = I²R (Equation 11.22)

P = V²/R (Equation 11.22)

These equations determine the rate at which energy is delivered by a current.

Units:

The SI unit of electric power is watt (W).

One watt (1 W) is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V (1 W = 1 volt × 1 ampere = 1 V A) (Equation 11.23).

The unit 'watt' is very small.

A much larger unit, kilowatt (kW), is used in practice: 1 kW = 1000 watts.

Electric Energy:

Electrical energy is the product of power and time.

The unit of electric energy is watt-hour (W h).

One watt-hour is the energy consumed when 1 watt of power is used for 1 hour.

The commercial unit of electric energy is kilowatt-hour (kW h), commonly known as a 'unit'.

Conversion: 1 kW h = 1000 watt × 3600 second = 3.6 × 10^6 watt second = 3.6 × 10^6 joule (J).

Clarification about Energy Consumption: It's a common misconception that electrons are consumed in an electric circuit. We pay electricity providers for the energy supplied to move electrons through electric gadgets, not for the electrons themselves.

Imagine an electric circuit as a water park's lazy river.

Electric current is like the flow of water itself.

Electric charge is like the individual water molecules moving along.

The potential difference (voltage) is like the pump that creates a height difference, pushing the water to flow from a higher point to a lower point. Without the pump, the water wouldn't flow on its own.

Resistance is like obstacles, narrow sections, or rough patches in the river that impede the smooth flow of water.

Ohm's Law tells you that if the pump is stronger (higher voltage), more water will flow (higher current), and if there are more obstacles (higher resistance), less water will flow, given the same pump strength.

Resistors in series are like multiple narrow sections placed one after another in a single path. All the water has to go through each narrow section, so the total resistance to flow is the sum of each narrow section's resistance. If one section breaks, the whole river stops.

Resistors in parallel are like multiple separate channels available for the water to flow through. The water can split up and take different paths. This effectively makes it easier for the water to flow overall, reducing the total resistance, and if one channel is blocked, water can still flow through the others.

The heating effect of electric current is like the water rubbing against the sides of the river and generating warmth due to friction, especially in the narrow or rough sections. This "friction" (resistance) converts some of the water's kinetic energy into heat.

Electric power is the rate at which the pump is moving water or the rate at which energy is being used to keep the water flowing through the system.

⚡ Class 10 Science – Chapter 12: Electricity

✅ NCERT Solutions

In-text Questions & Answers

Page 200

Q1. What does an electric circuit mean?
Answer:
An electric circuit is a continuous and closed path of an electric current. It typically consists of a source of electricity (cell or battery), wires, a switch, and a load (like a bulb).

Q2. Define the unit of current.
Answer:
The unit of electric current is ampere (A).
1 ampere = 1 coulomb/second
It is the current flowing when 1 coulomb of charge passes through a point in 1 second.

Q3. Calculate the number of electrons constituting one coulomb of charge.
Answer:
Charge of 1 electron = $1.6 \times 10^{-19}$ C
Number of electrons in 1 C = $\frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}$ electrons

Page 202

Q1. Name a device that helps to maintain a potential difference across a conductor.
Answer:
A cell or a battery helps maintain a potential difference across a conductor.

Q2. What is meant by saying that the potential difference between two points is 1 V?
Answer:
It means that 1 joule of work is done to move 1 coulomb of charge between two points.
$1 \text{V} = \frac{1 \text{J}}{1 \text{C}}$

Q3. How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Energy = Charge × Potential difference = 1 C × 6 V = 6 J

Page 209

Q1. On what factors does the resistance of a conductor depend?
Answer:
Resistance depends on:

Length of the conductor
Cross-sectional area
Material of the conductor
Temperature

Q2. Will current flow more easily through a thick wire or a thin wire of the same material and length?
Answer:
Through a thick wire, because it has a lower resistance (larger cross-sectional area).

Q3. Let the resistance of an electrical component remains constant. What happens to the current if the potential difference across it is doubled?
Answer:
From Ohm’s Law: $I = \frac{V}{R}$
If V is doubled, current also doubles.

Q4. Why are coils of electric toasters and electric irons made of an alloy rather than pure metal?
Answer:

Alloys have higher resistance than pure metals.
They do not oxidize or melt easily at high temperatures.

Q5. Use the data in Table 12.2 to answer the following —
(a) Which among iron and mercury is a better conductor?
Answer:
Iron, because its resistivity $(10^{-7} \Omega \cdot m)$ is lower than that of mercury.

(b) Which material is best for wire used for electrical transmission?
Answer:
Silver, as it has the lowest resistivity.

Page 213

Q1. Draw a schematic diagram of a circuit consisting of a battery, resistor, ammeter, and a plug key.
Answer:
👉 Refer to textbook Figure 12.7
(A drawing shows: Battery → Key → Resistor → Ammeter → back to battery)

Q2. Why is the ammeter connected in series?
Answer:
Because it measures the current through the component. In series, the current remains the same across all components.

Q3. Why is the voltmeter connected in parallel?
Answer:
Because it measures the potential difference across two points, and the voltage across parallel components is the same.

Page 216

Q1. What happens to the resistance of a circuit if the number of resistors is increased (a) in series (b) in parallel?
Answer:
(a) In series: Resistance increases.
(b) In parallel: Resistance decreases.

Q2. Why are resistors in parallel less than the least individual resistance?
Answer:
Because current gets multiple paths to flow, and overall resistance is reduced.

Page 218

Q1. Why does the cord of an electric heater not glow while the heating element does?
Answer:
The heater cord has very low resistance, so it doesn’t heat up. The element has high resistance, so it gets heated and glows.

Q2. Compute the heat generated in 10 seconds in a 10 Ω resistor carrying 2 A current.
Answer:
Using $H = I^2Rt$ :
$H = 2^2 × 10 × 10 = 400 \text{ J}$

Page 220

Q1. What determines the rate at which energy is delivered by a current?
Answer:
The electric power, which depends on voltage and current: $P = VI$

Q2. An electric motor takes 5 A from a 220 V line. Determine the power consumed and energy used in 2 hours.
Answer:
Power = $VI = 220 × 5 = 1100 \text{ W} = 1.1 \text{ kW}$
Energy = Power × Time = $1.1 × 2 = 2.2 \text{ kWh}$

✅ NCERT Back Exercise Questions – Page 221

Q1. A current of 1 A flows in a wire for 5 minutes. How much charge passes through the wire?
Answer:
$Q = I \times t = 1 × 300 = 300 \text{ C}$

Q2. Name a device that helps to maintain a potential difference across a conductor.
Answer:
Cell or battery

Q3. Calculate the resistance of a conductor if 0.1 A current flows through it when 0.5 V is applied.
Answer:
$R = \frac{V}{I} = \frac{0.5}{0.1} = 5 \Omega$

Q4. The resistance of a wire is 5 Ω. What would be the resistance of a similar wire twice as long?
Answer:
Resistance $R \propto l$ .
New resistance = $2 × 5 = 10 \Omega$

Q5. Why are electric bulbs filled with inert gases like nitrogen or argon?
Answer:
To prevent oxidation of the filament and to prolong bulb life.

Q6. Why is series arrangement not used for domestic circuits?
Answer:

All appliances get same current
If one device fails, the whole circuit breaks
No independent control possible

Here is a marks-wise list of past CBSE Board exam questions from Class 10 Science Chapter 12 – Electricity, including year-wise appearance and scoring-friendly answers, aligned with the CBSE 2025–26 pattern.

⚡ Electricity – CBSE Previous Year Questions with Answers

📘 Class 10 Science | Chapter 12 | Marks-wise | Updated for 2025

🟩 1-MARK QUESTIONS

Q1. What is the SI unit of electric charge?
📌 CBSE 2015
Answer:
The SI unit of electric charge is coulomb (C).

Q2. Define one volt.
📌 CBSE 2017
Answer:
One volt is the potential difference when 1 joule of work is done to move 1 coulomb of charge.
$1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}}$

Q3. What is the relation between 1 kilowatt hour and joule?
📌 CBSE 2019
Answer:
1 kWh = 3.6 × 10⁶ joules

Q4. What is meant by electric current?
📌 CBSE 2020
Answer:
Electric current is the rate of flow of electric charge through a conductor.
$I = \frac{Q}{t}$

🟨 2-MARK QUESTIONS

Q5. A bulb is rated 100 W, 220 V. Find the current and resistance.
📌 CBSE 2020
Answer:
Given: $P = 100 \text{ W}, V = 220 \text{ V}$

$I = \frac{P}{V} = \frac{100}{220} = 0.45 \text{ A}$
$R = \frac{V}{I} = \frac{220}{0.45} \approx 488.89 \Omega$

Q6. Define resistance. On what factors does it depend?
📌 CBSE 2018
Answer:
Resistance is the opposition offered by a conductor to the flow of current. It depends on:

Length (R ∝ L)
Area (R ∝ 1/A)
Material
Temperature

🟧 3-MARK QUESTIONS

Q7. State Ohm’s law. How is it verified?
📌 CBSE 2015, 2019
Answer:
Ohm’s Law: The current through a conductor is directly proportional to the potential difference across its ends, provided the temperature remains constant.
$V \propto I$ ⇒ $V = IR$

Verification:

Connect a circuit with a resistor, ammeter, voltmeter, battery.
Vary voltage using a variable resistor.
Record readings.
Plot V-I graph – it should be a straight line.

Q8. Two resistors of 6 Ω and 12 Ω are connected in (a) series and (b) parallel. Find total resistance in each case.
📌 CBSE 2018
Answer:
(a) Series: $R = 6 + 12 = 18 \Omega$
(b) Parallel: $\frac{1}{R} = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{3}{12} \Rightarrow R = 4 \Omega$

Q9. A wire of resistance 10 Ω is stretched to double its length. What is its new resistance?
📌 CBSE 2017
Answer:
When length is doubled,
$R' = R \left( \frac{l'}{l} \right)^2 = 10 \times 2^2 = 40 \Omega$

🟥 5-MARK QUESTIONS

Q10. What is the heating effect of electric current? Derive the expression for heat produced.
📌 CBSE 2017, 2019
Answer:
When current flows through a resistor, it generates heat.
Joule’s Law of Heating:
Heat $H = I^2Rt$

Derivation:
From Ohm’s Law: $V = IR$
Work done = $W = V \times I \times t = I^2Rt$

Applications: Electric heater, iron, fuse

Q11. State advantages of parallel arrangement over series in household circuits.
📌 CBSE 2016
Answer:

All appliances get the same voltage.
One appliance can be used independently of the others.
If one fails, others continue working.
Less total resistance ensures efficient power usage.

Q12. An electric iron draws a current of 4 A when connected to 220 V. Calculate: (a) Resistance (b) Energy consumed in 2 hrs.
📌 CBSE 2022
Answer:
(a) $R = \frac{V}{I} = \frac{220}{4} = 55 \Omega$
(b) $P = VI = 220 × 4 = 880 W$
Energy = $0.88 \text{ kW} × 2 = 1.76 \text{ kWh}$

Q13. A 100 W bulb is used for 5 hours daily. Find energy consumed in (a) kWh, (b) Joules in 30 days.
📌 CBSE 2020
Answer:
(a) Power = 0.1 kW
Time = 5 × 30 = 150 hours
Energy = 0.1 × 150 = 15 kWh
(b) $15 × 3.6 × 10^6 = 5.4 × 10^7 \text{ J}$