⚡ 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

⚡ 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) $15 × 3.6 × 10^6 = 5.4 × 10^7 \text{ J}$

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