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Class 10

CBSE Class 10 Science Chapter 12 Notes: Electricity

Complete local notes for CBSE Class 10 Science Electricity: electric current, potential difference, Ohm's law, resistance, series and parallel circuits, heating effect, electric power, formulas, solved numericals, diagrams, and board exam tips.

Updated 23 May 202614 min read

Chapter Overview

Electricity is one of the highest-value Physics chapters in CBSE Class 10 Science. It is formula-based, but it is not only about memorising formulas. You must understand what current, potential difference, resistance, and power mean, and then apply them correctly in circuits.

This chapter is important for:

  • Numericals
  • Circuit diagrams
  • Formula derivations
  • Reasoning questions
  • Series and parallel resistance problems
  • Everyday applications like fuse, heater, bulb, and electric power rating

1. Electric Charge and Electric Current

Electric charge is a basic property of matter. In metallic conductors, electric current is due to the flow of electrons.

Electric current is the rate of flow of electric charge.

I = Q / t

where:

SymbolMeaningSI Unit
IElectric currentampere (A)
QChargecoulomb (C)
tTimesecond (s)

One Ampere

If one coulomb of charge flows through a conductor in one second, the current is one ampere.

1 A = 1 C / 1 s

Solved Example

If 300 C of charge flows through a wire in 60 s, find the current.

I = Q / t I = 300 / 60 I = 5 A
Answer5 A.

2. Electric Potential Difference

Potential difference between two points is the work done to move a unit charge from one point to another.

V = W / Q

where:

SymbolMeaningSI Unit
VPotential differencevolt (V)
WWork donejoule (J)
QChargecoulomb (C)

One Volt

If one joule of work is done to move one coulomb of charge, the potential difference is one volt.

1 V = 1 J / 1 C

3. Circuit Symbols

You should be able to recognise and draw common symbols neatly.

ComponentPurpose
CellProvides potential difference
BatteryCombination of cells
Plug key or switchOpens or closes circuit
ResistorOffers resistance
AmmeterMeasures current
VoltmeterMeasures potential difference

Ammeter and Voltmeter Connection

InstrumentConnectionResistance
AmmeterSeriesVery low resistance
VoltmeterParallelVery high resistance

4. Ohm's Law

Ohm's law states that the current through a conductor is directly proportional to the potential difference across its ends, provided temperature and other physical conditions remain constant.

V proportional to I V = IR

where R is resistance.

Graph

For an ohmic conductor, the V-I graph is a straight line passing through the origin.

The slope of the V-I graph gives resistance:

R = V / I

Solved Example

A resistor has potential difference 12 V and current 3 A. Find resistance.

R = V / I R = 12 / 3 R = 4 ohm
Answer4 ohm.

5. Resistance

Resistance is the property of a conductor by which it opposes the flow of current.

R = V / I

SI unit: ohm.

Factors Affecting Resistance

Resistance of a conductor depends on:

1. Length of conductor 2. Area of cross-section 3. Nature of material 4. Temperature

The relation is:

R = rho l / A

where:

SymbolMeaning
rhoResistivity of material
lLength
AArea of cross-section

Key Ideas

  • Longer wire means higher resistance.
  • Thicker wire means lower resistance.
  • Different materials have different resistivity.
  • Resistance of metallic conductors generally increases with temperature.

6. Resistivity

Resistivity is a property of the material. It does not depend on length or thickness of the conductor.

SI unit of resistivity:

ohm metre

Material Use

MaterialWhy it is used
Copper and aluminiumLow resistivity, used in wires
NichromeHigh resistance, high melting point, used in heaters
TungstenHigh melting point, used in bulb filament

7. Resistors in Series

In series combination, resistors are connected end to end.

Properties

  • Same current flows through every resistor.
  • Total potential difference is divided among resistors.
  • Equivalent resistance is greater than individual resistances.

Formula:

Rs = R1 + R2 + R3

Example

If R1 = 2 ohm, R2 = 3 ohm, and R3 = 5 ohm, then:

Rs = 2 + 3 + 5 Rs = 10 ohm

8. Resistors in Parallel

In parallel combination, resistors are connected across the same two points.

Properties

  • Same potential difference acts across every resistor.
  • Current is divided among branches.
  • Equivalent resistance is less than the smallest individual resistance.

Formula:

1/Rp = 1/R1 + 1/R2 + 1/R3

For two resistors:

Rp = (R1 x R2) / (R1 + R2)

Example

Find equivalent resistance of 6 ohm and 3 ohm in parallel.

Rp = (6 x 3) / (6 + 3) Rp = 18 / 9 Rp = 2 ohm

9. Series vs Parallel

PointSeriesParallel
CurrentSame through all resistorsDivides in branches
VoltageDivides across resistorsSame across each resistor
Equivalent resistanceMore than individual resistorsLess than smallest resistor
Use in homesNot preferredPreferred

Why Household Circuits Use Parallel Combination

Household circuits use parallel combination because:

1. Each appliance gets the same voltage. 2. Appliances can be switched on or off independently. 3. If one appliance fails, others continue working.

10. Heating Effect of Electric Current

When current flows through a resistor, electrical energy is converted into heat energy.

Joule's law of heating:

H = I^2Rt

where:

SymbolMeaningUnit
HHeat producedjoule
ICurrentampere
RResistanceohm
tTimesecond

Applications

  • Electric heater
  • Electric iron
  • Electric kettle
  • Fuse
  • Bulb filament

11. Electric Fuse

An electric fuse is a safety device connected in series with the live wire.

It has:

  • Low melting point
  • Suitable resistance

When excessive current flows, the fuse wire melts and breaks the circuit. This prevents damage and fire.

12. Electric Power

Electric power is the rate at which electrical energy is consumed or converted.

P = VI

Using Ohm's law:

P = I^2R P = V^2 / R

SI unit: watt (W).

Commercial unit of electrical energy:

kilowatt hour (kWh)
1 kWh = 3.6 x 10^6 J

13. Power Rating of Appliances

If a bulb is rated 220 V, 100 W, it means:

  • It should be connected to 220 V supply.
  • It consumes 100 J of energy per second at 220 V.

Current drawn:

P = VI I = P / V I = 100 / 220 I = 0.45 A approximately

14. Formula Sheet

QuantityFormula
CurrentI = Q/t
Potential differenceV = W/Q
Ohm's lawV = IR
ResistanceR = V/I
Resistivity relationR = rho l/A
Series resistanceRs = R1 + R2 + R3
Parallel resistance1/Rp = 1/R1 + 1/R2 + 1/R3
Heat producedH = I^2Rt
PowerP = VI
PowerP = I^2R
PowerP = V^2/R
Electrical energyE = Pt

15. Solved Numericals

Example 1

A current of 0.5 A flows through a resistor of 20 ohm. Find the potential difference.

V = IR V = 0.5 x 20 V = 10 V
Answer10 V.

Example 2

Find the heat produced when 2 A current flows through a 10 ohm resistor for 5 minutes.

Given:

I = 2 A R = 10 ohm t = 5 min = 300 s

Formula:

H = I^2Rt H = 2^2 x 10 x 300 H = 4 x 10 x 300 H = 12000 J
Answer12000 J.

Example 3

Two resistors of 4 ohm and 6 ohm are connected in series. Find total resistance.

Rs = R1 + R2 Rs = 4 + 6 Rs = 10 ohm
Answer10 ohm.

Example 4

Two resistors of 4 ohm and 6 ohm are connected in parallel. Find total resistance.

Rp = (R1 x R2) / (R1 + R2) Rp = (4 x 6) / (4 + 6) Rp = 24 / 10 Rp = 2.4 ohm
Answer2.4 ohm.

16. Important Question Types

Question TypeWhat to Write
DefinitionExact meaning and SI unit
Ohm's lawStatement, formula, condition, graph idea
Series/parallelFormula and circuit behaviour
NumericalGiven, formula, substitution, answer with unit
Fuse questionSafety device, series connection, melting action
Household wiringParallel combination reasons

17. Practice Questions

1. Define electric current and write its SI unit. 2. What is meant by potential difference? 3. State Ohm's law. 4. Why is an ammeter connected in series? 5. Why is a voltmeter connected in parallel? 6. Find current when 120 C charge flows in 2 minutes. 7. Find resistance when 6 V produces 0.5 A current. 8. Three resistors 2 ohm, 4 ohm, and 6 ohm are connected in series. Find equivalent resistance. 9. Two resistors 3 ohm and 6 ohm are connected in parallel. Find equivalent resistance. 10. Why are household circuits connected in parallel? 11. A heater draws 5 A current from 220 V supply. Find its power. 12. Find heat produced by 3 A current through 5 ohm for 10 s.

18. Quick Answers

1. Rate of flow of charge; ampere. 2. Work done per unit charge. 3. V = IR, at constant temperature. 4. To measure the same current flowing through the circuit. 5. To measure potential difference across a component. 6. I = 120/120 = 1 A. 7. R = 6/0.5 = 12 ohm. 8. 2 + 4 + 6 = 12 ohm. 9. Rp = (3 x 6)/(3 + 6) = 2 ohm. 10. Same voltage, independent control, no total failure. 11. P = VI = 220 x 5 = 1100 W. 12. H = I^2Rt = 9 x 5 x 10 = 450 J.

Common Mistakes to Avoid

  • Forgetting to convert minutes into seconds.
  • Writing resistance unit as watt instead of ohm.
  • Connecting ammeter and voltmeter in the wrong way.
  • Adding resistances directly in parallel circuits.
  • Using P = V^2/R with wrong voltage.
  • Missing SI units in final numerical answers.

Final Revision Checklist

  • I can define current, potential difference, resistance, and power.
  • I can state and use Ohm's law.
  • I can solve series and parallel resistance questions.
  • I can explain why homes use parallel circuits.
  • I can use H = I^2Rt correctly.
  • I can calculate power and electrical energy.
  • I can write every numerical with formula and unit.

Internal Links

  • [CBSE Class 10 Science guide](/school/cbse/class/10/science)
  • [Chemical Reactions and Equations notes](/school/cbse/class/10/science/cbse-chemical-reactions-and-equations)
  • [CBSE Class 10 Maths guide](/school/cbse/class/10/maths)

NCERT Exercise Solutions

Intext Questions

Page 200, Q1: What does an electric circuit mean?

AnswerAn electric circuit is a continuous closed path through which electric current can flow. It consists of:

Current flows from the positive terminal of the source through the external circuit to the negative terminal. If the circuit is broken (open), no current flows.

  • A source of electrical energy (battery or cell)
  • Conducting wires connecting components
  • A load (device that uses electricity — bulb, motor, etc.)
  • A switch (to open or close the circuit)

Page 202, Q1: Define the unit of current.

AnswerThe unit of current is the Ampere (A). 1 Ampere is defined as the flow of 1 Coulomb of charge per second: 1 A = 1 C/s i.e., if 1 coulomb of charge passes through a cross-section of a conductor in 1 second, the current is 1 ampere.

Page 209, Q1: Judge the equivalent resistance when the following are connected in parallel: (a) 1 Ω and 10⁶ Ω (b) 1 Ω and 10³ Ω and 10⁶ Ω

AnswerFor parallel combination: 1/R = 1/R₁ + 1/R₂ + ...

(a) 1/R = 1/1 + 1/10⁶ = 1 + 0.000001 ≈ 1 R ≈ 1 Ω (slightly less than 1 Ω)

(b) 1/R = 1/1 + 1/1000 + 1/1000000 ≈ 1.001 R ≈ 0.999 Ω (approximately 1 Ω)

In parallel, the equivalent resistance is always less than the smallest resistor. Adding more resistors in parallel always reduces total resistance.


Chapter-End Exercise

Q1. A piece of wire of resistance R is cut into five equal parts. These parts are connected in parallel. What is the equivalent resistance of the combination?

AnswerEach part has resistance = R/5. Five parts in parallel: 1/Rₚ = 5 × (5/R) = 25/R Rₚ = R/25

The equivalent resistance is R/25 — much smaller than R.


Q2. Which uses more energy: a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?

AnswerEnergy = Power × Time

TV: E = 250 W × 1 h = 250 Wh = 250 × 3600 J = 900,000 J Toaster: E = 1200 W × (10/60) h = 200 Wh = 200 × 3600 J = 720,000 J

The TV set uses more energy (250 Wh vs 200 Wh for toaster).


Q3. An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery. Calculate: (a) total resistance of the circuit (b) current through the circuit (c) potential difference across the lamp

AnswerGiven: R_lamp = 20 Ω, R_conductor = 4 Ω, V = 6 V (series circuit)

(a) Total resistance: R = 20 + 4 = 24 Ω

(b) Current: I = V/R = 6/24 = 0.25 A

(c) Voltage across lamp: V_lamp = I × R_lamp = 0.25 × 20 = 5 V (Remaining 1 V is across the conductor: 0.25 × 4 = 1 V ✓ Total = 6 V ✓)


Q4. How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?

AnswerTotal resistance needed: R = V/I = 220/5 = 44 Ω

For n resistors of 176 Ω in parallel: 1/44 = n/176 n = 176/44 = 4

4 resistors of 176 Ω each must be connected in parallel.


Q5. Show how would you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (a) 9 Ω (b) 4 Ω.

Answer(a) 9 Ω: Connect two 6 Ω resistors in parallel, then connect the third in series. Parallel pair: 1/R_p = 1/6 + 1/6 = 2/6 → R_p = 3 Ω Series with third: R_total = 3 + 6 = 9 Ω ✓

(b) 4 Ω: Connect two 6 Ω resistors in series, then connect the third in parallel with the series pair. Series pair: R_s = 6 + 6 = 12 Ω Parallel with third: 1/R = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 → R = 4 Ω ✓


Q6. Several electric bulbs designed to be used on a 220 V electrical supply line are rated 10 W. How many lamps can be connected in parallel with each other across the supply so that the total power consumed does not exceed 1100 W?

AnswerPower per bulb = 10 W Maximum total power = 1100 W

Number of lamps = 1100 / 10 = 110 lamps

All are connected in parallel so each receives the full 220 V. Each consumes 10 W, and 110 × 10 = 1100 W ≤ 1100 W.

Frequently Asked Questions

The most important formulas are I = Q/t, V = W/Q, V = IR, R = rho l/A, H = I^2Rt, P = VI, P = I^2R, and P = V^2/R.

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