Electricity — Class 10 Notes

CBSE 2027 — dates, rules and official papers →

Physics · Ch 11Class 10 · CBSE 2026–27High weightage

Electricity

What the board actually asks: resistance combinations (series & parallel), Ohm's-law and $R = ρl / A$ numericals with correct units, electric power ( $P = V I = I^{2} R = V^{2} / R$ ) and energy in kW h, and Joule's-law heating — the numericals here carry almost every mark.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

The Big Idea AUTHORED

A potential difference pushes charge; resistance decides how much flows — and the leftover energy turns to heat.

One relation, $V = I R$ , ties together current, potential difference and resistance. Combine resistors (series adds, parallel reciprocals add), then track the energy: $P = V I$ and $H = I^{2} R t$ . Almost every board mark here is a numerical, so signs, units and the right formula choice are what score.

Formulae

Electric current

I = \frac{Q}{t}

Tip: I = Q/t; 1 A = 1 C/s.

Ohm's Law

Ohm's law

V = I R

Tip: Constant temperature is part of the law.

Resistance and Resistivity

Resistance from resistivity

R = ρ \frac{l}{A}

Tip: R ∝ l, R ∝ 1/A; ρ depends on material and temperature, not on size.

Resistors in Series

Resistors in series

R_{s} = R_{1} + R_{2} + R_{3}

Tip: Same current, voltages add.

Resistors in Parallel

Resistors in parallel

\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

Tip: Same voltage, currents add.

Electric Power

Electric power

P = V I = I^{2} R = \frac{V ^{2}}{R}

Tip: Pick the form that matches what is constant.

Heating Effect of Electric Current (Joule's Law)

Joule's heating

H = I^{2} R t

Tip: Also H = VIt.

Commercial energy unit

1 kW h = 3.6 \times 1 0^{6} J

Tip: 1 kWh = 1 unit.

Electric Charge and Number of Electrons

Q = n e

, where

n

is the number of electrons and the charge of one electron is

1.6 \times 1 0^{- 19}

coulomb.

Electric Potential Difference

V = \frac{W}{Q}

. One volt is the potential difference when 1 joule of work is done to move a charge of 1 coulomb:

1 V = 1 J C^{- 1}

. A voltmeter is connected in parallel across the component; an ammeter is connected in series.

Resistance, R = V/I, and the V–I Graph

R = \frac{V}{I}

, and

1 Ω = 1 V A^{- 1}

. The V–I graph of a conductor that obeys Ohm's law is a straight line through the origin.

Resistivity (

ρ = R A / l

)

ρ = \frac{R A}{l}

, measured in ohm-metre (

Ω m

). Alloys generally have higher resistivity than the metals they are made of.

Electric Circuit Symbols

Electric cell · battery · plug key (open) · plug key (closed) · wire joint · wires crossing without joining · electric bulb · resistor of resistance

R

· variable resistance (rheostat) · ammeter · voltmeter.

Electrical Energy and Cost

E = P \times t

. Electrical energy in kilowatt-hours = power in kW × time in hours. Cost = number of units (kWh) × rate per unit.

Practical Applications of the Heating Effect

Electric heaters and irons use the heating effect. A bulb filament is made of tungsten, which has a very high melting point (3380 °C). A fuse is a wire of low melting point placed in series; if the current becomes too high it melts and breaks the circuit.

Shortcuts & memory tricks →

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Electric current

NCERT verbatim — memorise this exact wording· Ch 11,

“A stream of electrons moving through a conductor constitutes an electric current. Conventionally, the direction of current is taken opposite to the direction of flow of electrons.”

What it means: Charge in motion. Quantitatively

I = \frac{Q}{t}

(charge per unit time); SI unit ampere (A). Conventional current runs from + to − — opposite to the electrons.

Electric potential difference

NCERT verbatim — memorise this exact wording· Ch 11,

“We define the electric potential difference between two points in an electric circuit carrying some current as the work done to move a unit charge from one point to the other”

What it means:

V = \frac{W}{Q}

— the work done per unit charge moved between two points. SI unit volt (V), where

1 V = 1 J C^{- 1}

Ohm's law

NCERT verbatim — memorise this exact wording· Ch 11,

“The potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. This is called Ohm's law.”

What it means:

V \propto I

, so

V = I R

with

R

constant at fixed temperature. On a

V

–

I

graph this is a straight line through the origin whose slope is

R

Resistance

NCERT verbatim — memorise this exact wording· Ch 11,

“R is a constant for the 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.”

What it means: How strongly a conductor opposes current:

R = \frac{V}{I}

. SI unit ohm (

Ω

). Double

R

and (at fixed

V

) the current halves.

Electrical resistivity

NCERT verbatim — memorise this exact wording· Ch 11,

“where ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is Ω m. It is a characteristic property of the material.”

What it means: The material's own opposition, independent of shape:

R = ρ \frac{l}{A}

. Metals/alloys

\sim 1 0^{- 8}

–

1 0^{- 6} Ω m

; insulators far higher.

Heating effect of electric current

NCERT verbatim — memorise this exact wording· Ch 11,

“if the electric circuit is purely resistive, that is, a configuration of resistors only connected to a battery; the source energy continually gets dissipated entirely in the form of heat. This is known as the heating effect of electric current.”

What it means: In a resistor-only circuit all the supplied energy becomes heat:

H = V I t

. This is what electric heaters, irons, toasters and bulb filaments use.

Joule's law of heating

NCERT verbatim — memorise this exact wording· Ch 11,

“This is known as Joule's law of heating. The law implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor.”

What it means:

H = I^{2} R t

. Heat rises with the square of current, linearly with resistance, and linearly with time.

Electric power

NCERT verbatim — memorise this exact wording· Ch 11,

“Equation (11.21) gives the rate at which electric energy is dissipated or consumed in an electric circuit. This is also termed as electric power.”

What it means: The rate of using electrical energy:

P = V I = I^{2} R = \frac{V ^{2}}{R}

. SI unit watt (W); the commercial unit of energy is the kilowatt hour (kW h).

Key terms — NCERT verbatim (state word-for-word)

Ampere: One ampere is constituted by the flow of one coulomb of charge per second (Ch 11, )
Volt: 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. (Ch 11, )
Ohm (unit): 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. (Ch 11, )
Rheostat: In an electric circuit, a device called rheostat is often used to change the resistance in the circuit. (Ch 11, )
Equivalent resistance in series: The equivalent resistance of several resistors in series is equal to the sum of their individual resistances. (Ch 11, )
Equivalent resistance in parallel: the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances. (Ch 11, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Electric current, charge & the circuit

Charge flows only around a continuous, closed path (an electric circuit). Current is the rate of that flow,

I = \frac{Q}{t}

, measured by an ammeter placed in series. The schematic below fixes the standard symbols and the +→− direction of conventional current.

NCERT Fig 11.1Schematic of a simple electric circuit — cell, electric bulb, ammeter and plug key

1-mark define-the-unit / direction-of-current

Ohm's law & resistance

V = I R

: at constant temperature the potential difference is proportional to the current, so a

V

–

I

plot is a straight line through the origin with slope

R

. A voltmeter (in parallel) reads

V

; an ammeter (in series) reads

I

2–3 mark Ohm's-law numerical / V–I graph

What resistance depends on — resistivity

R = ρ \frac{l}{A}

: resistance grows with length, falls with cross-sectional area, and depends on the material through its resistivity

ρ

. Alloys have higher

ρ

(and don't oxidise), so heating elements use them.

3-mark

R = ρl / A

numerical (watch the

A = π r^{2}

)

Combining resistors — series vs parallel

Series (same current everywhere):

R_{s} = R_{1} + R_{2} + R_{3}

, larger than any one resistor. Parallel (same voltage across each):

\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

, smaller than the smallest. Household wiring is parallel so each device gets full voltage and works independently.

NCERT Fig 11.6Resistors in series — three resistors R1, R2, R3 joined end to end between X and Y

3–5 mark combination numerical + derivation

Power, Joule heating & energy billed

Power is the rate of energy use:

P = V I = I^{2} R = \frac{V ^{2}}{R}

(watt). Heat dissipated in a resistor follows Joule's law

H = I^{2} R t

. Energy sold commercially is in kilowatt hours, where

1 kW h = 3.6 \times 1 0^{6} J

NCERT Fig 11.7Resistors in parallel — three resistors R1, R2, R3 connected between the common points X and Y

3–5 mark power/heating numerical + cost in kW h

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Example 11.2,

“How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?”

Given + relation.

Q = 2 C

V = 12 V

. From

V = \frac{W}{Q}

, rearrange to

W = V Q

.Concept: definition of potential difference

1 mark

Substitute + answer.

W = 12 V \times 2 C = 24 J

.Concept: answer with unit

1 mark

Mark-scheme logic (authored): 1 for the correct relation

W = V Q

from

V = W / Q

; 1 for the substitution and the answer stated with its unit (J).

Example 11.7,

“An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 11.9). Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.”

Example 11.8,

“In the circuit diagram given in Fig. 11.10, suppose the resistors R1, R2 and R3 have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.”

Example 11.11,

“100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor.”

Problem-solving strategies AUTHORED

Numerical?List every quantity with its SI unit before substituting — convert minutes to seconds, kW to W, mm diameter to m. Most lost marks are unit/conversion slips, not algebra.

Series or parallel?Series: same current, resistances add (

R_{s}

bigger than the largest). Parallel: same voltage, reciprocals add (

R_{p}

smaller than the smallest). Check your answer against those two bounds.

Which power formula?Pick the form matching the data:

P = V I

(given V and I),

P = I^{2} R

(given I and R),

P = V^{2} / R

(given V and R). All three are the same equation.

Energy or cost question?Work in kilowatt hours: energy (kW h) = power (kW) × time (h); cost = units × rate. Remember

1 kW h = 3.6 \times 1 0^{6} J

Resistivity numerical?In

R = ρl / A

the area is

A = π r^{2}

— halve the diameter to get the radius first, or you will be off by a factor of four.

Common mistakes that cost marks

Where students lose easy marks

Adding parallel resistances directly (

R_{1} + R_{2}

) instead of adding reciprocals — the #1 combination error.

Dropping units on the final answer (A, V, Ω, J, W) — a numerical answer without its unit loses a mark.

Using the diameter as the radius in

A = π r^{2}

when applying

R = ρl / A

— gives an area four times too big.

Not converting time or power units — minutes must become seconds, and kilowatts must become watts, before substituting.

Confusing where meters go — the ammeter is in series and the voltmeter is in parallel, not the reverse.

Formulae

Electric current

I = \frac{Q}{t}

Tip: I = Q/t; 1 A = 1 C/s.

Ohm's Law

Ohm's law

V = I R

Tip: Constant temperature is part of the law.

Resistance and Resistivity

Resistance from resistivity

R = ρ \frac{l}{A}

Tip: R ∝ l, R ∝ 1/A; ρ depends on material and temperature, not on size.

Resistors in Series

Resistors in series

R_{s} = R_{1} + R_{2} + R_{3}

Tip: Same current, voltages add.

Resistors in Parallel

Resistors in parallel

\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

Tip: Same voltage, currents add.

Electric Power

Electric power

P = V I = I^{2} R = \frac{V ^{2}}{R}

Tip: Pick the form that matches what is constant.

Heating Effect of Electric Current (Joule's Law)

Joule's heating

H = I^{2} R t

Tip: Also H = VIt.

Commercial energy unit

1 kW h = 3.6 \times 1 0^{6} J

Tip: 1 kWh = 1 unit.

Electric Charge and Number of Electrons

Q = n e

, where

n

is the number of electrons and the charge of one electron is

1.6 \times 1 0^{- 19}

coulomb.

Electric Potential Difference

V = \frac{W}{Q}

. One volt is the potential difference when 1 joule of work is done to move a charge of 1 coulomb:

1 V = 1 J C^{- 1}

. A voltmeter is connected in parallel across the component; an ammeter is connected in series.

Resistance, R = V/I, and the V–I Graph

R = \frac{V}{I}

, and

1 Ω = 1 V A^{- 1}

. The V–I graph of a conductor that obeys Ohm's law is a straight line through the origin.

Resistivity (

ρ = R A / l

)

ρ = \frac{R A}{l}

, measured in ohm-metre (

Ω m

). Alloys generally have higher resistivity than the metals they are made of.

Electric Circuit Symbols

Electric cell · battery · plug key (open) · plug key (closed) · wire joint · wires crossing without joining · electric bulb · resistor of resistance

R

· variable resistance (rheostat) · ammeter · voltmeter.

Electrical Energy and Cost

E = P \times t

. Electrical energy in kilowatt-hours = power in kW × time in hours. Cost = number of units (kWh) × rate per unit.

Practical Applications of the Heating Effect

Shortcuts & Tricks

Ammeter in series, voltmeter in parallel (across the component).

MCQ / check only: don't write this as working

Series total is larger than the largest resistor; parallel total is smaller than the smallest. Use to reject options.

MCQ / check only: don't write this as working

Two resistors in parallel: R₁R₂/(R₁+R₂). n equal resistors R in parallel: R/n. (Write 1/Rₚ in a written answer.)

MCQ / check only: don't write this as working

Same current (series) → P = I²R. Same voltage (parallel, household) → P = V²/R.

Safe in written answers

R ∝ l and R ∝ 1/A, so doubling length doubles R; halving the diameter makes R four times (A ∝ d²).

MCQ / check only: don't write this as working

Convert to kW and hours first, then units × rate.

Safe in written answers

Memory Tricks

Vikas ne Imli × Roti li: V = I × R.

Ohm's law. V = potential difference · I = current · R = resistance, in that order.

Ampere = coulomb ek second mein. Volt = joule per coulomb. Ohm = volt per ampere.

Units. 1 A = 1 C/s · 1 V = 1 J/C · 1 Ω = 1 V/A.

Series: current wahi, voltage baante, R jodo. Parallel: voltage wahi, current baante, 1/R jodo.

Series vs parallel. series: I same, V adds, R adds; parallel: V same, I adds, 1/R adds.

P = VI mein V = IR daalo to I²R; I = V/R daalo to V²/R.

Three power forms. P = VI → I²R → V²/R.

Garmi = I² R t: current ka square, resistance, time.

H = I²Rt. I² · R · t, in that order.

1 unit bijli = 36 lakh joule.

1 kWh = 3.6 × 10⁶ J. 1000 W × 3600 s = 36,00,000 J = 3.6 × 10⁶ J.

Mindmap

- - I = Q/t · unit ampere
  - V = W/Q · unit volt
  - Ammeter (series) · Voltmeter (parallel)
- - V = IR · straight-line V–I graph
  - R = V/I · unit ohm (Ω)
  - R = ρ l / A (resistivity)
- - Series: $R_{s} = R_{1} + R_{2} + R_{3}$
  - Parallel: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$
  - Series same I · Parallel same V
- - P = VI = I²R = V²/R (watt)
  - Joule: H = I²Rt
  - 1 kW h = 3.6 × 10⁶ J
- - Parallel added like series
  - Missing SI units
  - Diameter used as radius
  - Unconverted min / kW

The core skills here are wiring the meters correctly, reading circuit symbols, and knowing exactly how current, voltage and resistance behave in series vs parallel. These rules make every numerical and circuit-diagram question fall out.

Meters & current direction

AAmmeter measures current — always connected in series.
VVoltmeter measures potential difference — always connected in parallel across the two points.
→Conventional current flows from the + terminal to the − terminal (opposite to electron flow).

Series combination

ISame current through every resistor.
VPotential differences add: $V = V_{1} + V_{2} + V_{3}$ .
RResistances add: $R_{s} = R_{1} + R_{2} + R_{3}$ — bigger than the largest resistor.

Parallel combination

VSame potential difference across every resistor.
ICurrents add: $I = I_{1} + I_{2} + I_{3}$ .
RReciprocals add: $\frac{1}{R _{p}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}$ — smaller than the smallest resistor.

Reading a circuit diagram (Table 11.1 symbols)

Cell: long thin line is the + terminal, short thick line is the − terminal; several cells in a row make a battery.

Plug key / switch: open (○) means no current; closed (⊙) completes the circuit.

A resistor is a zig-zag (or a plain rectangle); a rheostat is a resistor with a sliding arrow for variable resistance.

A dot at a wire crossing means the wires are joined; no dot means they cross without connecting.

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

3 marksSection C

Kaveri connected a bulb A (100 Ω) to a 240 V supply. (a) How much energy will be consumed if it is kept ON for 4 hours each day for a week? Express in kJ. (b) She then connects another similar bulb B in series with A across 240 V. Will the brightness of A change? Explain mathematically.

Step-marked solution

[1.5 marks] (a) Power P = V^2/R = 240^2/100 = 576 W. Energy = P × t = 576 W × (4 h × 7 days × 3600 s) = 576 × 100800 = 58,060,800 J = 58,060.8 kJ.
[1 mark] (b) Series: Rnet = 100 + 100 = 200 Ω. Total power Ptot = V^2/Rnet = 240^2/200 = 288 W. Each bulb gets half: PA' = 144 W (since identical bulbs share equally).
[0.5 mark] Compare: PA' = 144 W < PA = 576 W ⟹ bulb A glows DIMMER in series.

3 marksSection C

(a) Vijaya connects three bulbs P, Q, R in series with a battery in two ways. In case I all three glow; in case II only P and R glow. What could be the reason for Q not glowing in case II? (b) Two resistances when connected in parallel give 10/3 Ω. When connected in series, they give 15 Ω. Calculate the individual resistances.

Step-marked solution

[1 mark] (a) In case II, the connecting wire between points across Q offers a much lower resistance path than bulb Q itself. Current bypasses Q through this short circuit ⟹ Q doesn't glow.
[1 mark] (b) Let resistances be R1, R2. Series: R1 + R2 = 15. Parallel: (R1·R2)/(R1+R2) = 10/3 ⟹ R1·R2 = 50.
[1 mark] From these: R1 and R2 are roots of x^2 − 15x + 50 = 0 ⟹ (x − 5)(x − 10) = 0 ⟹ R1 = 5 Ω, R2 = 10 Ω.

4 marksSection E

Four resistors (R1=15Ω, R2=30Ω, R3=15Ω, R4=15Ω), a voltmeter across R4, and a 20 V battery in a circuit. R2 and R3 in parallel; R1 and R4 in series with the parallel block. (a) Net resistance? (b) Voltmeter reading across R4? OR (b) Power dissipated by R1? (c) If R3 is removed, will net current increase, decrease, or remain same? Justify.

Step-marked solution

[1 mark] (a) R2 || R3 = (30 × 15)/(30 + 15) = 450/45 = 10 Ω. Net R = R1 + 10 + R4 = 15 + 10 + 15 = 40 Ω.
[1 mark] (b) Total I = V/R = 20/40 = 0.5 A. Voltmeter across R4: V4 = I × R4 = 0.5 × 15 = 7.5 V.
[1 mark] [OR] Power dissipated by R1 = I^2 × R1 = (0.5)^2 × 15 = 0.25 × 15 = 3.75 W.
[1 mark] (c) Removing R3: the parallel block becomes just R2 = 30 Ω (instead of 10 Ω). Net R goes UP (15 + 30 + 15 = 60 Ω). Since I = V/R and R increased, net current DECREASES.

Practise Electricity competency questions

Read NCERT Chapter 11 on LazyTopper (jesc111)

Official NCERT chapter PDF (jesc111) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Science, Ch 11. One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Electric current	verbatim	NCERT 2026-27 Ch11
Electric potential difference	verbatim	NCERT 2026-27 Ch11
Ohm's law	verbatim	NCERT 2026-27 Ch11
Resistance	verbatim	NCERT 2026-27 Ch11
Electrical resistivity	verbatim	NCERT 2026-27 Ch11
Heating effect of electric current	verbatim	NCERT 2026-27 Ch11
Joule's law of heating	verbatim	NCERT 2026-27 Ch11
Electric power	verbatim	NCERT 2026-27 Ch11
Ampere	verbatim	NCERT 2026-27 Ch11
Volt	verbatim	NCERT 2026-27 Ch11
Ohm (unit)	verbatim	NCERT 2026-27 Ch11
Rheostat	verbatim	NCERT 2026-27 Ch11
Equivalent resistance in series	verbatim	NCERT 2026-27 Ch11
Equivalent resistance in parallel	verbatim	NCERT 2026-27 Ch11
Work done to move a charge	example	NCERT 2026-27 Ch11 Ex 11.2
Resistors in series — total R, current, p.d.	example	NCERT 2026-27 Ch11 Ex 11.7
Resistors in parallel — branch currents, total R	example	NCERT 2026-27 Ch11 Ex 11.8
Joule's law — find p.d. from heat produced	example	NCERT 2026-27 Ch11 Ex 11.11
NCERT Fig 11.1	figure	NCERT 2026-27 Ch11 Fig 11.1
NCERT Fig 11.6	figure	NCERT 2026-27 Ch11 Fig 11.6
NCERT Fig 11.7	figure	NCERT 2026-27 Ch11 Fig 11.7

Electricity · spec-rendered note · NCERT Reprint 2026-27, Class 10 Science, Ch 11

Practise writing answers the way examiners mark them → Guide