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Electricity

Samacheer Kalvi (TN State Board) Class 10 Science — Physics Chapter 4, Electricity. Electric current and potential difference, Ohm's law, resistance and resistivity, resistors in series and parallel, electric power and energy, and the heating effect of current (Joule's law). Notes with formulas, worked numericals, common mistakes, practice and quick revision.

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Define current and potential difference
2Apply Ohm's law V = IR
3Find resistance of series and parallel combinations
4Calculate electric power and energy
5State and apply Joule's law of heating

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Why this chapter matters

Electricity underlies every appliance and circuit. Ohm's law, series–parallel combinations and power/energy calculations are among the most reliable scoring areas in the TN SSLC Science exam.

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A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 9 Science — Electric charge / circuits

Basic circuits, charge and simple connections.

Electricity — Class 10 Science (Samacheer Kalvi)

TN State Board (Samacheer Kalvi) Class 10 Science, Physics — Chapter 4. Current, voltage, resistance and the power delivered by electric circuits.

1. About this chapter

This chapter explains electric current and potential difference, Ohm's law, resistance and the factors affecting it, how resistors combine in series and parallel, and the power, energy and heating effect of current.

2. Current, voltage and Ohm's law

Electric current: rate of flow of charge, I = Q / t (ampere, A).
Potential difference: work done per unit charge, V = W / Q (volt, V).
Ohm's law: at constant temperature, V = I R (current ∝ voltage).
Resistance (R): opposition to current, unit ohm (Ω).

3. Resistivity and combinations

Resistance of a wire: R = ρ L / A (ρ = resistivity). R increases with length, decreases with area, and rises with temperature (for metals).
Series: R = R₁ + R₂ + R₃ (same current; voltages add).
Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃ (same voltage; currents add).

4. Power, energy and heating effect

Quantity	Formula	Unit
Electric power	P = V I = I²R = V²/R	watt (W)
Electric energy	E = P × t	joule; 1 kWh = 3.6×10⁶ J
Heat (Joule's law)	H = I²R t	joule

Joule's law of heating: heat produced ∝ I², R and time.
Domestic energy is billed in units = kilowatt-hour (kWh). A fuse (in the live wire) melts on excess current; earthing protects from shock.

5. Worked examples

Example 1. A 12 V battery drives 2 A through a resistor. Find R and power. R = V/I = 12/2 = 6 Ω; P = VI = 12×2 = 24 W.

Example 2. Two resistors 6 Ω and 3 Ω in parallel. Find the effective resistance. 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 → R = 2 Ω.

Example 3. A 1000 W heater runs for 2 hours. Find the energy in kWh. E = P×t = 1 kW × 2 h = 2 kWh.

6. Common mistakes

Mistake: Adding resistances directly in parallel. Fix: Use 1/R = 1/R₁ + 1/R₂; the parallel resistance is smaller than the smallest resistor.
Mistake: Mixing up power formulas. Fix: Choose P = VI, I²R or V²/R based on the given quantities.
Mistake: Confusing kWh with kW. Fix: kW is power; kWh (= kW × hour) is energy.

7. Practice (book-back style)

State Ohm's law and write its equation.
On what factors does the resistance of a wire depend?
Three 6 Ω resistors are in series. Find the total resistance.
A bulb draws 0.5 A at 220 V. Find its power.
State Joule's law of heating.

8. Answer key

At constant temperature, V ∝ I, so V = IR.
Length (R ∝ L), area of cross-section (R ∝ 1/A), material (ρ) and temperature.
R = 6 + 6 + 6 = 18 Ω.
P = VI = 220 × 0.5 = 110 W.
Heat H = I²Rt; heat produced is proportional to I², resistance and time.

9. Quick revision

Physics Ch 4 · current, voltage, resistance, power, heating.
I = Q/t; V = IR; R = ρL/A.
Series: R = R₁+R₂+…; Parallel: 1/R = 1/R₁+1/R₂+…
P = VI = I²R = V²/R; H = I²Rt; 1 kWh = 3.6×10⁶ J.
Fuse in live wire; earthing prevents shock.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Electric current

I = Q / t

Unit ampere (A).

Ohm's law

V = I R

At constant temperature.

Resistance of a wire

R = ρ L / A

ρ = resistivity.

Series / parallel

R = R₁+R₂ ; 1/R = 1/R₁+1/R₂

Series adds; parallel reduces.

Electric power

P = V I = I²R = V²/R

Unit watt.

Joule's heating

H = I²R t

1 kWh = 3.6×10⁶ J.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Adding resistances directly in parallel

✓ Use 1/R = 1/R₁ + 1/R₂; parallel resistance is smaller than the smallest resistor.

WATCH OUT

✗ Picking the wrong power formula

✓ Use P = VI, I²R or V²/R based on the quantities given.

WATCH OUT

✗ Confusing kW with kWh

✓ kW is power; kWh (= kW × hour) is energy.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Multiple choice questions

MCQs on Ohm's law, resistance, power and energy.

Questions

Short answer questions

Definitions, factors of resistance, Joule's law.

Questions

Numerical / circuit problems

Series–parallel resistance, power and energy sums.

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Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Concept

State Ohm's law.

▶ Show solution

At constant temperature, the current through a conductor is proportional to the potential difference across it; V = IR.

Q2EASY· Numerical

A 12 V battery drives 2 A through a resistor. Find R and the power.

▶ Show solution

R = V/I = 12/2 = 6 Ω; P = VI = 24 W.

Q3MEDIUM· Numerical

Find the effective resistance of 6 Ω and 3 Ω in parallel.

▶ Show solution

1/R = 1/6 + 1/3 = 3/6 → R = 2 Ω.

Q4EASY· Numerical

Three 6 Ω resistors are in series. Find the total resistance.

▶ Show solution

R = 6 + 6 + 6 = 18 Ω.

Q5MEDIUM· Numerical

A 1000 W heater runs for 2 hours. Find the energy used in kWh.

▶ Show solution

E = P×t = 1 kW × 2 h = 2 kWh.

Q6MEDIUM· Concept

State Joule's law of heating and give one application.

▶ Show solution

H = I²Rt; heat is proportional to I², resistance and time. Application: electric heater / iron / fuse.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Physics Chapter 4 of Samacheer Kalvi Class 10 Science.
•I = Q/t; V = IR; R = ρL/A.
•Series: R = R₁+R₂+…; Parallel: 1/R = 1/R₁+1/R₂+…
•P = VI = I²R = V²/R; H = I²Rt.
•1 kWh = 3.6×10⁶ J; energy is billed in units (kWh).
•Fuse goes in the live wire; earthing prevents shock.

Tamil Nadu (TNBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5-10 marks across MCQ, short answer, circuit and numerical questions

Question type	Marks each	Typical count	What it tests
MCQ	1	1-2	Ohm's law, units and definitions
Short Answer	2-3	1-2	Factors of resistance, Joule's law
Numerical / Circuit	2-5	1-2	Series–parallel, power and energy

Prep strategy

Master V = IR and the power formulas
Practise series and parallel resistance sums
Learn energy in kWh and Joule's law
Draw and label circuit diagrams

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Household wiring

Appliances are connected in parallel so each gets the full voltage.

Electric heating

Heaters, irons and geysers use Joule's heating effect.

Safety devices

Fuses and MCBs protect circuits from excess current.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write the relevant formula before substituting
Reduce series–parallel networks step by step
Keep units (A, V, Ω, W) consistent
Label circuit diagrams clearly

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Find the equivalent resistance of a mixed series–parallel network.
Show that P = I²R and P = V²/R follow from P = VI and Ohm's law.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

TN SSLC Class 10 Public ExamHigh

Foundation / NTSE PhysicsMedium

School unit testsHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Adding parallel paths gives the current more routes to flow, so the overall opposition (resistance) decreases below that of any single branch.

One unit is 1 kilowatt-hour (kWh) — the energy used by a 1 kW appliance running for 1 hour, equal to 3.6×10⁶ joules.

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