Chapter 11 of 13

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Rajasthan (RBSE)Class 10 Science★ 8–10 marks · RBSE Class 10 Science board (Unit — Effects of Current)

Electricity

The most numerical chapter of RBSE Class 10 Science. Current, potential difference and Ohm's law (V = IR); resistance, resistivity and the four factors that control it; series vs parallel combinations; electric power (P = VI = I²R = V²/R) and the heating effect of current; plus the commercial unit of energy (kWh) — with 12 fully solved board-pattern numericals.

⏱ 42 min readHard difficulty✓ Free · NCERT-aligned

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

1Define current, potential difference and resistance with their SI units and measuring instruments
2State and apply Ohm's law (V = IR), including interpreting the V–I graph
3Use R = ρl/A to explain how length, area, material and temperature affect resistance
4Calculate equivalent resistance of resistors in series and in parallel and choose the right combination
5Apply the heating effect H = I²Rt and explain fuses, heaters and bulb filaments
6Compute electric power (P = VI = I²R = V²/R) and energy in kWh, and read an electricity bill

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Why this chapter matters

Electricity is the most numerical chapter in Class 10 Science and the biggest single source of board marks in physics. Almost every RBSE paper carries a circuit or power numerical. Master V = IR and P = VI here and you secure both these marks and the foundation for Class 11–12 physics and JEE/NEET.

Before you start — revise these

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

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Atoms and Molecules (Class 9)

Charge of the electron; what a coulomb of charge means.

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Work and Energy (Class 9)

Work, energy and power — directly reused as electrical energy and power.

Electricity — RBSE Class 10 (Science)

Flip a switch in Jaipur and a fan 3 metres away spins instantly. Nothing visibly travelled — yet energy moved from the power station to your room in a blink. This chapter is the rulebook for that invisible flow: how much charge moves (current), what pushes it (voltage), what resists it (resistance), and how they lock together in one tidy equation, V = IR.

1. Electric current and circuit

Electric current is the rate of flow of electric charge:

$I = \frac{Q}{t}$

where Q is charge in coulombs (C) and t is time in seconds. The unit of current is the ampere (A): 1 A = 1 C/s.

One coulomb is the charge of about $6.25 \times 1 0^{18}$ electrons. By convention, current flows in the direction opposite to electron flow (from + to − outside the cell). A device that measures current is an ammeter, connected in series.

2. Potential difference — the push

To make charge flow, you need a "push" — a potential difference (voltage) provided by a cell or battery.

$V = \frac{W}{Q}$

Potential difference is the work done to move a unit charge between two points. Its unit is the volt (V): 1 V = 1 J/C. A voltmeter measures it, connected in parallel across the component.

3. Ohm's law — the central equation

At constant temperature, the current through a conductor is directly proportional to the potential difference across it.

$V = I R$

A graph of V (y-axis) against I (x-axis) is a straight line through the origin, and its slope is the resistance R.

Resistance (R) opposes the flow of current; its unit is the ohm (Ω): 1 Ω = 1 V/A. A good conductor has low resistance; an insulator very high. A rheostat is a variable resistor used to change the current in a circuit.

4. What decides resistance? — Resistivity

The resistance of a wire depends on four things:

$R = ρ \frac{l}{A}$

Length (l): R ∝ l — a longer wire has more resistance.
Area of cross-section (A): R ∝ 1/A — a thicker wire has less resistance.
Material: captured by resistivity ρ (rho), a property of the material. Unit: Ω·m.
Temperature: resistance generally rises with temperature.

Metals like copper and aluminium have very low resistivity → used for wires. Alloys like nichrome have high resistivity and don't oxidise easily at high temperature → used in heating elements (irons, toasters, geysers). Filaments are made of tungsten (very high melting point).

5. Combinations of resistors

Series — same current, voltages add

Resistors joined end to end carry the same current; the total (equivalent) resistance is the sum:

$R_{s} = R_{1} + R_{2} + R_{3} + \dots$

R_s is larger than any individual resistor. A drawback: if one component fails (a fused bulb), the whole circuit breaks.

Parallel — same voltage, currents add

Resistors joined across the same two points have the same voltage; the reciprocal of the total resistance is the sum of reciprocals:

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

R_p is smaller than the smallest individual resistor. Advantages — each device gets the full voltage, each can be switched independently, and one failing doesn't stop the rest. That's why home wiring uses parallel connections.

6. Heating effect of electric current

When current flows through a resistor, electrical energy turns into heat — Joule's law of heating:

$H = I^{2} R t$

This is useful (heaters, geysers, electric irons, the filament of an incandescent bulb, and the fuse — a thin wire that melts and breaks the circuit when current gets dangerously high) and sometimes wasteful (heat lost in transmission lines).

7. Electric power and the commercial unit

Electric power is the rate at which electrical energy is consumed:

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

The unit is the watt (W): 1 W = 1 J/s. A 60 W bulb uses 60 joules every second.

Energy = power × time. The SI unit is the joule, but electricity bills use the kilowatt-hour (kWh), the commercial unit of energy, also called 1 unit:

$1 kWh = 1000 W \times 3600 s = 3.6 \times 1 0^{6} J$

So a 1000 W (1 kW) geyser running for 1 hour uses exactly 1 unit of electricity.

8. Closing thought

Almost everything in this chapter flows from one relationship — V = IR — and one definition — power = energy per second. Combine them and you can:

predict the current any appliance draws,
choose between series and parallel wiring,
size a fuse safely, and
read your own electricity bill in kWh.

It is the most numerical chapter of Class 10 Science and one of the most rewarding for the RBSE board — if you can rearrange V = IR and P = VI confidently and watch your units, the marks are almost guaranteed. Write the formula, substitute with units, then solve.

Key formulas & results

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

Electric current

I = Q / t

Q in coulombs, t in seconds; unit ampere (A).

Potential difference

V = W / Q

Work per unit charge; unit volt (V) = J/C.

Ohm's law

V = IR

At constant temperature; R is resistance in ohm (Ω).

Resistivity

R = ρ l / A

ρ = resistivity (Ω·m); R ∝ length, R ∝ 1/area.

Series resistance

Rₛ = R₁ + R₂ + R₃ …

Same current; total bigger than any resistor.

Parallel resistance

1/Rₚ = 1/R₁ + 1/R₂ + …

Same voltage; total smaller than the smallest.

Heating effect

H = I²Rt

Joule's law; basis of heaters and fuses.

Electric power

P = VI = I²R = V²/R

Unit watt (W) = J/s.

Commercial energy unit

1 kWh = 3.6 × 10⁶ J = 1 'unit'

Energy = power × time; what the bill charges.

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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

✗ Connecting the ammeter in parallel and the voltmeter in series

✓ Ammeter ALWAYS in series (it measures the current through). Voltmeter ALWAYS in parallel (it measures the voltage across).

WATCH OUT

✗ Adding resistances directly in a parallel circuit

✓ In parallel use 1/Rₚ = 1/R₁ + 1/R₂. Forgetting to take the final reciprocal is the #1 error — Rₚ must come out SMALLER than the smallest resistor.

WATCH OUT

✗ Forgetting to convert minutes/hours to seconds in H = I²Rt

✓ Time must be in seconds for energy in joules. 5 minutes = 300 s. Mismatched units silently destroy the answer.

WATCH OUT

✗ Mixing up the dependence of R on length and area

✓ R ∝ length (longer = more resistance) but R ∝ 1/area (thicker = less resistance). They go opposite ways.

WATCH OUT

✗ Using watts × hours but reporting joules as the bill unit

✓ Bills use kWh, not joules. 1 kWh = 1 unit = 3.6 × 10⁶ J. Convert power to kW and time to hours for billing.

WATCH OUT

✗ Saying home wiring is in series

✓ Homes are wired in PARALLEL so each appliance gets the full 220 V and can be switched independently; one bulb failing doesn't cut the rest.

NCERT exercises (with solutions)

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

In-text Q (Current & p.d.)

Definitions, units, ammeter/voltmeter placement

Questions

In-text Q (Ohm's law & resistance)

V = IR problems, resistivity, factors affecting R

Questions

In-text Q (Combinations)

Series and parallel equivalent resistance

Questions

In-text Q (Power & heating)

Heating effect, power, energy in kWh

Questions

End-of-chapter Ex

End-of-chapter exercise

Mostly numericals: circuits, power, energy and cost

Questions

Practice problems

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

Q1EASY· Current

How much charge flows when a current of 2 A passes for 5 minutes?

▶ Show solution

Step 1 — t = 5 min = 300 s.

Step 2 — Q = I × t = 2 × 300 = 600 C.

✦ Answer: 600 coulombs.

Q2EASY· Ohm's law

A 12 V battery drives a current of 3 A through a resistor. Find its resistance.

▶ Show solution

Step 1 — R = V/I.

Step 2 — R = 12/3 = 4 Ω.

✦ Answer: 4 Ω.

Q3EASY· Instruments

How should an ammeter and a voltmeter be connected in a circuit?

▶ Show solution

Ammeter in series (measures current through); voltmeter in parallel (measures voltage across).

✦ Answer: ammeter — series; voltmeter — parallel.

Q4MEDIUM· Series

Three resistors of 2 Ω, 3 Ω and 5 Ω are joined in series across a 20 V battery. Find the total resistance and the current.

▶ Show solution

Step 1 — Rₛ = 2 + 3 + 5 = 10 Ω.

Step 2 — I = V/Rₛ = 20/10 = 2 A.

✦ Answer: 10 Ω; current = 2 A (same through each).

Q5MEDIUM· Parallel

Two resistors of 6 Ω and 3 Ω are connected in parallel. Find the equivalent resistance.

▶ Show solution

Step 1 — 1/Rₚ = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2.

Step 2 — Rₚ = 2 Ω.

✦ Answer: 2 Ω (smaller than both resistors, as expected).

Q6MEDIUM· Power

An electric bulb is rated 220 V, 100 W. Find the current it draws and its resistance.

▶ Show solution

Step 1 — P = VI ⇒ I = P/V = 100/220 = 0.45 A.

Step 2 — R = V²/P = (220)²/100 = 48400/100 = 484 Ω.

✦ Answer: current ≈ 0.45 A; resistance = 484 Ω.

Q7MEDIUM· Resistivity

How does the resistance of a wire change if its length is doubled and its area of cross-section is halved?

▶ Show solution

Step 1 — R = ρl/A.

Step 2 — Doubling l multiplies R by 2; halving A multiplies R by 2 again.

Step 3 — New R = 2 × 2 = 4 times the original.

✦ Answer: The resistance becomes 4 times the original.

Q8HARD· Heating effect

An electric iron of resistance 25 Ω draws a current of 4 A. Calculate the heat developed in 30 seconds.

▶ Show solution

Step 1 — H = I²Rt.

Step 2 — H = (4)² × 25 × 30 = 16 × 25 × 30.

Step 3 — H = 16 × 750 = 12000 J.

✦ Answer: 12000 J (12 kJ).

Q9HARD· Energy cost

A 1000 W heater runs for 2 hours a day. If electricity costs ₹6 per unit, find the monthly (30-day) cost.

▶ Show solution

Step 1 — Daily energy = power × time = 1 kW × 2 h = 2 kWh = 2 units.

Step 2 — Monthly energy = 2 × 30 = 60 units.

Step 3 — Cost = 60 × ₹6 = ₹360.

✦ Answer: ₹360 per month.

Q10HARD· Combination circuit

A 4 Ω and a 12 Ω resistor are connected in parallel, and this combination is joined in series with a 5 Ω resistor across a 12 V battery. Find (a) the equivalent resistance and (b) the total current.

▶ Show solution

Step 1 — Parallel part: 1/Rₚ = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3 ⇒ Rₚ = 3 Ω.

Step 2 — Add the series 5 Ω: R = 3 + 5 = 8 Ω.

Step 3 — Total current I = V/R = 12/8 = 1.5 A.

✦ Answer: (a) 8 Ω, (b) 1.5 A.

Q11HARD· Series vs parallel

Why is series connection not used for domestic wiring? Give two reasons and state which connection is used instead.

▶ Show solution

Step 1 — In series, the same current flows and the voltage is shared, so each appliance gets less than the full 220 V and may not work properly.

Step 2 — If one appliance/bulb fails, the circuit breaks and ALL appliances stop.

Step 3 — Also, appliances cannot be switched on/off independently in series.

Step 4 — Hence homes use PARALLEL connection: each device gets full voltage, works independently, and others keep running if one fails.

✦ Answer: series shares voltage and fails together; domestic wiring uses parallel.

5-minute revision

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

•I = Q/t (ampere); V = W/Q (volt); R = V/I (ohm). Ammeter in series, voltmeter in parallel.
•Ohm's law V = IR; V–I graph is a straight line through the origin with slope R.
•R = ρl/A: R ∝ length, R ∝ 1/area; nichrome (high ρ) for heaters, copper (low ρ) for wires, tungsten for filaments.
•Series: Rₛ = R₁+R₂+…, same current, bigger total. Parallel: 1/Rₚ = Σ1/R, same voltage, smaller total.
•Heating effect H = I²Rt; basis of heaters, irons and the safety fuse.
•Power P = VI = I²R = V²/R (watt).
•Energy = power × time; 1 kWh = 1 unit = 3.6×10⁶ J — what the electricity bill charges.
•Home wiring is parallel: full voltage to each device, independent switching, fault-tolerant.

Rajasthan (RBSE) marks blueprint

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

Typical chapter weightage: 8–10 marks

Question type	Marks each	Typical count	What it tests
MCQ / Assertion–Reason	1	1–2	Units, instrument placement, Ohm's law statement
Short answer / numerical	2	2	V = IR, series/parallel, power calculations
Short answer	3	1	Heating effect; energy cost; resistivity reasoning
Long / numerical	4–5	1	Combination circuit; series vs parallel justification

Prep strategy

Write the formula first, substitute with units, then solve — earns method marks even on a slip
Drill 15+ numericals mixing V=IR, series, parallel and power until automatic
Always convert time to seconds (for joules) or hours (for kWh) before substituting
Memorise the parallel rule and ALWAYS take the final reciprocal

Where this shows up in the real world

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

Your electricity bill

The 'units' on a Rajasthan discom bill are kilowatt-hours — exactly the energy = power × time you compute here.

Home wiring

Every socket in your house is wired in parallel so each device gets the full 220 V and can be switched on its own.

Safety fuses and MCBs

Fuses and miniature circuit breakers use the heating effect of current to cut off dangerous overloads.

Electric heaters and geysers

Nichrome elements convert electrical energy to heat via H = I²Rt — the science of your room heater and water geyser.

LED vs incandescent

Incandescent bulbs waste energy as heat in a tungsten filament; LEDs give the same light at a fraction of the wattage — lower kWh, lower bill.

Power transmission

Electricity is sent at high voltage to keep current low and minimise I²R heating losses over long lines across the state.

Exam strategy

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

For every numerical: write the formula, list the given values with units, substitute, then compute.
Convert all quantities to SI units BEFORE substituting (minutes→seconds, kW→W when needed).
In parallel problems, never forget the final reciprocal step; sanity-check Rₚ < smallest resistor.
State units in the final answer — a number without a unit can lose the last mark.
Memorise all three forms of power (VI, I²R, V²/R) and pick the one matching the given data.
For 'reasoning' questions (series vs parallel, nichrome vs copper) give two distinct reasons.

Going beyond the textbook

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

Kirchhoff's current and voltage laws for analysing complex multi-loop circuits.
Internal resistance of a cell and the difference between EMF and terminal voltage.
Temperature coefficient of resistance and superconductivity at very low temperatures.
Wheatstone bridge and the meter bridge for precise resistance measurement.

Where else this chapter is tested

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

RBSE Class 10 Board (BSER Ajmer)Very high — a circuit or power numerical almost every year

NTSE / state scholarshipHigh — Ohm's law and power MCQs

JEE FoundationVery high — direct base for Class 11–12 current electricity

Science Olympiad (NSO)High — circuit reasoning and numericals

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Questions students ask

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

Yes. RBSE prescribes the NCERT Science textbook, so the content and numericals are identical. RBSE (BSER Ajmer) sets the exam paper and marking scheme.

Resistance R = ρl/A. A larger area of cross-section A gives the electrons more 'room' to flow, so R ∝ 1/A — a thicker wire offers less resistance for the same length and material.

Nichrome has high resistivity (so it heats up strongly for a given current via H = I²Rt) and a high melting point, and it does not oxidise readily at high temperature. Copper has low resistivity, so it barely heats — perfect for wires, useless for heaters.

A kilowatt (kW) is a unit of POWER (rate of energy use). A kilowatt-hour (kWh) is a unit of ENERGY — the energy used by a 1 kW device in 1 hour. Your bill charges for kWh ('units'), not kW.

A fuse is a short, thin wire of low melting point placed in series. If the current rises dangerously (a short circuit or overload), the heating effect H = I²Rt melts the fuse, breaking the circuit before wiring or appliances are damaged.

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