Is it correct to say: Adding resistances in parallel directly (like series)?

For parallel, use 1/R_p = 1/R₁ + 1/R₂ + …. Invert at the end. R_p is smaller than the smallest R.

Is it correct to say: Confusing resistance R (depends on length/area) with resistivity ρ (material property only)?

Resistivity ρ is a fixed property of the material at a given temperature. Cutting a wire changes R, never ρ.

Is it correct to say: Connecting ammeter in parallel or voltmeter in series?

Ammeter goes in SERIES (low resistance). Voltmeter in PARALLEL (high resistance).

Is it correct to say: Forgetting to convert power to kW or time to hours when using kWh?

1 kWh = 1 kW × 1 h. Convert watts to kW (÷1000) and time to hours before using E = P × t.

Is it correct to say: Using V = IR with mismatched values (V across whole circuit but R of one resistor)?

V, I and R must all refer to the same component (or same complete loop).

Is it correct to say: Thinking a fatter wire has more resistance?

R ∝ 1/A. Thicker (larger area) means LOWER resistance.

Electricity

Ohm's law, resistance, series and parallel circuits, electric power and Joule's heating — NCERT Class 10 Physics Ch 12

📐 10 formulas✏️ 3 examples🎯 6 practice⚖️ 5-6 marks🏫 CBSE📚 Class 10✓ 2025–26 syllabus

💡

Board Exam Tips

→Ohm's law numerical (V = IR) with a series/parallel combination is asked almost every board.
→Always draw the circuit diagram before solving — 1 mark just for the neat diagram.
→State units next to every answer: V, A, Ω, W, J. One-mark loss otherwise.
→Learn the derivation of R = ρl/A — a favourite 3-mark question.
→Difference between resistance and resistivity carries 2 marks almost every year.

📊 Diagram

Series and parallel resistor networks with an ammeter and voltmeter

📐 Formulas(10)

Electric Current

I = \frac{Q}{t}

Symbol	Meaning
$I$	Current (ampere, A)
$Q$	Charge (coulomb, C)
$t$	Time (second, s)

Potential Difference

V = \frac{W}{Q}

Symbol	Meaning
$V$	Potential difference (volt, V)
$W$	Work done (joule, J)
$Q$	Charge moved (C)

Ohm's Law★ Board fav

V = I R

Resistance and Resistivity★ Board fav

R = ρ \frac{l}{A}

Symbol	Meaning
$R$	Resistance (ohm, Ω)
$ρ$	Resistivity of the material (Ω·m)
$l$	Length of the conductor (m)
$A$	Area of cross-section (m²)

Resistors in Series

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

Resistors in Parallel★ Board fav

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

Electric Power

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

Electrical Energy

E = P \times t = V I t

Joule's Law of Heating★ Board fav

H = I^{2} R t

Commercial Unit of Energy

1 kWh = 3.6 \times 1 0^{6} J

✏️ Solved Examples

1Solved Exampleeasy3 steps

A 5 Ω resistor is connected to a 10 V battery. Find the current through it.

List given quantities

V = 10 V, R = 5 Ω

2Solved Exampleboard3 steps

Three resistors of 4 Ω, 6 Ω and 12 Ω are connected in parallel across a 6 V battery. Find (a) the equivalent resistance, (b) the current drawn from the battery.

Reciprocal sum for parallel combination

\frac{1}{R _{p}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{3 + 2 + 1}{12} = \frac{6}{12}

3Solved ExampleHOTS4 steps

An electric bulb rated 100 W – 220 V is used for 5 hours daily. Calculate the electric energy consumed per month (30 days) and the cost at ₹5 per unit.

Power in kW

P = 100 W = 0.1 kW

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Adding resistances in parallel directly (like series)
✓For parallel, use 1/R_p = 1/R₁ + 1/R₂ + …. Invert at the end. R_p is smaller than the smallest R.
2
Confusing resistance R (depends on length/area) with resistivity ρ (material property only)
✓Resistivity ρ is a fixed property of the material at a given temperature. Cutting a wire changes R, never ρ.
3
Connecting ammeter in parallel or voltmeter in series
✓Ammeter goes in SERIES (low resistance). Voltmeter in PARALLEL (high resistance).
4
Forgetting to convert power to kW or time to hours when using kWh
✓1 kWh = 1 kW × 1 h. Convert watts to kW (÷1000) and time to hours before using E = P × t.
5
Using V = IR with mismatched values (V across whole circuit but R of one resistor)
✓V, I and R must all refer to the same component (or same complete loop).
6
Thinking a fatter wire has more resistance
✓R ∝ 1/A. Thicker (larger area) means LOWER resistance.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
A wire of resistance 10 Ω is stretched to double its length. Find its new resistance.
Q2
Two resistors of 6 Ω and 3 Ω are in series across a 9 V battery. Find the current and voltage across the 6 Ω resistor.
Q3
An electric iron of 750 W runs for 2 hours. Calculate the energy consumed in kWh and in joules.
Q4
State two differences between resistance and resistivity.
Q5
A 220 V heater draws 5 A of current. Find its resistance and the heat produced in 10 min.
Q6
Three 6 Ω resistors are connected in parallel. What is the equivalent resistance?

📝 Notes

Electricity

Electricity deals with electric charges in motion — the current that lights our bulbs and drives our fans.

Core quantities

Charge (Q): measured in coulombs. 1 electron carries 1.6 × 10⁻¹⁹ C.
Current (I): rate of flow of charge. Conventional current flows opposite to electron flow.
Potential difference (V): the "push" that drives current from high to low potential.
Resistance (R): opposition offered by a conductor. Depends on material (ρ), length (l), area (A) and temperature.

Ohm's law — when it holds

Ohm's law V = IR is valid only when:

Temperature is constant.
The conductor is metallic (ohmic).
No mechanical strain acts on the wire.

Non-ohmic examples: diode, filament bulb at high current, electrolyte.

Series vs Parallel — quick memory hook

Series: current same; voltages add; R_s LARGER; used for Christmas lights (all off if one fails).
Parallel: voltage same; currents add; R_p SMALLER; used in household wiring (each appliance independent).

Heating effect — everyday use

Joule's law H = I²Rt is the basis of:

Electric bulb (tungsten filament, high melting point)
Electric iron, geyser, heater (nichrome coil, high resistivity)
Fuse wire (low melting point, breaks on overload)

Quick sanity checks

Doubling voltage on a fixed R doubles the current and quadruples the power (P = V²/R).
Halving the wire's area doubles its resistance.
1 unit of electricity = 1 kWh = 3.6 MJ.