Last Updated: May 2026
CBSE Class 10 Science Chapter 12 — Electricity is a high-yield chapter, contributing 6–8 marks across short and long-answer questions in every CBSE board paper. The chapter covers electric current, potential difference, Ohm’s law, resistance, resistors in series and parallel, electric power, heating effect, and the practical formulas you need for numerical problems.
Quick Reference Table — Core Formulas
| Quantity | Formula / Definition | SI Unit |
|---|---|---|
| Electric Current (I) | I = Q / t | Ampere (A) = Coulomb / second |
| Potential Difference (V) | V = W / Q | Volt (V) = Joule / Coulomb |
| Resistance (R) | R = V / I | Ohm (Ω) = Volt / Ampere |
| Resistivity (ρ) | R = ρ × L / A | Ohm-metre (Ω·m) |
| Electric Power (P) | P = V × I = I²R = V²/R | Watt (W) |
| Energy (E) | E = P × t | Joule (J) or kilowatt-hour (kWh) |
| Heat (H) | H = I²Rt (Joule’s law of heating) | Joule |
Ohm’s Law and Its Verification
Statement: The current flowing through a conductor at constant temperature is directly proportional to the potential difference across its ends. V = IR.
Graph: A V-I graph for a metallic conductor at constant temperature is a straight line passing through the origin. The slope of the V-I graph gives R; the slope of the I-V graph gives 1/R.
Conditions for Ohm’s law: Constant temperature, constant physical conditions, metallic conductors. Filaments of incandescent bulbs, semiconductor junctions, and electrolytes do not obey Ohm’s law strictly.
Factors Affecting Resistance
R = ρ × L / A — resistance depends on:
- Length (L) — directly proportional. Doubling length doubles resistance.
- Area of cross-section (A) — inversely proportional. Doubling area halves resistance.
- Resistivity (ρ) — depends on material and temperature. Pure metals have low ρ; alloys higher.
- Temperature — for metals ρ increases with temperature; for semiconductors ρ decreases.
Practical material choice:
- Heating element of an iron / toaster: Nichrome (high resistivity, low oxidation, high melting point)
- Filament of bulb: Tungsten (very high melting point ~3380°C)
- Connecting wires: Copper or Aluminium (low resistivity)
- Fuse wire: Tin-Lead alloy (low melting point — melts to break circuit on overload)
Combination of Resistors
Series: Same current through all. Total R_s = R₁ + R₂ + R₃ + … . Used when resistance is to be increased or current limited.
Parallel: Same potential difference across all. Total 1/R_p = 1/R₁ + 1/R₂ + 1/R₃ + … . Used in household wiring so that each appliance gets the full mains voltage and one device’s failure does not break others.
Two-resistor parallel shortcut: R_p = (R₁ × R₂) / (R₁ + R₂)
Electric Power and Energy
Power formulas — choose the convenient one:
- P = V × I (when V and I are known)
- P = I²R (when I and R are known)
- P = V²/R (when V and R are known)
Commercial unit of energy: 1 kilowatt-hour (kWh) = 1000 W × 3600 s = 3.6 × 10⁶ J. The “1 unit” on your electricity bill = 1 kWh.
Joule’s Law of Heating
The heat produced in a conductor is directly proportional to:
- Square of the current (I²)
- Resistance (R)
- Time (t)
H = I² × R × t
Practical applications: Electric iron, toaster, geyser, electric heater, kettle. The element is made of a high-resistance alloy like nichrome to maximise heat. Fuse protects circuits because the fuse wire’s heating exceeds its melting point on overload current.
Sample Numerical — Series-Parallel Combination
Problem: Two resistors of 4 Ω and 6 Ω are connected in parallel. The combination is connected in series with a 2 Ω resistor across a 12 V battery. Find: (i) total resistance, (ii) current from the battery, (iii) potential difference across the parallel combination.
Solution:
- R_parallel = (4 × 6)/(4 + 6) = 24/10 = 2.4 Ω
- R_total = 2.4 + 2 = 4.4 Ω
- I = V/R = 12/4.4 ≈ 2.73 A
- V across parallel = I × R_parallel = 2.73 × 2.4 ≈ 6.55 V
10 Practice MCQs — Class 10 Electricity
Quiz data missing.
Frequently Asked Questions
Why is tungsten used as the filament of a bulb?
Tungsten has a very high melting point (~3380°C), which lets the filament heat to incandescence without melting. It also has high resistivity (so a small length glows brightly) and good ductility for fine-wire manufacture.
Why are appliances connected in parallel in household wiring?
In parallel, each appliance gets the full mains voltage (220 V in India) and operates at its rated power independently. If one device fails, the others continue working — series connection would shut everything down.
What is 1 kilowatt-hour in joules?
1 kWh = 1000 watts × 3600 seconds = 3.6 × 10⁶ joules. This is the commercial unit of electricity (“1 unit” on the meter).
A 100 W bulb operates for 5 hours. How much energy does it consume in kWh?
Energy = Power × Time = 100 W × 5 h = 500 Wh = 0.5 kWh. The bill therefore charges 0.5 units for that operation.
Continue Your Class 10 Science Prep
- CBSE Class 10 Science — Light: Reflection and Refraction
- CBSE Class 10 Science — Acids, Bases and Salts
- Ready For Boards Courses
- Board Exam FAQ
Bottom line: Memorise the seven core formulas, the conditions for Ohm’s law, the factors affecting resistance, the series-parallel shortcuts, and Joule’s law of heating. Solve at least ten numerical problems combining these — Electricity questions are almost entirely formula-driven and reward speed.
