Class 10 Science Notes — Chapter 11: Electricity
Current, potential difference, resistance, Ohm's law, and the heating effect of current.
Detailed NCERT notes
- Electric current I = charge Q / time t. Unit: ampere (A). 1 A = 1 C/s. Conventional current flows from + to − (opposite to electron flow).
- Potential difference V between two points = work done to move unit charge = W/Q. Unit: volt (V).
- Ohm's law: V = IR (at constant temperature). R is the resistance of the conductor; unit: ohm (Ω).
- Resistance depends on length (∝ L), cross-section (∝ 1/A), and material (resistivity ρ): R = ρL/A. Unit of ρ: Ω·m.
- Conductors have low resistivity (metals like Cu, Ag); insulators have very high resistivity (rubber, glass).
- Resistors in series: same current I; total V = V₁ + V₂ + ...; equivalent Rₛ = R₁ + R₂ + R₃ + …
- Resistors in parallel: same voltage V across each; total I = I₁ + I₂ + ...; equivalent 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ + …
- Parallel connection is preferred at home: appliances get full voltage, work independently, and total resistance decreases.
- Heating effect of current (Joule's law): H = I²Rt. Applications: bulb filament (tungsten, high m.p.), electric iron, heater, fuse.
- Electric power: P = VI = I²R = V²/R. Unit: watt (W). Commercial unit of energy: kilowatt-hour (kWh). 1 kWh = 3.6 × 10⁶ J.
- Electric fuse melts and breaks the circuit if current exceeds a safe value — protects against short circuits/overloading.
Formulas & key results
- I = Q/t
- V = IR (Ohm's law)
- R = ρL/A
- Series: R = R₁ + R₂ + …
- Parallel: 1/R = 1/R₁ + 1/R₂ + …
- P = VI = I²R = V²/R
- H = I²Rt (Joule's heating)
Mind map
- Charge → current → potential difference → resistance
- Series vs parallel
- Heating effect, power, energy (kWh)
Tricks & shortcuts
- Series: same current; parallel: same voltage.
- In parallel, equivalent resistance is less than the smallest resistor.
Common mistakes to avoid
- Adding parallel resistances directly.
- Confusing 1 kWh with 1 kJ (1 kWh = 3.6 × 10⁶ J).
- Ignoring the effect of temperature on R.
Competency-based questions & answers
- Q. Three resistors 2 Ω, 3 Ω, 6 Ω are connected in parallel. Find equivalent.A. 1/R = 1/2 + 1/3 + 1/6 = 6/6 = 1 ⇒ R = 1 Ω.
- Q. A 60 W bulb operates at 220 V. Calculate current and resistance.A. I = P/V = 60/220 ≈ 0.27 A; R = V/I ≈ 806.7 Ω (or V²/P = 806.7 Ω).
- Q. Why is the filament of a bulb made of tungsten?A. Very high melting point (~3380 °C), high resistivity, and can emit light at high temperatures without breaking.