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.
