🔬Science (Class 10)
Electricity
Ohm's law, resistors in series/parallel, heating effect, and electric power.
Class 10
What You Will Learn in Electricity
Electricity involves the flow of electric charge (current) through conductors, governed by Ohm's law, and described by circuits with resistance, voltage, and power relationships.
- Electric current I = Q/t (A); potential difference V = W/Q (V); Ohm's law: V = IR.
- Resistance R = ρL/A; ρ is resistivity (material property), L = length, A = cross-section area.
- Series: R_total = R₁+R₂+...; same current, voltage divides. Parallel: 1/R_total = 1/R₁+1/R₂+...; same voltage, current divides.
- Power P = VI = I²R = V²/R (Watts); Electrical energy E = Pt (joules or kWh).
- 1 unit (kWh) = 1000W × 3600s = 3.6×10⁶ J; cost of electricity in ₹ based on units consumed.
- Heating effect (Joule's law): H = I²Rt; used in electric heaters, irons, incandescent bulbs, fuses.
Key Formulas
V = IRP = VI = I²R = V²/RR_series = R₁+R₂1/R_parallel = 1/R₁+1/R₂
Example
A 100W bulb at 220V draws I=P/V=100/220≈0.45 A and has R=V²/P=484 Ω.
Solved Numerical Example
A wire 2m long, cross-section 0.5mm², resistivity 1.6×10⁻⁶ Ω·m. Find resistance. R = ρL/A = (1.6×10⁻⁶ × 2)/(0.5×10⁻⁶) = 3.2×10⁻⁶/5×10⁻⁷ = 6.4 Ω.
Expected Exam Questions — Electricity
Q1.Three resistors 2Ω, 3Ω, 6Ω are connected in parallel. Find equivalent resistance.
Answer: 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. R = 1 Ω.
Q2.An electric iron rated 1000W, 220V is used for 2 hours daily for 30 days. Find units consumed and cost at ₹6/unit.
Answer: Energy = P×t = 1000W × 2h/day × 30 days = 60,000 Wh = 60 kWh = 60 units. Cost = 60 × ₹6 = ₹360.
Q3.Why are household appliances connected in parallel?
Answer: In parallel: each appliance receives the full mains voltage (220V), operates independently (one failing doesn't affect others), can be switched individually, different currents flow based on power rating. Series connection would divide voltage and depend on individual resistances.
🔘 MCQ Practice — Electricity
MCQ 1.If a wire's length is doubled and diameter is halved, its resistance becomes:
A. 4R
B. 2R
C. 8R ✓
D. R/2
✓ Correct Answer: 8R
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