Electric Power

• Electric power: P = VI = I²R = V²/R.
• Energy: E = Pt; commercial unit kWh (kilowatt-hour).
• Heating effect: H = I²Rt (Joule's law).
• Rating of appliances — 60 W, 100 W bulb at 220 V.
• Fuse and safety (brief NCERT context).

1. Power — P = VI (watt = volt × ampere).
2. Alternate forms — P = I²R = V²/R (using V = IR).
3. Energy — E = P × t; joule when P in W, t in s.
4. kWh — energy used by 1 kW device in 1 hour; 1 kWh = 3.6 × 10⁶ J.
5. 100 W bulb — converts 100 J electrical energy per second to light + heat.
6. Heating effect — H = I²Rt; used in heater, iron, fuse.
7. Higher power — more current for same voltage: I = P/V.
8. Series bulbs — same I; higher R bulb glows brighter (P = I²R).
9. Parallel bulbs — same V; lower R draws more current, brighter.
10. Bill calculation — units = kW × hours; cost = units × rate.

Cost of running 2 kW heater for 3 hours at ₹5 per unit

Given: P = 2 kW, t = 3 h, rate = ₹5/kWh
Step 1 — Energy = P × t = 2 × 3 = 6 kWh (units)
Step 2 — Cost = 6 × 5 = ₹30
Step 3 — In joules: 6 × 3.6 × 10⁶ = 2.16 × 10⁷ J
Conclusion: 6 units consumed; cost ₹30.

• Confusing power (W) with energy (J or kWh).
• Using P = V²R instead of P = V²/R.
• Misconception: kWh is power unit (it's energy).
• Forgetting to convert minutes to hours in kWh problems.
• Assuming 100 W and 60 W bulbs in series equally bright.

• Write three formulas for electric power.
• How many joules in 1 kWh?
• 220 V, 11 A device — find power.
• Why does fuse wire melt on overload?

Open the Practice tab for graded questions on Electric Power.