What is electric power?
Electric power is the rate at which electrical energy is used (or supplied) in a circuit: P = W/t. Since W = VIt,
P = V × I
Using Ohm's law, P = I²R = V²/R. Its SI unit is the watt (W): 1 W is the power used when 1 A flows under a p.d. of 1 V, i.e. 1 W = 1 V × 1 A = 1 J/s. Bigger units: 1 kilowatt (kW) = 1000 W.
An appliance label such as "1000 W, 220 V" is its power rating: it uses 1000 W when run on 220 V. From this you can find its current (I = P/V) and resistance (R = V²/P).
Electrical energy and the kilowatt-hour
Energy used = power × time: E = P × t. In joules (W × s) the numbers become huge, so for homes we use the kilowatt-hour (kWh), the energy used by a 1 kW appliance in 1 hour. This is the commercial unit of electrical energy, and 1 kWh is called 1 unit on the bill.
1 kWh = 1000 W × 3600 s = 3.6 × 106 J.
Remember: the kWh is a unit of energy, not power.
How to calculate an electricity bill
For each appliance: units per day = power in kW × hours used per day. Add all appliances, multiply by the number of days, then multiply by the rate per unit.
Example (at ₹7 per unit, 30 days): LED bulb 10 W for 4 h = 0.04 kWh; fan 60 W for 4 h = 0.24 kWh; TV 100 W for 4 h = 0.4 kWh; iron 1000 W for 4 h = 4 kWh. Total 4.68 units a day, × 30 = 140.4 units, × ₹7 ≈ ₹983. (Real tariffs have slabs and fixed charges, but the idea is the same.)
Power and brightness: exam favourites
For bulbs on the same supply (parallel), the one with higher power rating is brighter and has lower resistance (R = V²/P). In series, the same current flows, so the bulb with higher resistance (the lower-rated one) gets more power (P = I²R) and glows brighter. Two identical bulbs in series on the same supply each get a quarter of their rated power.
Key formulas and definitions
- P = W / t = V I = I² R = V² / R
- 1 W = 1 V × 1 A = 1 J/s; 1 kW = 1000 W
- E = P × t; 1 kWh = 3.6 × 10⁶ J = 1 unit
- Bill = kW × hours per day × days × rate per unit
Worked examples
1. A bulb draws 0.5 A from a 220 V supply. Find its power.
P = VI = 220 × 0.5 = 110 W.
2. A 1100 W electric iron works on 220 V. Find the current and its resistance.
I = P/V = 1100/220 = 5 A. R = V/I = 220/5 = 44 Ω (or V²/P = 48400/1100 = 44 Ω).
3. How much energy in kWh does a 2 kW heater use in 3 hours? Express it in joules.
E = 2 × 3 = 6 kWh = 6 × 3.6 × 10⁶ = 2.16 × 10⁷ J.
4. A 60 W fan runs 10 hours a day for 30 days. Find the units and the cost at ₹6 per unit.
Units = 0.06 × 10 × 30 = 18 kWh. Cost = 18 × 6 = ₹108.
5. A current of 2 A through a resistor produces 80 W. Find the resistance.
P = I²R, so R = 80 / 4 = 20 Ω.
6. Two bulbs, 100 W and 60 W (both 220 V), are joined in parallel to 220 V. Find the total current.
Total P = 160 W. I = 160/220 ≈ 0.73 A.
7. A house has four 20 W LEDs for 6 h, a 1500 W AC for 5 h and a 200 W fridge for 24 h each day. Find the monthly (30-day) bill at ₹7 per unit.
Daily: 0.08 × 6 = 0.48; 1.5 × 5 = 7.5; 0.2 × 24 = 4.8. Total 12.78 units/day × 30 = 383.4 units. Bill = 383.4 × 7 ≈ ₹2684.
Common mistakes
- Calling the kWh a unit of power. It is a unit of energy; power is in watts.
- Forgetting to change watts to kilowatts before finding units (60 W = 0.06 kW).
- Using P = I²R with the wrong current in parallel circuits. In parallel V is the same, so P = V²/R is easier.
- Thinking a higher-watt bulb has more resistance. On the same voltage, R = V²/P, so higher power means lower resistance.