8 min readConcept sheetPart I (1 mark) on units; Part III (4 marks) power and energy-bill numericals

Electric power and electrical energy

Power is the rate of using electrical energy: P = VI. Bills count energy in kilowatt hours: 1 unit = 1 kWh = 3.6 × 10⁶ J.

PowerP = VI = I²R = V²/R; unit watt (W)

1 watt1 joule per second

EnergyE = P × t

Commercial unit1 kWh = 3.6 × 10⁶ J = 1 unit

Electric power

Electric power is the rate at which electrical energy is used, or the work done per second: P = W/t. Since W = VIt, P = VI. Using Ohm's law, P = I²R and P = V²/R. The SI unit is the watt (W). 1 watt is the power when 1 joule of work is done in 1 second. It is also the power when a current of 1 A flows under 1 V. Bigger unit: 1 kilowatt (kW) = 1000 W. Motors are often rated in horsepower: 1 HP = 746 W.

Electrical energy and the kilowatt hour

Electrical energy used = power × time, E = P × t. The joule is too small for household use. So electricity boards use the kilowatt hour (kWh), called one unit. 1 kWh is the energy used by a 1000 W appliance in 1 hour. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Quick conversions

1 kW1000 W

1 HP746 W

1 kWh (1 unit)3.6 × 10⁶ J

Energy in kWh(power in W × hours used) ÷ 1000

Worked example: current and resistance of a bulb

Problem

A bulb is rated 60 W, 220 V. Find the current through it and its resistance.

Working

Given P = 60 W, V = 220 V.
Formula: P = VI, so I = P/V = 60 ÷ 220 ≈ 0.27 A (rounded to 2 decimal places).
Formula: P = V²/R, so R = V²/P.
R = (220)² ÷ 60 = 48,400 ÷ 60 ≈ 806.7 Ω (rounded to 1 decimal place).

Answer

I ≈ 0.27 A; R ≈ 806.7 Ω

Worked example: the monthly energy bill

Problem

A home uses 4 LED bulbs of 10 W each for 5 hours a day. It also runs 2 fans of 60 W each for 10 hours a day. Find the energy used in 30 days and its cost at an example rate of ₹ 7 per unit.

Working

Bulbs per day: 4 × 10 W × 5 h = 200 Wh.
Fans per day: 2 × 60 W × 10 h = 1200 Wh.
Total per day: 200 + 1200 = 1400 Wh = 1.4 kWh.
For 30 days: 1.4 kWh × 30 = 42 kWh = 42 units.
Cost: 42 × ₹ 7 = ₹ 294.

Answer

42 units; ₹ 294 at the example rate

In real life

The electricity meter at home counts kWh, and the bill charges per unit. A 9 W LED bulb gives about the same light as a 60 W filament bulb, so it uses about 85% less energy.

Exam angle

1 marks: “1 kWh is equal to ___ joule.” 3.6 × 10⁶ J.

2 marks: “Define electric power. Give its SI unit.” The rate at which electrical energy is used: P = W/t = VI. SI unit: watt (W).

4 marks: “An electric heater of 1500 W is used for 2 hours a day. Find the energy used in 30 days and the cost at an example rate of ₹ 7 per unit.” E = 1.5 kW × 2 h × 30 = 90 kWh = 90 units. Cost = 90 × ₹ 7 = ₹ 630.

Remember it

Power = Voltage × I (P = VI). Units on the bill = kilowatts × hours.