10 min readConcept sheet2 and 3-mark numericals, and 4-mark case-based questions on an electricity bill

Electric power and the kilowatt hour

Electric power is the rate of using electrical energy: P = VI = I²R = V²/R. Bills charge for energy in kilowatt hours (units), not for power.

PowerP = VI = I²R = V²/R

Unit of powerwatt (W); 1 W = 1 V × 1 A; 1 kW = 1000 W

Commercial unit of energy1 kWh = 1 unit = 3.6 × 10⁶ J

Energy usedE = P × t

Power and its unit

Electric power is the rate at which electrical energy is used (or produced): P = VI. Using V = IR, we also get P = I²R and P = V²/R. The SI unit is the watt (W). 1 W is the power used when 1 A flows under a potential difference of 1 V. A bigger unit is the kilowatt: 1 kW = 1000 W. The rating printed on an appliance, such as 60 W, 220 V, tells you how fast it uses energy when connected to 220 V.

The kilowatt hour

Energy usedPower × time: E = P × t

1 kilowatt hour (kWh)Energy used by a 1 kW appliance in 1 hour

1 kWh in joules1000 W × 3600 s = 3.6 × 10⁶ J

1 unit on the electricity bill1 kWh

What we pay forEnergy (kWh), not power

Worked example: current and resistance of a bulb

Problem

A bulb is rated 100 W, 220 V. Find (a) the current it draws and (b) its resistance when glowing.

Working

Given: P = 100 W, V = 220 V.
(a) P = VI, so I = P/V = 100 W ÷ 220 V = 0.4545 A, which rounds to 0.45 A.
(b) P = V²/R, so R = V²/P = (220 V)² ÷ 100 W = 48,400 ÷ 100 = 484 Ω.
Check: R = V/I = 220 ÷ (100/220) = 220 × 220 ÷ 100 = 484 Ω.

Answer

(a) About 0.45 A (rounded to two decimal places); (b) 484 Ω

Worked example: the monthly electricity bill

Problem

A family uses four 10 W LED bulbs for 5 hours a day, two 75 W ceiling fans for 8 hours a day, and a 1000 W electric iron for 30 minutes a day. Find the energy used in 30 days and its cost at an example rate of ₹ 7 per unit.

Working

Bulbs: 4 × 10 W × 5 h = 200 Wh = 0.2 kWh per day.
Fans: 2 × 75 W × 8 h = 1200 Wh = 1.2 kWh per day.
Iron: 1000 W × 0.5 h = 500 Wh = 0.5 kWh per day.
Total per day = 0.2 + 1.2 + 0.5 = 1.9 kWh.
For 30 days: 1.9 kWh × 30 = 57 kWh = 57 units.
Cost = 57 units × ₹ 7 per unit = ₹ 399.

Answer

57 units; ₹ 399 at the example rate of ₹ 7 per unit

Power vs energy

PowerHow fast energy is used. Unit: watt (W) or kilowatt (kW). A 2 kW geyser uses energy twice as fast as a 1 kW iron, however long each runs.

EnergyHow much energy is used in all. Unit: joule (J) or kilowatt hour (kWh). It depends on the power and the time: E = P × t.

A 2 kW heater used for 3 hours a day for 30 days uses 2 × 3 × 30 = 180 kWh. At an example rate of ₹ 7 per unit, that costs ₹ 1260.

In real life

The star label on a fridge or AC and the wattage on an LED bulb help families pick appliances that use fewer units. Switching off fans and lights in empty rooms cuts the kWh on the monthly bill.

Exam angle

2 marks: “What is the commercial unit of electrical energy? Express it in joules.” The kilowatt hour (kWh). 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

3 marks: “An electric heater of resistance 44 Ω is connected to a 220 V supply. Find the current it draws and its power.” I = V/R = 220 ÷ 44 = 5 A. P = VI = 220 × 5 = 1100 W (1.1 kW).

2 marks: “Which uses more energy: a 200 W television used for 2 hours, or a 1500 W toaster used for 15 minutes?” TV: 0.2 kW × 2 h = 0.4 kWh. Toaster: 1.5 kW × 0.25 h = 0.375 kWh. The television uses more.

Remember it

P = VI (Power is 'VIP'). Energy = power × time. 1 unit = 1 kWh = 3.6 million joules.