PhysicsICSE Class 8 · Electricity

Kilowatt Hour and Electricity Bills: kWh, Units and Cost (ICSE Class 8 Physics)

Updated 17 October 2026 · Fact-checked

One kilowatt hour (1 kWh) is the energy used by a 1 kW appliance working for 1 hour, and it is the "1 unit" on an electricity bill. Energy in kWh = power in watt × time in hours ÷ 1000, and 1 kWh = 3.6 × 10⁶ J.

1 kW running for 1 hour is 1 kWh, called 1 unit on the bill and equal to 3.6 × 10⁶ J; multiply units by the rate to get the cost.

Illustration drawn for QuestionBuddy.

Understand Kilowatt Hour and Electricity Bills

Why a bigger unit than the joule? A 100 W bulb uses 100 J every second, which is 360 000 J in an hour. Household energy is therefore huge when counted in joules, so supply companies use a bigger unit, the kilowatt hour (kWh).

Definition. 1 kWh is the electrical energy used when an appliance of power 1 kilowatt (1000 W) works for 1 hour. On the bill it is called 1 unit.

Formula.

$E (kWh) = \frac{P ( W ) \times t ( h )}{1000}$ or $E (kWh) = P (kW) \times t (h)$ .

The first form has the power in watt and the second in kilowatt; both give the same answer if used consistently. Time must be in hours: change minutes to hours by dividing by 60.

kWh in joules. $1 kWh = 1000 W \times 3600 s = 3600000 J = 3.6 \times 1 0^{6} J$ .

joules to kWh: divide by $3.6 \times 1 0^{6}$ ;
kWh to joules: multiply by $3.6 \times 1 0^{6}$ .

Cost. $cost = units used \times rate per unit$ . The rate is always given in the question. Some bills use slabs (the first few units at a lower rate, the rest at a higher rate): work out each slab separately and add.

Several appliances or many days. Energy of one appliance for one day $\times$ number of appliances $\times$ number of days. Different appliances are calculated one by one and added.

Power is not energy. A heater of 2 kW has high power, but a 100 W bulb left on for 40 hours uses more energy ( $100 \times 40 = 4000$ Wh = 4 kWh) than the heater used for 1.5 hours ( $2 \times 1.5 = 3$ kWh). The bill depends on power × time.

Saving units. Switch off unused appliances, and replace high-power bulbs with low-power ones of the same brightness (for example LED instead of filament bulbs): the same light for far fewer units.

Kilowatt Hour and Electricity Bills: formulas and key facts

Energy in kWh: $E (kWh) = \frac{P ( W ) \times t ( h )}{1000}$; Time in hours; 1 kWh = 1 unit.
kWh in joules: $1 kWh = 3.6 \times 1 0^{6} J$; 1000 W × 3600 s.
Cost of electricity: $cost = units \times rate per unit$; Add slab by slab if rates differ.
Several appliances: $E = \sum P t (each appliance, each day, all days)$; Add the separate energies.

How to solve Kilowatt Hour and Electricity Bills questions

Write every power in the same unit, every time in hours, then multiply and divide by 1000 only once.

1List each appliance: its power (W), hours per day, number of appliances.
2Change minutes to hours (divide by 60).
3Find energy for one appliance for one day: $\frac{P \times t}{1000}$ kWh.
4Multiply by the number of appliances and days; add different appliances.
5Cost = total units × rate (slab by slab if needed).
6For joules, multiply kWh by $3.6 \times 1 0^{6}$ .

Kilowatt Hour and Electricity Bills in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Why the geyser costs more than the lights

Suppose a house has 6 LED bulbs of 9 W each, on for 5 h a day, and a geyser of 2 kW used for 30 minutes a day.

Which uses more energy in one day, and by how many times?

Try it first, then show the solutionSolution

Bulbs: $6 \times 9 = 54$ W; $54 \times 5 = 270$ Wh = 0.27 kWh.
Geyser: $2 \times 0.5 = 1$ kWh.
Ratio $= \frac{1}{0.27} \approx 3.7$ .

Answer: The geyser uses 1 kWh against 0.27 kWh: about 3.7 times more.

Memory hook: High power, even for a short time, can beat low power for a long time.

Reading the bill

Imagine the reading on the supply unit is 5 214 on 1 June and 5 394 on 1 July, and the rate is ₹6 per unit.

Find the bill for the month, ignoring any fixed charge.

Try it first, then show the solutionSolution

Units used $= 5394 - 5214 = 180$ .
Cost $= 180 \times 6 = ₹1080$ .

Answer: 180 units; ₹1080.

Memory hook: Units used = later reading minus earlier reading.

Tricks to remember Kilowatt Hour and Electricity Bills

Short memory aids. Say each one aloud once, then use it on a question the same day.

Watt-hours then thousand
Multiply watts by hours, then divide by 1000: "W-H over a thousand".
Use it when: Converting to kWh.
Kilo means 1000 twice
1 kWh = 1000 W for 3600 s: the 1000 and the 3600 give 3.6 million.
Use it when: Remembering 3.6 × 10⁶ J.
One unit, one hour of 1 kW
A 1 kW heater for 1 h, a 100 W bulb for 10 h, a 10 W LED for 100 h: all are 1 unit.
Use it when: Checking your answer.

Kilowatt Hour and Electricity Bills: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Joules in one unit · 1 mark · MCQBoard level

One kilowatt hour is equal to:
(a) $3.6 \times 1 0^{3}$ J
(b) $3.6 \times 1 0^{5}$ J
(c) $3.6 \times 1 0^{6}$ J
(d) $3.6 \times 1 0^{9}$ J

Try it first, then show the solutionSolution

$1 kWh = 1000 \times 3600 = 3600000$ J $= 3.6 \times 1 0^{6}$ J.

Answer: (c) $3.6 \times 1 0^{6}$ J

Example 2: Bulbs for a month · 2 marks · Short answerBoard level

Five bulbs of 40 W each burn for 6 hours every day. Find the electrical energy used in 30 days, in kWh.

Try it first, then show the solutionSolution

Total power $= 5 \times 40 = 200$ W. (½)
Energy $= \frac{200 \times 6 \times 30}{1000}$ (1)
$= 36$ kWh. (½)

Answer: 36 kWh (36 units)

Example 3: A month of household use · 4 marks · Long answerHarder board level

In a 30-day month a house uses a 2 kW heater for 1.5 h daily, a 100 W TV for 5 h daily and four 10 W lamps for 6 h daily. Find (a) the units used by each, (b) the total, (c) the bill at ₹6 per unit.

Try it first, then show the solutionSolution

(a) Heater: $2 \times 1.5 \times 30 = 90$ kWh. (1)
TV: $\frac{100 \times 5 \times 30}{1000} = 15$ kWh. (1)
Lamps: $\frac{4 \times 10 \times 6 \times 30}{1000} = 7.2$ kWh. (1)
(b) Total $= 90 + 15 + 7.2 = 112.2$ units; (c) cost $= 112.2 \times 6 = ₹673.20$ . (1)

Answer: (a) 90, 15 and 7.2 units (b) 112.2 units (c) ₹673.20

Example 4: Joules to units · 3 marks · Short answerHarder board level

A geyser gives $1.26 \times 1 0^{7}$ J of energy to water in a day. (a) Express this in kWh. (b) Find the daily cost at ₹7 per unit.

Try it first, then show the solutionSolution

(a) $E = \frac{1.26 \times 1 0 ^{7}}{3.6 \times 1 0 ^{6}}$ (1)
$= 3.5$ kWh. (1)
(b) Cost $= 3.5 \times 7 = ₹24.50$ . (1)

Answer: (a) 3.5 kWh (b) ₹24.50

Example 5: Is the swap worth it? · 4 marks · Case-basedHOTS / competency

A family uses four 100 W filament bulbs for 5 h daily, and plans to replace them by four 12 W LED lamps giving the same light. For 30 days and ₹8 per unit, find (a) units used before, (b) units used after, (c) money saved in the month.

Try it first, then show the solutionSolution

(a) $\frac{4 \times 100 \times 5 \times 30}{1000} = 60$ units. (1)
(b) $\frac{4 \times 12 \times 5 \times 30}{1000} = 7.2$ units. (1)
(c) Units saved $= 60 - 7.2 = 52.8$ . (1)
Money saved $= 52.8 \times 8 = ₹422.40$ . (1)

Answer: (a) 60 units (b) 7.2 units (c) ₹422.40 saved

Where students lose marks in Kilowatt Hour and Electricity Bills

The slips examiners see most often, and what to write instead.

Mistake: Using minutes in the formula
The unit needs power in kW and time in hours.
Wrong working
$\frac{2000 \times 30}{1000} = 60$ kWh for a 2 kW heater used 30 minutes.
Right working
$\frac{2000 \times 0.5}{1000} = 1$ kWh.
Do this: Convert time to hours before substituting.
Mistake: Forgetting the number of appliances or days
The formula gives energy for one appliance for the time stated.
Wrong working
"3 bulbs of 60 W for 5 h: $\frac{60 \times 5}{1000} = 0.3$ kWh."
Right working
$\frac{3 \times 60 \times 5}{1000} = 0.9$ kWh.
Do this: Underline "each", "all" and the number of days in the question.
Mistake: Saying a unit is a unit of power
Kilowatt hour is energy; kilowatt is power.
Wrong working
"1 kWh is the power used in 1 hour."
Right working
"1 kWh is the energy used by a 1 kW appliance in 1 hour."
Do this: Say "energy" with kWh and "power" with kW.

Exam tips

Write "1 kWh = 1 unit = 3.6 × 10⁶ J" at the top of every bill question.
Show the steps: formula, substitution, answer with unit (kWh or units, rupees).
In an "add the energies" question, finish each appliance separately and add at the end.
Quick check: 1 unit is 1 kW for 1 h, so a 100 W lamp needs 10 h per unit.

Practice questions on Kilowatt Hour and Electricity Bills

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Very short answerBoard level
A refrigerator motor of 150 W runs for 8 hours in a day. How many units does it use that day?
Hint
Power in W times hours, divided by 1000.
Answer
$\frac{150 \times 8}{1000} = 1.2$ units (1.2 kWh).
Q2 · 2 marks · Short answerHarder board level
An electric iron of 1.2 kW is used for 45 minutes. At ₹6.50 per unit, find the cost.
Hint
Change 45 min to hours first.
Answer
$t = 0.75$ h; $E = 1.2 \times 0.75 = 0.9$ kWh; cost $= 0.9 \times 6.5 = ₹5.85$ .
Q3 · 3 marks · Short answerHarder board level
(a) Convert 2.5 kWh into joules. (b) For how long could that energy run a 250 W appliance?
Hint
Multiply by $3.6 \times 1 0^{6}$ ; then use $t = \frac{E}{P}$ .
Answer
(a) $2.5 \times 3.6 \times 1 0^{6} = 9 \times 1 0^{6}$ J. (b) $t = \frac{9 \times 1 0 ^{6}}{250} = 36000$ s $= 10$ h.
Q4 · 3 marks · Short answerHarder board level
A house uses 270 units in a 30-day month. If it were used at a constant rate, what steady power would that be? Find also the bill at ₹6 per unit.
Hint
Hours in 30 days = $30 \times 24$ .
Answer
Time $= 720$ h; $P = \frac{270}{720} = 0.375$ kW $= 375$ W. Bill $= 270 \times 6 = ₹1620$ .
Q5 · 4 marks · Long answerHOTS / competency
Geyser A is rated 3 kW and runs 20 minutes a day. Geyser B is rated 2 kW and runs 40 minutes a day. In 30 days, which uses more energy and by how many units? What is the difference in cost at ₹7 per unit?
Hint
Do not decide by power alone.
Answer
A: $3 \times \frac{1}{3} = 1$ kWh per day, so 30 units. B: $2 \times \frac{2}{3} = \frac{4}{3}$ kWh per day, so 40 units. B uses 10 units more; extra cost $= 10 \times 7 = ₹70$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is 1 kWh in joules?Show answer
3.6 × 10⁶ J.
2.What is one "unit" on the bill?Show answer
1 kilowatt hour.
3.Formula for energy in kWh from watts and hours?Show answer
P × t ÷ 1000.
4.Is kWh a unit of power or energy?Show answer
Energy.
5.How many watts in 1 kW?Show answer
1000 W.

Kilowatt Hour and Electricity Bills: frequently asked questions

What is a kilowatt hour?

A kilowatt hour is the energy used by a 1 kW (1000 W) appliance working for 1 hour. It is the "unit" of electrical energy on household bills, and 1 kWh = 3.6 × 10⁶ J.

How do I calculate the electricity bill?

Find the units used for each appliance as power in W × time in h ÷ 1000, add them, and multiply the total units by the rate per unit given in the question.

Why is the kilowatt hour used instead of the joule?

Because household energy in joules is a very large number. One kWh is 3.6 million joules, so bills use a more convenient size.