PhysicsICSE Class 10 · Electrical Power and Energy

Kilowatt Hour and Electricity Bills: Units Consumed and Cost (ICSE Class 10)

Updated 16 October 2026 · Fact-checked

The kilowatt hour (kW h) is the commercial unit of electrical energy: the energy a 1 kW appliance uses in one hour: 1 kW h = 1000 W × 3600 s = 3.6 × 10⁶ J, one 'unit' on a bill. Energy in kW h = power in kW × time in hours, W = Pt.

Energy is power × time, so a tall short bar and a flat long bar can have the same area: a 2 kW heater and a 100 W lamp can each use 1 kW h.

Illustration drawn for QuestionBuddy.

Understand Kilowatt Hour and Electricity Bills

Why a bigger unit? The joule is far too small for household use. A 1000 W iron used for just one hour consumes $1000 \times 3600 = 3600000$ J. Electricity boards therefore measure and charge for energy in a larger unit.

The kilowatt hour. One kilowatt hour is the electrical energy consumed by an appliance of power 1 kW working for 1 hour:

$1 kW h = 1000 W \times 3600 s = 3.6 \times 1 0^{6} J .$

It is called the commercial unit of electrical energy; one kW h is 1 unit on an electricity bill.
It is a unit of energy, not of power. 'Kilowatt' is power; 'kilowatt hour' is power × time.
Conversions: kW h → J, multiply by $3.6 \times 1 0^{6}$ ; J → kW h, divide by $3.6 \times 1 0^{6}$ .

Energy from power and time. From $P = \frac{W}{t}$ :

$W = P t .$

To get energy in	Use power in	And time in
joules (SI)	watts	seconds
kilowatt hours (commercial)	kilowatts	hours

The common error is to multiply watts by hours and call the answer kW h. Divide watts by 1000 first.

Same energy, different appliances. Energy depends on both power and time. A 2 kW heater for half an hour and a 100 W lamp for 10 hours each use 1 kW h. A low-power appliance left on for a long time can cost as much as a high-power one used briefly.

The electricity meter. The meter where the supply enters a house is a kilowatt hour meter (energy meter). Its reading goes up continuously while power is drawn; it shows the total energy used, in kW h, not the power at a given moment.

$Units consumed = present reading - previous reading .$

Calculating a household's consumption and bill.
1. For each appliance: (number of appliances) × (power in kW) × (hours used per day) gives kW h per day.
2. Add the daily energy of all appliances.
3. Multiply by the number of days in the billing period.
4. Multiply the total units by the rate per unit (₹ per kW h) for the cost. If the rate changes with the slab (for example, a lower rate for the first 100 units), charge each slab at its own rate.

Saving energy. Because cost ∝ power × time, savings come from using lower-power appliances (an LED lamp instead of a filament lamp giving the same light), switching off when not in use, and avoiding standby power. A small saving per hour adds up over many hours, days and appliances.

Kilowatt Hour and Electricity Bills: formulas and key facts

Energy from power and time: $W = P t$; W in J with P in W and t in s; in kW h with P in kW and t in h.
Kilowatt hour: $1 kW h = 3.6 \times 1 0^{6} J$; 1 kW h = 1 unit.
Units consumed: $units = \sum (n \times P_{k W} \times h_{p er d a y}) \times days$; Then cost = units × rate.
Meter reading: $U ni t sco n s u m e d = p r ese n t r e a d in g - p r e v i o u sr e a d in g ∣ M e t er u ni t : k W h (e n er g y)$; Not kW.

How to solve Kilowatt Hour and Electricity Bills questions

Set up a table of appliances, convert everything to kW and hours, and keep a running total.

1Convert each power to kW (divide watts by 1000) and each time to hours (divide minutes by 60).
2For each appliance: number × power (kW) × hours per day = kW h per day.
3Add for the daily total, then multiply by the number of days.
4Multiply by the rate per unit; for slab rates, split the units first.
5For joules, multiply kW h by $3.6 \times 1 0^{6}$ ; for time, use $t = \frac{W}{P}$ (h = kW h ÷ kW).
6Write the unit with every answer: kW h (units), J, h or ₹.

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.

The cost of standby

Suppose your television, switched off with the remote but left on standby, still draws 5 W all day and night.

How many units does it waste in 30 days, and what does that cost at ₹8 per unit?

Try it first, then show the solutionSolution

Energy $= 0.005 kW \times 24 h \times 30 = 3.6$ kW h.
Cost = 3.6 × 8 = ₹28.80.

Answer: 3.6 units, about ₹28.80 a month, for nothing.

Memory hook: Small watts × many hours = real money.

Reading the meter yourself

Imagine you note your home's electricity meter reading as 2385.6 one morning and 2392.8 the next morning.

How much energy did your home use in that day, and what was the average power?

Try it first, then show the solutionSolution

Units $= 2392.8 - 2385.6 = 7.2$ kW h.
Average power $= \frac{7.2 kW h}{24 h} = 0.3$ kW $= 300$ W.

Answer: 7.2 kW h; an average of 300 W.

Memory hook: Meter shows energy; divide by time for power.

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.

3.6 million
1 kW h = 1000 × 3600 = 3.6 MJ: 'kilo times an hour of seconds'.
Use it when: Converting kW h and joules.
Match the pairs
W with s gives J; kW with h gives kW h. Never mix W with h.
Use it when: Avoiding the factor-1000 error.
N-P-H-D-R
Number × Power (kW) × Hours × Days × Rate = bill.
Use it when: Household bill questions.
kW is speed, kW h is distance
Power is how fast energy is used, as speed is how fast distance is covered.
Use it when: kW versus kW h.

Kilowatt Hour and Electricity Bills: solved examples

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

Example 1: kW h in joules · 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

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

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

Example 2: Defining the kilowatt hour · 3 marks · Short answerBoard level

(a) Define one kilowatt hour. (b) Show that 1 kW h = $3.6 \times 1 0^{6}$ J. (c) Is the kilowatt hour a unit of power or of energy?

Try it first, then show the solutionSolution

(a) The electrical energy consumed by an appliance of power 1 kW in 1 hour. (1)
(b) $1 kW h = 1000 W \times 3600 s = 3600000$ W s $= 3.6 \times 1 0^{6}$ J. (1)
(c) Energy (power × time). (1)

Answer: (a) Energy used by 1 kW in 1 h (b) 1000 × 3600 J (c) Energy

Example 3: Units used by an iron · 2 marks · Short answerHarder board level

A 750 W iron is used for 40 minutes every day. Find the energy it uses (a) per day, (b) in 30 days, in kW h.

Try it first, then show the solutionSolution

(a) $W = 0.75 kW \times \frac{40}{60} h = 0.5$ kW h. (1)
(b) $0.5 \times 30 = 15$ kW h. (1)

Answer: (a) 0.5 kW h (b) 15 kW h

Example 4: A month's bill · 4 marks · Long answerHarder board level

A household uses four 12 W LED lamps for 5 h a day, two 60 W fans for 10 h a day, a 120 W television for 4 h a day and a 2 kW geyser for 30 minutes a day. Find the units consumed in 30 days and the bill at ₹7.50 per unit.

Try it first, then show the solutionSolution

Lamps: $4 \times 0.012 \times 5 = 0.24$ kW h; fans: $2 \times 0.06 \times 10 = 1.2$ kW h. (1)
Television: $0.12 \times 4 = 0.48$ kW h; geyser: $2 \times 0.5 = 1.0$ kW h. (1)
Daily total $= 0.24 + 1.2 + 0.48 + 1.0 = 2.92$ kW h; in 30 days $= 87.6$ kW h. (1)
Bill = 87.6 × 7.50 = ₹657. (1)

Answer: 87.6 units; ₹657

Example 5: Joules to units and time · 2 marks · Short answerHarder board level

(a) An appliance uses $7.2 \times 1 0^{6}$ J. Express this in kW h and find the cost at ₹6 per unit. (b) For how long must a 400 W appliance run to use 1 unit?

Try it first, then show the solutionSolution

(a) $\frac{7.2 \times 1 0 ^{6}}{3.6 \times 1 0 ^{6}} = 2$ kW h; cost = 2 × 6 = ₹12. (1)
(b) $t = \frac{W}{P} = \frac{1 kW h}{0.4 kW} = 2.5$ h. (1)

Answer: (a) 2 kW h, ₹12 (b) 2.5 h

Example 6: Is switching to LED worth it? · 3 marks · Short answerHOTS / competency

A shop has six 60 W filament lamps lit for 6 h a day. They are replaced by six 9 W LED lamps that give the same light, costing ₹100 each. Electricity costs ₹8 per unit. (a) Find the units saved in 30 days. (b) Find the money saved in 30 days. (c) In about how many months will the LED lamps pay for themselves?

Try it first, then show the solutionSolution

(a) Saving per day $= 6 \times (0.060 - 0.009) \times 6 = 1.836$ kW h; in 30 days $= 55.08$ kW h. (1)
(b) 55.08 × 8 = ₹440.64. (1)
(c) Cost of lamps = 6 × 100 = ₹600; $\frac{600}{440.64} \approx 1.4$ , so within 2 months. (1)

Answer: (a) 55.08 kW h (b) ₹440.64 (c) About 1.4 months

Example 7: Case study: the meter and a slab tariff · 4 marks · Case-basedHOTS / competency

A family's energy meter read 4562.4 at the start of a 30-day month and 4718.4 at the end. The tariff is ₹5 per unit for the first 100 units and ₹7 per unit after that.
(a) What does the meter measure, and in what unit?
(b) Find the units consumed and the bill.
(c) Express the energy consumed in joules.
(d) A 2 kW heater was used for 1 hour every day of the month. What percentage of the units did it account for?

Try it first, then show the solutionSolution

(a) Electrical energy, in kilowatt hours (not power in kW). (1)
(b) $4718.4 - 4562.4 = 156$ units; bill = 100 × 5 + 56 × 7 = 500 + 392 = ₹892. (1)
(c) $156 \times 3.6 \times 1 0^{6} = 5.616 \times 1 0^{8}$ J. (1)
(d) Heater: $2 \times 1 \times 30 = 60$ kW h; $\frac{60}{156} \times 100 \approx 38.5%$ . (1)

Answer: (a) Energy in kW h (b) 156 units, ₹892 (c) 5.616 × 10⁸ J (d) ≈ 38.5%

Where students lose marks in Kilowatt Hour and Electricity Bills

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

Mistake: Wrong unit for the meter reading
The meter adds up energy, so its unit is the kilowatt hour; the kilowatt is a unit of power.
Wrong working
'The meter reading is in kW.'
Right working
'The meter reading is in kW h (kilowatt hours).'
Do this: Remember: meter, bill and units are all energy.
Usually costs: Reported by CISCE examiners for 2025 Q8(i).
Mistake: Watts multiplied by hours
W × h gives watt hours, 1000 times smaller than kW h.
Wrong working
60 W × 10 h = 600 kW h.
Right working
0.06 kW × 10 h = 0.6 kW h.
Do this: Convert watts to kilowatts in the first line.
Usually costs: Usually costs the answer mark.
Mistake: Minutes not changed to hours
The kW h needs time in hours.
Wrong working
0.75 kW × 40 = 30 kW h.
Right working
0.75 kW × (40/60) h = 0.5 kW h.
Do this: Divide minutes by 60.
Usually costs: Usually costs the answer mark.
Mistake: Forgetting the number of appliances or days
Bills add every appliance over every day of the period.
Wrong working
Using one fan when the question says two, or one day's units as the bill.
Right working
number × kW × h per day × days.
Do this: Make a table with a column for each factor.
Usually costs: Usually costs the method and answer marks.

Exam tips

Set out a bill question as a table (appliance, number, kW, h per day, kW h per day) before adding.
State '1 unit = 1 kW h' when you switch from energy to money.
Show the conversion 1 kW h = 1000 W × 3600 s when asked to derive 3.6 × 10⁶ J.
Round money to two decimal places and give the ₹ sign.

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 · MCQBoard level
The meter installed where the electricity supply enters a house measures:
(a) power in kW
(b) energy in kW h
(c) current in A
(d) voltage in V
Hint
What is the bill based on?
Answer
(b) energy in kW h
Q2 · 1 mark · Very short answerBoard level
Express 5 kW h in joules.
Hint
Multiply by $3.6 \times 1 0^{6}$ .
Answer
$5 \times 3.6 \times 1 0^{6} = 1.8 \times 1 0^{7}$ J.
Q3 · 1 mark · Very short answerBoard level
How many units does a 100 W lamp use in 10 hours?
Hint
0.1 kW × 10 h.
Answer
$0.1 \times 10 = 1$ kW h, that is, 1 unit.
Q4 · 1 mark · Very short answerHarder board level
For how long must a 2.5 kW geyser run to use 5 units?
Hint
$t = \frac{W}{P}$ .
Answer
$t = \frac{5}{2.5} = 2$ h.
Q5 · 1 mark · Very short answerHarder board level
Convert $9 \times 1 0^{5}$ J into kW h.
Hint
Divide by $3.6 \times 1 0^{6}$ .
Answer
$\frac{9 \times 1 0 ^{5}}{3.6 \times 1 0 ^{6}} = 0.25$ kW h.
Q6 · 2 marks · Short answerHarder board level
A 150 W television is used for 6 h a day for 31 days. Find the units consumed and the cost at ₹6.50 per unit.
Hint
0.15 kW × 6 h × 31.
Answer
$0.15 \times 6 \times 31 = 27.9$ kW h; cost = 27.9 × 6.50 = ₹181.35.
Q7 · 2 marks · Short answerHarder board level
A 1 kW iron is used for 15 minutes a day for 30 days. Find the energy used in kW h and the cost at ₹8 per unit.
Hint
15 minutes = 0.25 h.
Answer
$1 \times 0.25 \times 30 = 7.5$ kW h; cost = 7.5 × 8 = ₹60.
Q8 · 2 marks · Short answerHarder board level
A 2 kW heater is used for 30 minutes and a 100 W lamp for 10 hours. Which uses more energy? Explain.
Hint
Energy = power × time.
Answer
Heater: $2 \times 0.5 = 1$ kW h; lamp: $0.1 \times 10 = 1$ kW h. They use the same energy: the lamp's lower power is made up for by its longer time.
Q9 · 3 marks · Short answerHOTS / competency
A family's monthly (30-day) bill is ₹1200 at ₹8 per unit. Apart from a 1.5 kW air conditioner, their other appliances use 2 units a day. For how many hours a day, on average, did the air conditioner run?
Hint
Find the total units, subtract the rest.
Answer
Total $= \frac{1200}{8} = 150$ units; others $= 2 \times 30 = 60$ units; AC $= 90$ units $= 3$ units a day; $t = \frac{3}{1.5} = 2$ h a day.
Q10 · 2 marks · Short answerHOTS / competency
A student writes, 'My fan's power is 60 kW h.' Point out the error and correct the statement.
Hint
Power versus energy.
Answer
kW h is a unit of energy, not power. The fan's power is 60 W (0.06 kW); it uses 0.06 kW h of energy each hour.

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.Commercial unit of electrical energy?Show answer
Kilowatt hour (kW h).
2.1 kW h in joules?Show answer
3.6 × 10⁶ J.
3.What does the household meter measure?Show answer
Electrical energy, in kW h.
4.Units used by a 500 W appliance in 4 h?Show answer
2 kW h.
5.Formula linking energy, power and time?Show answer
W = Pt.

Kilowatt Hour and Electricity Bills: frequently asked questions

What is one kilowatt hour?

One kilowatt hour is the electrical energy consumed by an appliance of power 1 kW in one hour. It equals 1000 W × 3600 s = 3.6 × 10⁶ J and is called one 'unit' on an electricity bill.

How do I calculate an electricity bill?

For each appliance multiply number × power in kW × hours used per day; add them for the daily units, multiply by the number of days, then multiply by the rate per unit.

Is kW h a unit of power?

No. The kilowatt (kW) is a unit of power; the kilowatt hour (kW h) is a unit of energy, because it is power multiplied by time.