Electrical Energy, Power and the Cost of Electricity – Basic Electrical Wiring and Installation (Senior 3 Physics) | Ganzaa

Basic Electrical Wiring and Installation

Lessons

Senior 3 · Physics

Basic Electrical Wiring and Installation

5 lessons

Lesson 3 of 5

Electrical Energy, Power and the Cost of Electricity

25 min

What does a token buy?

A family in Kigali buys an electricity token with mobile money and types the 20-digit code into its prepaid meter, which adds a number of "units". A unit is one kilowatt-hour. How long those units last depends on which appliances the family uses, how powerful they are and for how long, and the price of each unit depends on how many the family has already used that month. This lesson puts numbers on all of it.

Power: three forms of one formula

An appliance with a current $I$ and a potential difference $V$ converts electrical energy at the rate

P = V I

Using $V = I R$ from Ohm's law, the same power can be written in the two other forms that are useful for resistors and heaters:

P = I^{2}R \qquad\qquad P = \frac{V^{2}}{R}

The energy converted in a time $t$ is $E = P t = V I t$ . In joules, $P$ is in watts and $t$ in seconds. For electricity bills the unit is the kilowatt-hour (S1): the energy used by a 1 kW appliance in 1 hour, $1\ \text{kWh} = 3.6 \times 10^{6}\ \text{J}$ .

Definition

Electrical power: the rate at which electrical energy is converted into other forms, $P = V I$ . Unit: the watt ( $1\ \text{W} = 1\ \mathrm{J\,s^{-1}}$ ).

Power rating: the power an appliance uses when connected to its rated voltage, printed on its label, e.g. "230 V, 2000 W". Kilowatt-hour (kWh, the "unit" on a meter): energy = power in kW × time in hours.

Worked example 1: reading a kettle's label

A kettle is rated 230 V, 2000 W. Find (a) the current it draws; (b) the resistance of its element; (c) the energy it uses in 4.0 minutes, in J and in kWh.

\text{(a) } I = \frac{P}{V} = \frac{2000}{230} \approx 8.7\ \text{A} \qquad \text{(b) } R = \frac{V^{2}}{P} = \frac{230^{2}}{2000} \approx 26\ \Omega

\text{(c) } E = Pt = 2000 \times 240 = 4.8 \times 10^{5}\ \text{J} \qquad E = 2.0\ \text{kW} \times \frac{4.0}{60}\ \text{h} \approx 0.13\ \text{kWh}

Worked example 2: two lamps in series

Two identical 230 V, 100 W filament lamps are connected in series across the 230 V mains. How much power does each lamp use? (Assume each lamp's resistance stays the same.)

R = \frac{V^{2}}{P} = \frac{230^{2}}{100} = 529\ \Omega \qquad I = \frac{230}{529 + 529} \approx 0.217\ \text{A}

P = I^{2}R = 0.217^{2} \times 529 \approx 25\ \text{W each}

Each lamp gets half the voltage, so a quarter of its rated power: another reason why appliances are always connected in parallel.

Working out a month's electricity

To estimate a household's use, add up power × hours for every appliance. The figures below are illustrative, for a small home.

Appliance	Power	Hours per day	Energy per day (kWh)
5 LED lamps	10 W each	5	$5 \times 0.010 \times 5 = 0.25$
Television	80 W	4	0.32
Refrigerator (switches itself on and off)	about 1.0 kWh per day		1.0
3 phone chargers	10 W each	2	0.06
Electric iron	1000 W	0.5	0.50
Total			2.13

In 30 days: $2.13 \times 30 \approx 64\ \text{kWh}$ . Notice that the iron, used for only half an hour, uses twice as much energy as all five lamps together.

REG's block tariff

Rwanda charges households in blocks: the first units of each month are cheap, so that small users pay little, and further units cost more. The tariff approved by RURA and published by REG, in force from 1 October 2025, is shown below (prices exclude VAT and the regulatory fee).

Units used in the month	Price per kWh
0 to 20 kWh	89 Frw
more than 20, up to 50 kWh	310 Frw
more than 50 kWh	369 Frw

Worked example 3: a month's bill

A household uses 65 kWh in a month. Find the cost at REG's household tariff, before VAT and the regulatory fee.

\text{first 20 kWh: } 20 \times 89 = 1780\ \text{Frw}

\text{next 30 kWh: } 30 \times 310 = 9300\ \text{Frw}

\text{last 15 kWh: } 15 \times 369 = 5535\ \text{Frw}

\text{total} = 1780 + 9300 + 5535 = 16\,615\ \text{Frw}

Efficiency: getting more for each unit

The efficiency of an appliance is the fraction of the energy it takes in that comes out in the useful form:

\eta = \frac{\text{useful energy (or power) output}}{\text{energy (or power) input}} \times 100\%

A filament lamp is very inefficient: typically only about 5% of its input becomes light, and the rest heats the room. An LED lamp gives the same light for a much smaller power, typically about a sixth or less, which is why replacing filament lamps is the cheapest way for most families to cut their bill.

LED lamps rated from 4.9 W to 20 W beside a 100 W incandescent (filament) lamp, third from the left. Typically, an LED lamp of about 15 W gives roughly as much light as the 100 W filament lamp, which turns most of its energy into heat. Photo: Bech, CC BY 4.0, via Wikimedia Commons

Worked example 4: changing to LED lamps

A family replaces five 60 W filament lamps with five 9 W LED lamps giving the same light (typical values). The lamps are on 5 hours a day. How much energy is saved in 30 days, and what is it worth if it would have been charged at 369 Frw per kWh?

\text{power saved} = 5 \times (60 - 9) = 255\ \text{W} = 0.255\ \text{kW}

\text{energy saved} = 0.255 \times 5 \times 30 \approx 38\ \text{kWh} \qquad \text{saving} = 38.25 \times 369 \approx 14\,100\ \text{Frw}

Warning

Watts and hours, not watts and seconds. For kWh, the power must be in kilowatts and the time in hours: a 1500 W iron used for 20 minutes uses $1.5 \times \frac{20}{60} = 0.5\ \text{kWh}$ , not $1500 \times 20$ . And remember that the block tariff makes the last units of the month the dearest.

Tip

Big power or long hours? The energy is power × time. Appliances that heat (kettle, iron, cooker, water heater) have the largest power; appliances that run all day (fridge, security lights) have the longest time. Those two groups decide most bills.

Calculating the cost of electric power (Step by Step Science). Watch how power in kilowatts and time in hours give the energy in kWh, and how the cost follows; the video uses a US price, so in Rwanda use REG's tariff instead.

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Practice

Write down three equations for electrical power, and say when each is most useful.
An iron is rated 230 V, 1500 W. Find the current it draws and the resistance of its element.
How many kWh does a 9 W LED lamp use if it is on for 6 hours a day for 30 days?
A household uses 18 kWh in a month. Find the cost at REG's household tariff, before VAT.
Another household uses 45 kWh. Find its cost, before VAT.
A 60 W filament lamp has an efficiency of 5%. What power is given out as light, and what power is wasted as heat?
Replacing one 60 W filament lamp with a 9 W LED lamp, used 5 hours a day for 30 days, saves how many kWh? What is that worth at 369 Frw per kWh?
A family has used 48 kWh this month. Ironing will use 5 kWh more. A neighbour says, "That will cost 5 × 89 = 445 Frw." Evaluate this claim and find the true extra cost.

Show answersHide answers

$P = V I$ (when V and I are known); $P = I^{2} R$ (when the current through a resistance is known, e.g. losses in a cable); $P = \frac{V^2}{R}$ (when the voltage across a resistance is known, e.g. a heater on the mains).
$I = \frac{1500}{230} \approx 6.5\ \text{A}$ ; $R = \frac{230^2}{1500} \approx 35\ \Omega$ .
$0.009 \times 6 \times 30 = 1.62\ \text{kWh}$ .
$18 \times 89 = 1602\ \text{Frw}$ .
$20 \times 89 + 25 \times 310 = 1780 + 7750 = 9530\ \text{Frw}$ .
Light: $0.05 \times 60 = 3\ \text{W}$ ; wasted: $57\ \text{W}$ .
$0.051 \times 5 \times 30 = 7.65\ \text{kWh}$ ; $7.65 \times 369 \approx 2820\ \text{Frw}$ .
Wrong: the cheap 89 Frw block covers only the first 20 kWh, which are already used. The next 2 kWh (48 to 50) cost 310 Frw each and the last 3 kWh cost 369 Frw each: $2 \times 310 + 3 \times 369 = 620 + 1107 = 1727\ \text{Frw}$ .

Key points

$P = VI = I^2R = \frac{V^2}{R}$ ; $E = P t$ ; 1 kWh = $3.6 \times 10^{6}$ J.
Energy in kWh = power in kW × time in hours; add up all appliances for a month.
REG household tariff (from 1 October 2025): 89 Frw per kWh up to 20 kWh, 310 Frw from 20 to 50 kWh, 369 Frw above 50 kWh, before VAT.
Efficiency = useful output ÷ input; LED lamps use a fraction of a filament lamp's power for the same light.

Sources and credits

20.4: Electric Power and Energy, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Consulted for the physics; no text was copied.
Tariff: Tariffs, Rwanda Energy Group: electricity end-user tariff effective 1 October 2025, VAT and regulatory fee excluded (page read 1 October 2026). Appliance powers and hours, and the LED and filament lamp figures, are typical or illustrative values.
Photo: Taille lampes LED.jpg by Bech, Wikimedia Commons, licensed under CC BY 4.0.
Video: Electric Power (3 of 3), Calculating the Cost of Electric Power, Step by Step Science, embedded from YouTube.

Written by Ganzaa: the lesson text, the tariff graph, the tables, worked examples and practice exercises.

1What does a token buy?

2Power: three forms of one formula

3Working out a month's electricity

4REG's block tariff

5Efficiency: getting more for each unit

What does a token buy?

Power: three forms of one formula

Working out a month's electricity

REG's block tariff

Efficiency: getting more for each unit