Electrical energy, power and the cost of electricity: WASSCE Physics

Every month, the electricity bill (or the units you top up on an EDSA prepaid meter) depends on how much electrical energy your appliances use. This lesson shows how to calculate energy and power — and how to work out what an appliance costs to run.

Learning objectives

By the end of this lesson you should be able to:

calculate electrical energy and power using $W = VIt$ and $P = VI$
use the alternative forms $P = I^2R$ and $P = \dfrac{V^2}{R}$
interpret power ratings of appliances
calculate the cost of electrical energy in kilowatt-hours

Electrical energy

When a charge $Q$ moves through a p.d. $V$ , the energy converted is $W = QV$ . Since $Q = It$ :

W = VIt

Using $V = IR$ , this can also be written

W = I^2Rt = \dfrac{V^2t}{R}

The energy is in joules (J) when $V$ is in volts, $I$ in amperes and $t$ in seconds.

Electrical power

Definition

Electrical power is the rate at which electrical energy is converted into other forms:

P = \dfrac{W}{t} = VI = I^2R = \dfrac{V^2}{R}

\begin{aligned}P &= \dfrac{W}{t} \\ &= VI \\ &= I^2R \\ &= \dfrac{V^2}{R}\end{aligned}

Its unit is the watt (W): 1 W = 1 J s⁻¹ = 1 V A.

Worked example

An electric iron draws 4 A from a 240 V supply. Find (a) its power, (b) its resistance, (c) the energy it uses in 30 minutes.

(a) $P = VI = 240 \times 4 = 960$ W.

(b) $R = \dfrac{V}{I} = \dfrac{240}{4} = 60$ Ω.

Power ratings

An appliance labelled "240 V, 60 W" converts 60 J of energy every second when connected to 240 V. From the rating you can find its working current and resistance:

I = \dfrac{P}{V} = \dfrac{60}{240} = 0.25\ \text{A} \qquad R = \dfrac{V^2}{P} = \dfrac{240^2}{60} = 960\ \Omega

\begin{gathered}\begin{aligned}I &= \dfrac{P}{V} \\ &= \dfrac{60}{240} \\ &= 0.25\ \text{A}\end{aligned} \\[6pt] \begin{aligned}R &= \dfrac{V^2}{P} \\ &= \dfrac{240^2}{60} \\ &= 960\ \Omega\end{aligned}\end{gathered}

Worked example

A 240 V, 100 W lamp is connected to a 120 V supply. What power does it now use (assume its resistance stays constant)?

$R = \dfrac{240^2}{100} = 576$ Ω. On 120 V: $P = \dfrac{120^2}{576} = 25$ W.

Halving the voltage reduces the power to one quarter, because $P \propto V^2$ .

Common mistake

A lamp's rated power applies only at its rated voltage. On a different voltage, first find its resistance from the rating, then calculate the new power.

The kilowatt-hour

The joule is too small for household bills, so electricity is sold in kilowatt-hours (kWh), called units.

Definition

One kilowatt-hour is the energy used by an appliance of power 1 kW working for 1 hour:

1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^{6}\ \text{J}

\begin{aligned}1\ \text{kWh} &= 1000\ \text{W} \times 3600\ \text{s} \\ &= 3.6 \times 10^{6}\ \text{J}\end{aligned}

\text{energy (kWh)} = \text{power (kW)} \times \text{time (h)} \qquad \text{cost} = \text{energy (kWh)} \times \text{price per kWh}

\begin{gathered}\begin{aligned}&\text{energy (kWh)} \\ &= \text{power (kW)} \times \text{time (h)}\end{aligned} \\[6pt] \begin{aligned}\text{cost} &= \text{energy (kWh)} \times \text{price per kWh}\end{aligned}\end{gathered}

Worked example

A household uses: four 60 W bulbs for 6 hours a day, a 1.2 kW iron for 30 minutes a day, and a 150 W fridge for 24 hours a day. If one unit costs NLe 2.50, find the cost for 30 days.

Appliance	Power (kW)	Hours per day	kWh per day
4 bulbs	0.24	6	1.44
Iron	1.2	0.5	0.60
Fridge	0.15	24	3.60
Total			5.64

Energy in 30 days $= 5.64 \times 30 = 169.2$ kWh. Cost $= 169.2 \times 2.50 =$ NLe 423.00.

In Sierra Leone

With an EDSA prepaid meter, the "units" you buy are kilowatt-hours. Replacing four 60 W filament bulbs with 9 W LED bulbs that give the same light cuts the lighting energy by about 85%, so the same units last much longer.

Key points

$W = VIt = I^2Rt = \dfrac{V^2t}{R}$ (J).
$P = VI = I^2R = \dfrac{V^2}{R}$ (W).
From a rating "V, P": $I = P/V$ , $R = V^2/P$ .
1 kWh = 3.6 × 10⁶ J; cost = kWh × price per unit.

WASSCE examination focus

WAEC tests this

Define the kilowatt-hour and show it equals 3.6 MJ, calculate energy, power, current and resistance from appliance ratings, and work out an electricity bill from a table of appliances. Convert watts to kilowatts and minutes to hours carefully.

Quick check

How many joules are there in 1 kWh?
A 2 kW heater runs for 3 hours. How many units of electricity does it use?
Find the resistance of a lamp rated 240 V, 40 W.

Answers: 1. $3.6 \times 10^{6}$ J. 2. 6 kWh. 3. $\dfrac{240^2}{40} = 1440$ Ω.

Electrical energy, power and the cost of electricity

Learning objectives

Electrical energy

Electrical power

Power ratings

The kilowatt-hour

Key points

WASSCE examination focus

Quick check

Keep going with electric energy and power

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