Electrical Power and Energy

Every electrical appliance has a power rating stamped on it — a kettle might be $3 \text{ kW}$ , a phone charger only a few watts. This number tells you how fast the appliance transfers energy, and it decides both how quickly it does its job and how much it costs to run. In this lesson, part of Topic P3 (Electricity) of OCR Gateway Science A, you will learn the equations that link power to current, voltage and resistance, calculate the energy an appliance transfers, and work out the cost of electrical energy using the kilowatt-hour.

By the end of this lesson you should be able to define electrical power, recall and use $P = VI$ and $P = I^2R$ , calculate energy transferred with $E = Pt$ and $E = QV$ , explain and use the kilowatt-hour to calculate the cost of electrical energy, and carry out worked calculations.

What Is Electrical Power?

Power is the rate at which energy is transferred — that is, the energy transferred per second. Power is given the symbol $P$ and measured in watts ( $\text{W}$ ), where one watt is one joule per second ( $1 \text{ W} = 1 \text{ J/s}$ ). A high-power appliance transfers a lot of energy each second; a low-power one only a little.

For electrical appliances, the power is the rate at which the appliance transfers electrical energy into other forms — heat and light in a lamp, kinetic energy in a motor, heat in a kettle. A $2000 \text{ W}$ (or $2 \text{ kW}$ ) kettle transfers $2000 \text{ J}$ of energy every second, which is why it boils water far faster than a $500 \text{ W}$ travel kettle.

Larger powers are often given in kilowatts ( $\text{kW}$ ), where $1 \text{ kW} = 1000 \text{ W}$ . Typical household values are worth a feel for: an LED lamp is around $5{-}10 \text{ W}$ , a television about $100 \text{ W}$ , a microwave around $800 \text{ W}$ , a hairdryer or toaster about $1{-}2 \text{ kW}$ , and a kettle or an electric heater around $2{-}3 \text{ kW}$ . The high-power appliances are nearly always the ones that produce heat, because heating water or air requires energy to be delivered very quickly.

Appliance	Typical power
LED lamp	$5{-}10 \text{ W}$
Television	$100 \text{ W}$
Microwave oven	$800 \text{ W}$
Toaster / hairdryer	$1{-}2 \text{ kW}$
Kettle / electric heater	$2{-}3 \text{ kW}$

Exam Tip: Power is the energy transferred per second, in watts ( $1 \text{ W} = 1 \text{ J/s}$ ). Watch the prefix: a power in kilowatts must be converted to watts (×1000) before using it in an energy calculation in joules.

The Power Equations: $P = VI$ and $P = I^2R$

The electrical power transferred by a component depends on the current through it and the potential difference across it:

$P = VI$

where $P$ is the power in watts ( $\text{W}$ ), $V$ is the potential difference in volts ( $\text{V}$ ), and $I$ is the current in amperes ( $\text{A}$ ). In words: power equals voltage times current.

Because $V = IR$ , we can substitute for $V$ to get a second, very useful form. Replacing $V$ with $IR$ in $P = VI$ gives $P = (IR) \times I$ , so:

$P = I^2R$

where $R$ is the resistance in ohms ( $\Omega$ ). This form is handy when you know the current and resistance but not the voltage. Notice the current is squared: doubling the current through a component quadruples the power it transfers — which is why a small increase in current causes a large increase in heating.

$P = VI \qquad P = I^2R$

Worked Example 1 — Power from voltage and current

A lamp operates at $230 \text{ V}$ and draws a current of $0.26 \text{ A}$ . Calculate its power.

Step 1 — write the equation: $P = VI$ .

Step 2 — substitute: $P = 230 \times 0.26$ .

Step 3 — calculate: $P = 59.8 \text{ W}$ (about $60 \text{ W}$ ).

Answer: the lamp's power is about $60 \text{ W}$ .

Worked Example 2 — Power from current and resistance

A heating element has a resistance of $50\ \Omega$ and carries a current of $4 \text{ A}$ . Calculate the power it transfers.

Step 1 — write the equation: $P = I^2R$ .

Step 2 — substitute: $P = 4^2 \times 50 = 16 \times 50$ .

Step 3 — calculate: $P = 800 \text{ W}$ .

Answer: the element transfers $800 \text{ W}$ .

Worked Example 3 — Rearranging $P = VI$ for current

An appliance is rated at $1380 \text{ W}$ when connected to the $230 \text{ V}$ mains. What current does it draw?

Step 1 — rearrange for current: $I = \dfrac{P}{V}$ .

Step 2 — substitute: $I = \dfrac{1380}{230}$ .

Step 3 — calculate: $I = 6 \text{ A}$ .

Answer: it draws a current of $6 \text{ A}$ . (This kind of calculation is used to choose the right fuse — covered in the next lesson.)

Worked Example 3b — Choosing between the two power equations

A resistor of $20\ \Omega$ has a current of $3 \text{ A}$ flowing through it. Find the power dissipated using $P = I^2R$ , and confirm it using $P = VI$ .

Step 1 — using $P = I^2R$ : square the current first, $3^2 = 9$ , then $P = 9 \times 20 = 180 \text{ W}$ .

Step 2 — to confirm, first find the p.d. with $V = IR = 3 \times 20 = 60 \text{ V}$ .

Step 3 — then $P = VI = 60 \times 3 = 180 \text{ W}$ .

Answer: $180 \text{ W}$ by both methods — a useful check that the two power equations agree.

Exam Tip: Use $P = VI$ when you know the voltage and current, and $P = I^2R$ when you know the current and resistance. In $P = I^2R$ , square the current first, then multiply by the resistance — a common slip is to multiply before squaring.

Calculating Energy Transferred: $E = Pt$ and $E = QV$

Since power is energy per second, the energy transferred in a time $t$ is simply the power multiplied by the time:

$E = Pt$

where $E$ is the energy transferred in joules ( $\text{J}$ ), $P$ is the power in watts ( $\text{W}$ ), and $t$ is the time in seconds ( $\text{s}$ ). The longer an appliance runs and the higher its power, the more energy it transfers.

You also met, in the earlier lesson, the equation $E = QV$ (energy = charge × p.d.). Both give the energy transferred, and which you use depends on the information given: use $E = Pt$ when you know the power and time, and $E = QV$ when you know the charge and voltage. The two are connected, because $P = VI$ and $Q = It$ together give $Pt = VI \times t = V \times (It) = QV$ .

$E = Pt \qquad E = QV$

Worked Example 4 — Energy from power and time

A $2000 \text{ W}$ kettle is switched on for $90 \text{ s}$ . How much energy does it transfer?

Step 1 — write the equation: $E = Pt$ .

Step 2 — substitute (time in seconds): $E = 2000 \times 90$ .

Step 3 — calculate: $E = 180\,000 \text{ J}$ (= $180 \text{ kJ}$ ).

Answer: $180\,000 \text{ J}$ of energy is transferred.

Worked Example 5 — Energy from power and time in minutes

A $1500 \text{ W}$ electric heater is on for $10 \text{ minutes}$ . Calculate the energy it transfers in joules.

Electrical Power and Energy

Electrical Power and Energy

What Is Electrical Power?

The Power Equations: P=VIP = VIP=VI and P=I2RP = I^2RP=I2R

Worked Example 1 — Power from voltage and current

Worked Example 2 — Power from current and resistance

Worked Example 3 — Rearranging P=VIP = VIP=VI for current

Worked Example 3b — Choosing between the two power equations

Calculating Energy Transferred: E=PtE = PtE=Pt and E=QVE = QVE=QV

Worked Example 4 — Energy from power and time

Worked Example 5 — Energy from power and time in minutes

More in Physics

The Power Equations: $P = VI$ and $P = I^2R$

Worked Example 3 — Rearranging $P = VI$ for current

Calculating Energy Transferred: $E = Pt$ and $E = QV$