Electrical Power and Energy Transfers

Power is the rate at which energy is transferred, measured in watts (W). One watt is one joule of energy transferred every second, so $1\text{ W}=1\text{ J/s}$.

In a circuit, energy is transferred as charge flows through components. A component with a high power transfers a lot of energy each second; a low-power component transfers little. The power depends on two things you can measure directly: the potential difference (pd) across the component and the current through it.

$P=VI$

• $P$ = power, in watts (W)
• $V$ = potential difference, in volts (V)
• $I$ = current, in amps (A)

You must recall and apply $P=VI$. It is not given on the Physics equation sheet.

Worked example. A lamp has a pd of 12 V across it and a current of 2 A through it. Find its power.

$P=VI=12\times 2=24\text{ W}$

The lamp transfers 24 J of energy every second.

Sometimes you know the current and the resistance but not the pd. A second power equation lets you work directly from those two quantities.

$P=I^{2}R$

• $P$ = power, in watts (W)
• $I$ = current, in amps (A)
• $R$ = resistance, in ohms (Ω)

You must recall and apply $P=I^{2}R$. It is not given on the Physics equation sheet.

This equation follows from the first. Since $V=IR$, substituting into $P=VI$ gives $P=(IR)\times I=I^{2}R$, so the two formulas always agree.

Worked example. A resistor of 5 Ω carries a current of 3 A. Find the power dissipated.

$P=I^{2}R=3^2\times 5=9\times 5=45\text{ W}$

Notice the current is squared first: $3^2=9$, then multiply by 5. Doubling the current would quadruple the power, because power depends on the square of the current. This is why thick, low-resistance cables are used to carry large currents with little wasted heating.

Both equations give power in watts; you pick the one that matches the quantities you are given.

You are given	Use	Why
pd and current	$P=VI$	Both quantities appear directly
current and resistance	$P=I^{2}R$	No pd needed
pd and resistance	First find $I=V/R$, then $P=VI$	Or combine to $P=V^2/R$ (not required, but follows from the two)

Worked example. A heater of resistance 20 Ω is connected to the 230 V mains. Find its power.

First find the current using $V=IR$ (recalled from resistance work), rearranged to $I=V/R$:

$I=\frac{V}{R}=\frac{230}{20}=11.5\text{ A}$

Then use $P=VI$:

$P=VI=230\times 11.5=2645\text{ W}\approx 2.6\text{ kW}$

Check with $P=I^{2}R=11.5^2\times 20=132.25\times 20=2645\text{ W}$. Both routes agree.

The total energy an appliance transfers depends on its power and how long it runs. A high-power appliance used briefly can transfer the same energy as a low-power one left on for hours.

$E=Pt$

• $E$ = energy transferred, in joules (J)
• $P$ = power, in watts (W)
• $t$ = time, in seconds (s)

You must recall and apply $E=Pt$. It is not given on the Physics equation sheet.

The time must be in seconds for the answer to come out in joules.

Worked example. A 2000 W kettle is switched on for 1 minute 30 seconds. How much energy does it transfer?

Convert the time: $1\text{ min }30\text{ s}=90\text{ s}$.

$E=Pt=2000\times 90=180000\text{ J}=180\text{ kJ}$

The single most common slip here is leaving the time in minutes. Always convert to seconds first.

Whenever charge is pushed through a component by a potential difference, work is done and energy is transferred. This gives a second energy equation, written in terms of charge rather than time.

$E=QV$

• $E$ = energy transferred, in joules (J)
• $Q$ = charge, in coulombs (C)
• $V$ = potential difference, in volts (V)

You must recall and apply $E=QV$. It is not given on the Physics equation sheet.

The potential difference tells you how much energy each coulomb of charge carries: a pd of 1 V means 1 joule is transferred per coulomb.

Worked example. A charge of 6 C passes through a bulb with a pd of 12 V across it. How much energy is transferred?

$E=QV=6\times 12=72\text{ J}$

If you are given a current and a time instead of a charge, first find the charge with $Q=It$ (recalled from charge and current), then use $E=QV$.

An appliance transfers energy electrically from the mains supply into the appliance's own energy stores. Naming the useful and wasted transfers is a standard exam task.

Appliance	Energy transferred usefully	Energy wasted
Kettle	Thermal (heating the water)	Thermal to the surroundings, sound
Electric motor / fan	Kinetic	Thermal (heating), sound
Filament lamp	Light	Thermal (the bulb gets hot)
Toaster	Thermal (heating the bread)	Thermal to the surroundings

The power rating printed on an appliance tells you how much energy it transfers each second. A higher power rating means more energy transferred per second, so for the same running time a higher-rated appliance transfers more energy in total.

Worked example — comparing two appliances. A 60 W lamp is on for 5 hours; a 900 W microwave runs for 4 minutes. Which transfers more energy?

Lamp: $t=5\times 3600=18000\text{ s}$, so $E=Pt=60\times 18000=1080000\text{ J}$.

Microwave: $t=4\times 60=240\text{ s}$, so $E=Pt=900\times 240=216000\text{ J}$.

The lamp transfers about five times as much energy, despite its far lower power, because it runs for far longer. Power and time both matter.

1. Leaving time in minutes or hours

$E=Pt$ gives joules only when $t$ is in seconds. Convert 1 minute to 60 s, 1 hour to 3600 s, before substituting.

2. Forgetting to square the current in P = I²R

The equation is $P=I^{2}R$, not $P=IR$. Work out $I^2$ first, then multiply by $R$. Squaring after multiplying by $R$ gives the wrong answer.

3. Mixing up the two power equations

Use $P=VI$ when you have pd and current; use $P=I^{2}R$ when you have current and resistance. Both are recall-and-apply, so you must have both memorised.

4. Confusing E = Pt with E = QV

Use $E=Pt$ when you know power and time. Use $E=QV$ when you know charge and pd. If given current and time, find charge first with $Q=It$, then apply $E=QV$.

5. Treating power rating as total energy

A "2000 W" kettle does not transfer 2000 J in total; it transfers 2000 J every second. Multiply by the time in seconds to get the total energy.