Electric Power and Energy at Home

Beginner

Every appliance has a power rating, in watts (W). Power tells you how fast the appliance uses energy: one watt is one joule every second. An LED lamp uses about 10 W, a television about 100 W and a kettle about 2000 W, which is 2 kilowatts (kW).

Step 1: energy = power × time. This energy is counted in kilowatt-hours (kWh): one kWh is what a 1 kW appliance uses in an hour, a 2 kW kettle in half an hour, or a 10 W LED lamp in 100 hours.

Step 2: the bill. The meter adds up the kWh, and you pay for energy, not for power: the number of kWh times the price of one kWh. A 2 kW heater left on for 3 hours uses $2\times 3=6\mathrm{kWh}$; at €0.20 per kWh, that costs €1.20.

Step 3: current and safety. Every appliance you switch on draws more current through the same supply wires, and current warms wires. With too many powerful appliances on one circuit the wires could overheat, so a fuse (a thin wire that melts) or a circuit breaker (a switch that clicks off) cuts the power first.

In the animation a lamp, a kettle and a heater switch on one after another. With each one the red current bar rises and the meter's disc spins faster. When the heater joins the kettle, the bar passes the limit line and the breaker trips.

Step 4: efficiency. An old filament bulb turns most of its energy into heat; an LED gives the same light with about a sixth of the power.

Check yourself: which uses more energy, a 100 W lamp on for 10 hours or a 2000 W kettle on for 15 minutes? (Answer: the lamp, 1 kWh against 0.5 kWh.)

Intermediate

Power: energy per second

An LED lamp left on long enough uses as much energy as a kettle switched on briefly. Power is the rate at which energy is transferred, $P=E/t$, in watts: $1\mathrm{W}=1J/s$. A 2000 W kettle takes 2000 joules from the mains every second, a 10 W LED lamp only 10.

The voltage $V$ across an appliance is the energy that each coulomb of charge gives up in passing through it, $1\mathrm{V}=1J/C$, and the current $I$ is the charge passing each second, $1\mathrm{A}=1C/s$. Their product is joules per second:
$P=VI$

Ohm's law, $V=IR$, from DC Circuits and Ohm's Law, gives two more forms:
$P=I^{2}R=\frac{V^2}{R}$

On the mains the voltage is fixed, so by $V^2/R$ a smaller resistance takes more power: the kettle element has only $230^2/2000\approx 26\Omega$, a hot 100 W lamp filament 529 Ω. For a fixed current, $I^{2}R$ says the larger resistance heats more, which is why supply cables are thick.

Appliances in parallel on the mains

Every socket sits between the same two wires, live and neutral, so household appliances are in parallel. Each gets the full mains voltage (230 V in Europe, 120 V in North America) whatever else is on, and the branch currents add up in the supply cable:
$I=\frac{P_1+P_2+\dots }{V}$

A circuit protected at $I_{\max }$ can therefore deliver at most $V{I}_{\max }$: 2300 W for 10 A at 230 V, 3680 W for 16 A. That is why a kitchen gets several circuits.

Energy, the kilowatt-hour and efficiency

The energy an appliance uses is its power multiplied by the time it runs:
$E=Pt$
With $P$ in watts and $t$ in seconds it comes out in joules; suppliers count larger units, kilowatt-hours, the energy of 1 kW for an hour:
$1\mathrm{kWh}=3.6\mathrm{MJ}$
since $1000\mathrm{W}\cdot 3600\mathrm{s}=3.6\times 10^6\mathrm{J}$. The meter adds up kilowatt-hours, and you pay for this energy, not for power: a 2000 W kettle used for 6 minutes and a 10 W LED left on for 20 hours both take 0.2 kWh.

The efficiency compares the useful output with the input:
$\eta =\frac{P_{\mathrm{useful}}}{P_{\mathrm{in}}}$
A kettle puts 80 to 90 % of its energy into the water, a filament lamp only about 5 % into visible light, which is why an LED needs about a sixth of the power for the same light.

Worked example: the circuit in the animation

The animation's 230 V circuit is protected by a 10 A breaker. The 100 W lamp draws $100/230=0.43\mathrm{A}$ and the 2000 W kettle $2000/230=8.70\mathrm{A}$, so together
$I=\frac{100+2000}{230}=9.13\mathrm{A}$
just under the limit, with $2300-2100=200\mathrm{W}$ to spare. The 2000 W heater then raises the current to
$I=\frac{100+2000+2000}{230}=17.83\mathrm{A}$
almost twice the rating.

These are the values on the $I$ readout, while the $P$ readout steps from 100 to 2100 to 4100 W and the red arrow $I$ on the supply wire lengthens; the branch labels show 0.43, 8.70 and 8.70 A. The red column in the current bar stays just below the 10 A line with the kettle and shoots past it with the heater. 1.2 s later the breaker trips: $P$ and $I$ drop to zero, and $E$ stops climbing.

With 1 s standing for 1 minute, the lamp has run 5.5 minutes by then, the kettle 3.5 and the heater 1.2, so in watt-hours
$E=\frac{100\cdot 5.5+2000\cdot 3.5+2000\cdot 1.2}{60}\mathrm{Wh}=165.8\mathrm{Wh}$
the 0.166 kWh of the $E$ readout (166 Wh in the meter window). At €0.25 per kWh this costs about 4 cents.

Worked example: a heater on an extension reel

A 1500 W fan heater runs on a 25 m extension reel with 1.0 mm² copper conductors. Its $1500/230=6.52\mathrm{A}$ goes out and back through 50 m of copper of resistivity $1.7\times 10^{-8}\Omega \cdot \mathrm{m}$. With $R=\rho L/A$ from Electric Current and Resistance,
$R_c=\frac{\rho L}{A}=\frac{1.7\times 10^{-8}\cdot 50}{1.0\times 10^{-6}}=0.85\Omega$
$P_c=I^2{R}_c=6.52^2\cdot 0.85\approx 36\mathrm{W}$

Spread along the unwound cable, 36 W is harmless; wound on the drum, the same heat, about that of a small soldering iron, is trapped and can overheat the insulation, hence the lower rating of a wound reel. The reel also drops $I{R}_c\approx 5.5\mathrm{V}$: by $P=V^2/R$, 2.4 % less voltage costs about 5 % of the power, leaving roughly 1430 W.

At full rating, 4 hours an evening for 30 days uses $E=1.5\cdot 4\cdot 30=180\mathrm{kWh}$, €45 at €0.25 per kWh; on the reel the meter shows about 176 kWh, 4 of them released in the reel.

Fuses, breakers and the current bar

A cable is rated for the largest current at which its Joule heating, $I^{2}R$, keeps the insulation safe, around 70 °C for PVC. A fuse (a thin wire that melts) or a circuit breaker (a switch that can be reset) on the live wire cuts the current beyond that rating: it protects the cable, not the appliance.

Heating grows as the square of the current, so a moderate overload is dangerous only if it lasts, a short circuit at once. A breaker's bimetal strip, warmed like the cable, trips after seconds to minutes; its electromagnet trips within about 10 ms when a short circuit drives hundreds of amperes.

In the animation the red bar is the current through the breaker and the line is its 10 A rating: below it, the wiring could run for ever. The 17.8 A with the heater on is an overload, so the breaker waits 1.2 s, standing for 1.2 minutes, before opening.

Common mistakes

• Appliances do not use up current: the 8.70 A entering the kettle through the live wire leaves through the neutral. What is used up is energy, 2000 joules a second.
• A kilowatt-hour is not "kilowatts per hour" but a power multiplied by a time, an energy of 3.6 MJ: a 3 kW heater uses 1 kWh in 20 minutes.
• Switching on more appliances in parallel does not raise the total resistance but lowers it: 529 Ω for the lamp alone, 25.2 Ω with the kettle, 12.9 Ω with all three, which is why the current grows.
• More resistance does not always mean more heat: at fixed voltage, $P=V^2/R$ gives the 26 Ω kettle element 2000 W and the 529 Ω filament 100 W. $I^{2}R$ favours the larger resistance only for equal currents, as in series.
• A bigger fuse is not safer: a 10 A cable fused at 30 A could carry three times its rated current, heating $3^2=9$ times as fast, without the fuse blowing.

Where you meet it

Every electricity bill is this lesson. A fridge averaging 40 W uses about 350 kWh a year, and a device on standby at 1 W still takes 8.76 kWh; an electric car's 60 kWh battery needs about 8 hours on a 7.4 kW home charger; a 1 GW power station matches half a million kettles boiling at once.

The magnetic field around every current, which opens a breaker, is the subject of Magnetic Fields of Magnets and Currents. Motors, Generators and Transformers explains why power lines run at hundreds of kilovolts to cut $I^{2}R$ losses, and Semiconductors, Diodes and LEDs why an LED beats a filament.

Advanced

Power as work done on moving charge

Every appliance carries a power rating in watts, and the whole lesson rests on one formula for it, $P=VI$. This section derives it from the work the electric field does on the charge that passes through a component, which shows why it holds for every kind of device.

Legend

$I$	Current entering at the end with the higher potential.	$\mathrm{A}$
$dq$	Charge that crosses a section of the conductor in a short time $dt$: $dq=Idt$.	$\mathrm{C}$
$\vec{E}$	Electric field inside the conductor, along the current; its size is $V/L$.	$V/m$
$L$	Length of the conductor.	$\mathrm{m}$
$V_a$	Potential at the end where the current enters.	$\mathrm{V}$
$V_b$	Potential at the end where the current leaves.	$\mathrm{V}$
$V$	Drop in potential along the conductor, $V=V_a-V_b$.	$\mathrm{V}$

Figure 1 shows the simplest component, a uniform conductor of length $L$. The current $I$ enters at the left end, where the potential is $V_a$, and leaves at the right end, where it is $V_b$. Below the conductor, on the same horizontal scale, the potential is plotted as a height. Look at the slanting line: the potential falls evenly from $V_a$ to $V_b$, and the total drop is $V=V_a-V_b$. The electric field $\vec{E}$ inside points down that slope, in the direction of the current, and because the fall is even, its size is the same everywhere: $E=V/L$.

Now look at the shaded slab $dq$. It is the charge that crosses one cross-section of the conductor in a short time $dt$. Current is charge per unit time, so $dq=Idt$. In that time a charge $dq$ enters at the left and an equal charge leaves at the right, because charge does not pile up inside. The net effect is the same as if the slab $dq$ had been carried the whole way from the potential $V_a$ to the potential $V_b$.

A difference in potential is work per unit charge: when a charge $q$ moves through a drop $V$, the field does the work $qV$ on it. So in the time $dt$ the field does the work $dW=Vdq=VIdt$, and the power, which is work per unit time, is
$P=\frac{dW}{dt}=V\frac{dq}{dt}=VI$

Only the potentials of the two ends entered this argument. Inside a component that is not a straight wire the potential may fall unevenly, or even rise in places, but the work done on a charge carried from one terminal to the other still depends only on the end values. Nor does the argument care what becomes of the energy: a motor turns it into mechanical work, a battery on charge into chemical energy, an LED partly into light.

The sign needs one convention. Take $V$ as the potential where the current enters minus the potential where it leaves. Then $P=VI>0$ means that the component absorbs energy from the circuit. A source drives current out of its positive terminal, so the current enters it at the lower potential, $V$ is negative and $P<0$: the source delivers energy to the circuit.

In a resistor all of the work becomes heat. The current is the same at both ends, so the charges drift out as fast as they drift in and gain no kinetic energy overall. The energy the field gives them is passed on, in collisions, to the vibrating atoms of the metal.

The same result can be found point by point inside the metal. Let there be $n$ charge carriers per unit volume, each with charge $q$ and drift velocity $\vec{v}$. The field pushes each one with the force $q\vec{E}$, so it does work on it at the rate $q\vec{E}\cdot \vec{v}$. Per unit volume that is $nq\vec{v}\cdot \vec{E}$, and $nq\vec{v}$ is the current density $\vec{j}$, the current per unit area of cross-section. The power released per unit volume is therefore $\vec{j}\cdot \vec{E}$. Ohm's law in its local form, $\vec{j}=\sigma \vec{E}$, where $\sigma$ is the conductivity of the material, turns this into $j^2/\sigma$. A wire of length $L$ and cross-section $A$ has the current density $j=I/A$, the volume $AL$ and the resistance $R=L/(\sigma A)$, so
$P=\frac{j^2}{\sigma }AL=I^2\frac{L}{\sigma A}=I^{2}R$

With $V=IR$ this also gives $P=V^2/R$, the convenient form when the voltage is fixed, as at a socket.

Where the energy really flows

The kettle takes 2000 W from the socket. It is natural to picture that energy travelling inside the copper of the cable, carried along by the electrons. It does not, and seeing where it does travel makes the idea of a field concrete. The electrons are far too slow to carry it. Copper has about $n=8.5\times 10^{28}$ free electrons per cubic metre, each with the charge $e=1.6\times 10^{-19}\mathrm{C}$, so the kettle's 8.70 A in a 1.5 mm² wire needs a drift speed of only $v=I/(neA)\approx 0.43mm/s$. With alternating current, which reverses 100 times a second, they merely shuffle back and forth by about 2 µm. Yet the element starts to warm the moment the switch closes. The energy travels in the electric and magnetic fields around the wires.

Legend

$I$	Current along the axis; in the end view it comes towards you (the dot).	$\mathrm{A}$
$\vec{E}$	Electric field along the axis, of size $V/L$ inside the metal and just outside it.	$V/m$
$L$	Length of the cylinder.	$\mathrm{m}$
$a$	Radius of the cylinder.	$\mathrm{m}$
$\vec{B}$	Magnetic field of the current, circling the cylinder; at the surface its size is $I/(2\pi {\epsilon }_0{c}^{2}a)$.	$\mathrm{T}$
$\vec{S}$	Energy flux of the field, power per unit area: it points from outside into the resistor.	$W/{\mathrm{m}}^2$

Figure 2 models the kettle's element as a plain cylindrical resistor of radius $a$ and length $L$, seen from the side on the left and end-on on the right. Three fields are drawn. The first is the electric field $\vec{E}$, which runs along the axis in the direction of the current $I$ that it drives. Its size inside is $V/L$. The part of an electric field that lies along a surface is the same on both sides of it, so just outside the metal there is also a field $V/L$ along the axis.

The second is the magnetic field $\vec{B}$ of the current, which circles the cylinder. In the end view the current comes towards you (the dot), and $\vec{B}$ goes round anticlockwise: point your right thumb along the current, and your fingers curl the way $\vec{B}$ goes. Its size at the surface follows from Ampère's law, $B\cdot 2\pi a=I/({\epsilon }_0{c}^2)$, so $B=I/(2\pi {\epsilon }_0{c}^{2}a)$. Since ${\epsilon }_0{c}^2=1/{\mu }_0$, this is the familiar $B={\mu }_{0}I/(2\pi a)$.

The third is the energy flux $\vec{S}$, the power that crosses a unit area, in watts per square metre. Wherever an electric and a magnetic field are both present, energy flows at the rate given by the Poynting vector, named after John Henry Poynting:
$\vec{S}={\epsilon }_0{c}^2\vec{E}\times \vec{B}$

Its direction follows from the figure. $\vec{E}$ points along the axis and $\vec{B}$ goes round it, so the two are at right angles, and their cross product, at right angles to both, lies along a radius. The right-hand rule decides between inward and outward. At the top of the end view $\vec{E}$ points towards you and $\vec{B}$ points to the left, and towards you crossed with leftward points down, into the cylinder. The same holds all round, so at every point of the side $\vec{S}$ points into the resistor, as the arrows $\vec{S}$ show in both views. On the flat end faces a radial $\vec{S}$ runs along the face, so no energy enters there.

Because $\vec{E}$ and $\vec{B}$ are at right angles, the size of $\vec{S}$ is simply the product of the sizes:
$S={\epsilon }_0{c}^2\cdot \frac{V}{L}\cdot \frac{I}{2\pi {\epsilon }_0{c}^{2}a}=\frac{VI}{2\pi aL}$
It is the same at every point of the side, whose area is $2\pi aL$, so the total power flowing in is $S\cdot 2\pi aL=VI$: exactly the rate at which the resistor turns energy into heat.

So the energy enters the resistor from the space around it, through its surface. Along the supply cable, the electric field reaches across the gap between the live and the neutral wire and the magnetic field circles each wire; there $\vec{S}$ runs mostly parallel to the wires, through the insulation and the air between them, from the socket towards the kettle. Only a small inward part enters the copper, and it is the cable's own $I^{2}R$ loss. When the current reverses, $\vec{E}$ and $\vec{B}$ both reverse and their cross product keeps its direction, so the energy always flows towards the load. The wires guide the energy rather than carry it, and its conservation is local: the energy in any region changes only by what flows across its surface.

Alternating current: rms values

The mains supply is alternating: the voltage and the current change all the time. What, then, do the 230 V and the 2000 W on a label mean?

The voltage is sinusoidal, $v=V_0\sin \omega t$, with the amplitude $V_0$ and $\omega =2\pi \cdot 50{\mathrm{s}}^{-1}$ in Europe. In a resistor the current $i=v/R$ follows the voltage exactly, $i=I_0\sin \omega t$ with $I_0=V_0/R$; the two are in phase. The instantaneous power is $p=vi$, and with ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$,
$p(t)=V_0{I}_0{\sin }^2\omega t=\frac{1}{2}{V}_0{I}_0(1-\cos 2\omega t)$

Read this term by term. The power never goes negative: it pulses at twice the mains frequency, 100 Hz, between 0 and $V_0{I}_0$. The cosine term averages to zero over each cycle, so the average power is $P=\frac{1}{2}{V}_0{I}_0$.

A steady current $I$ would heat the same resistor at the rate $I^{2}R$. It matches the alternating one when $I^2$ equals the average of $i^2$, which is $\frac{1}{2}{I}_0^2$, because ${\sin }^2$ averages to one half. That steady value, the square root of the mean of the square, is the rms current, $I_{\mathrm{rms}}=I_0/\sqrt{2}$; the rms voltage is likewise $V_{\mathrm{rms}}=V_0/\sqrt{2}$. Then $\frac{1}{2}{V}_0{I}_0=V_{\mathrm{rms}}{I}_{\mathrm{rms}}$, and all the direct-current formulas, $P=VI=I^{2}R=V^2/R$, hold for the average power when rms values are used. That is why ratings are given as rms values: an rms current heats a resistor exactly like an equal steady current.

For the kettle, the 230 V on the label is rms, so the voltage peaks at $230\sqrt{2}\approx 325\mathrm{V}$. Its rms current of 8.70 A peaks at $8.70\sqrt{2}\approx 12.3\mathrm{A}$, and its power swings between 0 and $V_0{I}_0=4000\mathrm{W}$, averaging 2000 W.

Phase and the power factor

In a kettle the current rises and falls together with the voltage. In devices with coils or capacitors, such as motors and transformers, the current is shifted in time against the voltage, and then $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ overstates the power actually taken. Figure 3 shows how to find the real power.

Legend

$V_0$	Amplitude of the voltage; its arrow is the reference direction.	$\mathrm{V}$
$I_0$	Amplitude of the current.	$\mathrm{A}$
$\phi$	Phase angle by which the current trails the voltage.	$\mathrm{rad}$
$I_0\cos \phi$	Part of the current in step with the voltage: the only part that delivers power on average.	$\mathrm{A}$
$I_0\sin \phi$	Part of the current a quarter of a cycle behind the voltage: its power averages to zero.	$\mathrm{A}$

Write the current as $i=I_0\sin (\omega t-\phi )$, where $\phi$ is the phase angle. A coil makes the current lag behind the voltage ($\phi >0$, as drawn); a capacitor makes it lead ($\phi <0$); the result below holds for both. Figure 3 draws each quantity as an arrow whose length is its amplitude and whose direction gives its phase. The voltage arrow $V_0$ is the reference, drawn horizontal. The current arrow $I_0$ is turned away from it by the angle $\phi$, clockwise, because the current trails. The two lengths are in different units, volts and amperes, so only the angle between the arrows carries meaning.

Now split the current arrow into two parts, as the figure does: its part along the voltage, $I_0\cos \phi$, and its part at right angles to it, $I_0\sin \phi$. In formulas this is the identity for the sine of a difference:
$I_0\sin (\omega t-\phi )=I_0\cos \phi \sin \omega t-I_0\sin \phi \cos \omega t$

The first part rises and falls exactly with the voltage. It behaves like the current in a resistor, so by the last section it delivers the average power $\frac{1}{2}{V}_0{I}_0\cos \phi$. The second part goes as $-\cos \omega t$, which is a quarter of a cycle behind the voltage. Its product with $v=V_0\sin \omega t$ is $-V_0{I}_0\sin \phi \sin \omega t\cos \omega t=-\frac{1}{2}{V}_0{I}_0\sin \phi \sin 2\omega t$. This is positive half of the time and negative the other half, so it averages to zero. It only moves energy back and forth: for a quarter of a cycle the device stores energy in the magnetic field of its coil (or the electric field of its capacitor), and for the next quarter it hands it back to the supply.

Adding the two products, and using $\cos \phi \cos 2\omega t+\sin \phi \sin 2\omega t=\cos (2\omega t-\phi )$, gives
$p(t)=\frac{1}{2}{V}_0{I}_0(\cos \phi -\cos (2\omega t-\phi ))$
For part of each cycle this is negative, while the device returns stored energy. Its average is the constant part:
$P=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \phi$

The factor $\cos \phi$ is called the power factor, and the product $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$, measured in volt-amperes (V·A), the apparent power. A motor drawing 5.0 A at 230 V with $\cos \phi =0.8$ has an apparent power of 1150 V·A but takes only 920 W. A load with a power factor of 1 would need only 4.0 A for the same 920 W, yet the cable and the breaker still carry the motor's full 5.0 A and heat as $I^{2}R$ accordingly. That is why suppliers charge industrial users for a poor power factor, and why factories add capacitors to cancel the lag of their motors.

The energy graph is an integral

The bill counts energy, not power. When the power changes with time, the energy is the area under the graph of power against time, its time integral, and the power is the slope of the energy:
$E=\int Pdt$

In the animation at the Advanced level, $P(t)$ is a staircase. The lamp brings it to 100 W at 0.5 s, the kettle to 2100 W at 2.5 s and the heater to 4100 W at 4.8 s. The dashed line marks $V{I}_{\max }=230\cdot 10=2300\mathrm{W}$, the most the circuit can deliver at the breaker's rated current $I_{\max }=10\mathrm{A}$. The heater's step crosses this line, the breaker trips at 6.0 s and the power drops to zero.

$E(t)$ is the running area under the staircase, with 1 s of animation standing for 1 minute. Each step adds its height times its width: $100\mathrm{W}\cdot 2\min =200\mathrm{W}\min \approx 3.3\mathrm{Wh}$ up to 2.5 s, then $2100\mathrm{W}\cdot 2.3\min =80.5\mathrm{Wh}$, reaching 83.8 Wh at 4.8 s, then $4100\mathrm{W}\cdot 1.2\min =82\mathrm{Wh}$, reaching 165.8 Wh at the trip. After that the graph stays flat. So the 1.2 minutes of overload add more energy than the 2.3 minutes before them. Each straight segment of $E(t)$ climbs with a slope equal to the power at that time, and each corner marks a step of $P(t)$; a smoothly varying power would make the graph curve.

A meter computes this integral. The older induction meter turns an aluminium disc with a torque proportional to the average of $vi$, that is to $P$, against a magnetic brake whose drag is proportional to the disc's speed. The two balance when the speed is proportional to $P$, so the number of turns counts $E$. An electronic meter samples $v$ and $i$ thousands of times a second, multiplies each pair and adds up the products. Both work with the instantaneous product $vi$, so the power factor is included automatically: they count the real power, not $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$.

Switch-on surges and distorted currents

So far every load has been a fixed resistance. Real loads are not, and two common effects make the current larger than $P/V$ suggests, which matters for fuses and cables.

A tungsten filament's resistance grows tenfold or more as it heats from room temperature to about 2700 K. A 100 W lamp has $R=V^2/P=230^2/100=529\Omega$ when hot but only about 50 Ω cold. At switch-on it therefore draws $230/50\approx 4.6\mathrm{A}$, over 1 kW, for a fraction of a second until the filament heats up. That sudden stress is why such bulbs usually fail at the moment they are switched on.

Many phone chargers draw current only in short pulses near the peaks of the voltage, so their current is not a sine wave at all. Heating goes as $i^2$, which weights tall, narrow pulses heavily, so their $I_{\mathrm{rms}}$ exceeds $P/V_{\mathrm{rms}}$ even without any phase shift.

Cable heating and the circuit breaker

A circuit breaker protects the cable, whose insulation is damaged if it runs too hot. So the rule for when it must trip follows from how a cable warms up.

Model the cable as a body of heat capacity $C$ whose temperature is $\Delta T$ above its surroundings. The current heats it at the rate $I^{2}R$, and it loses heat to its surroundings at a rate $k\Delta T$, proportional to how much warmer it is, where $k$ is a heat-loss coefficient in watts per kelvin. The rate of change of its stored heat, $Cd\Delta T/dt$, is the difference:
$C\frac{d\Delta T}{dt}=I^{2}R-k\Delta T$

At first $\Delta T$ is small, the loss is negligible and the temperature climbs at the rate $I^{2}R/C$. As it rises, the loss catches up, until gain and loss balance at the final rise $\Delta T=I^{2}R/k$. The solution approaches this value as $1-e^{-t/\tau }$, with the time constant $\tau =C/k$, and is practically there after a few time constants, typically minutes for house wiring.

The final rise grows as $I^2$. Cables are sized so that their rated current keeps them below the safe limit of their insulation. The overload in the animation, $17.83/10=1.78$ times the rating, gives $1.78^2\approx 3.2$ times the temperature rise, far beyond that limit.

The bimetal strip in the breaker is heated by the same current and has its own heat capacity and heat loss, so it is a small model of the cable: it bends and trips the breaker when its own temperature reaches a set value. The larger the overload, the higher its final temperature would be and the earlier on its rise it passes the trip point, so its delay shortens steeply as the overload grows. In a short circuit the current might reach 300 A, which heats $(300/10{)}^2=900$ times as fast as 10 A. That is too fast to wait for a strip to warm, so a magnetic coil, pulled by the current itself, opens the contacts within about 10 ms, half a mains cycle.

Worked problem: the overload in a real cable

The animation assumes the full 230 V at every socket, as if the cable had no resistance. Suppose instead that the circuit's cable runs 15 m out and 15 m back, in 1.5 mm² copper. With all three appliances on, what current flows, what power does the cable take, and how much could it warm in the 1.2 minutes before the trip?

Legend

$V$	Voltage of the supply (rms).	$\mathrm{V}$
$I$	Current, the same all round the loop.	$\mathrm{A}$
$R_w$	Resistance of the whole cable, both conductors, drawn in one wire.	$\Omega$
$R_L$	Combined resistance of the three appliances.	$\Omega$
$P_w$	Power released as heat in the cable.	$\mathrm{W}$
$P_L$	Power the appliances take.	$\mathrm{W}$

Figure 4 draws the situation as a series circuit. The supply $V$ drives one current $I$ round the loop: through the cable, whose two conductors together have the resistance $R_w$ (drawn in a single wire), and through the three appliances, which act together as one resistance $R_L$. Because the same $I$ flows through both, the heat they release, $P_w=I^2{R}_w$ and $P_L=I^2{R}_L$, is in the ratio of their resistances. Treat each appliance as a fixed resistance equal to its value at the rated 230 V, which is a good model for heating elements.

First the two resistances. Copper has the resistivity $\rho =1.7\times 10^{-8}\Omega \mathrm{m}$, and the cable has a total length of 30 m and the cross-section $A=1.5\times 10^{-6}{\mathrm{m}}^2$, so $R_w=\rho L/A=1.7\times 10^{-8}\cdot 30/(1.5\times 10^{-6})=0.34\Omega$. The appliances take 4100 W together at 230 V, so $R_L=V^2/P=230^2/4100=12.90\Omega$.

The two are in series, so their resistances add, and the current and the cable's heating are
$I=\frac{V}{R_L+R_w}=\frac{230}{12.90+0.34}=17.37\mathrm{A}$
$P_w=I^2{R}_w=17.37^2\cdot 0.34\approx 103\mathrm{W}$

The cable takes part of the voltage, $17.37\cdot 0.34\approx 5.9\mathrm{V}$, so the sockets get $230-5.9=224.1\mathrm{V}$ and the appliances $P_L=I^2{R}_L\approx 3890\mathrm{W}$, not 4100 W. The cable's 103 W is 3.4 W per metre, three times the 34 W it releases at the rated 10 A, since $(17.37/10{)}^2=3.0$.

Now the warming. In the 72 s before the trip the cable releases $103\cdot 72\approx 7.4\mathrm{kJ}$. Its copper has the volume $30\cdot 1.5\times 10^{-6}=4.5\times 10^{-5}{\mathrm{m}}^3$; at 8960 kg/m³ that is a mass of 0.40 kg, and at 385 J/(kg·K) a heat capacity of 155 J/K. If no heat escaped, the copper would warm by
$\Delta T=\frac{7400}{155}\approx 48\mathrm{K}$

In reality the insulation and the wall absorb part of the heat, so the rise is smaller. Against that, copper's resistance grows by about 0.4 % per kelvin, so by up to about 19 % over a 48 K rise, which raises the heating a little. The order of magnitude is the point: tens of kelvin in about a minute, still rising, so an overload of this size must be cut within minutes.

Limits of validity

Everything above rests on simplifications. This section says where they hold.

Circuit theory treats the wires as ideal links that carry the same current all along their length at every instant. That requires a change at one end to reach the other end quickly compared with a cycle. Across a house such a change takes about a tenth of a microsecond, against a period of 20 ms at 50 Hz; put another way, the circuit must be much smaller than the wavelength $c/f$, which is 6000 km at 50 Hz. At radio frequencies, with wavelengths of metres, the wires radiate and the field picture takes over completely.

At 50 Hz the current also fills the conductor evenly. An alternating current is pushed towards the surface of a conductor, within a layer called the skin depth, and in copper at 50 Hz that depth is 9.2 mm, more than the 0.7 mm radius of a 1.5 mm² wire. At 1 MHz it shrinks to 65 µm, and the wire's resistance rises.

The supply is not ideal either: European mains may stray 10 % from 230 V, and by $P=V^2/R$ a kettle on 207 V gives only $2000\cdot 0.9^2=1620\mathrm{W}$. Ohm's law fails for filaments, whose resistance depends on their temperature, and for LEDs and electronics. Even copper's resistance rises by about 0.4 % per kelvin, and only superconducting cables, cooled with liquid nitrogen to 77 K, carry direct current with no resistive loss at all.

Where this leads

The magnetic field around every current is developed in Magnetic Fields of Magnets and Currents, and Electromagnetic Induction shows how the 50 Hz voltage is generated. Motors, Generators and Transformers explains why power crosses countries at 400 kV: for the same power, a thousand times the voltage means a thousandth of the current and so a millionth of the $I^{2}R$ loss.

The Poynting vector returns in Electromagnetic Waves, carrying sunlight's roughly 1 kW/m². Blackbody Radiation and the Quantum explains why a filament at 2700 K radiates mostly infrared, Semiconductors, Diodes and LEDs why an LED does far better, and First Law of Thermodynamics and Heat Engines why a heat pump gives three or four times more heat per kilowatt-hour than a heater.