Motors, Generators and Transformers

Beginner

Put a loop of wire between the poles of a magnet and send a current round it. One side of the loop is pushed up, the other side down, so the loop twists. Left alone, the loop swings round until it lines up with the magnet and stops. To keep it spinning, the current in the loop must be reversed every half turn; a split ring turning with the loop, rubbing on two fixed contacts, does exactly that. You now have an electric motor, the heart of a drill, a fan or an electric car.

Run the machine backwards and it becomes a generator. In the animation a coil keeps spinning between the north and south poles of a magnet, as a turbine would turn it in a power station. Its sides sweep through the magnetism and push a current round the circuit. Watch the red arrow on the wire: it grows, shrinks and points the other way every half turn. This is alternating current (AC), the kind in your sockets, and the lamp flashes twice per turn.

The current then reaches a transformer: two coils on one iron ring. The changing current in the first coil makes changing magnetism in the iron, which drives a current in the second coil. More turns on the second coil give a higher voltage, but never more power: ten times the voltage means a tenth of the current. That is why electricity travels at very high voltage: a small current in the cables wastes far less energy as heat.

Check yourself: a transformer has 50 turns in and 500 turns out. With 12 V AC going in, what comes out? Answer: 120 V, with a tenth of the current.

Intermediate

Force on a wire, torque on a coil

A magnetic field pushes sideways on moving charges, and so on a wire that carries a current. For a straight wire of length $L$ carrying a current $I$ through a uniform field $B$, the force is
$F=BIL\sin \theta$
where $\theta$ is the angle between the wire and the field: the force is largest when the wire crosses the field at right angles and zero along the field lines. It is perpendicular to both, and Fleming's left-hand rule gives its direction: first finger along the field, second finger along the current, thumb along the force. So 1 T pushes with 1 N on each metre of wire carrying 1 A across it.

Bend the wire into a rectangular coil of $N$ turns and area $A$ on an axle between the magnet's poles. Its two long sides carry opposite currents and are pushed opposite ways, so the coil feels a torque
$\tau =NBIA\sin \theta$
where $\theta$ is now the angle between the field and the normal to the coil. The torque is largest when the plane of the coil lies along the field and zero when the coil faces the field, where a coil left alone comes to rest. A DC motor keeps turning thanks to a commutator, a split ring on the axle touched by two brushes, which reverses the current every half turn, just as the torque is about to reverse.

The generator: a motor run backwards

Remove the battery and turn the coil at a steady angular velocity $\omega$. Measuring the angle from the position where the coil faces the field, the flux through each turn is $\Phi =BA\cos \omega t$, and Faraday's law, $\epsilon =-Nd\Phi /dt$, gives the EMF and its peak value:

• $\epsilon =NBA\omega \sin \omega t$
• ${\epsilon }_0=NBA\omega$

This alternating EMF has frequency $f=\omega /2\pi$. It is largest when the plane of the coil lies along the field: the sides then cut straight across the field lines, and the flux, passing through zero, changes fastest. It is zero when the coil faces the field, where the flux is largest but momentarily constant. Slip rings carry the EMF out without twisting the wires; a commutator in their place would give pulsing DC. Meters quote the root-mean-square (rms) value, the peak divided by $\sqrt{2}$ for a sine wave: a 230 V supply peaks at about 325 V.

Transformers: two coils, one flux

A transformer has no moving parts: a primary coil of $N_p$ turns and a secondary of $N_s$ turns share one closed iron core. An alternating current in the primary makes an alternating flux that the iron guides through the secondary. Each turn of either coil links the same changing flux and gets the same EMF, so the voltages are in the ratio of the turns, and without losses the power in equals the power out:

• $\frac{V_s}{V_p}=\frac{N_s}{N_p}$
• $V_p{I}_p=V_s{I}_s$

A step-up transformer ($N_s>N_p$) raises the voltage and lowers the current by the same factor, so no energy is created. A steady direct current makes a steady flux, which induces nothing, so a transformer needs AC. Large transformers are over 99 % efficient.

Worked example: the machine in the animation

The coil has $N=200$ turns of area $A=0.020{\mathrm{m}}^2$ (20 cm by 10 cm) in a field $B=0.50\mathrm{T}$, and it makes one turn every 2 s, so $\omega =2\pi /2=\pi \approx 3.14rad/s$. Then
${\Phi }_{\max }=BA=0.50\cdot 0.020=0.010\mathrm{Wb}$
${\epsilon }_0=NBA\omega =200\cdot 0.50\cdot 0.020\cdot \pi \approx 6.28\mathrm{V}$

At $t=0$ the coil faces the field: the $\Phi$ readout shows 10.0 mWb and the $\epsilon$ readout zero. An eighth of a turn later, at $t=0.25\mathrm{s}$, the flux is $10\cos 45^{\circ }\approx 7.1\mathrm{mWb}$ and the EMF $6.28\sin 45^{\circ }\approx 4.44\mathrm{V}$. At $t=0.5\mathrm{s}$ the plane of the coil lies along the field, $\Phi$ passes through zero and $\epsilon$ peaks at 6.28 V. At $t=1\mathrm{s}$ the flux is −10.0 mWb and the EMF is zero again.

With $N_p=100$ and $N_s=200$, the secondary gives $V_s=2\epsilon$, peaking at 12.57 V (about 12.6 V) on the $V_s$ readout, in step with $\epsilon$. The lamp's power grows as $V_s^2$, so it flashes at both peaks, positive and negative: twice per turn.

Reading the arrows: why the generator resists

The blue $v$ arrows show the long sides moving round their circle: at $t=0$ they slide along the field $B$ and cut nothing; at $t=0.5\mathrm{s}$ they cross it at right angles. Only the motion across the field drives the charges, hence the $\sin \omega t$. The purple $F$ arrows are the motor effect inside the generator: the field pushes on the induced current with $F=BIL$, always against the motion. That is Lenz's law; if the forces helped instead, the coil would speed itself up and make energy from nothing.

How hard must the coil be turned to keep the lamp lit? Treat the lamp as a 25 Ω resistor. At the peak it carries $12.57/25\approx 0.50\mathrm{A}$, so the ideal transformer needs twice that, about 1.0 A, in the coil. The torque is then $\tau =NBIA=200\cdot 0.50\cdot 1.0\cdot 0.020=2.0\mathrm{N}\cdot \mathrm{m}$, and turning against it at $\pi rad/s$ takes $\tau \omega \approx 6.3\mathrm{W}$, exactly the lamp's peak power, $12.57\cdot 0.50\approx 6.3\mathrm{W}$. Disconnect the lamp and the coil turns freely.

Worked example: high-voltage power lines

A wind farm sends 20 MW through lines with a total resistance of 4.0 Ω. They heat up at the rate $I^{2}R$, and for a given power the current is $I=P/V$, so
$P_{\mathrm{loss}}=I^{2}R={\left(\frac{P}{V}\right)}^{2}R$

At the generators' own 11 kV the current would be $20\times 10^6/(11\times 10^3)\approx 1818\mathrm{A}$, and the loss $1818^2\cdot 4.0\approx 13.2\mathrm{MW}$: two thirds of the power would only heat the wires. A step-up transformer with a turns ratio of 1 : 12 raises the voltage to 132 kV. The current falls to 151.5 A and the loss to $151.5^2\cdot 4.0\approx 92\mathrm{kW}$, under half a percent: twelve times the voltage, $12^2=144$ times less loss. Step-down transformers near homes then bring it back to 230 V.

Common mistakes

• A transformer does not work on steady DC: a 12 V battery gives 0 V on the secondary after a brief kick at switch-on.
• The EMF is not largest when the coil faces the magnets: there the flux, 10 mWb in the animation, is largest but momentarily constant, so $\epsilon =0$. The EMF peaks at 6.28 V when $\Phi =0$.
• A step-up transformer does not create energy: the lamp gets twice the coil's voltage but half its current, 0.50 A against 1.0 A, so both sides carry 6.3 W.
• The two angles called $\theta$ differ: in $F=BIL\sin \theta$ it lies between wire and field, in $\tau =NBIA\sin \theta$ between field and normal, so a coil lying along the field has the largest torque.

Where you meet it

Electric motors drive fans, lifts, pumps and electric cars, and use close to half of the world's electricity. An electric car's motor even works as a generator when it brakes. A two-pole power-station generator turning at 3000 rpm gives the 50 Hz of the European grid, and transformers sit at both ends of every power line and inside phone chargers. The 230 V met in Electric Power and Energy at Home is an rms value, and the changing fields behind every transformer lead on to Electromagnetic Waves.

Advanced

Two routes to the same EMF

A generator is a coil of $N$ turns, turned at a steady angular speed $\omega$ in a uniform magnetic field $\vec{B}$. Faraday's flux rule gives its EMF in one line, but it hides what actually pushes the charges round the circuit. So we first find the EMF from the magnetic force on the charges inside the moving wire, and then check that the flux rule gives the same answer. Let each turn be a rectangle with two long sides of length $L$, parallel to the axle, and two short ends of length $w$, the width of the coil, so that its area is $A=Lw$.

Legend

$\vec{B}$	Magnetic field, uniform and horizontal in the plane of the figure.	$\mathrm{T}$
$\omega$	Angular speed of the coil, which turns clockwise here.	$rad/s$
$\frac{w}{2}$	Radius of the circle on which each long side moves: half the width $w$ of the coil.	$\mathrm{m}$
$\vec{v}$	Velocity of a long side, tangent to its circle, of size $\omega w/2$.	$m/s$
$\theta$	Angle between $\vec{v}$ and $\vec{B}$, and also between $\vec{A}$ and $\vec{B}$: $\theta =\omega t$.	$\mathrm{rad}$
$\vec{A}$	Area vector of the coil: size $A=Lw$, at right angles to the coil, so parallel to $\vec{v}$ of the upper side.	${\mathrm{m}}^2$
$\vec{F}$	Magnetic force $q\vec{v}\times \vec{B}$ on a positive charge $q$ in a long side. It points along the wire: out of the page on the upper side (dot), into it on the lower side (cross).	$\mathrm{N}$

Figure 1 looks along the axle, which points into the page. Each long side then appears as a small disc, and the coil itself as the line joining the two discs. The field $\vec{B}$ runs horizontally across the page, and the coil turns clockwise at $\omega$.

Start with the motion of one long side. It lies a distance $w/2$ from the axle, so it moves on the dashed circle of radius $w/2$ with speed $v=\omega w/2$. Its velocity $\vec{v}$ is tangent to that circle. Measure time from the moment when the sides move along the field lines, with $\vec{v}$ parallel to $\vec{B}$. The velocity turns with the coil, so by time $t$ it has turned through the angle $\theta =\omega t$, marked in the figure between $\vec{v}$ and the dashed horizontal line at the upper side.

Next, the force. A positive charge $q$ carried along by the side feels the magnetic force $\vec{F}=q\vec{v}\times \vec{B}$. Both $\vec{v}$ and $\vec{B}$ lie in the plane of the page, so their cross product is perpendicular to the page, that is, along the wire. On the upper side it points out of the page (the dot); on the lower side, whose velocity is reversed, it points into the page (the cross). Because $\theta$ is the angle between $\vec{v}$ and $\vec{B}$, the force has size $qvB\sin \theta =qvB\sin \omega t$.

Now the work. The EMF is the work done on a unit charge carried once round the circuit. Along a long side the force per unit charge, $vB\sin \omega t$, points along the wire over the whole length $L$, so it does the work $BLv\sin \omega t$ per unit charge. The short ends lie in the plane of the page; the force on their charges still points along the axle, straight across the wire, so it does no work along them.

Finally, go once round a turn. Along the upper side we travel out of the page, the same way as the force there. Along the lower side we come back into the page, and the force there also points into the page, so it again pushes the way we travel. The two sides therefore add rather than cancel, and each turn gives $2BLv\sin \omega t$. The $N$ turns are connected in series, so their EMFs add too:
$\epsilon =2NBL\frac{\omega w}{2}\sin \omega t=NBA\omega \sin \omega t$
where the last step uses $A=Lw$.

The flux route is shorter. The area vector $\vec{A}$ has size $A$ and stands at right angles to the plane of the coil. In Figure 1 it is parallel to the velocity of the upper side, because that velocity is also at right angles to the coil. So $\vec{A}$ turns with $\vec{v}$ and makes the same angle $\theta =\omega t$ with $\vec{B}$. The flux through one turn is therefore $\Phi =BA\cos \omega t$, largest at $t=0$, when the coil faces the field. Faraday's law gives
$\epsilon =-N\frac{d\Phi }{dt}=-NBA\frac{d}{dt}\cos \omega t=NBA\omega \sin \omega t$
the same result.

Put in the animation's numbers. Its coil has $L=0.20\mathrm{m}$ along the axle and $w=0.10\mathrm{m}$ across, so $A=0.020{\mathrm{m}}^2$, with $N=200$ turns in $B=0.50\mathrm{T}$, turning at $\omega =\pi rad/s$ (one turn every 2 s). Each long side moves at $v=\pi \cdot 0.05\approx 0.157m/s$ and contributes at most $BLv=0.50\cdot 0.20\cdot 0.157\approx 15.7\mathrm{mV}$. The 200 turns have 400 long sides, all in series, so the peak EMF is $400\cdot 15.7\mathrm{mV}\approx 6.28\mathrm{V}$, the same as $NBA\omega =200\cdot 0.50\cdot 0.020\cdot \pi$.

The two routes agree, but they describe different physics. In the generator the wire moves, and the magnetic force on its moving charges does the pushing; the flux rule is a compact summary of that force. In a transformer nothing moves, so there is no force $q\vec{v}\times \vec{B}$ at all. Instead, the changing flux creates an electric field that circulates round the core,
$\oint \vec{E}\cdot d\vec{l}=-\frac{d\Phi }{dt}$
and this electric field drives the charges in the secondary coil. The fact that one rule covers two such different mechanisms hints that electric and magnetic fields are two aspects of a single field, which relativity makes precise.

Reading the graph: flux leads the EMF

The animation's graph shows the flux and the EMF together. Reading it well means seeing Faraday's law as a statement about slopes, which is the quickest way to sketch any induced EMF from its flux.

Legend

$\Phi$	Magnetic flux through one turn, $BA\cos \omega t$, drawn scaled to the height of ${\epsilon }_0$.	$\mathrm{Wb}$
$\epsilon$	EMF of the coil, $-Nd\Phi /dt$.	$\mathrm{V}$
${\epsilon }_0$	Peak EMF of the coil, $NBA\omega$.	$\mathrm{V}$
$V_s$	Voltage of the transformer's secondary coil; its peak is $2{\epsilon }_0$ for $N_s/N_p=2$.	$\mathrm{V}$
$t$	Time.	$\mathrm{s}$
$T$	Period: the time of one turn of the coil.	$\mathrm{s}$
$\frac{T}{4}$	A quarter period: the time by which the flux leads the EMF, a phase of $90^{\circ }$.	$\mathrm{s}$

Figure 2 redraws that graph over one period $T$, the time of one turn: here $T=2\pi /\omega =2\mathrm{s}$. The green curve is the flux $\Phi (t)$ through one turn, scaled to the height of the EMF so that the two are easy to compare. The red curve is the EMF $\epsilon (t)$. The gold curve, which we come to below, is the output $V_s(t)$ of a transformer fed by the generator.

First compare where the curves peak. The flux $\Phi =BA\cos \omega t$ has a crest at $t=0$, and the EMF has its first crest a quarter period later, at $T/4=0.5\mathrm{s}$. The flux leads the EMF by a quarter cycle, a phase of $90^{\circ }$. The pattern repeats every period, so the flux crests again at 2 s, 4 s and so on.

Next see why. Faraday's law, $\epsilon =-Nd\Phi /dt$, says that the EMF is minus $N$ times the slope of the green curve. At a crest or a trough of $\Phi$ the curve is flat for an instant, so $\epsilon =0$ there, even though the flux itself is at its largest. Where $\Phi$ falls through zero, at $T/4$, it is at its steepest; the dashed tangent shows that slope, and there $\epsilon$ reaches its peak $+{\epsilon }_0=+6.28\mathrm{V}$. Half a period later $\Phi$ rises through zero just as steeply, and $\epsilon =-6.28\mathrm{V}$.

The numbers confirm it. The flux through one turn has amplitude $BA=0.50\cdot 0.020=10\mathrm{mWb}$, so the flux linkage $N\Phi$ has amplitude $200\cdot 10\mathrm{mWb}=2.0\mathrm{Wb}$. A cosine of amplitude $a$ and angular frequency $\omega$ has a steepest slope of $a\omega$, here $2.0\cdot \pi \approx 6.28Wb/s$. Since $1Wb/s=1\mathrm{V}$, that is the peak EMF ${\epsilon }_0$.

The gold curve $V_s(t)$ has the same zeros and crests as $\epsilon (t)$ but is twice as tall, peaking at $2{\epsilon }_0$. An ideal transformer copies the waveform, scales it by its turns ratio, here $N_s/N_p=2$, and adds no phase shift; Figure 4 shows why.

Last, what single number to quote for such a voltage. A resistor $R$ driven by ${\epsilon }_0\sin \omega t$ takes the power $({\epsilon }_0^2/R){\sin }^2\omega t$. Over a cycle ${\sin }^2\omega t$ averages $\frac{1}{2}$, because ${\sin }^2$ and ${\cos }^2$ have the same average and add up to 1. So the mean power is half the peak power, the same as a steady voltage ${\epsilon }_0/\sqrt{2}$ would give. This is the root-mean-square (rms) value, here $6.28/\sqrt{2}\approx 4.44\mathrm{V}$.

Energy balance: the generator pays in torque

A generator turns mechanical work into electrical energy, so something must resist the rotation; otherwise the coil would deliver power for nothing. That something is the magnetic force on the very current the coil delivers.

Let the coil deliver a current $I$ in phase with its EMF, as it does into a resistor. In each turn, a long side carries $I$ along a length $L$ at right angles to $\vec{B}$, so it feels a force $BIL$, directed at right angles to $\vec{B}$ in the plane of Figure 1. Only the part of this force along the side's velocity $\vec{v}$ turns the coil. Since $\vec{v}$ makes the angle $\omega t$ with $\vec{B}$, that part is $BIL\sin \omega t$, and by Lenz's law it acts against $\vec{v}$, at the lever arm $w/2$. Two sides per turn and $N$ turns give the counter-torque
$\tau =2N(BIL\sin \omega t)\frac{w}{2}=NBIA\sin \omega t$
To keep the coil turning at a steady $\omega$, whoever drives it must supply this torque, so the mechanical power put in is
$\tau \omega =(NBIA\sin \omega t)\omega =(NBA\omega \sin \omega t)I=\epsilon I$
exactly the electrical power delivered, at every instant.

Which force does the work? Not the magnetic one: it is always perpendicular to each charge's total velocity, the sum of the coil's motion and the charge's drift along the wire, so it does no work at all. It only redirects work. The part of the force due to the coil's motion pushes the charges along the wire and drives the current; the part due to their drift along the wire pushes sideways on the wire and resists the rotation. These two works are equal and opposite, so all the electrical energy comes from whoever turns the shaft.

The motor: back EMF and the torque line

Run the same machine as a motor and the coupling works the other way round. The supply drives a current, the force on that current turns the coil, and the turning coil, being a generator too, makes an EMF that opposes the supply. This back EMF is what sets a motor's current and speed, and it decides how much torque the motor can give at each speed.

Legend

$\omega$	Speed of the motor shaft.	$rad/s$
$\tau$	Torque the motor delivers; it falls linearly as $\omega$ rises.	$\mathrm{N}\cdot \mathrm{m}$
${\tau }_0$	Stall torque, $kV/R$: the torque with the shaft held, $\omega =0$.	$\mathrm{N}\cdot \mathrm{m}$
${\omega }_0$	No-load speed, $V/k$: the back EMF equals the supply voltage, no current flows and the torque is zero.	$rad/s$
$\frac{{\omega }_0}{2}$	Half the no-load speed, where the mechanical power is largest.	$rad/s$
$\frac{{\tau }_0}{2}$	Half the stall torque: the torque at ${\omega }_0/2$.	$\mathrm{N}\cdot \mathrm{m}$
$P$	Mechanical power $\tau \omega$, the area of the shaded rectangle; at ${\omega }_0/2$ it has its largest value, $V^2/(4R)$.	$\mathrm{W}$

Figure 3 plots the torque $\tau$ that a motor delivers against the speed $\omega$ of its shaft, for a fixed supply voltage $V$ applied through the resistance $R$ of its winding.

To see why the graph is a straight line, model the motor. With many coils on the rotor and a commutator that switches each coil in at the right moment, the back EMF ${\epsilon }_b$ is nearly steady and proportional to the speed. Three equations then describe the motor:
$V={\epsilon }_b+IR,{\epsilon }_b=k\omega ,\tau =kI$
The first is the circuit: the supply voltage is shared between the back EMF and the drop across the winding. The second and third contain the same constant $k$, and they must: the energy balance above, ${\epsilon }_{b}I=\tau \omega$, holds only if the constant linking ${\epsilon }_b$ to $\omega$ equals the one linking $\tau$ to $I$.

The first equation gives the current, $I=(V-k\omega )/R$, and the third turns it into a torque:
$\tau =\frac{k(V-k\omega )}{R}$
This is the falling line of the figure. At its left end the shaft is held still, $\omega =0$, so there is no back EMF and only the winding limits the current, to $V/R$; the torque is then the stall torque ${\tau }_0=kV/R$. As the speed rises the back EMF grows, so the current falls and the torque with it. At the right end, the no-load speed ${\omega }_0=V/k$, the back EMF has risen to the supply voltage, no current is left, and the torque is zero.

The mechanical power is $P=\tau \omega$. On the graph that is the area of a rectangle with one corner at the origin and the opposite corner on the line, like the shaded one. Writing the line as $\tau ={\tau }_0(1-\omega /{\omega }_0)$ gives
$P={\tau }_0\omega \left(1-\frac{\omega }{{\omega }_0}\right)$
which is zero at both ends of the line. It is a parabola in $\omega$, symmetric about the midpoint of its two zeros, so it is largest at $\omega ={\omega }_0/2$, where the torque is ${\tau }_0/2$. The peak power is therefore $({\tau }_0/2)({\omega }_0/2)=V^2/(4R)$.

How much of the input becomes work? The motor takes the electrical power $VI$ and turns ${\epsilon }_{b}I$ of it into mechanical power, so its efficiency, ignoring friction, is ${\epsilon }_b/V=\omega /{\omega }_0$. At the speed of peak power ${\epsilon }_b=V/2$: only half the power taken in becomes mechanical power, and the other half heats the winding. Near the no-load speed, where the rectangle is wide but low, the efficiency is high.

Take a 12 V motor with $R=0.50\Omega$. At start-up $\omega =0$, so there is no back EMF and $I=12/0.50=24\mathrm{A}$. Running at 3000 rpm, that is $\omega =3000\cdot 2\pi /60\approx 314rad/s$, its back EMF is ${\epsilon }_b=11\mathrm{V}$, and it draws only $(12-11)/0.50=2.0\mathrm{A}$. Of the 24 W taken in, ${\epsilon }_{b}I=22\mathrm{W}$ becomes mechanical work and $I^{2}R=2\mathrm{W}$ heats the winding, an efficiency of 92 %.

The running point also gives the motor constant, $k={\epsilon }_b/\omega =11/314\approx 0.035\mathrm{V}\cdot s/rad$, so the torque is $\tau =kI=0.035\cdot 2.0=0.070\mathrm{N}\cdot \mathrm{m}$; as a check, $\tau \omega =0.070\cdot 314\approx 22\mathrm{W}$.

Now place this motor on Figure 3. Its stall torque is ${\tau }_0=kV/R=0.035\cdot 12/0.50\approx 0.84\mathrm{N}\cdot \mathrm{m}$ and its no-load speed is ${\omega }_0=V/k\approx 343rad/s$, near 3270 rpm. So 3000 rpm lies close to the right-hand end of the line, where the torque is small and the efficiency $\omega /{\omega }_0$ is high. The torque falls linearly with speed from 0.84 N·m to zero, and the mechanical power peaks at half the no-load speed, about 1640 rpm, at $V^2/(4R)=72\mathrm{W}$. There the motor draws $(12-6)/0.50=12\mathrm{A}$, or 144 W, so only 50 % of it becomes work.

Inside the transformer

A transformer changes the size of an alternating voltage with no moving parts, which is why power grids run on AC. All its behaviour follows from one fact: the iron core makes the same flux pass through every turn of both coils.

Legend

$V_p$	Voltage of the source across the primary coil.	$\mathrm{V}$
$I_p$	Current in the primary coil.	$\mathrm{A}$
$N_p$	Number of turns of the primary coil.
$\Phi$	Magnetic flux in the iron core, the same through every turn of both coils.	$\mathrm{Wb}$
$N_s$	Number of turns of the secondary coil.
$I_s$	Current in the secondary coil and the load.	$\mathrm{A}$
$V_s$	Voltage across the secondary coil and the load.	$\mathrm{V}$
$R$	Resistance of the load on the secondary.	$\Omega$

Figure 4 shows a transformer with a load. The source applies the voltage $V_p$ across the $N_p$ turns of the primary coil and drives the current $I_p$ through them. The iron core carries the flux $\Phi$ round its closed loop, along the green arrows, so that the same $\Phi$ threads all $N_s$ turns of the secondary coil. The secondary applies the voltage $V_s$ to the load $R$ and drives the current $I_s$ through it.

Since every turn of either coil links the same flux, every turn has the same EMF, $d\Phi /dt$ in size. Turns in series add, and with the resistance of the windings neglected each coil's terminal voltage equals its EMF, so in size
$V_p=N_p\frac{d\Phi }{dt},V_s=N_s\frac{d\Phi }{dt}$
Dividing one by the other cancels $d\Phi /dt$ and leaves the turns ratio, $V_s/V_p=N_s/N_p$. Both voltages follow the same $d\Phi /dt$, so they also rise and fall together, with no phase shift, as Figure 2 showed.

Now read the first equation the other way round: the primary voltage fixes the flux, whatever the load. For a sinusoidal voltage of peak $V_0$ the flux is a sinusoid too, of peak ${\Phi }_0=V_0/(N_p\omega )$, because differentiating a sinusoid multiplies its amplitude by $\omega$. A 230 V, 50 Hz primary of 1000 turns has $V_0\approx 325\mathrm{V}$ and $\omega \approx 314rad/s$, so ${\Phi }_0\approx 1.04\mathrm{mWb}$. To keep the iron below about 1.2 T the core then needs a cross-section of ${\Phi }_0/B\approx 8.6\times 10^{-4}{\mathrm{m}}^2$, some 8.6 cm². The area needed falls as $1/\omega$: at the 400 Hz used on aircraft an eighth of it would do, and chargers switching at tens of kilohertz use tiny ferrite cores.

Why does the primary current rise when a load is connected? The load current $I_s$ flows round the secondary turns and, on its own, would magnetise the core the opposite way to $I_p$; in Figure 4 the two currents circulate in opposite senses round the core. But the flux is fixed by $V_p$. So the primary must draw an extra current whose magnetising effect, measured by current times number of turns, cancels that of the secondary: $N_p{I}_p=N_s{I}_s$, apart from the small magnetising current that sets up the flux itself. Multiplying by the voltage ratio gives $V_p{I}_p=V_s{I}_s$: power in equals power out.

A transformer therefore also changes resistances. Seen from the primary, the load is $V_p/I_p$. Substituting $V_p=(N_p/N_s)V_s$ and $I_p=(N_s/N_p)I_s$ gives
$\frac{V_p}{I_p}={\left(\frac{N_p}{N_s}\right)}^{2}R$
which is how a transformer matches a loudspeaker to an amplifier.

Two limiting cases remain: an open secondary and a direct current. With the secondary open, no current $I_s$ flows, and the secondary voltage is set by the primary current alone: in size $V_s=Md{I}_p/dt$, where $M$ is the mutual inductance of the two coils. On DC, $d\Phi /dt=0$: nothing is induced in either coil, nothing crosses to the secondary, and only the copper's small resistance limits the primary current.

Worked problem: loading the generator

Connect a 25 Ω lamp, treated as a plain resistor, to the secondary of the animation's transformer ($N_p=100$, $N_s=200$), whose primary is fed by the generator, and neglect the resistance of the coils. What current does the generator coil carry, how hard must it be turned, and what would change without the transformer?

First, the load as the coil sees it. From the primary, the lamp looks like $(N_p/N_s{)}^{2}R=(100/200{)}^2\cdot 25=6.25\Omega$. The coil drives its EMF, of peak 6.28 V, into this resistance, so the coil current is a sine wave in phase with the EMF, of peak $6.28/6.25\approx 1.01\mathrm{A}$. The secondary doubles the voltage and halves the current, so the lamp carries a peak of about 0.50 A at a peak of 12.57 V.

Next, the torque. The energy balance gave the counter-torque $NBAI\sin \omega t$, where $I$ is now the primary current $I_p$, itself proportional to $\sin \omega t$. With $NBA=200\cdot 0.50\cdot 0.020=2.0\mathrm{Wb}$, and $1\mathrm{Wb}\cdot \mathrm{A}=1\mathrm{N}\cdot \mathrm{m}$,
$\tau =NBA{I}_p\sin \omega t\approx 2.0{\sin }^2\omega t\mathrm{N}\cdot \mathrm{m}$
It peaks at 2.0 N·m when the plane of the coil lies along the field, where the EMF and the current are largest, and averages half of that, 1.0 N·m. As a check, the mean mechanical power, $\bar{\tau }\omega =1.005\cdot \pi \approx 3.16\mathrm{W}$, equals the lamp's mean power, half its peak voltage times its peak current: $\frac{1}{2}\cdot 12.57\cdot 0.503\approx 3.16\mathrm{W}$.

Without the transformer the lamp would sit directly on the slip rings, a 25 Ω load instead of 6.25 Ω. It would take only ${\epsilon }_0^2/(2R)=6.28^2/50\approx 0.79\mathrm{W}$, a quarter as much, and the coil would be four times easier to turn. The transformer creates no energy: by making the lamp look four times smaller in resistance, it lets the coil drive four times the current at the same EMF, and so deliver four times the power, all of it supplied by whoever turns the coil.

Notice, finally, that the torque varies as ${\sin }^2\omega t=\frac{1}{2}(1-\cos 2\omega t)$: it throbs at twice the rotation frequency, which shakes a single-phase machine. Power stations avoid this with three sets of coils set $120^{\circ }$ apart. The $\cos 2\omega t$ parts of their three torques are then a third of a cycle apart and cancel, since
${\sin }^2\theta +{\sin }^2(\theta -120^{\circ })+{\sin }^2(\theta +120^{\circ })=\frac{3}{2}$
so a balanced three-phase generator gives a perfectly steady torque and power.

Limits of validity

The ideal transformer assumes three things: all the flux links both coils, the windings have no resistance, and the iron is magnetised for free and without loss. Real cores lose energy in two ways. Hysteresis costs a fixed amount of energy each time the iron's magnetisation is reversed, so its power loss grows roughly in proportion to the frequency. Eddy currents are currents induced in the iron itself; they grow as the square of both the frequency and the thickness of the iron sheets, which is why cores are built from insulated laminations about 0.3 mm thick. Large transformers reach 99.5 % efficiency, yet a 500 MVA unit then sheds 2.5 MW of heat, hence its oil tank and radiators.

Saturation sets a further limit. Silicon steel saturates near 1.5 to 2 T: beyond that, more current adds almost no flux. Since the flux is fixed by ${\Phi }_0=V_0/(N_p\omega )$, a transformer run below its design frequency or above its design voltage is driven into saturation and draws huge currents.

The treatment is also quasi-static: the fields are taken to act everywhere at once. At 50 Hz the wavelength is $c/f=6000\mathrm{km}$, so any machine is tiny by comparison, but a 1000 km line is a sixth of a wavelength and must be analysed as a transmission line. The flux rule also needs care when conducting material slides past a fixed contact, as in a spinning-disc generator, where it is unclear which loop the flux should be counted through. The force $q\vec{v}\times \vec{B}$ is then the reliable guide.

AC is not always best. A long cable has a large capacitance, which AC must keep charging and discharging. Beyond roughly 50 to 100 km of undersea cable, or several hundred kilometres of overhead line, high-voltage DC with electronic converters wins. Insulation, and corona discharge in the air round the conductors, cap the highest transmission voltages at around 1 MV.

Where this leads

The force on currents from Magnetic Force on Moving Charges and the flux rule from Electromagnetic Induction are all the physics a power grid needs. The induced electric field of every transformer, joined by its partner, a magnetic field made by a changing electric field, gives the self-sustaining waves of Electromagnetic Waves. In the frame moving with a side of the coil, the generator's magnetic push becomes an electric field: a first glimpse of Special Relativity. The rms values and power factor of AC supplies were treated in Electric Power and Energy at Home, and the rectifiers and inverters behind brushless motors and HVDC links come from Semiconductors, Diodes and LEDs.