ISEGORIA / MATH ENCYCLOPEDIA
Electric motors: turning current into torque
Every electric motor uses the same two facts: a current in a magnetic field feels a force, and a conductor moving through a field has a voltage induced in it. In a DC motor the commutator keeps the current in each conductor pointing the right way, and the induced back EMF sets the speed. In an AC motor three stationary windings, fed with currents a third of a cycle apart, produce a magnetic field that rotates by itself, and an induction motor’s rotor is dragged along by the currents that field induces in it.
Before you begin: Electromagnetic induction, phasors and rotational mechanics
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. A field that rotates by itself
Put three windings round a stator with their axes 120° apart and feed them three-phase currents, each lagging the last by a third of a cycle. Each winding on its own makes a field that only pulses back and forth along its axis, but the three pulses add up to a field of constant strength that turns steadily round the bore. With p pairs of poles it turns once every p cycles of the supply. Run the supply, unbalance one phase, swap two leads, or take two phases away and see what is left.
Worked example. A four-pole motor (\(p=2\)) on a 50 Hz supply has a field turning at \(60\times50/2=1500\) rpm; a two-pole motor on 60 Hz, 3600 rpm. If phase c carries only half the current of the others, the field splits into a forward part of 1.25 and a backward part of 0.25 (in units of one phase’s peak), so its tip traces an ellipse with semi-axes 1.5 and 1.0 instead of a circle of radius 1.5.
Watch out. Each winding is idealised as producing a field that varies as the cosine of angle around the gap, which real distributed windings approximate closely but not exactly; the space harmonics they add are left out. The field strength is in arbitrary units, and the central arrows are in electrical degrees, which equal mechanical degrees only for one pole pair.
Where does the factor 3/2 come from?
Write each pulsating field as two half-size fields turning in opposite directions: \(\cos(\omega t-\phi)\,\hat{\mathbf u}\) is half a forward-turning vector plus half a backward-turning one. With the windings and currents both 120° apart, the three forward halves line up and add to \(3\times\frac12=\frac32\), while the three backward halves point 120° apart from each other and cancel exactly.
2. The DC motor: torque, back EMF and efficiency
In a permanent-magnet DC motor the torque is proportional to the armature current, and the spinning armature generates a back EMF proportional to its speed, with the same constant k in both. Ohm’s law then ties them together: the applied voltage is shared between the back EMF and the resistance drop. That makes speed fall in a straight line as load rises, and it makes power and efficiency peak at very different loads. Change the voltage and the winding resistance, drag the load, and watch the motor start.
Worked example. With \(k=0.02\) N·m/A, \(R=0.6\ \Omega\) and 0.01 N·m of friction at 12 V, the motor runs unloaded at 5590 rpm and stalls at 0.39 N·m. Loaded with 0.1 N·m it turns at 4150 rpm, draws 5.5 A and delivers 43.5 W from 66 W, an efficiency of 66%. Its maximum power, 57 W, comes at half the stall torque, but its best efficiency, 71%, comes at only 0.053 N·m, where the current is \(\sqrt{0.5\times20}=3.2\) A.
Watch out. The model has one constant friction torque and ignores iron loss, brush drop and armature reaction, and the resistance is taken at room temperature; a real winding heats up and its resistance rises by about 0.4% per degree. The transient is the exact solution of the two coupled equations above, integrated numerically, with the rotor held still until the motor’s torque exceeds the load.
Why does a stalled motor burn out, but a running one does not?
Because the only thing limiting the current at standstill is the winding resistance: \(i=V/R\), here 20 A, and the heat \(i^2R\) is 240 W, all of it in the winding. Once the motor turns, the back EMF \(k\omega\) opposes the supply and the current falls to what the load needs. A running motor is protected by its own speed.
3. The induction motor’s torque–speed curve
An induction motor’s rotor has no connection to the supply. The rotating field sweeps past its bars, induces currents in them, and the force on those currents drags the rotor round. If the rotor ever caught up with the field nothing would be induced, so it always runs slightly slow; that shortfall, the slip, sets everything. From Steinmetz’s equivalent circuit comes the whole torque–speed curve, from braking through motoring to generating. Change the load, the rotor resistance and the supply frequency.
Worked example. The motor here, star-connected at 400 V and 50 Hz with \(R_1=0.4\), \(X_1=X_2=1.0\), \(X_m=30\) and \(R_2=0.35\ \Omega\), gives 74.7 N·m at 3% slip, 1455 rpm: 11.4 kW at 20.2 A and a power factor of 0.87. Its breakdown torque is 200 N·m, 2.7 times that, at 17% slip. Starting straight on the line it gives 76 N·m but draws 110 A, 5.4 times its running current. Tripling the rotor resistance moves the breakdown to 52% slip with exactly the same 200 N·m.
Watch out. The parameters are representative of a motor of this size, not a particular product. The circuit ignores iron loss, friction and windage, and saturation, so the efficiency shown counts only copper losses and is optimistic by a few percent. Real squirrel-cage rotors also change their effective resistance with slip because of the skin effect in deep bars, which raises the starting torque above this curve.
Why does the breakdown torque not depend on the rotor resistance?
Because the rotor resistance only ever appears divided by the slip, as \(R_2/s\). Doubling \(R_2\) and doubling s gives the same circuit, the same currents and the same air-gap power, so every point on the curve simply moves to twice the slip at the same torque. The peak moves sideways but keeps its height, which is why wound-rotor motors add resistance to start heavy loads and remove it once running.