Back emf and energy conversion

Hold a motor's shaft still and the armature is only a coil of wire: the supply sees R and nothing else, and the current is V/R. Let it turn and Faraday's law takes over. The flux through the rotating armature changes, an emf appears, and Lenz's law fixes its polarity so that it opposes the current that produced the motion — hence back emf. The loop rule now reads V = ε + IR, so I = (V − kω)/R. For a 24 V motor with R = 0.50 Ω and k = 0.24 V s rad⁻¹, the switch-on current is 48 A; at 86.1 rad s⁻¹ the emf is 20.7 V, the winding sees only 3.3 V, and the current is 6.7 A, smaller by a factor of 7.2. The motor limits its own current by running, which is why a large machine needs a starting resistor or a soft starter, and why the lights dim for a moment when the compressor kicks in.

Nothing in the circuit tells the motor what load it is carrying; the speed does. Add load torque and the shaft slows, so ε = kω falls, so V − ε grows, so I rises, and τ = kI rises with it until torque again matches the load. Solving V = kω + (τ/k)R for ω gives the torque–speed line, ω = V/k − τR/k², or equivalently τ = (kV/R)(1 − ω/ω₀) with ω₀ = V/k the no-load speed. For the same 24 V motor, ω₀ = 100 rad s⁻¹ and the stall torque is kV/R = 11.5 N m. A 1.6 N m load pulls the speed to 86.1 rad s⁻¹ and draws 6.67 A; double the load to 3.2 N m and the speed falls to 72.2 rad s⁻¹ while the current doubles to 13.3 A. The armature current is a torque meter, never a speedometer.

The emf constant in ε = kω carries units V s rad⁻¹; the torque constant in τ = kI carries N m A⁻¹. Those are the same SI combination, since V s = (J C⁻¹) s = J A⁻¹ = N m A⁻¹, and in a machine whose only internal loss is resistive the two constants are not merely alike but equal. The argument is one line of energy accounting. Multiply V = ε + IR by I to get VI = I²R + εI. The first term is copper heating, so all the rest, εI, is the power crossing the air gap into mechanical form, which is τω. Then kₑ ωI = kₜ Iω at every speed and every current, which forces kₑ = kₜ. Measure one and you have the other: a motor generating 0.24 V per rad s⁻¹ delivers 0.24 N m per ampere. That equality is not a design choice but a statement that induction and the magnetic force are two faces of one interaction.

Take VI = I²R + τω seriously and the central design tension appears. At the operating point above — 24 V, 6.67 A — the supply delivers 160 W, the armature wastes I²R = 22.2 W as heat, and 137.8 W leaves through the shaft, which is exactly τω = 1.6 N m × 86.1 rad s⁻¹. The conversion efficiency of this idealised machine is η = εI/VI = ε/V = ω/ω₀, here 86%. But the mechanical output P = kω(V − kω)/R is a downward parabola in ω: zero at stall, zero at ω₀, and peaking at ω₀/2 where ε = IR = V/2. That maximum is V²/4R = 288 W for this motor — reached at 50% efficiency, with 288 W simultaneously cooking the winding. Maximum power and good efficiency are different operating points, and every traction drive lives with the choice.

Nothing in V = ε + IR assumes the machine is a motor. Drive the shaft from outside and ω climbs; once kω exceeds V, the sign of V − kω flips and so does the current. Take the same motor on a 24 V line, with a descending load turning it at 125 rad s⁻¹: ε = 30 V, so I = (24 − 30)/0.50 = −12 A, meaning 12 A flows out of the armature and into the supply. The torque kI reverses too — 2.88 N m now braking — and the 360 W leaving the shaft, τω = 2.88 × 125, splits into 12² × 0.50 = 72 W of copper heat and 24 × 12 = 288 W returned to the line. That is regenerative braking: the same hardware, the same k, the same equation, read with the other sign. The changeover speed is exactly ω₀.

This model carries one loss, I²R, and a real machine carries several more. Bearing friction and windage take a torque τ_f before anything reaches the shaft, so the useful output is (kI − τ_f)ω: a modest τ_f = 0.25 N m at 86.1 rad s⁻¹ removes 21.5 W, turning 137.8 W of conversion into 116 W of output and dropping the efficiency from 86% to 73%. The same friction is why the no-load speed never quite reaches ω₀ — the motor settles where kI matches τ_f, at 97.8 rad s⁻¹ rather than 100. Hysteresis and eddy-current loss in the turning iron add more drag, brushes drop a volt or two that is not part of IR, the armature inductance makes the switch-on current rise over a time L/R instead of instantly, and at large current the armature's own field distorts the main field so that k itself droops.