Nice — I’ll walk you through a clear, practical step-by-step method for designing a motor

winding and taking it “up to the circuit” (i.e., estimating turns, wire size, resistance, copper loss
and showing the per-phase equivalent). I’ll give the generic formulas you’ll use and a worked
numeric example so you can follow the math end-to-end.

I’ll assume you want a 3-phase induction motor example. If you need single-phase or different
ratings, tell me and I’ll adapt (but I won’t wait — below is a full worked example).

1) Overview — the main design steps

1. Pick the motor rating / electrical specs (power, voltage, frequency, poles).
2. Compute rated line & phase currents.
3. Choose design flux (Bp) and pole core area (Ac) → compute flux per pole Φ.
4. From required phase EMF and Φ compute turns per phase (use winding factor).
5. Choose conductor cross-section using current density J → choose wire gauge.
6. Compute mechanical layout: slots, coils per phase, turns per coil.
7. Compute winding length and DC resistance (R) per phase.
8. Estimate copper losses, finalize conductor choice.
9. Produce electrical connection (star/delta) and draw per-phase equivalent circuit (R1, X1,
magnetizing branch, R2′/s, X2′).

2) Useful formulas (keep these handy)

 Phase voltage (star): Vph=Vline3V_{ph} = \dfrac{V_{line}}{\sqrt{3}}
 Input electrical power (approx): Pin=PoutηP_{in} = \dfrac{P_{out}}{\eta}
 Rated line current: Iline=Pin3 Vline pfI_{line} = \dfrac{P_{in}}{\sqrt{3}\,V_{line}\,\
text{pf}}
 Flux per pole: Φ=Bpeak⋅Ac\Phi = B_{peak}\cdot A_c (B in T, Ac in m² → Φ in Wb)
 Induced RMS EMF per phase (approx sinusoidal): Eph=4.44 f Nph Φ kwE_{ph} =
4.44\,f\,N_{ph}\,\Phi\,k_w
→ rearranged: Nph=Eph4.44 f Φ kwN_{ph} = \dfrac{E_{ph}}{4.44\,f\,\Phi\,k_w}
where kwk_w = winding factor (slot + pitch distribution factor), ~0.85–0.98 depending
on winding.
 Conductor cross-section from current density: Acond=IJA_{cond} = \dfrac{I}{J}
(A/mm²)
 DC resistance: R=ρ⋅LAR = \rho\cdot \dfrac{L}{A} (ρ for copper ≈ 1.724×10−8
Ω⋅m1.724\times10^{-8}\,\Omega\cdot m at 20°C; L in m; A in m²)
 Copper loss (per phase): Pcu,phase=Iphase2⋅RphaseP_{cu,phase} = I_{phase}^2 \cdot
R_{phase} ; total copper loss = 3× that (for 3-phase)
 Synchronous speed ns=120 fpolesn_s = \dfrac{120\,f}{\text{poles}}
3) Worked example (3-phase induction
motor)
Given / assumed design data

 Output power Pout=5 000 WP_{out} = 5\,000\ \text{W} (5 kW)

 Line voltage VLL=400 VV_{LL} = 400\ \text{V} (3-phase)
 Frequency f=50 Hzf = 50\ \text{Hz}
 Poles = 4 (4-pole motor)
 Efficiency η=90%=0.90\eta = 90\% = 0.90 (assumed)
 Power factor pf = 0.85 (assumed)
 Peak flux density Bpeak=1.2 TB_{peak} = 1.2\ \text{T} (typical)
 Core cross-sectional area Ac=0.005 m2A_c = 0.005\ \text{m}^2 (5,000 mm² — example
small motor)
 Winding factor kw=0.95k_w = 0.95 (reasonable for semi-distributed winding)
 Current density J=4 A/mm2J = 4\ \text{A/mm}^2 (practical enamel wire rule of thumb)
 Mean length per turn (MLT) ≈ 0.25 m (assumed stator geometry)
 Copper resistivity ρ=1.724×10−8 Ω⋅m\rho = 1.724\times 10^{-8}\ \Omega\cdot\text{m}

Now compute step by step (I’ll show the arithmetic):

1. Input power PinP_{in}

Pin=Poutη=50000.90=5555.555… W≈5555.56 W.P_{in}=\dfrac{P_{out}}{\eta}=\dfrac{5000}

{0.90}=5555.555\ldots\ \text{W}\approx 5555.56\ \text{W}.

2. Phase voltage VphV_{ph}

Vph=VLL3=4001.73205=230.940 V (approx).V_{ph}=\dfrac{V_{LL}}{\sqrt{3}}=\dfrac{400}
{1.73205}=230.940\ \text{V} \ (\text{approx}).

3. Rated line (and phase) current II

I=Pin3 VLL pf=5555.555…1.73205×400×0.85.I=\dfrac{P_{in}}{\sqrt{3}\,V_{LL}\,\text{pf}}

= \dfrac{5555.555\ldots}{1.73205\times 400\times 0.85}.

Calculate denominator: 1.73205×400=692.8201.73205\times 400 = 692.820. Then

692.820×0.85=589.897692.820\times 0.85 = 589.897.
So

I=5555.555…589.897=9.417 A (approx).I = \dfrac{5555.555\ldots}{589.897}=9.417\ \

text{A}\ \text{(approx)}.
(rounded to 3 decimals: 9.417 A). So choose conductor sized for ≈9.4 A.

4. Conductor area from current density

Acond=IJ=9.4174=2.3543 mm2.A_{cond} = \dfrac{I}{J}=\dfrac{9.417}{4}=2.3543\ \

text{mm}^2.

→ Pick standard conductor 2.5 mm² enamelled copper.

5. Flux per pole Φ

Φ=Bpeak⋅Ac=1.2×0.005=0.006 Wb.\Phi = B_{peak}\cdot A_c = 1.2\times 0.005 = 0.006\ \

text{Wb}.

6. Required turns per phase NphN_{ph}

Use Eph≈VphE_{ph} \approx V_{ph} (neglect small drops): Eph=230.94 VE_{ph}=230.94\ \
text{V}.

Nph=Eph4.44 f Φ kw=230.944.44×50×0.006×0.95.N_{ph}=\dfrac{E_{ph}}{4.44\,f\,\Phi\,k_w}
=\dfrac{230.94}{4.44\times 50\times 0.006\times 0.95}.

Compute denominator step by step:

 4.44×50=222.04.44\times 50 = 222.0
 222.0×0.006=1.332222.0\times 0.006 = 1.332
 1.332×0.95=1.26541.332\times 0.95 = 1.2654

So

Nph=230.941.2654=182.5 turns (per phase).N_{ph}=\dfrac{230.94}{1.2654}=182.5\ \text{turns

(per phase)}.

7. Map turns to coils / slots

Suppose you choose 36 stator slots (common small motor choice). For 3 phases:

 slots per phase = 36/3 = 12 slots/phase.

 If you decide coils per phase = 6 (two slots per coil), then turns per coil =
Ncoil=Nph6=182.56≈30.4N_{coil} = \dfrac{N_{ph}}{6} = \dfrac{182.5}{6} \approx
30.4.
Round to integer: use 30 turns/coil (or 31 depending on distribution).

8. Winding length and DC resistance per phase

 Total turns per phase (use exact chosen integer; use 182.5 for estimate).
 Mean length per turn (MLT) = 0.25 m (assumed).
Total copper length per phase:
Lphase=Nph×MLT=182.5×0.25=45.625 m.L_{phase} = N_{ph}\times \text{MLT} = 182.5\
times 0.25 = 45.625\ \text{m}.

Convert chosen conductor area to m²: A=2.5 mm2=2.5×10−6 m2.A = 2.5\ \text{mm}^2 = 2.5\
times 10^{-6}\ \text{m}^2.

DC resistance per phase:

Rphase=ρ⋅LphaseA=1.724×10−8×45.6252.5×10−6.R_{phase} = \rho\cdot\dfrac{L_{phase}}
{A} =1.724\times10^{-8}\times\dfrac{45.625}{2.5\times10^{-6}}.

Compute numerator: 1.724×10−8×45.625=7.8715×10−71.724\times10^{-8}\

times45.625=7.8715\times10^{-7}.
Divide by 2.5×10−62.5\times10^{-6}: 7.8715×10−7/2.5×10−6=0.31486 Ω.7.8715\times10^{-
7} / 2.5\times10^{-6} = 0.31486\ \Omega.

So Rphase≈0.315 ΩR_{phase}\approx 0.315\ \Omega (DC resistance, 20°C).

9. Copper loss
Phase current ≈9.417 A, so

Pcu,phase=I2Rphase=9.4172×0.31486.P_{cu,phase} = I^2 R_{phase} = 9.417^2\times 0.31486.

Compute 9.4172=88.6949.417^2 = 88.694. Multiply: 88.694×0.31486=27.93 W88.694\times

0.31486 = 27.93\ \text{W} per phase.
Total copper loss (3 phases) ≈ 3×27.93=83.79 W.3\times 27.93 = 83.79\ \text{W}.

(These are estimates — in practice you’ll adjust turns/wire size to target copper loss, slot fill,
thermal limits.)

4) Winding layout & practical notes

 Slot/coil selection: choose number of slots such that winding is reasonably distributed
(e.g., 36, 42, 48 slots), and coil span (short-pitch or full-pitch) to control harmonics.
 Winding factor kwk_w: product of pitch and distribution factors; typical 0.9–0.98.
Using kwk_w in the EMF equation adjusts the turns computed above.
 Parallel paths: If you put parallel conductor paths in a phase (to reduce resistance),
reduce turns per coil accordingly. For example, 2 parallel paths halves turns per
conductor string and halves conductor area per string (or double cross section).
 Thermal & insulation: enamel thickness, slot insulation, thermal class of insulation
determine allowable current density — adjust J accordingly (2–6 A/mm² common range:
lower J for higher temp rating).
  Skin/strand: for high frequency or harmonic concerns or to ease bending, use Litz or
multiple strands; but typical line freq ≤ 60 Hz uses solid enamelled conductors.

5) Per-phase equivalent circuit (up to

“circuit”)
For an induction motor the standard per-phase equivalent (referred to stator) is:

----R1----jX1----+----jXm----+
| |
| Rm (core loss)
| |
+----(R2'/s + jX2')--- to rotor branch

More commonly drawn as:

 R1 = stator resistance (ohmic, we estimated ~0.315 Ω/phase DC)

 X1 = stator leakage reactance (depends on geometry; compute from leakage inductance)
 Magnetizing branch: Xm (magnetizing reactance) in parallel with core loss resistor Rm
 Rotor referred: R2′/s (where s is slip) in series with X2′

How to use this circuit:

 At startup (s≈1) rotor resistance referred is R2′/1 -> produce starting torque.
 At near-sync (s small) the rotor branch behaves differently; the torque and current follow
standard induction motor formulas.
 You estimate X1, X2′, Xm using either measured tests (no-load & locked-rotor) or from
magnetic geometry (complex).

6) Quick checklist you can run through for

any motor design
1. Specify rating: P, V, f, poles, connection (Y/Δ).
2. Choose Bpeak and Ac → compute Φ.
3. Choose kw and compute Nph = Vph/(4.44 f Φ kw).
4. Decide slots, coils/phase → turns per coil = Nph / (coils/phase).
5. Choose J → get conductor area → choose nearest standard wire size.
6. Compute MLT and Rphase (ρ L / A).
7. Compute copper loss and check thermal limit /slot fill.
 8. If copper loss too high → increase conductor area (parallel conductors or larger gauge) or
reduce number of turns (but must keep voltage).
9. Draw winding diagram and assign slot numbers and phasing (A, B, C).
10. Model equivalent circuit (R1, X1, Xm, R2′, X2′) and perform performance checks
(starting current/torque, full load slip, efficiency).

7) If you want exact winding diagram & slot

mapping
Tell me:

 exact motor frame/stator bore size or number of slots you want (common choices: 12, 18,
24, 36, 48)
 whether you want a concentrated or distributed winding
 star (Y) or delta connection requirement
With that I’ll produce: slot-by-slot coil assignment (e.g., slot 1: +30 turns A+, slot 7: −30
turns A−, etc.), coil span, and a neat ASCII wiring map or a downloadable image.

8) Want me to adapt this to your case?

If you have a particular motor rating (e.g., 1.5 kW, 220 V, 60 Hz, 6-pole) or a target number of
slots, tell me the numbers and I’ll redo these calculations specifically for that motor and produce:

 concrete turns per coil, turns table, recommended wire gauge, estimated R and copper
loss, and a per-slot winding map right now.

Which motor (rating / voltage / poles / slots) should I compute next — or shall I produce a slot-
by-slot winding diagram for the 36-slot example I used above?