PracticeVideosBlogFlashcards

16.3 Three-Phase & Single-Phase Induction Motors

Key Takeaways

  • Fractional slip s=ns−nrnss = \frac{n_s - n_r}{n_s} dictates induction motor operation, where rotor frequency is fr=sff_r = s f, standstill induced EMF E2E_2 scales to sE2s E_2, and standstill rotor reactance X2X_2 scales to sX2s X_2.

  • Power flow and loss distribution follow the fundamental ratio Pag:Pcu2:Pmech=1:s:(1−s)P_{\text{ag}} : P_{\text{cu2}} : P_{\text{mech}} = 1 : s : (1-s), where air-gap power is Pag=3I2′2R2′sP_{\text{ag}} = 3 I_2'^2 \frac{R_2'}{s}, rotor copper loss is Pcu2=sPagP_{\text{cu2}} = s P_{\text{ag}}, and converted mechanical power is Pmech=(1−s)PagP_{\text{mech}} = (1-s) P_{\text{ag}}.

  • Developed electromagnetic torque equation is Te=3ωsV12(R2′/s)(R1+R2′/s)2+(X1+X2′)2T_e = \frac{3}{\omega_s} \frac{V_1^2 (R_2' / s)}{(R_1 + R_2'/s)^2 + (X_1 + X_2')^2}. Peak breakdown torque Tmax≈32ωsV122(X1+X2′)T_{\text{max}} \approx \frac{3}{2 \omega_s} \frac{V_1^2}{2(X_1 + X_2')} is independent of rotor resistance R2′R_2', but the slip at maximum torque smax=R2′R12+(X1+X2′)2s_{\text{max}} = \frac{R_2'}{\sqrt{R_1^2 + (X_1 + X_2')^2}} scales directly with R2′R_2'.

  • Starting methods for 3-phase motors include Direct-on-line (DOL), Star-Delta (reducing line current and starting torque to 13\frac{1}{3} of DOL values), Autotransformer (scaling current and torque by k2k^2), Soft Starter, and Rotor Resistance (Wound-Rotor).

  • Single-phase induction motors lack inherent starting torque per double-revolving field theory (equal forward and backward torques at s=1s=1). Starting is enabled via auxiliary windings or phase-shifting capacitors (split-phase, capacitor-start, capacitor-run, and shaded-pole configurations).

Last updated: August 2026

16.3 Three-Phase & Single-Phase Induction Motors

Operating Principles & Slip Mechanics

The three-phase induction motor is the most widely utilized electric motor in modern industrial plants due to its rugged self-starting construction, high reliability, and low cost. It operates on the principle of electromagnetic induction without requiring electrical connections to the rotor.

Creation of Stator Rotating Magnetic Field (RMF)

When balanced 3-phase AC currents flow through stator windings physically displaced by 120∘120^\circ electrical around the stator frame, they establish a rotating magnetic field of constant magnitude Bnet=1.5BmB_{\text{net}} = 1.5 B_m rotating at synchronous speed:

ns=120fP[rpm]n_s = \frac{120 f}{P} \quad [\text{rpm}]

Induction of Rotor Current & Slip Definition

The relative motion between the stator RMF (nsn_s) and the stationary rotor conductors (nr=0n_r = 0) cuts rotor bars, inducing an electromotive force (e2e_2) by Faraday's law. Because rotor bars are short-circuited by end-rings, heavy rotor currents (i2i_2) flow. Interaction between i2i_2 and the RMF produces electromagnetic torque (TeT_e) accelerating the rotor in the direction of the RMF (Lenz's law).

The fractional difference between synchronous speed nsn_s and actual rotor speed nrn_r is defined as slip (ss):

s=ns−nrns  ⟹  nr=(1−s)nss = \frac{n_s - n_r}{n_s} \quad \implies \quad n_r = (1 - s) n_s
  • At standstill (nr=0n_r = 0): Slip s=1.0s = 1.0 (100%100\% slip).
  • At synchronous speed (nr=nsn_r = n_s): Slip s=0s = 0. (Flux cutting ceases, e2=0,i2=0,Te=0e_2 = 0, i_2 = 0, T_e = 0; hence an induction motor can never reach nsn_s).
  • Rotor frequency: fr=sff_r = s f.
  • Running induced rotor EMF: E2r=sE2E_{2r} = s E_2.
  • Running rotor reactance: X2r=sX2X_{2r} = s X_2.

Per-Phase Equivalent Circuit & Power Flow Balance

To analyze performance, the rotor circuit operating at frequency frf_r is transformed to the primary stator frequency ff by dividing the rotor parameters by ss:

I2′=E2′(R2′s)2+(X2′)2\mathbf{I}_2' = \frac{E_2'}{\sqrt{\left(\frac{R_2'}{s}\right)^2 + (X_2')^2}}

The total equivalent rotor resistance is split into two components:

R2′s=R2′+(1−ss)R2′\frac{R_2'}{s} = R_2' + \left( \frac{1 - s}{s} \right) R_2'
  • R2′R_2' represents actual physical rotor winding I2RI^2 R copper losses.
  • (1−ss)R2′\left( \frac{1 - s}{s} \right) R_2' represents the electrical equivalent of the mechanical shaft load.
                  INDUCTION MOTOR PER-PHASE EQUIVALENT CIRCUIT

        I1 ───►   R1         X1                  R'2        X'2       ───► I'2
     ───────────█████──────UUUUUU───────┬───────█████──────UUUUUU──────────────┐
                                        │                                      │
                                   I0   │                                     ███  [(1-s)/s] R'2
                                    ┌───┴───┐                                 ███  (Mechanical
                                 Ic │       │ Im                               │    Load)
                                   ███     UUU                                 │
                                Rc ███  Xm UUU                                 │
                                    │       │                                  │
     ───────────────────────────────┴───────┴──────────────────────────────────┴┘

Stages of Power Flow in 3-Phase Induction Motors

  1. Stator Electrical Input Power: Pin=3VLILcos⁡θP_{\text{in}} = \sqrt{3} V_L I_L \cos \theta
  2. Stator Losses: Stator Copper Loss Pcu1=3I12R1P_{\text{cu1}} = 3 I_1^2 R_1, Stator Core Loss PiP_i
  3. Air-Gap Power (PagP_{\text{ag}}): Power transferred electromagnetically from stator to rotor across the air gap: Pag=Pin−Pcu1−Pi=3(I2′)2R2′sP_{\text{ag}} = P_{\text{in}} - P_{\text{cu1}} - P_i = 3 (I_2')^2 \frac{R_2'}{s}
  4. Rotor Copper Loss (Pcu2P_{\text{cu2}}): Internal ohmic heating in rotor conductors: Pcu2=3(I2′)2R2′=sPagP_{\text{cu2}} = 3 (I_2')^2 R_2' = s P_{\text{ag}}
  5. Gross Mechanical Power Developed (PmechP_{\text{mech}}): Power converted into rotational mechanical form: Pmech=Pag−Pcu2=(1−s)Pag=3(I2′)2R2′(1−ss)P_{\text{mech}} = P_{\text{ag}} - P_{\text{cu2}} = (1 - s) P_{\text{ag}} = 3 (I_2')^2 R_2' \left( \frac{1 - s}{s} \right)
  6. Net Shaft Power (PshaftP_{\text{shaft}}): Output power available at the mechanical shaft coupling: Pshaft=Pmech−Protational(Protational=Pfriction+Pwindage+Pstray)P_{\text{shaft}} = P_{\text{mech}} - P_{\text{rotational}} \quad (P_{\text{rotational}} = P_{\text{friction}} + P_{\text{windage}} + P_{\text{stray}})

Fundamental Power Ratio Mandate: Pag:Pcu2:Pmech=1:s:(1−s)P_{\text{ag}} : P_{\text{cu2}} : P_{\text{mech}} = 1 : s : (1 - s) Rotor Conversion Efficiency: ηrotor=PmechPag=1−s\eta_{\text{rotor}} = \frac{P_{\text{mech}}}{P_{\text{ag}}} = 1 - s


Torque-Slip Characteristics & NEMA Motor Design Classes

The developed electromagnetic torque as a function of slip ss is:

Te=Pagωs=3ωs[V12(R2′s)(R1+R2′s)2+(X1+X2′)2]T_e = \frac{P_{\text{ag}}}{\omega_s} = \frac{3}{\omega_s} \left[ \frac{V_1^2 \left(\frac{R_2'}{s}\right)}{\left(R_1 + \frac{R_2'}{s}\right)^2 + (X_1 + X_2')^2} \right]
                    3-PHASE INDUCTION MOTOR TORQUE-SLIP CURVE

    Torque T ▲
             │                   Breakdown / Peak Torque Tmax
             │                        ┌───────┐
             │                       ╱         ╲
             │                      ╱           ╲
             │                     ╱             ╲
   Tstart ───┼─► █                ╱               ╲
             │   █               ╱                 ╲   Low-Slip Normal
             │   █              ╱                   ╲  Operating Region (Linear)
             │   █             ╱                     ───────────
             └───┴────────────┴──────────────────────────┴──────────────► Slip s
             s=1.0          s = s_max                  s=0
          (Standstill)    (Breakdown Slip)         (Synchronous Speed)

Peak Breakdown (Pull-Out) Torque (TmaxT_{\text{max}})

By differentiation dTeds=0\frac{dT_e}{ds} = 0, maximum torque and its corresponding breakdown slip smaxs_{\text{max}} are:

smax=R2′R12+(X1+X2′)2≈R2′X1+X2′s_{\text{max}} = \frac{R_2'}{\sqrt{R_1^2 + (X_1 + X_2')^2}} \approx \frac{R_2'}{X_1 + X_2'} Tmax=32ωs[V12R1+R12+(X1+X2′)2]≈32ωs[V122(X1+X2′)]T_{\text{max}} = \frac{3}{2 \omega_s} \left[ \frac{V_1^2}{R_1 + \sqrt{R_1^2 + (X_1 + X_2')^2}} \right] \approx \frac{3}{2 \omega_s} \left[ \frac{V_1^2}{2(X_1 + X_2')} \right]

Critical Engineering Principle: Peak breakdown torque magnitude TmaxT_{\text{max}} is completely independent of rotor resistance R2′R_2'. However, the slip smaxs_{\text{max}} at which peak torque occurs is directly proportional to R2′R_2'. Increasing rotor resistance shifts the peak torque toward standstill (s=1.0s = 1.0), augmenting starting torque without altering maximum torque capacity.

NEMA Motor Design Classifications

NEMA ClassStarting Torque (% Rated)Starting Current (% Rated)Breakdown Torque (% Rated)Full-Load SlipRotor Construction & Target Applications
Class A150%−170%150\% - 170\%High (500%−800%500\% - 800\%)200%−250%200\% - 250\%Low (<5%< 5\%)Standard single cage; machine tools, blowers, centrifugal pumps
Class B130%−160%130\% - 160\%Medium (400%−500%400\% - 500\%)200%−250%200\% - 250\%Low (<5%< 5\%)Deep-bar rotor (skin effect increases start R2R_2); Standard Industrial Workhorse
Class CHigh (200%−250%200\% - 250\%)Medium (400%−500%400\% - 500\%)190%−225%190\% - 225\%Low (<5%< 5\%)Double-cage rotor (outer high-RR bar for start, inner high-XX bar for run); crushers, conveyors
Class DExtra High (275%+275\%+)Low (300%−400%300\% - 400\%)275%+275\%+ at s=1s=1High (5%−13%5\% - 13\%)High-resistance brass rotor bars; punch presses, shears, hoists, elevators

3-Phase Motor Starting Methods

Across starting, locked-rotor currents reach 500%−800%500\% - 800\% of rated full-load current. Standard starting schemes reduce line voltage or adjust impedance:

  1. Direct-On-Line (DOL) Starting: 100%100\% full voltage applied. Max starting torque (100%Tstart,DOL100\% T_{\text{start,DOL}}), but severe inrush current.
  2. Star-Delta (Y-Δ\text{Y-}\Delta) Starter: Motor windings are connected in Star during starting, then switched to Delta during running. Vphase,Y=Vline3  ⟹  Iline,start,Y=13Iline,start,DOL,Tstart,Y=13Tstart,DOLV_{\text{phase,Y}} = \frac{V_{\text{line}}}{\sqrt{3}} \quad \implies \quad I_{\text{line,start,Y}} = \frac{1}{3} I_{\text{line,start,DOL}}, \quad T_{\text{start,Y}} = \frac{1}{3} T_{\text{start,DOL}}
  3. Autotransformer Starter: Taps provide reduced fraction kk (e.g. k=0.80k = 0.80 or 80%80\%) of line voltage. Imotor,start=kIDOL,Iline,start=k2IDOL,Tstart=k2Tstart,DOLI_{\text{motor,start}} = k I_{\text{DOL}}, \quad I_{\text{line,start}} = k^2 I_{\text{DOL}}, \quad T_{\text{start}} = k^2 T_{\text{start,DOL}}
  4. Rotor Resistance Starting (Wound-Rotor Motors): External resistors connected to rotor slip rings add Rext′R_{\text{ext}}' to rotor circuit, setting smax=1.0s_{\text{max}} = 1.0 to achieve Tstart=TmaxT_{\text{start}} = T_{\text{max}} while maintaining low starting current.

Single-Phase Induction Motors & Starting Topologies

Single-phase induction motors are widely used in home appliances, HVAC, and commercial equipment where 3-phase power is unavailable.

Double-Revolving Field Theory

A single-phase pulsating magnetic flux B(t)=Bmcos⁡(ωt)B(t) = B_m \cos(\omega t) does not rotate naturally. According to Ferrari's double-revolving field theory, a pulsating field resolves into two counter-rotating fields of equal magnitude Bf=Bb=Bm2B_f = B_b = \frac{B_m}{2} rotating at +ns+n_s and −ns-n_s.

sf=s,sb=2−ss_f = s, \quad s_b = 2 - s

At standstill (nr=0,s=1.0n_r = 0, s = 1.0), forward slip sf=1.0s_f = 1.0 and backward slip sb=1.0s_b = 1.0. The forward torque and backward torque are exactly equal and opposite (Tf=TbT_f = T_b), resulting in zero net starting torque (Tstart=0T_{\text{start}} = 0).

                  SINGLE-PHASE INDUCTION MOTOR STARTING TOPOLOGIES

   (A) Split-Phase               (B) Capacitor-Start            (C) Permanent-Split Capacitor
     Main Winding                  Main Winding                    Main Winding
   ┌───███████───┐               ┌───███████───┐                 ┌───███████───┐
   │             │               │             │                 │             │
 L ├─████─┬─  ──┤ N            L ├─████─┬─  ──┤ N              L ├─████─┬─  ──┤ N
   │ Aux  │ Centrifugal          │ Aux  │ Centrifugal            │ Aux  │
   │ Wind └─o/ o─┘               │ Wind ├──┤├───o/ o─┘           │ Wind ├──┤├──┐
   │                             │      │ C_start                │      │ C │
   └─────────────────────────────┴──────┴────────────────────────┴──────┴───┴──┘

Single-Phase Starting Methods

  1. Split-Phase Motor: Aux winding (high resistance, low reactance) in parallel with main winding (low resistance, high reactance) creates a ∼30∘\sim 30^\circ phase shift. Centrifugal switch opens aux winding at 75%75\% speed. Moderate starting torque.
  2. Capacitor-Start Motor: Heavy-duty electrolytic capacitor in series with aux winding creates a near 90∘90^\circ phase shift. Centrifugal switch disconnects aux circuit at 75%75\% speed. High starting torque (300%−450%300\% - 450\%).
  3. Permanent-Split Capacitor (PSC) Motor: Run capacitor remains permanently in series with aux winding. No centrifugal switch. Quiet operation, low starting torque, high running power factor.
  4. Two-Value (Capacitor-Start, Capacitor-Run) Motor: Employs large start capacitor for initial high torque and small run capacitor for high running efficiency.
  5. Shaded-Pole Motor: Salient poles fitted with a copper shorting ring (shading coil) across part of each pole face. Delays flux in shaded segment, creating a weak rotating field. Lowest starting torque and efficiency (20%−40%20\% - 40\%); used in small desk fans and blowers.
Loading diagram...
Power Flow and Loss Conversion Stages in a Three-Phase Induction Motor
NEMA Class B Induction Motor Developed Torque (% Rated Torque) vs Slip s
Test Your Knowledge

A 460 V, 60 Hz, 4-pole, wye-connected 3-phase induction motor runs at 1728 rpm while taking 48 kW from the line. Total stator losses equal 2.4 kW, and total mechanical friction and windage rotational loss equals 1.2 kW. What are the percentage slip, air-gap power Pag, rotor copper loss Pcu2, output shaft power in horsepower (hp), and overall motor efficiency?

A

s = 5.0%, Pag = 48.0 kW, Pcu2 = 2.400 kW, 54.20 hp, η = 84.2%

B

s = 4.0%, Pag = 45.6 kW, Pcu2 = 1.200 kW, 59.50 hp, η = 92.5%

C

s = 4.0%, Pag = 45.6 kW, Pcu2 = 1.824 kW, 57.07 hp, η = 88.7%

D

s = 3.5%, Pag = 46.2 kW, Pcu2 = 1.617 kW, 58.10 hp, η = 90.1%

Test Your Knowledge

A 480 V, 60 Hz, 6-pole wound-rotor 3-phase induction motor has a standstill rotor resistance per phase of R2' = 0.20 Ω and a standstill rotor reactance of X2' = 1.0 Ω. Stator impedance is negligible. What is the normal slip at maximum breakdown torque, and what external resistance per phase Rext' must be inserted into the rotor circuit to develop maximum breakdown torque right at starting (s = 1.0)?

A

s_max = 10%, R_ext = 0.90 Ω

B

s_max = 25%, R_ext = 0.75 Ω

C

s_max = 20%, R_ext = 1.00 Ω

D

s_max = 20%, R_ext = 0.80 Ω

Test Your Knowledge

A 400 V, 60 Hz, delta-connected 3-phase squirrel-cage induction motor draws a line starting current of 300 A and produces a starting torque equal to 1.8 times full-load torque under Direct-On-Line (DOL) starting. If a Star-Delta (Y-Δ) starter is installed, what are the line starting current and starting torque under star starting?

A

100.0 A starting current, 0.60 T_fl starting torque

B

173.2 A starting current, 1.04 T_fl starting torque

C

150.0 A starting current, 0.90 T_fl starting torque

D

100.0 A starting current, 0.90 T_fl starting torque

Sections you finish are checked off in the contents.