Lecture 5B: Equivalent Circuit & Power Flow | Polyphase Induction Machines

RECAP

Key Results from Lecture 5A

✓Established Results

Rotating field at synchronous speed: n_s = 120f/P
Slip: s = (n_s−n_r)/n_s, rotor frequency f_r = s·f
Rotor speed: ω_r = ω_s(1−s)
Rotor must slip to sustain torque
Typical rated slip: 0.5%–8%

Today's Objectives

Transformer analogy & turns-ratio referral
Per-phase equivalent circuit (full & simplified)
Physical meaning of every circuit element
The critical R_r′/s power split
Complete power flow analysis & torque expression
Efficiency & power factor vs. load

SECTION 01

Why Do We Need an Equivalent Circuit?

🎯Objectives

Predict torque, current, PF, efficiency at any operating point — no prototype needed
Provide the mathematical basis for drive control design
Enable parameter extraction from standard motor tests (Lecture 5C)

What the Model Gives Us

Quantity	From Circuit
Stator current I_s	KVL
Rotor current I_r′	KVL
Input power P_in	3V_sI_scosφ
Air-gap power P_ag	3I_r′²R_r′/s
Torque T_em	P_ag/ω_s
Efficiency η	P_out/P_in

📋 Scope: The per-phase steady-state equivalent circuit assumes balanced 3-phase supply, symmetrical machine, constant rotor speed, and sinusoidal air-gap flux distribution.

SECTION 02

The Transformer Analogy

Table 1 — Transformer vs. Induction Motor
Transformer	Induction Motor
Primary winding	Stator winding
Secondary winding	Rotor winding
Magnetic core	Air-gap flux path
Open/loaded secondary	Short-circuited rotor
Fixed secondary	Rotating secondary
Fixed frequency	Slip frequency f_r = sf

⚡

The Crucial Difference

The rotating secondary introduces a slip-dependent rotor impedance

This feature — the slip-dependent impedance R_r′/s — separates induction motor analysis from ordinary transformer analysis and is the key to understanding all operating characteristics.

Transformer analogy diagram showing stator as primary and rotor as rotating secondary winding — Transformer analogy: stator (primary) vs. rotor (rotating short-circuited secondary).

SECTION 03

Turns-Ratio Referral — Rotor to Stator

Stator and rotor have different numbers of turns. To place them in one circuit, rotor quantities are scaled to the stator side using the effective turns ratio:

Turns Ratio & Referral Equations

a = N_s / N_r

R_r′ = a² R_r    (referred rotor resistance)
L_r′ = a² L_r    (referred rotor leakage inductance)
X_r′ = a² X_r    (referred rotor leakage reactance)
I_r′ = I_r / a    (referred rotor current)
E_r′ = a E_r    (referred rotor voltage)

🔑 Prime (′) Notation: All rotor quantities in the equivalent circuit (R_r′, X_r′, I_r′) are referred to the stator side. After referral: E_ag = E_r′, allowing physical isolation between stator and rotor circuits to be removed.

SECTION 04

Slip-Dependent Rotor Impedance

At rotor frequency f_r = sf, the per-phase rotor voltage equation is:

Rotor Circuit at Slip Frequency → Referred Impedance

s·E_ag = I_r(R_r + j·s·X_r)

Dividing by s: E_ag = I_r(R_r/s + jX_r)

After referral: E_ag = I_r′ (R_r′/s + jX_r′)

Table 2 — Impedance vs. Slip
Slip s	R_r′/s	Behaviour
s = 1 (standstill)	R_r′	High current
s = 0.05 (rated)	20 R_r′	Moderate current
s → 0 (synchronous)	→ ∞	No current, no torque

💡 Key Insight: X_r′ is evaluated at stator frequency f. Slip appears only in the resistance term R_r′/s. As the motor accelerates (slip ↓), R_r′/s ↑ — rotor current naturally decreases even at constant supply voltage.

SECTION 05

Full Per-Phase Equivalent Circuit

Fig. 1 — Full per-phase equivalent circuit: stator (R_s, jX_s), magnetising branch (R_c ∥ jX_m), rotor (R_r′/s, jX_r′).

SECTION 06

Physical Meaning of Circuit Elements

Table 3 — Circuit Element Reference
Region	Symbol	Name	Physical Meaning
Stator	R_s	Stator resistance	Copper losses in stator windings: P_s,Cu = 3I_s²R_s
Stator	jX_s	Stator leakage reactance	Flux linking only the stator; X_s = ω_eL_ls
Magnetising	R_c	Core-loss resistance	Iron losses (hysteresis + eddy current); P_core = 3E_ag²/R_c
Magnetising	jX_m	Magnetising reactance	Main mutual flux; I_m ≈ 25–40% I_rated
Rotor	R_r′/s	Referred rotor resistance	Key element — contains both rotor Cu loss and mechanical output power
Rotor	jX_r′	Referred rotor leakage	Flux linking only rotor; X_r′ = ω_eL_lr′; evaluated at stator frequency

📌 Note on R_c: R_c is very large and absorbs negligible current. It is often neglected (lumped into no-load losses) or treated as a constant fixed loss. The simplified circuit retains only jX_m.

SECTION 07

The Critical Power Split of R_r′/s

Resistance Split: Copper Loss + Mechanical Output

R_r′/s = R_r′ + R_r′·(1−s)/s

← Rotor Cu loss → ← Mechanical output →

🔥R_r′ — Real Resistor

Heat dissipated in rotor bars. This is an actual physical resistor and represents true copper loss.

Rotor Copper Loss

P_r,Cu = 3I_r′² R_r′ = s·P_ag

⚙️R_r′(1−s)/s — Fictitious Resistor

Circuit representation of shaft work. Not a physical component — it is the circuit proxy for mechanical work done on the load.

Mechanical Output

P_conv = 3I_r′² R_r′(1−s)/s = (1−s)P_ag

SECTION 08

Simplified Per-Phase Equivalent Circuit

In most calculations R_c is very large and absorbs negligible current. The magnetising branch reduces to a single shunt element jX_m.

KCL at the Air-Gap Node

I⃗_s = I⃗_m + I⃗_r′

Simplified per-phase equivalent circuit of the induction motor with core-loss resistance R_c neglected — Fig. 3 — Simplified equivalent circuit (R_c neglected); most commonly used in practice.

Typical Current Magnitudes

I_m ≈ 25–40% of I_rated
I_r′ ≈ 90–100% of I_rated at full load

SECTION 09

Phasor Diagram

📐Step-by-Step Construction

Take E_ag as the reference phasor
I_m lags E_ag by 90° (purely inductive jX_m)
I_r′ lags E_ag by θ_r = arctan(sX_r′/R_r′) — small at rated slip
I⃗_s = I⃗_m + I⃗_r′ (phasor sum)
V⃗_s = E⃗_ag + I⃗_s(R_s + jX_s)

φ φ = angle between V_s and I_s = power factor angle

Phasor diagram of the induction motor showing stator voltage, rotor current, magnetising current, and power factor angle — Fig. 4 — Phasor diagram at rated load: V_s, I_s, I_m, I_r′ and power factor angle φ.

Why Power Factor is Always Lagging

I_m is always 90° lagging regardless of load. Even at full load, I_s carries a substantial reactive component — the "magnetising current penalty."

Condition	PF Angle φ	Power Factor	Notes
No-load	80°–85°	0.1–0.3	I_s ≈ I_m; purely reactive
Rated load	25°–40°	0.75–0.90	I_r′ significant; I_m still present

SECTION 10

Power Flow Analysis — The Complete Chain

Power flow chain diagram from three-phase electrical input through stator, air gap, and rotor to mechanical shaft output — Fig. 5 — Power flow chain: P_in → P_ag → P_conv → P_out.

PInput Power (3-phase)

P_in = 3 V_s I_s cosφ

CStator Copper Loss

P_s,Cu = 3 I_s² R_s

AAir-Gap Power

P_ag = P_in − P_s,Cu − P_core = 3 I_r′² R_r′/s

RRotor Copper Loss

P_r,Cu = 3 I_r′² R_r′ = s · P_ag

MConverted Power

P_conv = (1−s) P_ag

OShaft Output Power

P_out = P_conv − P_fw

P_fw = friction & windage losses

⚖️

Fundamental Power Ratio

P_ag : P_r,Cu : P_conv = 1 : s : (1−s)

At s = 0.03: only 3% of air-gap power is lost as rotor heat. Large motors use very small rated slip (0.5–2%) to achieve high efficiency — at s = 0.01, only 1% of air-gap power is wasted.

Graph of air-gap power, rotor copper loss, and converted power fractions versus slip for an induction motor — Fig. 6 — Power ratio vs. slip: P_r,Cu/P_ag = s and P_conv/P_ag = 1−s.

SECTION 11

Electromagnetic Torque

Torque–Power Linkage

P_conv = T_em · ω_r = T_em · ω_s(1−s)
P_ag = T_em · ω_s

T_em = P_ag / ω_s = 3 I_r′² R_r′ / (s · ω_s)

Table 4 — Three Equivalent Forms of Torque
Approach	Formula
From P_ag	T_em = P_ag / ω_s
From P_conv	T_em = P_conv / ω_r
From circuit	T_em = 3I_r′² R_r′ / (s·ω_s)

📌 Torque is set by air-gap power and synchronous (not rotor) speed. At constant V_s and ω_s, torque is controlled via slip (which controls I_r′). This is the foundation for the torque-slip equation in Lecture 5C.

SECTION 12

Efficiency & Power Factor vs. Load

ηEfficiency

η = P_out/P_in = (P_in − ΣP_loss)/P_in

Variable losses: P_s,Cu, P_r,Cu (rise with load)
Fixed losses: P_core, P_fw (approx. constant)

φPower Factor

PF = cosφ = P_in / (3 V_s I_s)

Always lagging due to I_m component. Poor at light loads.

Table 5 — Typical Values at Rated Load
Motor Size	Efficiency η	Power Factor
Small (<5 kW)	75–88%	0.70–0.80
Medium (5–100 kW)	88–94%	0.80–0.87
Large (>100 kW)	93–97%	0.85–0.92

At Light Loads — Both η and PF Degrade Significantly

Fixed losses dominate P_in
I_s ≈ I_m ⇒ very low PF
Running oversized motors at light load wastes energy
Peak efficiency occurs at approximately 75% rated load where variable losses equal fixed losses

Graph of efficiency and power factor versus per-unit load for a typical polyphase induction motor — Fig. 7 — Efficiency η and power factor cosφ vs. load; peak η ≈ 75% rated load.

SECTION 13

Lecture Summary

1. Transformer Analogy

Stator = primary; Rotor = short-circuited rotating secondary
Referred quantities: R_r′ = a²R_r, X_r′ = a²X_r, a = N_s/N_r

2. Equivalent Circuit

Full: R_s, jX_s, R_c∥jX_m, R_r′/s, jX_r′
Simplified: drop R_c; use jX_m shunt only

3. Power Split of R_r′/s

R_r′/s = R_r′ + R_r′(1−s)/s (Cu loss + mechanical output)

4. Power Flow

P_ag : P_r,Cu : P_conv = 1 : s : (1−s)
T_em = P_ag/ω_s = 3I_r′²R_r′/(s·ω_s)

5. Phasor Diagram

I_m always 90° lagging ⇒ PF always lagging
No-load: I_s ≈ I_m, very low PF
Rated load: PF ≈ 0.75–0.90

Next: Lecture 5C

Torque-slip characteristics via Thévenin equivalent, maximum torque analysis, NEMA design classes, and parameter measurement from standard motor tests.