15.1 Single-Phase & Three-Phase Transformers

Key Takeaways

  • The RMS induced EMF in a transformer winding is $E = 4.44 f N \Phi_m = 4.44 f N B_m A_c$, establishing that induced voltage is directly proportional to frequency, turn count, and maximum magnetic flux.
  • Equivalent circuit parameters referred to the primary side are $R_{e1} = R_1 + a^2 R_2$ and $X_{e1} = X_1 + a^2 X_2$, where the transformation turns ratio is $a = N_1 / N_2$.
  • Voltage regulation measures terminal voltage variation from no-load to full-load: zero regulation occurs at a leading power factor when $\tan \phi_2 = R_{e2} / X_{e2}$, while maximum regulation occurs at a lagging power factor when $\tan \phi_2 = X_{e2} / R_{e2}$.
  • Maximum operating efficiency occurs at the load condition where variable copper losses equal constant core losses ($P_{\text{cu}} = P_i$), with the load fraction given by $x_{\text{max}} = \sqrt{P_i / P_{\text{cu,fl}}}$.
  • Current Transformers (CTs) must never have their secondary circuits opened while the primary is energized, as high induced voltage ($V = L \frac{di}{dt}$) and core saturation create fatal electric shock and equipment destruction hazards.
Last updated: August 2026

15.1 Single-Phase & Three-Phase Transformers

Executive Summary & Fundamental Operating Principles

A transformer is a static electromagnetic device that transfers electrical energy between two or more coupled circuits through the medium of a time-varying magnetic field without any change in frequency. Transformers form the critical backbone of electric power transmission and distribution grids, enabling high-voltage transmission to minimize line losses ($I^2 R$) over long distances and low-voltage distribution for end-user safety.

The operation of a transformer relies fundamentally on Faraday's Law of Electromagnetic Induction and Lenz's Law. When a sinusoidal alternating voltage $v_1(t) = V_{1,m} \sin(\omega t)$ is applied to the primary winding of $N_1$ turns, it establishes a mutual alternating magnetic flux $\Phi(t) = \Phi_m \sin(\omega t)$ in the high-permeability ferromagnetic core. This alternating flux links both the primary winding and the secondary winding of $N_2$ turns.

According to Faraday's law, the instantaneous induced counter-EMF in the primary and induced EMF in the secondary are:

e1(t)=N1dΦdt=N1ddt[Φmsin(ωt)]=N1ωΦmcos(ωt)e_1(t) = -N_1 \frac{d\Phi}{dt} = -N_1 \frac{d}{dt} \left[ \Phi_m \sin(\omega t) \right] = -N_1 \omega \Phi_m \cos(\omega t)

e2(t)=N2dΦdt=N2ddt[Φmsin(ωt)]=N2ωΦmcos(ωt)e_2(t) = -N_2 \frac{d\Phi}{dt} = -N_2 \frac{d}{dt} \left[ \Phi_m \sin(\omega t) \right] = -N_2 \omega \Phi_m \cos(\omega t)

The maximum peak value of the induced EMF is $E_{m} = 2 \pi f N \Phi_m$. Consequently, the RMS Induced EMF Equation for both primary and secondary windings is derived as:

E1=2π2fN1Φm=4.44fN1Φm=4.44fN1BmAcE_1 = \frac{2 \pi}{\sqrt{2}} f N_1 \Phi_m = 4.44 f N_1 \Phi_m = 4.44 f N_1 B_m A_c

E2=2π2fN2Φm=4.44fN2Φm=4.44fN2BmAcE_2 = \frac{2 \pi}{\sqrt{2}} f N_2 \Phi_m = 4.44 f N_2 \Phi_m = 4.44 f N_2 B_m A_c

where:

  • $f$ = system frequency in Hertz (Hz)
  • $N_1, N_2$ = number of primary and secondary turns
  • $\Phi_m$ = maximum core flux in Webers (Wb)
  • $B_m$ = maximum core flux density in Tesla (T)
  • $A_c$ = effective magnetic cross-sectional core area in square meters (m$^2$)

Transformation Ratio ($a$)

The ratio of primary to secondary induced voltage defines the ideal turns ratio $a$:

a=N1N2=E1E2V1V2=I2I1a = \frac{N_1}{N_2} = \frac{E_1}{E_2} \approx \frac{V_1}{V_2} = \frac{I_2}{I_1}

  • If $a > 1$ ($N_1 > N_2$, $V_1 > V_2$), the transformer is a step-down transformer.
  • If $a < 1$ ($N_1 < N_2$, $V_1 < V_2$), the transformer is a step-up transformer.

Core Construction & Core Losses

To ensure maximum magnetic coupling ($k \approx 1$) and minimize reluctance, transformer cores are constructed from high-grade, cold-rolled grain-oriented (CRGO) silicon steel. Adding 3% to 4.5% silicon increases electrical resistivity, reducing stray eddy currents.

Core-Type vs. Shell-Type Topologies

Construction ParameterCore-Type TransformerShell-Type Transformer
Magnetic PathsSingle continuous magnetic circuit framing the windingsDouble magnetic circuit surrounding the windings (central limb)
Winding ArrangementConcentric windings divided on both vertical limbs (LV inside, HV outside)Sandwich or interleaved pancake windings placed on the central limb
Mechanical ProtectionExposed windings on outer limbsSurroundings shell protects internal coils against mechanical impact
Cooling & RepairSuperior natural air/oil cooling; easier coil maintenanceMore complex coil assembly; higher mechanical bracing against short-circuit forces
ApplicationHigh-voltage transmission and large power transformer applicationsLow-voltage, high-current industrial and distribution applications

Core Loss Breakdown ($P_i$)

No-load core loss (iron loss $P_i$) remains essentially constant from zero load to full load at constant rated voltage and frequency. It consists of two physical components:

Core Loss: Pi=Ph+Pe\text{Core Loss: } P_i = P_h + P_e

  1. Hysteresis Loss ($P_h$): Caused by continuous reversal of magnetic domains in the ferromagnetic material during each AC cycle. Governed by Steinmetz's empirical formula: Ph=khfBmn(where n1.6 for silicon steel)P_h = k_h f B_m^n \quad \text{(where } n \approx 1.6 \text{ for silicon steel)}
  2. Eddy Current Loss ($P_e$): Induced circulation currents in the conductive core body driven by alternating flux $\frac{d\Phi}{dt}$. Governed by: Pe=kef2Bm2t2P_e = k_e f^2 B_m^2 t^2 where $t$ is the individual lamination thickness (typically $0.35\text{ mm}$ to $0.50\text{ mm}$). Laminating the core and insulating adjacent sheets with thin lacquer or varnish restricts eddy current loops, dramatically reducing thermal losses ($P_e \propto t^2$).

Practical Transformer Equivalent Circuit & Test Procedures

An ideal transformer is lossless and infinitely permeable. A real transformer exhibits primary and secondary winding resistances ($R_1, R_2$), leakage reactances ($X_1, X_2$), core hysteresis/eddy losses ($R_c$), and finite core permeability ($X_m$).

                  PRACTICAL TRANSFORMER EQUIVALENT CIRCUIT (PRIMARY REFERRED)
        I1 ───►   R1         X1                  R'2        X'2       ───► I'2
     ───────────█████──────UUUUUU───────┬───────█████──────UUUUUU──────────────┐
                                        │                                      │
                                   I0   │                                      │
                                    ┌───┴───┐                                 ███  R_Load
                                 Ic │       │ Im                              ███
                                   ███     UUU                                 │
                                Rc ███  Xm UUU                                 │
                                    │       │                                  │
     ───────────────────────────────┴───────┴──────────────────────────────────┴┘

Reflection of Impedances Across Turns Ratio ($a$)

To simplify circuit calculations, secondary parameters are transferred to the primary side:

R2=a2R2,X2=a2X2,ZL=a2ZL,V2=aV2,I2=I2aR_2' = a^2 R_2, \quad X_2' = a^2 X_2, \quad Z_L' = a^2 Z_L, \quad V_2' = a V_2, \quad I_2' = \frac{I_2}{a}

Primary-Referred Equivalent Parameters:

Re1=R1+R2=R1+a2R2R_{e1} = R_1 + R_2' = R_1 + a^2 R_2 Xe1=X1+X2=X1+a2X2X_{e1} = X_1 + X_2' = X_1 + a^2 X_2 Ze1=Re12+Xe12Z_{e1} = \sqrt{R_{e1}^2 + X_{e1}^2}

Secondary-Referred Equivalent Parameters:

Re2=R2+R1a2,Xe2=X2+X1a2,Ze2=Re22+Xe22R_{e2} = R_2 + \frac{R_1}{a^2}, \quad X_{e2} = X_2 + \frac{X_1}{a^2}, \quad Z_{e2} = \sqrt{R_{e2}^2 + X_{e2}^2}

Open-Circuit (OC) and Short-Circuit (SC) Standard Tests

Transformer equivalent circuit parameters are experimentally determined using two standardized non-destructive laboratory tests:

Test ParameterOpen-Circuit (OC) TestShort-Circuit (SC) Test
Energized WindingLow-Voltage (LV) Side (HV left open)High-Voltage (HV) Side (LV shorted with thick busbar)
Applied VoltageRated Nominal Voltage ($V_{1,\text{rated}}$)Reduced Voltage ($V_{\text{sc}} \approx 5% - 10%$ of rated $V_1$)
Current FlowingSmall No-Load Current ($I_0 \approx 2% - 5%$ of rated $I$)Rated Full-Load Current ($I_{\text{sc}} = I_{\text{fl}}$)
Primary Measurement$V_0, I_0, P_0$ (Wattmeter reading)$V_{\text{sc}}, I_{\text{sc}}, P_{\text{sc}}$ (Wattmeter reading)
Extracted ParametersCore loss $P_i = P_0$; Excitation branch $R_c, X_m$Full-load copper loss $P_{\text{cu,fl}} = P_{\text{sc}}$; Series impedance $R_{eq}, X_{eq}, Z_{eq}$

Calculation Steps from Test Data:

  1. From OC Test ($V_0, I_0, P_0$): cosϕ0=P0V0I0,Ic=I0cosϕ0,Im=I0sinϕ0\cos \phi_0 = \frac{P_0}{V_0 I_0}, \quad I_c = I_0 \cos \phi_0, \quad I_m = I_0 \sin \phi_0 Rc=V0Ic,Xm=V0ImR_c = \frac{V_0}{I_c}, \quad X_m = \frac{V_0}{I_m}
  2. From SC Test ($V_{\text{sc}}, I_{\text{sc}}, P_{\text{sc}}$): Zeq,HV=VscIsc,Req,HV=PscIsc2,Xeq,HV=Zeq,HV2Req,HV2Z_{\text{eq,HV}} = \frac{V_{\text{sc}}}{I_{\text{sc}}}, \quad R_{\text{eq,HV}} = \frac{P_{\text{sc}}}{I_{\text{sc}}^2}, \quad X_{\text{eq,HV}} = \sqrt{Z_{\text{eq,HV}}^2 - R_{\text{eq,HV}}^2}

Voltage Regulation Analysis

Voltage Regulation (VR) measures the percentage change in secondary terminal voltage magnitude when full rated load at a specified power factor is reduced to zero, assuming primary supply voltage remains constant.

%VR=V2,no-loadV2,full-loadV2,full-load×100%\% \text{VR} = \frac{V_{2,\text{no-load}} - V_{2,\text{full-load}}}{V_{2,\text{full-load}}} \times 100\%

Using secondary-referred parameters, $V_{2,\text{nl}} = \frac{V_1}{a}$. The phasor expression yields the Approximate Voltage Regulation Formula:

%VR=I2Re2cosϕ2±I2Xe2sinϕ2V2,fl×100%\% \text{VR} = \frac{I_2 R_{e2} \cos \phi_2 \pm I_2 X_{e2} \sin \phi_2}{V_{2,\text{fl}}} \times 100\%

or in per-unit (pu) terms:

VRpu=Rpucosϕ2±Xpusinϕ2\text{VR}_{\text{pu}} = R_{\text{pu}} \cos \phi_2 \pm X_{\text{pu}} \sin \phi_2

  • Plus Sign ($+$): Used for lagging power factor (inductive load). Terminal voltage drops under load.
  • Minus Sign ($-$): Used for leading power factor (capacitive load). Terminal voltage can rise under load.

Critical Voltage Regulation Conditions

  1. Maximum Voltage Regulation (Worst Voltage Drop): Occurs at lagging power factor when: tanϕ2=Xe2Re2    cosϕ2=Re2Ze2\text{Occurs at lagging power factor when: } \tan \phi_2 = \frac{X_{e2}}{R_{e2}} \implies \cos \phi_2 = \frac{R_{e2}}{Z_{e2}} %VRmax=I2Ze2V2,fl×100%=Zpu×100%\% \text{VR}_{\text{max}} = \frac{I_2 Z_{e2}}{V_{2,\text{fl}}} \times 100\% = Z_{\text{pu}} \times 100\%
  2. Zero Voltage Regulation (Perfect Terminal Voltage Stability): Occurs at leading power factor when: tanϕ2=Re2Xe2    cosϕ2=Xe2Ze2\text{Occurs at leading power factor when: } \tan \phi_2 = \frac{R_{e2}}{X_{e2}} \implies \cos \phi_2 = \frac{X_{e2}}{Z_{e2}} %VR=0    Re2cosϕ2=Xe2sinϕ2\% \text{VR} = 0 \implies R_{e2} \cos \phi_2 = X_{e2} \sin \phi_2

Efficiency Optimization & All-Day Efficiency

Transformer commercial efficiency $\eta$ is the ratio of active power output to active power input:

η=PoutPin×100%=PoutPout+Pi+Pcu×100%\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{P_{\text{out}}}{P_{\text{out}} + P_i + P_{\text{cu}}} \times 100\%

Let $S_{\text{fl}}$ be the rated full-load kVA capacity, $x = \frac{S}{S_{\text{fl}}} = \frac{I_2}{I_{2,\text{fl}}}$ be the fractional load factor, and $\cos \phi$ be the load power factor. Copper loss scales quadratically with fractional load ($P_{\text{cu}} = x^2 P_{\text{cu,fl}}$):

η=xSflcosϕxSflcosϕ+Pi+x2Pcu,fl×100%\eta = \frac{x S_{\text{fl}} \cos \phi}{x S_{\text{fl}} \cos \phi + P_i + x^2 P_{\text{cu,fl}}} \times 100\%

Condition for Maximum Efficiency

To determine the load fraction $x$ that maximizes efficiency for a given power factor, differentiate $\eta$ with respect to $x$ and set $\frac{d\eta}{dx} = 0$. This proves that maximum efficiency occurs when variable copper loss equals constant core loss:

x2Pcu,fl=Pi    xmax=PiPcu,flx^2 P_{\text{cu,fl}} = P_i \implies x_{\text{max}} = \sqrt{\frac{P_i}{P_{\text{cu,fl}}}}

The kVA load corresponding to maximum efficiency is:

Smax efficiency=SflPiPcu,flS_{\text{max efficiency}} = S_{\text{fl}} \sqrt{\frac{P_i}{P_{\text{cu,fl}}}}

All-Day Efficiency (Distribution Transformers)

Unlike power transformers (which operate near 100% full load continuously in utility substations), distribution transformers supply fluctuating residential/commercial loads throughout 24 hours. Their cores remain energized 24/7 (constant 24-hour core energy loss), while copper loss varies with hourly loading. Distribution transformers are designed with $P_i \ll P_{\text{cu,fl}}$ ($x_{\text{max}} \approx 50% - 70%$) to maximize All-Day Energy Efficiency:

ηall-day=Total Energy Output in 24 hours (kWh)Total Energy Input in 24 hours (kWh)=k=1n(Pout,k×tk)k=1n(Pout,k×tk)+24Pi+k=1n(xk2Pcu,fl×tk)\eta_{\text{all-day}} = \frac{\text{Total Energy Output in 24 hours (kWh)}}{\text{Total Energy Input in 24 hours (kWh)}} = \frac{\sum_{k=1}^n (P_{\text{out},k} \times t_k)}{\sum_{k=1}^n (P_{\text{out},k} \times t_k) + 24 P_i + \sum_{k=1}^n (x_k^2 P_{\text{cu,fl}} \times t_k)}


Parallel Operation of Single-Phase & Three-Phase Transformers

Connecting transformers in parallel increases system capacity and reliability. To prevent circulating currents, overheating, and unequal load division, five essential conditions must be satisfied:

  1. Equal Voltage Ratings / Turns Ratio: Primary and secondary nominal voltages must match exactly ($a_A = a_B$). A ratio mismatch causes a continuous circulating current $\dot{I}_{\text{circ}} = \frac{\dot{E}_A - \dot{E}_B}{\dot{Z}_A + \dot{Z}_B}$ even at no-load.
  2. Identical Polarity & Phase Sequence: For 3-phase transformers, vector group phase displacement must match (e.g., both Dy11). Reversing polarity results in destructive dead short-circuits.
  3. Equal Per-Unit Impedances ($Z_{\text{pu}}$): Transformers divide total load in inverse proportion to their per-unit impedances. If $Z_{\text{pu,A}} = Z_{\text{pu,B}}$, each unit shares load strictly proportional to its kVA rating.
  4. Equal $X/R$ Ratios: Matching $X/R$ ratios ensures that both transformer currents are in phase with each other and with the total load current, operating at identical power factors.
  5. Identical Frequency Rating: Standard $60\text{ Hz}$ utility rating.

Complex Load Sharing Equation

For two transformers A and B connected in parallel supplying total complex load $\dot{S}_T$:

S˙A=S˙T(Z˙BZ˙A+Z˙B),S˙B=S˙T(Z˙AZ˙A+Z˙B)\dot{S}_A = \dot{S}_T \left( \frac{\dot{Z}_B}{\dot{Z}_A + \dot{Z}_B} \right), \quad \dot{S}_B = \dot{S}_T \left( \frac{\dot{Z}_A}{\dot{Z}_A + \dot{Z}_B} \right)

where $\dot{Z}_A, \dot{Z}_B$ are equivalent ohmic impedances referred to the same voltage level.


Instrument Transformers: Potential Transformers (PTs) & Current Transformers (CTs)

Instrument transformers scale down dangerous high voltages and high currents to standardized safe levels for protective relays, digital meters, and control instruments.

                  INSTRUMENT TRANSFORMER SCHEMATIC & CT SAFETY RULE

      High-Voltage Line (I_primary) ───────────█████───────────────► Load
                                                CT
                                              ──┬──
                                                │ Primary (1-2 turns)
                                              ══╧══ Core
                                                │ Secondary (N2 turns)
                                              ──┬──
                                                │   ◄── MUST NEVER BE OPEN-CIRCUITED!
                                              ┌─┴─┐     ALWAYS SHORTED OR
                                              │ A │     CONNECTED TO LOW-Z BURDEN
                                              └─┬─┘
                                                │
                                                ▼ Ground

Potential Transformers (PTs / VTs)

  • Function: Precision step-down voltage transformer connected in parallel across power lines.
  • Secondary Rating: Standardized at $120\text{ V}$ or $115\text{ V}$ line-to-line (or $69.3\text{ V}$ line-to-neutral).
  • Operation: Operates under near open-circuit conditions with very high secondary impedance (metering voltage coils). Designed with extremely small leakage reactance to minimize ratio error and phase angle error.

Current Transformers (CTs)

  • Function: Step-down current transformer (step-up voltage ratio) connected in series with the power conductor.
  • Secondary Rating: Standardized at $5\text{ A}$ or $1\text{ A}$.
  • Operation: Primary winding consists of one or a few heavy turns (or a bar-type conductor passing through a toroidal core). Secondary consists of many fine turns wound on a high-permeability core.

CRITICAL REE EXAM SAFETY MANDATE: CT SECONDARY OPEN-CIRCUIT HAZARD

Mandatory Safety Rule: The secondary circuit of an energized Current Transformer must NEVER be opened while primary current is flowing.

Physical Mechanism: Under normal operation, primary MMF ($N_1 I_1$) is opposed and almost completely neutralized by secondary MMF ($N_2 I_2$), leaving a tiny net magnetizing MMF ($N_1 I_0$). If the secondary is opened ($I_2 = 0$), the opposing secondary MMF vanishes instantly. The entire heavy primary line current ($I_1$) becomes pure magnetizing current.

Consequences:

  1. Core flux density $\Phi_m$ spikes violently into deep saturation.
  2. Extreme rate-of-change of flux $\frac{d\Phi}{dt}$ induces lethal peak voltage spikes ($V_{\text{peak}} = L \frac{di}{dt}$, often thousands of Volts) across open secondary terminals, presenting immediate fatal electric shock hazards to personnel and destroying winding insulation.
  3. Severe core hysteresis heating causes catastrophic thermal breakdown. Always short-circuit CT secondary terminals using a testing switch before disconnecting downstream ammeters or relays.
Loading diagram...
Transformer Parameter Extraction Workflow from OC & SC Tests
Transformer Commercial Efficiency (%) vs. Percentage Full Load (at 0.8 Lagging Power Factor)
Test Your Knowledge

A 20 kVA, 2000/200 V single-phase transformer underwent open-circuit and short-circuit tests. The open-circuit test yielded a core loss of 120 W, while the short-circuit test yielded a full-load copper loss of 300 W. At what kVA load does maximum operating efficiency occur?

A
B
C
D
Test Your Knowledge

A single-phase distribution transformer has an equivalent resistance of 1.0% and an equivalent leakage reactance of 4.0%. What is the percentage voltage regulation at full rated load with a power factor of 0.80 lagging?

A
B
C
D
Test Your Knowledge

What primary operational hazard occurs if the secondary winding of an energized Current Transformer (CT) is accidentally open-circuited while primary line current is flowing?

A
B
C
D