4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing

PE Electrical and Computer Exam

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1 Chapter 1: Exam Overview & Power Engineering Fundamentals

1.1 PE Power Exam Structure, Official Specification & NCEES Reference Strategy 1.2 Units, Conversions, Engineering Constants & Fundamental Power Equations 1.3 Measurement & Metering Principles (Power, Energy, Power Factor) 1.4 Instrument Testing & Diagnostics (Insulation Megger & Ground Resistance)

2 Chapter 2: Circuit Analysis Fundamentals

2.1 DC Circuit Laws, Network Theorems (Thevenin, Norton, Superposition) & Transients 2.2 Single-Phase AC Steady-State Analysis, Phasors & Impedance 2.3 AC Power Relationships (Real, Reactive, Apparent Power & Power Factor Correction) 2.4 Resonance, Frequency Response & Harmonic Distortion

3 Chapter 3: Three-Phase Systems & Symmetrical Components

3.1 Three-Phase Circuit Analysis (Wye and Delta Configurations & Transformations) 3.2 Balanced Three-Phase Power Calculations & Phasor Relationships 3.3 The Per-Unit (pu) System: Bases, Conversions & System Modeling 3.4 Symmetrical Components (Fortescue Transformation, Sequence Networks & Unbalance)

4 Chapter 4: Electric Power Transformers

4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing 4.2 Transformer Voltage Regulation, Efficiency & Loss Calculations 4.3 Three-Phase Transformer Connections, Phase Shifts & Paralleling 4.4 Autotransformers, Instrument Transformers (CTs & VTs) & Special Transformers

5 Chapter 5: Rotating Machines: Induction & Synchronous

5.1 Three-Phase Induction Motors: Principles, Equivalent Circuits & Slip-Torque Characteristics 5.2 Induction Motor Starting, Speed Control & Braking Methods 5.3 Synchronous Generators: Operation, Excitation, Power-Angle Equations & V-Curves 5.4 Synchronous Motors & Condensers for Power Factor Correction

6 Chapter 6: DC Machines & Power Devices

6.1 DC Generators & Motors: Types (Shunt, Series, Compound), Characteristics & Speed Control 6.2 Power Capacitors, Shunt Reactors & Harmonic Filter Design 6.3 Battery Energy Storage Systems (BESS) & Ultra-Capacitors 6.4 Renewable Generation Integration: Solar Photovoltaics (PV) & Wind Power Systems

7 Chapter 7: Power Electronics & Industrial Controls

7.1 Power Semiconductor Switching Devices (Diodes, Thyristors, MOSFETs, IGBTs) 7.2 AC-to-DC Converters (Uncontrolled & Controlled Rectifiers) 7.3 DC-to-DC Switch-Mode Converters (Buck, Boost, Buck-Boost) & DC-AC Inverters 7.4 Variable Frequency Drives (VFDs), Motor Starters & Ladder Logic / PLC Controls

8 Chapter 8: Transmission & Distribution Systems

8.1 Transmission Line Parameters (Resistance, Inductance, Capacitance & GMD/GMR) 8.2 Line Models (Short, Medium Nominal-π, Long Lines) & ABCD Two-Port Parameters 8.3 Surge Impedance, Surge Impedance Loading (SIL) & Voltage Regulation 8.4 Distribution System Architectures, Voltage Drops & Feeder Loading 8.5 Power Flow Studies: Bus Classification, Y-Bus & Newton-Raphson Solutions 8.6 Power System Stability: Swing Equation, Equal-Area Criterion & Critical Clearing Time

9 Chapter 9: Power System Fault Analysis

9.1 Fault Analysis Fundamentals & Subtransient/Transient Machine Reactances 9.2 Three-Phase Symmetrical Fault Calculations 9.3 Single Line-to-Ground (SLG) & Line-to-Line (LL) Fault Analysis 9.4 Double Line-to-Ground (DLG) Faults & Bus Impedance Matrix (Zbus) Methods

10 Chapter 10: Power System Protection & Coordination

10.1 Protective Relaying Principles, IEEE Device Designations & Instrument Transformers 10.2 Overcurrent Protection (50/51), Time-Current Curves (TCC) & Relay Coordination 10.3 Differential Protection (87) for Transformers, Busbars & Generators 10.4 Distance / Impedance Protection (21) for Transmission Lines & Directional Relays (67)

11 Chapter 11: National Electrical Code (NEC) & Installation Design

11.1 NEC Fundamentals, Branch Circuit & Feeder Conductor Sizing (Ampacity & Derating) 11.2 Overcurrent Protection Sizing, Raceway Fill & Box Fill Calculations 11.3 Motor Branch Circuits, Controllers & Feeder Protection (NEC Article 430) 11.4 Grounding and Bonding Systems per NEC Article 250

12 Chapter 12: Electrical Safety, Standards & Facility Applications

12.1 Hazardous (Classified) Locations: Classes, Divisions, Zones, T-Codes & Sealing (NEC Articles 500-506) 12.2 Electrical Safety in the Workplace: NFPA 70E & Arc Flash Hazard Calculations 12.3 National Electrical Safety Code (NESC) Clearance & Utility Safety Requirements 12.4 Facility Applications: Illumination Design, Demand/Connected Loads & Lightning/Surge Protection 12.5 Special Occupancies & Systems: Health Care, Fire Pumps, Pools, Emergency & Standby (NEC Chapters 5-7)

4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing

Key Takeaways Explain This Topic AI

Ideal transformers enforce strict voltage, current, and impedance scaling: V1/V2 = N1/N2 = a, I1/I2 = 1/a, and Z1/Z2 = a^2, where a is the turns ratio N1/N2.

Faraday's Law of Induction determines the root-mean-square induced EMF in a magnetic core: E_rms = (2*pi / sqrt(2)) * f * N * Φ_max ≈ 4.44 * f * N * Φ_max = 4.44 * f * N * B_max * A_c.

The exact practical transformer equivalent circuit includes primary and secondary winding resistances (R1, R2'), leakage reactances (X1, X2'), core loss resistance (Rc), and magnetizing reactance (Xm); shifting the shunt excitation branch to the input terminals creates the cantilever approximate equivalent circuit.

The Open-Circuit (No-Load) test is performed at rated voltage typically on the low-voltage (LV) winding with the HV winding open, measuring Voc, Ioc, Poc to extract shunt excitation parameters (Rc, Xm).

The Short-Circuit test is performed at rated current typically on the high-voltage (HV) winding with the LV winding shorted, measuring Vsc, Isc, Psc to extract series equivalent resistance (Req), leakage reactance (Xeq), and percent impedance (%Z = (Vsc / Vrated) * 100%).

Last updated: August 2026

4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing

Executive Overview: Electric power transformers are static electromagnetic devices that transfer electrical energy between two or more circuits through mutual magnetic induction. Operating at efficiencies routinely exceeding 98 % 98\% 98% to 99.5 % 99.5\% 99.5% , transformers constitute the fundamental enabling technology of the modern AC power grid, stepping up voltages at generating stations to minimize transmission I 2 R I^2 R I 2 R line losses and stepping down voltages at distribution substations for safe end-use utilization. On the NCEES PE Electrical and Computer: Power examination, transformer mastery requires rapid extraction of equivalent circuit parameters from Open-Circuit (OC) and Short-Circuit (SC) test data, precise impedance reflection across turns ratios, and flawless per-unit parameter conversions.

1. Physics of Magnetic Induction & The Ideal Transformer

A transformer functions on the principle of Faraday's Law of Electromagnetic Induction. When a sinusoidal alternating voltage v 1 ( t ) = V m 1 cos ⁡ ( ω t ) v_1(t) = V_{m1} \cos(\omega t) v 1 ( t ) = V m 1 cos ( ω t ) is applied to a primary winding of N 1 N_1 N 1 turns wrapped around a ferromagnetic core of cross-sectional area A c A_c A c , it establishes an alternating magnetic flux ϕ ( t ) = Φ max ⁡ sin ⁡ ( ω t ) \phi(t) = \Phi_{\max} \sin(\omega t) ϕ ( t ) = Φ m a x sin ( ω t ) within the core. IDEAL TWO-WINDING TRANSFORMER SCHEMATIC:

Primary Winding (N1) Ferromagnetic Core Secondary Winding (N2)
I1 -> Flux Φ(t) -> I2
+-------( ( (-------------------------+------------------------( ( (-------+
| ( ( ( | ( ( ( |
| ( ( ( +---+---+ ( ( ( |
+ | ( ( ( | | ( ( ( | +
V1 (~) ( ( ( | Φ | ( ( ( [Z_L] V2
| ( ( ( | | ( ( ( | -
| ( ( ( +---+---+ ( ( ( |
| ( ( (-------------------------+------------------------( ( (-------+
+--------------------------------------------------------------------------+
<----------------- Core Mean Path Length lc ----------------->
Mathematical Derivation of Induced EMF (Faraday's Law)

The instantaneous induced electromotive force (EMF) e ( t ) e(t) e ( t ) across a winding of N N N turns linking time-varying magnetic flux ϕ ( t ) \phi(t) ϕ ( t ) is given by Faraday's Law: ≠ ( t ) = − N d ϕ ( t ) d t = − N d d t [ Φ max ⁡ sin ⁡ ( ω t ) ] = − N ω Φ max ⁡ cos ⁡ ( ω t ) = N ω Φ max ⁡ sin ⁡ ( ω t − 90 ∘ ) \ne(t) = -N \frac{d\phi(t)}{dt} = -N \frac{d}{dt}\left[ \Phi_{\max} \sin(\omega t) \right] = -N \omega \Phi_{\max} \cos(\omega t) = N \omega \Phi_{\max} \sin\left(\omega t - 90^\circ\right)  = ( t ) = − N d t d ϕ ( t ) = − N d t d [ Φ m a x sin ( ω t ) ] = − N ω Φ m a x cos ( ω t ) = N ω Φ m a x sin ( ω t − 9 0 ∘ )

The peak induced EMF is E max ⁡ = N ω Φ max ⁡ = 2 π f N Φ max ⁡ E_{\max} = N \omega \Phi_{\max} = 2\pi f N \Phi_{\max} E m a x = N ω Φ m a x = 2 π f N Φ m a x . Converting to the root-mean-square (RMS) value: E r m s = E max ⁡ 2 = 2 π 2 f N Φ max ⁡ = 2 π f N Φ max ⁡ ≈ 4.44288 f N Φ max ⁡ E_{rms} = \frac{E_{\max}}{\sqrt{2}} = \frac{2\pi}{\sqrt{2}} f N \Phi_{\max} = \sqrt{2}\pi f N \Phi_{\max} \approx 4.44288 f N \Phi_{\max} E r m s = 2 E m a x = 2 2 π f N Φ m a x = 2 π f N Φ m a x ≈ 4.44288 f N Φ m a x

Expressing peak flux in terms of peak core magnetic flux density B max ⁡ B_{\max} B m a x (Tesla) and core effective cross-sectional area A c A_c A c ( m 2 m^2 m 2 ): E r m s = 4.44 f N B max ⁡ A c E_{rms} = 4.44 f N B_{\max} A_c E r m s = 4.44 f N B m a x A c

The Ideal Transformer Equations & Impedance Reflection

An ideal transformer assumes: (1) zero winding resistance ( R 1 = R 2 = 0 R_1 = R_2 = 0 R 1 = R 2 = 0 ), (2) zero leakage flux (all flux is confined to the core, X 1 = X 2 = 0 X_1 = X_2 = 0 X 1 = X 2 = 0 ), (3) infinite core permeability ( μ r → ∞ \mu_r \to \infty μ r → ∞ , requiring zero magnetizing current, X m → ∞ X_m \to \infty X m → ∞ ), and (4) zero core losses ( R c → ∞ R_c \to \infty R c → ∞ ).

Defining the turns ratio a a a : a = N 1 N 2 = V 1 V 2 = I 2 I 1 a = \frac{N_1}{N_2} = \frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{\mathbf{I}_2}{\mathbf{I}_1} a = N 2 N 1 = V 2 V 1 = I 1 I 2

Applying Ampere's Law around the closed magnetic path yields zero net magnetomotive force (MMF): F n e t = N 1 I 1 − N 2 I 2 = 0 ⟹ N 1 I 1 = N 2 I 2 ⟹ I 1 I 2 = N 2 N 1 = 1 a \mathcal{F}_{net} = N_1 \mathbf{I}_1 - N_2 \mathbf{I}_2 = 0 \implies N_1 \mathbf{I}_1 = N_2 \mathbf{I}_2 \implies \frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_2}{N_1} = \frac{1}{a} F n e t = N 1 I 1 − N 2 I 2 = 0 ⟹ N 1 I 1 = N 2 I 2 ⟹ I 2 I 1 = N 1 N 2 = a 1

When a load impedance Z L = V 2 / I 2 \mathbf{Z}_L = \mathbf{V}_2 / \mathbf{I}_2 Z L = V 2 / I 2 is connected to the secondary winding, the input impedance Z i n = V 1 / I 1 \mathbf{Z}_{in} = \mathbf{V}_1 / \mathbf{I}_1 Z in = V 1 / I 1 seen looking into the primary terminals is: Z i n = V 1 I 1 = a V 2 I 2 / a = a 2 ( V 2 I 2 ) = a 2 Z L \mathbf{Z}_{in} = \frac{\mathbf{V}_1}{\mathbf{I}_1} = \frac{a \mathbf{V}_2}{\mathbf{I}_2 / a} = a^2 \left(\frac{\mathbf{V}_2}{\mathbf{I}_2}\right) = a^2 \mathbf{Z}_L Z in = I 1 V 1 = I 2 / a a V 2 = a 2 ( I 2 V 2 ) = a 2 Z L +---------------------------------------------------------------------------------------------------+

| IDEAL TRANSFORMER SCALING LAWS (TURNS RATIO a = N1 / N2) |
+---------------------------------------------------------------------------------------------------+
| Voltage Transformation: | V1 / V2 = a ==> V1 = a * V2 |
| Current Transformation: | I1 / I2 = 1 / a ==> I1 = I2 / a |
| Impedance Transformation: | Z1 / Z2 = a^2 ==> Z_referred_to_primary = a^2 * Z2 |
| Admittance Transformation: | Y1 / Y2 = 1 / a^2 ==> Y_referred_to_primary = Y2 / a^2 |
| Complex Power Invariance: | S1 = V1 * I1* = (a*V2) * (I2*/a) = V2 * I2* = S2 |
+---------------------------------------------------------------------------------------------------+
2. Practical Transformer Modeling & Equivalent Circuits

Real-world power transformers deviate from ideal behavior due to physical non-idealities:

Winding Resistances ( R 1 , R 2 R_1, R_2 R 1 , R 2 ): Finite conductivity of copper or aluminum conductors causes Joule heating ( I 2 R I^2 R I 2 R losses).

Leakage Fluxes ( X 1 = ω L l 1 , X 2 = ω L l 2 X_1 = \omega L_{l1}, X_2 = \omega L_{l2} X 1 = ω L l 1 , X 2 = ω L l 2 ): Magnetic flux lines that link only one winding without traversing the entire core create series leakage inductances.

Core Magnetizing Reactance ( X m X_m X m ): Finite permeability of the ferromagnetic steel core requires a finite magnetizing current I m \mathbf{I}_m I m to establish core flux Φ \Phi Φ .

Core Loss Resistance ( R c R_c R c or R f e R_{fe} R f e ): Hysteresis loop energy dissipation and eddy currents induced in core laminations dissipate active power, represented by a shunt resistance across the induced EMF.

The Exact T-Equivalent Circuit EXACT T-EQUIVALENT CIRCUIT (REFERRED TO PRIMARY WINDING):

Primary Winding Ideal Core Ratio 1:1 Secondary Winding (Referred)
R1 X1 I1' -> R2' X2'
o---------[ R1 ]------[ X1 ]---------+----------------------------+-------------[ R2' ]-----[ X2' ]---------o
+ | | +
[ Rc ] [ Xm ]
V1 | | V2'
+----------------------------+ -
| Shunt Branch | |
o------------------------------------+----------------------------+------------------------------------o
<------- Core Excitation ---->

Where secondary parameters are referred to the primary side: V 2 ′ = a V 2 , I 2 ′ = I 2 a , R 2 ′ = a 2 R 2 , X 2 ′ = a 2 X 2 , Z L ′ = a 2 Z L \mathbf{V}_2' = a \mathbf{V}_2, \quad \mathbf{I}_2' = \frac{\mathbf{I}_2}{a}, \quad R_2' = a^2 R_2, \quad X_2' = a^2 X_2, \quad \mathbf{Z}_L' = a^2 \mathbf{Z}_L V 2 ′ = a V 2 , I 2 ′ = a I 2 , R 2 ′ = a 2 R 2 , X 2 ′ = a 2 X 2 , Z L ′ = a 2 Z L

The Cantilever (Approximate) Equivalent Circuit

Because the no-load excitation current I 0 = I c + I m \mathbf{I}_0 = \mathbf{I}_c + \mathbf{I}_m I 0 = I c + I m is typically very small ( 1 % 1\% 1% to 3 % 3\% 3% of rated full-load current in power transformers), the voltage drop across the primary series impedance Z 1 = R 1 + j X 1 \mathbf{Z}_1 = R_1 + jX_1 Z 1 = R 1 + j X 1 caused by excitation current is negligible. Shifting the shunt branch ( R c ∥ j X m ) (R_c \parallel jX_m) ( R c ∥ j X m ) directly to the primary input terminals yields the Cantilever Approximate Equivalent Circuit : CANTILEVER (APPROXIMATE) EQUIVALENT CIRCUIT (REFERRED TO PRIMARY):

+----------------------------+ Req1 Xeq1
| | +-----[ Req1 ]----[ Xeq1 ]-----+
| | | |
[ Rc ] [ Xm ] | |
+ | | | | +
V1 +----------------------------+ | | V2' = a*V2
| Shunt Branch | | | -
+----------------------------+ | |
o-----------------------------------------------------+------------------------------+------o

Where the total series equivalent parameters referred to the primary winding are: R e q 1 = R 1 + R 2 ′ = R 1 + a 2 R 2 R_{eq1} = R_1 + R_2' = R_1 + a^2 R_2 R e q 1 = R 1 + R 2 ′ = R 1 + a 2 R 2 X e q 1 = X 1 + X 2 ′ = X 1 + a 2 X 2 X_{eq1} = X_1 + X_2' = X_1 + a^2 X_2 X e q 1 = X 1 + X 2 ′ = X 1 + a 2 X 2 Z e q 1 = R e q 1 + j X e q 1 = R e q 1 2 + X e q 1 2 ∠ θ e q \mathbf{Z}_{eq1} = R_{eq1} + j X_{eq1} = \sqrt{R_{eq1}^2 + X_{eq1}^2} \angle \theta_{eq} Z e q 1 = R e q 1 + j X e q 1 = R e q 1 2 + X e q 1 2 ∠ θ e q

If referred to the secondary winding ( a = N 1 / N 2 a = N_1 / N_2 a = N 1 / N 2 ): R e q 2 = R e q 1 a 2 = R 1 a 2 + R 2 , X e q 2 = X e q 1 a 2 = X 1 a 2 + X 2 , Z e q 2 = Z e q 1 a 2 R_{eq2} = \frac{R_{eq1}}{a^2} = \frac{R_1}{a^2} + R_2, \quad X_{eq2} = \frac{X_{eq1}}{a^2} = \frac{X_1}{a^2} + X_2, \quad \mathbf{Z}_{eq2} = \frac{\mathbf{Z}_{eq1}}{a^2} R e q 2 = a 2 R e q 1 = a 2 R 1 + R 2 , X e q 2 = a 2 X e q 1 = a 2 X 1 + X 2 , Z e q 2 = a 2 Z e q 1

3. Open-Circuit (No-Load) Test Characterization

The Open-Circuit (OC) Test determines the shunt excitation branch parameters: core loss resistance R c R_c R c and magnetizing reactance X m X_m X m . OPEN-CIRCUIT (NO-LOAD) TEST INSTRUMENTATION SETUP:

Wattmeter (P_oc)
+---[ ± M ]---+
| |
o--------+---( A )-----+--------------------( ( ( LV ) ) )------------------o (Open)
Rated I_oc ( ( ( ) ) )
V_rated,LV ( V ) V_oc ( ( ( ) ) ) HV Terminals
o----------------+--------------------------( ( ( ) ) )------------------o (Open)
Transformer Core
Standard Testing Procedure & Justification

Side Selection: The test is almost universally performed by energizing the Low-Voltage (LV) winding with rated voltage while leaving the High-Voltage (HV) winding completely open-circuited. Energizing the LV side requires standard, readily available utility test voltages (e.g., 120 V , 240 V , 480 V 120\text{ V}, 240\text{ V}, 480\text{ V} 120 V , 240 V , 480 V ) and ensures personnel safety by avoiding kilovolt potentials on the meter connections.

Physical Assumptions: Because the secondary is open ( I 2 = 0 I_2 = 0 I 2 = 0 ), the current drawn from the source is solely the no-load excitation current ( I o c = I 0 ≈ 0.01 − 0.03 I r a t e d I_{oc} = I_0 \approx 0.01 - 0.03 I_{rated} I oc = I 0 ≈ 0.01 − 0.03 I r a t e d ). Because I o c I_{oc} I oc is minuscule, the series winding copper loss I o c 2 R 1 I_{oc}^2 R_1 I oc 2 R 1 is negligible ( < 0.1 % < 0.1\% < 0.1% of total loss). Therefore, all active power measured by the wattmeter P o c P_{oc} P oc represents core losses ( P c o r e P_{core} P core ) .

Mathematical Parameter Extraction (LV Referred)

Given test measurements V o c , I o c , P o c V_{oc}, I_{oc}, P_{oc} V oc , I oc , P oc :

Core Loss Resistance ( R c R_c R c ): P o c = V o c 2 R c , L V ⟹ R c , L V = V o c 2 P o c P_{oc} = \frac{V_{oc}^2}{R_{c,LV}} \implies R_{c,LV} = \frac{V_{oc}^2}{P_{oc}} P oc = R c , L V V oc 2 ⟹ R c , L V = P oc V oc 2

Apparent Power and Power Factor Angle ( θ o c \theta_{oc} θ oc ): S o c = V o c I o c , P F o c = cos ⁡ θ o c = P o c V o c I o c ⟹ θ o c = arccos ⁡ ( P o c V o c I o c ) S_{oc} = V_{oc} I_{oc}, \quad PF_{oc} = \cos\theta_{oc} = \frac{P_{oc}}{V_{oc} I_{oc}} \implies \theta_{oc} = \arccos\left(\frac{P_{oc}}{V_{oc} I_{oc}}\right) S oc = V oc I oc , P F oc = cos θ oc = V oc I oc P oc ⟹ θ oc = arccos ( V oc I oc P oc )

Current Components (In-Phase Core Loss Current I c I_c I c and Quadrature Magnetizing Current I m I_m I m ): I c = I o c cos ⁡ θ o c = V o c R c , L V , I m = I o c sin ⁡ θ o c = I o c 2 − I c 2 I_c = I_{oc} \cos\theta_{oc} = \frac{V_{oc}}{R_{c,LV}}, \quad I_m = I_{oc} \sin\theta_{oc} = \sqrt{I_{oc}^2 - I_c^2} I c = I oc cos θ oc = R c , L V V oc , I m = I oc sin θ oc = I oc 2 − I c 2

Magnetizing Reactance ( X m X_m X m ): X m , L V = V o c I m = V o c I o c sin ⁡ θ o c X_{m,LV} = \frac{V_{oc}}{I_m} = \frac{V_{oc}}{I_{oc} \sin\theta_{oc}} X m , L V = I m V oc = I oc sin θ oc V oc

Alternative Admittance Formulation: Y m = I o c V o c , G c = P o c V o c 2 = 1 R c , L V , B m = Y m 2 − G c 2 = 1 X m , L V Y_m = \frac{I_{oc}}{V_{oc}}, \quad G_c = \frac{P_{oc}}{V_{oc}^2} = \frac{1}{R_{c,LV}}, \quad B_m = \sqrt{Y_m^2 - G_c^2} = \frac{1}{X_{m,LV}} Y m = V oc I oc , G c = V oc 2 P oc = R c , L V 1 , B m = Y m 2 − G c 2 = X m , L V 1

Exam Key Note: If parameters are required on the HV side, scale the LV-derived shunt values by a 2 a^2 a 2 : R c , H V = a 2 R c , L V R_{c,HV} = a^2 R_{c,LV} R c , H V = a 2 R c , L V and X m , H V = a 2 X m , L V X_{m,HV} = a^2 X_{m,LV} X m , H V = a 2 X m , L V , where a = V H V / V L V a = V_{HV} / V_{LV} a = V H V / V L V .

4. Short-Circuit Test Characterization

The Short-Circuit (SC) Test determines the series equivalent parameters: total winding resistance R e q R_{eq} R e q , leakage reactance X e q X_{eq} X e q , and percent impedance % Z \%Z % Z . SHORT-CIRCUIT TEST INSTRUMENTATION SETUP:

Wattmeter (P_sc)
+---[ ± M ]---+
| |
o--------+---( A )-----+--------------------( ( ( HV ) ) )------------------o--+
Variable I_sc ( ( ( ) ) ) |
AC Supply ( V ) V_sc ( ( ( ) ) ) Solid Short
o----------------+--------------------------( ( ( ) ) )------------------o--+
(0 - 10% V_rated) Transformer Core (LV Terminals)
Standard Testing Procedure & Justification

Side Selection: The test is performed by applying a reduced AC voltage to the High-Voltage (HV) winding while the Low-Voltage (LV) winding is solidly short-circuited with a zero-impedance copper bar. Energizing the HV side reduces the required test current to nominal full-load HV levels ( I r a t e d , H V = S r a t e d / V r a t e d , H V I_{rated,HV} = S_{rated} / V_{rated,HV} I r a t e d , H V = S r a t e d / V r a t e d , H V ), allowing standard laboratory meters to measure the current accurately.

Physical Assumptions: The applied voltage V s c V_{sc} V sc required to circulate rated current is very low (typically only 2 % 2\% 2% to 8 % 8\% 8% of rated V r a t e d , H V V_{rated,HV} V r a t e d , H V ). Because core loss is proportional to V 2 V^2 V 2 (or B max ⁡ 2 B_{\max}^2 B m a x 2 ), the core loss at V s c V_{sc} V sc is less than ( 0.05 ) 2 = 0.25 % (0.05)^2 = 0.25\% ( 0.05 ) 2 = 0.25% of normal no-load core loss. Therefore, the shunt branch draws zero significant current and all power measured by the wattmeter P s c P_{sc} P sc represents full-load series copper loss ( P c u , F L P_{cu,FL} P c u , F L ) .

Mathematical Parameter Extraction (HV Referred)

Given test measurements V s c , I s c , P s c V_{sc}, I_{sc}, P_{sc} V sc , I sc , P sc :

Total Equivalent Series Resistance ( R e q , H V R_{eq,HV} R e q , H V ): P s c = I s c 2 R e q , H V ⟹ R e q , H V = P s c I s c 2 P_{sc} = I_{sc}^2 R_{eq,HV} \implies R_{eq,HV} = \frac{P_{sc}}{I_{sc}^2} P sc = I sc 2 R e q , H V ⟹ R e q , H V = I sc 2 P sc

Total Equivalent Series Impedance Magnitude ( Z e q , H V Z_{eq,HV} Z e q , H V ): Z e q , H V = V s c I s c Z_{eq,HV} = \frac{V_{sc}}{I_{sc}} Z e q , H V = I sc V sc

Total Equivalent Series Leakage Reactance ( X e q , H V X_{eq,HV} X e q , H V ): X e q , H V = Z e q , H V 2 − R e q , H V 2 X_{eq,HV} = \sqrt{Z_{eq,HV}^2 - R_{eq,HV}^2} X e q , H V = Z e q , H V 2 − R e q , H V 2

Power Factor Angle of Equivalent Impedance ( θ e q \theta_{eq} θ e q ): θ e q = arctan ⁡ ( X e q , H V R e q , H V ) = arccos ⁡ ( P s c V s c I s c ) \theta_{eq} = \arctan\left(\frac{X_{eq,HV}}{R_{eq,HV}}\right) = \arccos\left(\frac{P_{sc}}{V_{sc} I_{sc}}\right) θ e q = arctan ( R e q , H V X e q , H V ) = arccos ( V sc I sc P sc )

Percent Impedance ( % Z \%Z % Z ): % Z = ( V s c V r a t e d , H V ) × 100 % = ( I r a t e d , H V Z e q , H V V r a t e d , H V ) × 100 % \%Z = \left(\frac{V_{sc}}{V_{rated,HV}}\right) \times 100\% = \left(\frac{I_{rated,HV} Z_{eq,HV}}{V_{rated,HV}}\right) \times 100\% % Z = ( V r a t e d , H V V sc ) × 100% = ( V r a t e d , H V I r a t e d , H V Z e q , H V ) × 100%

5. Per-Unit System Formulation for Transformers

In power system analysis, transformers are modeled in the Per-Unit (pu) system. A major computational advantage of the per-unit system is that the per-unit series impedance of a transformer is identical whether calculated from the primary or secondary side : Z b a s e , H V = V b a s e , H V 2 S b a s e , Z b a s e , L V = V b a s e , L V 2 S b a s e Z_{base,HV} = \frac{V_{base,HV}^2}{S_{base}}, \quad Z_{base,LV} = \frac{V_{base,LV}^2}{S_{base}} Z ba se , H V = S ba se V ba se , H V 2 , Z ba se , L V = S ba se V ba se , L V 2 Z p u = Z e q , H V Z b a s e , H V = Z e q , H V V b a s e , H V 2 / S b a s e = Z e q , L V V b a s e , L V 2 / S b a s e = % Z 100 Z_{pu} = \frac{Z_{eq,HV}}{Z_{base,HV}} = \frac{Z_{eq,HV}}{V_{base,HV}^2 / S_{base}} = \frac{Z_{eq,LV}}{V_{base,LV}^2 / S_{base}} = \frac{\%Z}{100} Z p u = Z ba se , H V Z e q , H V = V ba se , H V 2 / S ba se Z e q , H V = V ba se , L V 2 / S ba se Z e q , L V = 100 % Z

Per-Unit Base Conversion

When a transformer nameplate impedance Z p u , o l d Z_{pu,old} Z p u , o l d (defined on its self-cooled rating S b a s e , o l d S_{base,old} S ba se , o l d and voltage V b a s e , o l d V_{base,old} V ba se , o l d ) is integrated into a system-wide study with new base values S b a s e , n e w S_{base,new} S ba se , n e w and V b a s e , n e w V_{base,new} V ba se , n e w : Z p u , n e w = Z p u , o l d × ( V b a s e , o l d V b a s e , n e w ) 2 × ( S b a s e , n e w S b a s e , o l d ) Z_{pu,new} = Z_{pu,old} \times \left(\frac{V_{base,old}}{V_{base,new}}\right)^2 \times \left(\frac{S_{base,new}}{S_{base,old}}\right) Z p u , n e w = Z p u , o l d × ( V ba se , n e w V ba se , o l d ) 2 × ( S ba se , o l d S ba se , n e w ) +---------------------------------------------------------------------------------------------------+

| OPEN-CIRCUIT VS. SHORT-CIRCUIT TEST COMPARISON SUMMARY |
+----------------------------+------------------------------------+---------------------------------+
| Parameter | Open-Circuit (No-Load) Test | Short-Circuit Test |
+----------------------------+------------------------------------+---------------------------------+
| **Terminals Energized** | Typically Low-Voltage (LV) Side | Typically High-Voltage (HV) Side|
| **Opposite Terminals** | Open-Circuited | Solidly Short-Circuited |
| **Applied Voltage Level** | 100% Rated Voltage ($V_{rated}$) | Reduced Voltage ($2 - 8\% V_r$) |
| **Current Drawn** | No-Load Current ($1 - 3\% I_r$) | 100% Rated Current ($I_{rated}$)|
| **Wattmeter Measures** | Core / Iron Losses ($P_{core}$) | Full-Load Copper Loss ($P_{cu}$)|
| **Parameters Extracted** | Shunt Branch: $R_c, X_m$ | Series Branch: $R_{eq}, X_{eq}$ |
+----------------------------+------------------------------------+---------------------------------+
6. Comprehensive Step-by-Step Worked Mathematical Example
Problem Scenario

A single-phase, 50 kVA 50\text{ kVA} 50 kVA , 2400 V / 240 V 2400\text{ V} / 240\text{ V} 2400 V /240 V , 60 Hz 60\text{ Hz} 60 Hz distribution transformer underwent standard factory open-circuit and short-circuit testing. The laboratory test logs record the following data:

Open-Circuit Test (Instruments connected on the Low-Voltage 240 V 240\text{ V} 240 V side, HV open): V o c = 240.0 V , I o c = 5.40 A , P o c = 180.0 W V_{oc} = 240.0\text{ V}, \quad I_{oc} = 5.40\text{ A}, \quad P_{oc} = 180.0\text{ W} V oc = 240.0 V , I oc = 5.40 A , P oc = 180.0 W

Short-Circuit Test (Instruments connected on the High-Voltage 2400 V 2400\text{ V} 2400 V side, LV shorted): V s c = 120.0 V , I s c = 20.833 A , P s c = 650.0 W V_{sc} = 120.0\text{ V}, \quad I_{sc} = 20.833\text{ A}, \quad P_{sc} = 650.0\text{ W} V sc = 120.0 V , I sc = 20.833 A , P sc = 650.0 W

Calculate:

The turns ratio a a a .

The shunt excitation parameters R c R_c R c and X m X_m X m referred to the Low-Voltage (LV) side and High-Voltage (HV) side.

The series equivalent parameters R e q , X e q R_{eq}, X_{eq} R e q , X e q , and Z e q Z_{eq} Z e q referred to the High-Voltage (HV) side and Low-Voltage (LV) side.

The transformer percent impedance ( % Z \%Z % Z ) and per-unit equivalent impedance Z p u \mathbf{Z}_{pu} Z p u . =========================================================================================

CALCULATION WORKFLOW & DETAILED STEP-BY-STEP SOLUTION:

=========================================================================================

Step 1: Compute Rated Currents and Turns Ratio
Rated HV Current: I_rated,HV = S_rated / V_rated,HV = 50,000 VA / 2400 V = 20.833 A
Rated LV Current: I_rated,LV = S_rated / V_rated,LV = 50,000 VA / 240 V = 208.33 A
Turns Ratio: a = V_rated,HV / V_rated,LV = 2400 / 240 = 10.0
Impedance Scaling Factor: a^2 = 10.0^2 = 100.0

Step 2: Extract Shunt Parameters from Open-Circuit Test Data (LV Side)

Core Loss Resistance (LV Side):

R_c,LV = (V_oc)^2 / P_oc
= (240.0)^2 / 180.0
= 57,600 / 180.0
= 320.00 ohms

Core Loss Current Component:

I_c = V_oc / R_c,LV = 240.0 / 320.0 = 0.750 A

Magnetizing Current Component:

I_m = sqrt( (I_oc)^2 - (I_c)^2 )
= sqrt( (5.40)^2 - (0.750)^2 )
= sqrt( 29.16 - 0.5625 )
= sqrt( 28.5975 )
= 5.34766 A

Magnetizing Reactance (LV Side):

X_m,LV = V_oc / I_m
= 240.0 / 5.34766
= 44.8795 ohms approx 44.88 ohms

Scale Shunt Parameters to High-Voltage (HV) Side:

R_c,HV = a^2 * R_c,LV = 100.0 * 320.00 ohms = 32,000 ohms (32.0 kohms)
X_m,HV = a^2 * X_m,LV = 100.0 * 44.8795 ohms = 4,487.95 ohms (4.488 kohms)

Step 3: Extract Series Parameters from Short-Circuit Test Data (HV Side)

Equivalent Series Resistance (HV Side):

R_eq,HV = P_sc / (I_sc)^2
= 650.0 / (20.8333)^2
= 650.0 / 434.028
= 1.49760 ohms approx 1.498 ohms

Equivalent Series Impedance Magnitude (HV Side):

Z_eq,HV = V_sc / I_sc
= 120.0 / 20.8333
= 5.7600 ohms

Equivalent Series Leakage Reactance (HV Side):

X_eq,HV = sqrt( (Z_eq,HV)^2 - (R_eq,HV)^2 )
= sqrt( (5.7600)^2 - (1.4976)^2 )
= sqrt( 33.1776 - 2.2428 )
= sqrt( 30.9348 )
= 5.56191 ohms approx 5.562 ohms

HV Impedance Phasor:

Z_eq,HV = 1.498 + j5.562 ohms = 5.760 /_ 74.93° ohms

Scale Series Parameters to Low-Voltage (LV) Side:

R_eq,LV = R_eq,HV / a^2 = 1.49760 / 100.0 = 0.014976 ohms (14.98 mohms)
X_eq,LV = X_eq,HV / a^2 = 5.56191 / 100.0 = 0.055619 ohms (55.62 mohms)
Z_eq,LV = Z_eq,HV / a^2 = 5.76000 / 100.0 = 0.057600 ohms (57.60 mohms)

Step 4: Compute Percent Impedance and Per-Unit Model

HV Base Impedance:

Z_base,HV = (V_rated,HV)^2 / S_rated = (2400)^2 / 50,000 = 5,760,000 / 50,000 = 115.20 ohms

LV Base Impedance:

Z_base,LV = (V_rated,LV)^2 / S_rated = (240)^2 / 50,000 = 57,600 / 50,000 = 1.1520 ohms

Per-Unit Impedance (HV calculation):

Z_pu = Z_eq,HV / Z_base,HV = 5.7600 / 115.20 = 0.0500 pu (5.00%)
R_pu = R_eq,HV / Z_base,HV = 1.4976 / 115.20 = 0.0130 pu (1.30%)
X_pu = X_eq,HV / Z_base,HV = 5.5619 / 115.20 = 0.04828 pu (4.83%)

Per-Unit Impedance (LV calculation check):

Z_pu = Z_eq,LV / Z_base,LV = 0.0576 / 1.1520 = 0.0500 pu (CONFIRMED IDENTICAL)

Percent Impedance directly from SC Test Voltage:

%Z = (V_sc / V_rated,HV) * 100% = (120.0 / 2400.0) * 100% = 5.00%
=========================================================================================
7. Common Exam Traps & Tactical Pitfalls

Impedance Scaling Factor Direction Inversion ( a 2 a^2 a 2 vs 1 / a 2 1/a^2 1/ a 2 ): Multiplying by a 2 a^2 a 2 instead of dividing by a 2 a^2 a 2 when transferring an impedance from the high-voltage side to the low-voltage side. Always remember the fundamental physical rule: high-voltage windings have high impedances and low currents; low-voltage windings have low impedances and high currents .

Test Side Misattribution: Assuming the Open-Circuit test gives HV parameters or that the Short-Circuit test gives LV parameters. Always check which physical terminals the meters were wired to before scaling by a 2 a^2 a 2 .

The Base Voltage Discrepancy in Percent Impedance: Calculating % Z \%Z % Z using V s c V_{sc} V sc measured on the HV side divided by the LV rated voltage (e.g., 120 / 240 = 50 % 120 / 240 = 50\% 120/240 = 50% instead of 120 / 2400 = 5.0 % 120 / 2400 = 5.0\% 120/2400 = 5.0% ). The test voltage must always be compared to the rated voltage of the energized winding .

Neglecting Core Loss Negligibility in Short-Circuit Calculations: Attempting to incorporate R c R_c R c and X m X_m X m into short-circuit test analysis. Under short-circuit conditions, the excitation branch is completely bypassed because applied voltage is < 10 % < 10\% < 10% of nominal.

4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing

PE Electrical and Computer Exam

Track your progress as you read

1 Chapter 1: Exam Overview & Power Engineering Fundamentals

2 Chapter 2: Circuit Analysis Fundamentals

3 Chapter 3: Three-Phase Systems & Symmetrical Components

4 Chapter 4: Electric Power Transformers

5 Chapter 5: Rotating Machines: Induction & Synchronous

6 Chapter 6: DC Machines & Power Devices

7 Chapter 7: Power Electronics & Industrial Controls

8 Chapter 8: Transmission & Distribution Systems

9 Chapter 9: Power System Fault Analysis

10 Chapter 10: Power System Protection & Coordination

11 Chapter 11: National Electrical Code (NEC) & Installation Design

12 Chapter 12: Electrical Safety, Standards & Facility Applications

4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing

Key Takeaways Explain This Topic AI

Last updated: August 2026

4.1 Transformer Theory, Equivalent Circuits & Open/Short-Circuit Testing

1. Physics of Magnetic Induction & The Ideal Transformer

The Ideal Transformer Equations & Impedance Reflection

The Cantilever (Approximate) Equivalent Circuit

3. Open-Circuit (No-Load) Test Characterization

Mathematical Parameter Extraction (LV Referred)

4. Short-Circuit Test Characterization

Mathematical Parameter Extraction (HV Referred)

5. Per-Unit System Formulation for Transformers

Per-Unit Base Conversion

=========================================================================================

Step 2: Extract Shunt Parameters from Open-Circuit Test Data (LV Side)

I_c = V_oc / R_c,LV = 240.0 / 320.0 = 0.750 A

Step 3: Extract Series Parameters from Short-Circuit Test Data (HV Side)

Z_eq,HV = 1.498 + j5.562 ohms = 5.760 /_ 74.93° ohms

Step 4: Compute Percent Impedance and Per-Unit Model

Z_base,HV = (V_rated,HV)^2 / S_rated = (2400)^2 / 50,000 = 5,760,000 / 50,000 = 115.20 ohms

Z_base,LV = (V_rated,LV)^2 / S_rated = (240)^2 / 50,000 = 57,600 / 50,000 = 1.1520 ohms

Z_pu = Z_eq,LV / Z_base,LV = 0.0576 / 1.1520 = 0.0500 pu (CONFIRMED IDENTICAL)

Loading diagram... Transformer Parameter Extraction Workflow from OC and SC Tests