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

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 $98%$ to $99.5%$, transformers constitute the fundamental enabling technology of the modern AC power grid, stepping up voltages at generating stations to minimize transmission $I^2 R$ 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.

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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 $v_1(t) = V_{m1} \cos(\omega t)$ is applied to a primary winding of $N_1$ turns wrapped around a ferromagnetic core of cross-sectional area $A_c$, it establishes an alternating magnetic flux $\phi(t) = \Phi_{\max} \sin(\omega t)$ 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) $e(t)$ across a winding of $N$ turns linking time-varying magnetic flux $\phi(t)$ is given by Faraday's Law:

$$e(t) = -N \frac{d\phi(t)}{dt} = -N \frac{d}{dt}\left[ \Phi_{\max} \sin(\omega t) \right] = -N \omega \Phi_{\max} \cos(\omega t) = N \omega \Phi_{\max} \sin\left(\omega t - 90^\circ\right)$$

The peak induced EMF is $E_{\max} = N \omega \Phi_{\max} = 2\pi f N \Phi_{\max}$. Converting to the root-mean-square (RMS) value:

$$E_{rms} = \frac{E_{\max}}{\sqrt{2}} = \frac{2\pi}{\sqrt{2}} f N \Phi_{\max} = \sqrt{2}\pi f N \Phi_{\max} \approx 4.44288 f N \Phi_{\max}$$

Expressing peak flux in terms of peak core magnetic flux density $B_{\max}$ (Tesla) and core effective cross-sectional area $A_c$ ($m^2$):

$$E_{rms} = 4.44 f N B_{\max} A_c$$

The Ideal Transformer Equations & Impedance Reflection

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

Defining the turns ratio $a$:

$$a = \frac{N_1}{N_2} = \frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{\mathbf{I}_2}{\mathbf{I}_1}$$

Applying Ampere's Law around the closed magnetic path yields zero net magnetomotive force (MMF):

$$\mathcal{F}_{net} = N_1 \mathbf{I}_1 - N_2 \mathbf{I}_2 = 0 \implies N_1 \mathbf{I}_1 = N_2 \mathbf{I}_2 \implies \frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_2}{N_1} = \frac{1}{a}$$

When a load impedance $\mathbf{Z}_L = \mathbf{V}_2 / \mathbf{I}2$ is connected to the secondary winding, the input impedance $\mathbf{Z}{in} = \mathbf{V}_1 / \mathbf{I}_1$ seen looking into the primary terminals is:

$$\mathbf{Z}_{in} = \frac{\mathbf{V}_1}{\mathbf{I}_1} = \frac{a \mathbf{V}_2}{\mathbf{I}_2 / a} = a^2 \left(\frac{\mathbf{V}_2}{\mathbf{I}_2}\right) = a^2 \mathbf{Z}_L$$

+---------------------------------------------------------------------------------------------------+
| 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                    |
+---------------------------------------------------------------------------------------------------+

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2. Practical Transformer Modeling & Equivalent Circuits

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

1. Winding Resistances ($R_1, R_2$): Finite conductivity of copper or aluminum conductors causes Joule heating ($I^2 R$ losses).
2. Leakage Fluxes ($X_1 = \omega L_{l1}, X_2 = \omega L_{l2}$): Magnetic flux lines that link only one winding without traversing the entire core create series leakage inductances.
3. Core Magnetizing Reactance ($X_m$): Finite permeability of the ferromagnetic steel core requires a finite magnetizing current $\mathbf{I}_m$ to establish core flux $\Phi$.
4. Core Loss Resistance ($R_c$ or $R_{fe}$): 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:

$$\mathbf{V}_2' = a \mathbf{V}_2, \quad \mathbf{I}_2' = \frac{\mathbf{I}_2}{a}, \quad R_2' = a^2 R_2, \quad X_2' = a^2 X_2, \quad \mathbf{Z}_L' = a^2 \mathbf{Z}_L$$

The Cantilever (Approximate) Equivalent Circuit

Because the no-load excitation current $\mathbf{I}_0 = \mathbf{I}_c + \mathbf{I}_m$ is typically very small ($1%$ to $3%$ of rated full-load current in power transformers), the voltage drop across the primary series impedance $\mathbf{Z}_1 = R_1 + jX_1$ caused by excitation current is negligible. Shifting the shunt branch $(R_c \parallel jX_m)$ 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:

$$R_{eq1} = R_1 + R_2' = R_1 + a^2 R_2$$ $$X_{eq1} = X_1 + X_2' = X_1 + a^2 X_2$$ $$\mathbf{Z}_{eq1} = R_{eq1} + j X_{eq1} = \sqrt{R_{eq1}^2 + X_{eq1}^2} \angle \theta_{eq}$$

If referred to the secondary winding ($a = N_1 / N_2$):

$$R_{eq2} = \frac{R_{eq1}}{a^2} = \frac{R_1}{a^2} + R_2, \quad X_{eq2} = \frac{X_{eq1}}{a^2} = \frac{X_1}{a^2} + X_2, \quad \mathbf{Z}_{eq2} = \frac{\mathbf{Z}_{eq1}}{a^2}$$

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

The Open-Circuit (OC) Test determines the shunt excitation branch parameters: core loss resistance $R_c$ and magnetizing reactance $X_m$.

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., $120\text{ V}, 240\text{ V}, 480\text{ V}$) and ensures personnel safety by avoiding kilovolt potentials on the meter connections.
• Physical Assumptions: Because the secondary is open ($I_2 = 0$), the current drawn from the source is solely the no-load excitation current ($I_{oc} = I_0 \approx 0.01 - 0.03 I_{rated}$). Because $I_{oc}$ is minuscule, the series winding copper loss $I_{oc}^2 R_1$ is negligible ($< 0.1%$ of total loss). Therefore, all active power measured by the wattmeter $P_{oc}$ represents core losses ($P_{core}$).

Mathematical Parameter Extraction (LV Referred)

Given test measurements $V_{oc}, I_{oc}, P_{oc}$:

1. Core Loss Resistance ($R_c$): $P_{oc} = \frac{V_{oc}^2}{R_{c,LV}} \implies R_{c,LV} = \frac{V_{oc}^2}{P_{oc}}$
2. Apparent Power and Power Factor Angle ($\theta_{oc}$): $S_{oc} = V_{oc} I_{oc}, \quad PF_{oc} = \cos\theta_{oc} = \frac{P_{oc}}{V_{oc} I_{oc}} \implies \theta_{oc} = \arccos\left(\frac{P_{oc}}{V_{oc} I_{oc}}\right)$
3. Current Components (In-Phase Core Loss Current $I_c$ and Quadrature Magnetizing Current $I_m$): $I_c = I_{oc} \cos\theta_{oc} = \frac{V_{oc}}{R_{c,LV}}, \quad I_m = I_{oc} \sin\theta_{oc} = \sqrt{I_{oc}^2 - I_c^2}$
4. Magnetizing Reactance ($X_m$): $X_{m,LV} = \frac{V_{oc}}{I_m} = \frac{V_{oc}}{I_{oc} \sin\theta_{oc}}$
5. Alternative Admittance Formulation: $Y_m = \frac{I_{oc}}{V_{oc}}, \quad G_c = \frac{P_{oc}}{V_{oc}^2} = \frac{1}{R_{c,LV}}, \quad B_m = \sqrt{Y_m^2 - G_c^2} = \frac{1}{X_{m,LV}}$

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

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4. Short-Circuit Test Characterization

The Short-Circuit (SC) Test determines the series equivalent parameters: total winding resistance $R_{eq}$, leakage reactance $X_{eq}$, and percent impedance $%Z$.

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 ($I_{rated,HV} = S_{rated} / V_{rated,HV}$), allowing standard laboratory meters to measure the current accurately.
• Physical Assumptions: The applied voltage $V_{sc}$ required to circulate rated current is very low (typically only $2%$ to $8%$ of rated $V_{rated,HV}$). Because core loss is proportional to $V^2$ (or $B_{\max}^2$), the core loss at $V_{sc}$ is less than $(0.05)^2 = 0.25%$ of normal no-load core loss. Therefore, the shunt branch draws zero significant current and all power measured by the wattmeter $P_{sc}$ represents full-load series copper loss ($P_{cu,FL}$).

Mathematical Parameter Extraction (HV Referred)

Given test measurements $V_{sc}, I_{sc}, P_{sc}$:

1. Total Equivalent Series Resistance ($R_{eq,HV}$): $P_{sc} = I_{sc}^2 R_{eq,HV} \implies R_{eq,HV} = \frac{P_{sc}}{I_{sc}^2}$
2. Total Equivalent Series Impedance Magnitude ($Z_{eq,HV}$): $Z_{eq,HV} = \frac{V_{sc}}{I_{sc}}$
3. Total Equivalent Series Leakage Reactance ($X_{eq,HV}$): $X_{eq,HV} = \sqrt{Z_{eq,HV}^2 - R_{eq,HV}^2}$
4. Power Factor Angle of Equivalent Impedance ($\theta_{eq}$): $\theta_{eq} = \arctan\left(\frac{X_{eq,HV}}{R_{eq,HV}}\right) = \arccos\left(\frac{P_{sc}}{V_{sc} I_{sc}}\right)$
5. Percent Impedance ($%Z$): $\%Z = \left(\frac{V_{sc}}{V_{rated,HV}}\right) \times 100\% = \left(\frac{I_{rated,HV} Z_{eq,HV}}{V_{rated,HV}}\right) \times 100\%$

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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:

$$Z_{base,HV} = \frac{V_{base,HV}^2}{S_{base}}, \quad Z_{base,LV} = \frac{V_{base,LV}^2}{S_{base}}$$ $$Z_{pu} = \frac{Z_{eq,HV}}{Z_{base,HV}} = \frac{Z_{eq,HV}}{V_{base,HV}^2 / S_{base}} = \frac{Z_{eq,LV}}{V_{base,LV}^2 / S_{base}} = \frac{\%Z}{100}$$

Per-Unit Base Conversion

When a transformer nameplate impedance $Z_{pu,old}$ (defined on its self-cooled rating $S_{base,old}$ and voltage $V_{base,old}$) is integrated into a system-wide study with new base values $S_{base,new}$ and $V_{base,new}$:

$$Z_{pu,new} = Z_{pu,old} \times \left(\frac{V_{base,old}}{V_{base,new}}\right)^2 \times \left(\frac{S_{base,new}}{S_{base,old}}\right)$$

+---------------------------------------------------------------------------------------------------+
| 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: $R_{eq}, X_{eq}$ |
+----------------------------+------------------------------------+---------------------------------+

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6. Comprehensive Step-by-Step Worked Mathematical Example

Problem Scenario

A single-phase, $50\text{ kVA}$, $2400\text{ V} / 240\text{ V}$, $60\text{ Hz}$ 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 $240\text{ V}$ side, HV open): $V_{oc} = 240.0\text{ V}, \quad I_{oc} = 5.40\text{ A}, \quad P_{oc} = 180.0\text{ W}$
• Short-Circuit Test (Instruments connected on the High-Voltage $2400\text{ V}$ side, LV shorted): $V_{sc} = 120.0\text{ V}, \quad I_{sc} = 20.833\text{ A}, \quad P_{sc} = 650.0\text{ W}$

Calculate:

1. The turns ratio $a$.
2. The shunt excitation parameters $R_c$ and $X_m$ referred to the Low-Voltage (LV) side and High-Voltage (HV) side.
3. The series equivalent parameters $R_{eq}, X_{eq}$, and $Z_{eq}$ referred to the High-Voltage (HV) side and Low-Voltage (LV) side.
4. The transformer percent impedance ($%Z$) and per-unit equivalent impedance $\mathbf{Z}_{pu}$.

=========================================================================================
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%
=========================================================================================

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7. Common Exam Traps & Tactical Pitfalls

• Impedance Scaling Factor Direction Inversion ($a^2$ vs $1/a^2$): Multiplying by $a^2$ instead of dividing by $a^2$ 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 $a^2$.
• The Base Voltage Discrepancy in Percent Impedance: Calculating $%Z$ using $V_{sc}$ measured on the HV side divided by the LV rated voltage (e.g., $120 / 240 = 50%$ instead of $120 / 2400 = 5.0%$). 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 $R_c$ and $X_m$ into short-circuit test analysis. Under short-circuit conditions, the excitation branch is completely bypassed because applied voltage is $< 10%$ of nominal.