Chapter 13 — Transformer models and parameters

A transformer is two coils that share a magnetic circuit. Everything an EMT program does with one follows from that sentence, and this chapter walks the whole distance: from the mutual-inductance description of a coupled pair, through the reason no serious program uses that description directly, to the trapezoidal companion the solver actually stamps, and finally to where the numbers on the parameter form come from — which is a pair of factory tests you can reproduce in the simulator.

Saturation, the one genuinely nonlinear thing a transformer does, is deferred to Chapter 14. This chapter builds the linear frame it hangs on.

Which component to use

NumaSim's single-phase Ideal Transformer is exactly what its name says: it enforces $v_2/v_1=1/n$ and $i_2/i_1=-n$ and nothing else. No leakage, no winding resistance, no core. It is the right tool for level-shifting inside a control or measurement chain, and the wrong tool for anything magnetic. Every lab in this module uses the three-phase Transformer (3-phase, 2-winding) component, which carries leakage, a magnetizing branch, and an optional saturation curve.

Learning objectives

By the end of this chapter you should be able to:

• Write the coupled-inductance description of a two-winding transformer and convert it exactly into the leakage-plus-magnetizing T-equivalent.
• Explain why the mutual-inductance matrix is a poor parameterization for a program, and why the T-equivalent is the one worth discretizing.
• Derive the trapezoidal companion of the leakage and magnetizing branches and say how the ideal ratio enters the nodal system.
• Recover $r$ and $x$ from a short-circuit test and the magnetizing current from an open-circuit test, and convert both to henries and ohms.
• Read a vector group, predict its phase shift, and say what each winding connection does to zero-sequence current.

13.1 Two coupled coils

Two windings on a common core, with $N_1$ and $N_2$ turns, obey

$$\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{cc}{R}_1& 0\\ 0& R_2\end{array}\right]\left[\begin{array}{c}{i}_1\\ i_2\end{array}\right]+\left[\begin{array}{cc}{L}_{11}& L_{12}\\ L_{12}& L_{22}\end{array}\right]\frac{d}{dt}\left[\begin{array}{c}{i}_1\\ i_2\end{array}\right],$$

where $L_{11}$ and $L_{22}$ are the self-inductances and $L_{12}$ the mutual. The matrix is symmetric because the coupling is reciprocal. Two derived quantities name what the numbers mean physically: the coupling coefficient

$$K_{12}=\frac{L_{12}}{\sqrt{L_{11}{L}_{22}}}$$

and the turns ratio $a=N_1/N_2$, which for a shared core flux is very nearly $\sqrt{L_{11}/L_{22}}$. A power transformer is built to couple as tightly as steel allows, so $K_{12}$ lands around $0.9995$ — a number that will shortly cause trouble.

The two-winding transformer. Primary terminals $p_1$, $p_2$ and secondary $s_1$, $s_2$ link only through the flux in the iron core between them.

The matrix form is not how anyone thinks about a transformer, so trade it for the classical T-equivalent. Define, all referred to winding 1,

$$L_m=a{L}_{12},L_1=L_{11}-a{L}_{12},L_2=a^2{L}_{22}-a{L}_{12}.$$

Then $L_{11}=L_1+L_m$ and $a^2{L}_{22}=L_2+L_m$, so the three new numbers reproduce the original three exactly. This is a change of variables, not an approximation: $L_1$ and $L_2$ are the leakage inductances, the flux each winding produces that misses the other, and $L_m$ is the magnetizing inductance, the flux they share. The circuit is a series leakage on each side, a shunt magnetizing branch in the middle, and an ideal $a:1$ ratio to move between the two voltage levels.

The T-equivalent, referred to winding 1. Series leakage $R_1+L_1$ and $R_2+L_2$, a shunt magnetizing branch $L_m$ beside the core-loss conductance $G_c$, and an ideal $a:1$ ratio block that carries the voltage level change and nothing else.

13.2 Why not just invert the inductance matrix

An EMT program could discretize the coupled pair directly. Applying the trapezoidal rule to $\mathbf{v}=\mathbf{L}d\mathbf{i}/dt$ gives a $2\times 2$ companion conductance $\mathbf{G}=\frac{\mathrm{\Delta }t}{2}{\mathbf{L}}^{-1}$, which is a perfectly respectable thing to stamp. It is also a bad idea, and the running example shows why.

Take the transformer used throughout this module: 100 MVA, 230/115 kV, leakage $x=0.10$ pu split evenly between the windings, magnetizing current 1% of rated. Work in per unit, where the turns ratio is 1 by construction. An open-circuit test measures $X_{11}=1/0.01=100$ pu, so

$$\mathbf{L}=\left[\begin{array}{cc}100.00& 99.95\\ 99.95& 100.00\end{array}\right]\text{pu}.$$

Now ask this matrix for the leakage — the number that fixes short-circuit current, voltage regulation, and the fault duty of everything downstream:

$$L_{11}-\frac{L_{12}^2}{L_{22}}=100-\frac{{99.95}^2}{100}=0.09998\text{pu}.$$

The answer, 0.1, arrives as the difference of two numbers that agree to four significant figures. The matrix's eigenvalues are $100\pm 99.95$, that is $199.95$ and $0.05$, so its condition number is about 4000. Push the magnetizing current down to 0.1%, which a modern grain-oriented core easily manages, and $L_{11}$ becomes 1000 while the small eigenvalue stays at 0.05: the condition number goes to 40 000. The better the transformer, the worse the matrix. In the limit of a perfect core the mutual equals the self, $\mathbf{L}$ is singular, and the most conceptually simple transformer of all — the ideal one — is the one you cannot represent.

Double precision absorbs a condition number of ${10}^4$ without complaint, so this is an argument about parameterization rather than a prediction of numerical failure. But it is a decisive one, for three reasons that together explain why NumaSim stores leakage and magnetizing separately:

1. It is what you are given. Nobody publishes $L_{11}$ and $L_{12}$. A datasheet gives $x$ and a magnetizing percentage, which are the T-equivalent.
2. The ideal limit stays representable. Sending $L_m\to \mathrm{\infty }$ removes the shunt branch and leaves a finite leakage. Nothing is singular; the branch simply disappears.
3. The nonlinearity gets a home. Saturation lives in the core, which in the T-equivalent is one identified branch. That lets Chapter 14 handle it with a scalar current injection instead of re-inverting a matrix every step — and re-inverting per step would undo the factorize-once economics of Chapter 3.

13.3 Discretizing the T-equivalent

With the model in T form, the discretization is Chapter 2 applied three times per phase leg.

The two leakage branches are each a series $R+L$, and a series $R+L$ has its own companion. Trapezoidal integration of $Ldi/dt=v-Ri$ gives

$$i(t)=G_{RL}v(t)+I^{\text{hist}},G_{RL}=\frac{1}{R+2L/\mathrm{\Delta }t},$$$$I^{\text{hist}}=\alpha i(t-\mathrm{\Delta }t)+G_{RL}v(t-\mathrm{\Delta }t),\alpha =\frac{2L/\mathrm{\Delta }t-R}{2L/\mathrm{\Delta }t+R}.$$

Setting $R=0$ recovers $G_L=\mathrm{\Delta }t/2L$ and $\alpha =1$, the plain inductor of Chapter 2. The damping factor $\alpha$ is how much of the previous current survives one step; for a leakage branch it is a hair under 1.

The magnetizing branch is a plain inductor companion, $G_m=\mathrm{\Delta }t/2L_m$, in parallel with a constant core-loss conductance derived from the no-load loss,

$$G_c=\frac{S_{\text{rated}}{p}_{\text{NL}}}{V_{\text{HV}}^2}.$$

The ideal ratio cannot be a conductance — it fixes a voltage relationship and passes whatever current that demands, exactly like the ideal voltage source of Chapter 3. It is handled the same way, by modified nodal analysis: the branch current becomes an extra unknown and the constraint $v_{\text{HV}}-n{v}_{\text{LV}}=0$ becomes an extra row. A three-phase two-winding transformer adds three such branches, one per leg, which is why the component always demands an MNA-capable solver.

One leg of the discretized transformer. Every reactive element has become a conductance in parallel with a known history source, so the leg is a resistive network the nodal solve handles directly. The only element that is not a conductance is the ideal ratio, which enters as an MNA constraint row.

Put numbers on it. The base impedance of the running example is $Z_b=V_{LL}^2/S_{3\varphi }=230{000}^2/{10}^8=529\mathrm{\Omega }$. Half the leakage on the HV side is $x=0.05$ pu, so $X_1=26.45\mathrm{\Omega }$ and $L_1=26.45/(2\pi \cdot 60)=70.2\text{mH}$, with $R_1=0.0025\times 529=1.32\mathrm{\Omega }$. At $\mathrm{\Delta }t=50\mu \text{s}$,

$$G_{RL}=\frac{1}{1.32+2(0.0702)/50\times {10}^{-6}}=\frac{1}{2807.7}=3.56\times {10}^{-4}\text{S}.$$

The magnetizing reactance is $100$ pu, i.e. $52900\mathrm{\Omega }$, so $L_m=140.3\text{H}$ and $G_m=\mathrm{\Delta }t/2L_m=1.78\times {10}^{-7}$ S. The magnetizing branch is two thousand times weaker than the leakage it hangs beside. Hold that ratio; it is the whole reason the next section's tests work.

13.4 Where the numbers come from

Two factory tests fill in the entire parameter form, and both are simple enough to run in the simulator.

Short-circuit test. Short the LV winding and raise the HV voltage until rated current flows. Because the shorted secondary presents a path thousands of times stiffer than the magnetizing branch, the exciting current is negligible and the whole applied voltage falls across the leakage. So the applied voltage, in per unit, is the leakage impedance:

$$|Z{|}_{pu}=\frac{V_{\text{applied}}}{V_{\text{rated}}}\text{at rated current},r_{pu}=\frac{P_{sc}}{S_{\text{rated}}},x_{pu}=\sqrt{|Z{|}_{pu}^2-r_{pu}^2}.$$

These are the schema's x_hv_lv_pu and r_hv_lv_pu. Typical values are $x$ between 0.08 and 0.12 and $r$ between 0.002 and 0.01, so $x_{pu}\approx |Z{|}_{pu}$ to within a fraction of a percent and the distinction rarely matters for magnitude — but it matters a great deal for damping, as §13.6 shows.

Open-circuit test. Leave the LV winding open and apply rated voltage to the HV winding. Now only the magnetizing branch is in the circuit:

$$\mathtt{\text{i\_mag\_hv\_pct}}=100\frac{I_{oc}}{I_{\text{rated}}},\mathtt{\text{no\_load\_loss\_pu}}=\frac{P_{oc}}{S_{\text{rated}}}.$$

Nameplate to ohms. The transformer's own base is $S_{\text{rated}}$ with the rated line-to-line voltage of the side in question. From there,

$$Z_b=\frac{V_{LL}^2}{S_{3\varphi }},L=\frac{x_{pu}{Z}_b}{2\pi f},I_{\text{rated}}=\frac{S_{3\varphi }}{\sqrt{3}{V}_{LL}}.$$

For the running example, $I_{\text{rated}}={10}^8/(\sqrt{3}\cdot 230000)=251.0$ A on the HV side and 502.0 A on the LV side.

One convention deserves scrutiny. The short-circuit test measures the sum of the two leakages and can never separate them, yet the model needs a value for each because the magnetizing branch sits between them. NumaSim splits evenly, $R_{HV}=R_{LV}=r_{pu}/2$ and $X_{HV}=X_{LV}=x_{pu}/2$, each referred to its own winding base. For a linear model the split is invisible at the terminals — any split summing to $x$ gives the same short-circuit impedance. It stops being invisible the moment the core saturates, because it decides how much of the terminal voltage actually reaches the magnetizing branch. Keep that in mind through Chapter 14.

A three-winding transformer is the case where the split is measurable: three short-circuit tests give $x_{HL}$, $x_{HT}$, $x_{LT}$, and the star-point inversion

$$x_H=\frac{1}{2}(x_{HL}+x_{HT}-x_{LT})$$

with its two cyclic partners recovers all three arms uniquely. NumaSim's transformer_3w_3ph takes exactly those three pairs. Note that one of the three arms often comes out slightly negative — a real result of the inversion, not an error, and harmless in the network solution.

13.5 Three-phase: connections, vector groups, and zero sequence

A three-phase transformer is three legs of the model above, plus a decision about how the three windings on each side are joined. The schema offers Y (wye with a floating neutral), YN (wye with an accessible neutral, either brought out to a port or grounded internally), and D (delta), independently per side.

The choice changes what each winding sees. A wye winding spans line to neutral and carries $V_{LL}/\sqrt{3}$; a delta winding spans line to line and carries $V_{LL}$. Same core, same line voltage, different winding volts — so the connection is part of the turns ratio, not a detail on top of it.

It also creates a phase shift. A delta winding's voltage is a line-to-line quantity, and $V_{ab}$ leads $V_{an}$ by 30°, so putting a delta on one side and a wye on the other shifts the secondary by an odd multiple of 30°. That shift is what a vector group records, in units of 30° on a clock face: hv_lv_vector_clock = 1 means the LV side lags the HV side by 30°, 11 means it leads by 30°, 6 means 180°, and 0 means in phase. It maps one-for-one onto the load-flow angle column, $\text{ANG}=-30°\times \text{clock}$.

Where the 30° comes from. The line-to-line phasor $V_{ab}$ is the difference of two line-to-neutral phasors 120° apart; it is $\sqrt{3}$ times longer and leads $V_{an}$ by exactly 30°. A delta winding is wound across that phasor, a wye winding across $V_{an}$.

Be aware of a modelling limit here: NumaSim implements clocks 0 and 6 exactly, as a polarity flip on the ideal-ratio constraint. Clocks 1, 5, 7, and 11 are approximated to the nearest exact clock with a runtime warning, on the grounds that a well-formed Dy1 or YNd1 already gets its 30° from the delta connection itself — the clock parameter would be double-counting it. Set the connections correctly and the phase shift takes care of itself.

Zero sequence is where the connection matters most, and the three cases are cleanly different:

• A grounded wye (YN) offers a path: zero-sequence current flows in all three phases together and returns through the neutral.
• An ungrounded wye (Y) offers none. The star point floats, so the three currents must sum to zero and zero-sequence current simply cannot exist in that winding.
• A delta offers a path that goes nowhere. Zero-sequence current circulates around the closed triangle and never appears on the lines. This is why a delta winding is the standard way to give a system a zero-sequence sink without exporting it.

Neutral grounding is a further choice. hv_grounding / lv_grounding select solid bonding or an $R_n+j{X}_n$ impedance; because the zero-sequence current in the neutral is three times the per-phase value, an impedance $Z_n$ adds $3Z_n$ to the terminal zero-sequence impedance — the standard trick for limiting ground-fault current without floating the system.

Finally, the component assumes a three-limb core, which is a magnetic statement, not an electrical one: the three legs share a return path, so the three fluxes must sum to zero and zero-sequence flux has to escape through the tank and the air. The model captures this by giving every leg the same grounded core shunt at the star point. Five-limb and shell-form cores, which do provide a zero-sequence flux path, behave differently and are not represented.

13.6 Lab: the short-circuit and open-circuit test

Open the transformer short-circuit test in simulator →

The circuit is the whole factory test: an ideal source at 23 kV — one tenth of the 230 kV rating, the level §13.4 says should drive rated current through a 0.1 pu impedance — feeding the HV terminals of the running example, with all three LV terminals shorted. Saturation is off; at 0.1 pu applied voltage the core never comes near the knee.

Recover the leakage. Run it and read I_HVa after the transient settles, past about 0.3 s. It sits at 250.8 A RMS, against a rated 251.0 A. Now do what an engineer with only the waveform would do:

$$|Z{|}_{pu}=\frac{V_{\text{applied,pu}}}{I_{pu}}=\frac{0.100}{0.999}=0.1001\text{pu},$$

which is $\sqrt{r^2+x^2}$ for $r=0.005$, $x=0.100$. Check the turns ratio too: I_LVa settles at 501.3 A, exactly 1.999 times the HV current, recovering 230/115 from two ammeters.

Confirm the assumption the test rests on. Look at Imag_a: 0.217 A RMS, under a tenth of a percent of rated. The shorted secondary really has bypassed the core, exactly as the 2000:1 conductance ratio of §13.3 predicted.

Get $r$ and $x$ separately. The first cycles are not the answer, but they carry the extra information. The source closes at a voltage zero and current through an inductive branch cannot jump, so the waveform starts with a DC offset that lifts the first peak to 658.4 A at 8.1 ms — 1.855 times the settled peak of 354.9 A. That factor is $1+e^{-\pi R/X}$, which for $X/R=20$ is 1.855. The offset then decays with the loop's own time constant: it halves in 36.8 ms, so $\tau =36.8/\ln 2=53$ ms $=X/(\omega R)=0.1/(377\times 0.005)$. One run, both parameters — magnitude from the settled current, ratio from the decay.

Run the other half. Convert the circuit into an open-circuit test: delete the LV short, terminate LV through a large resistance instead (1 MΩ per phase is ample), and raise the source to the full 230 kV. The magnetizing current now swings 3.55 A peak, i.e. 2.51 A RMS, which is 1.00% of rated — the i_mag_hv_pct you typed in, measured back out.

Read the amplitude, though, not the peak. The magnetizing current arrives sitting on a 3.53 A DC offset that barely moves across the whole run (3.545 A at 0.1 s, 3.532 A at 0.5 s), because with the secondary open the flux loop is 140 H over 1.3 Ω and its time constant is close to two minutes. That stubborn offset is not an artefact. It is the same mechanism that makes transformer inrush decay so slowly, and it is the subject of the next chapter.

Two more things worth trying. Set x_hv_lv_pu = 0.06 and the settled current rises to 1.66 pu — the same applied voltage across a smaller impedance. Set the source phase = 90 so the circuit closes near a voltage crest, where the steady-state current is already passing through zero and no offset is needed: the first peak falls from 658 A to 366 A and the waveform is symmetric from the start. That is point-on-wave closing, and Chapter 14 turns it into the dominant variable of the whole inrush problem.

13.7 Summary

• A transformer is a coupled inductor pair, $\mathbf{v}=\mathbf{R}\mathbf{i}+\mathbf{L}d\mathbf{i}/dt$, and the T-equivalent ($L_1$, $L_2$, $L_m$, ideal ratio $a$) is an exact change of variables on that matrix, not an approximation.
• The matrix form is a bad parameterization: leakage appears as the difference of two nearly equal numbers, the conditioning gets worse as the core gets better, and the ideal-core limit is singular. The T-equivalent has none of those problems, matches the datasheet, and gives saturation a single branch to live in.
• Discretization reuses Chapter 2 verbatim: a series $R+L$ companion for each leakage, an inductor companion plus a loss conductance for the core, and an MNA constraint row for the ideal ratio.
• A short-circuit test gives $|Z|$ and $r$; an open-circuit test gives the magnetizing current and no-load loss. The 50/50 leakage split is a convention, invisible in the linear model and consequential once the core saturates.
• Wye and delta windings see different voltages and shift the phase by odd multiples of 30°; a grounded wye passes zero-sequence current, an ungrounded wye blocks it, and a delta circulates it internally.

13.8 Problems

Problem 13.1. A 50 MVA, 138/13.8 kV transformer has $x=0.09$ pu at 60 Hz. Compute the HV-side base impedance, the total leakage reactance in ohms, and the per-winding leakage inductance in henries under NumaSim's even split.

Solution 13.1

$Z_b=V_{LL}^2/S_{3\varphi }=138{000}^2/50\times {10}^6=380.9\mathrm{\Omega }$. Total leakage $X=0.09\times 380.9=34.28\mathrm{\Omega }$. The HV half is $0.045$ pu $=17.14\mathrm{\Omega }$, so $L_{HV}=17.14/(2\pi \cdot 60)=45.5\text{mH}$. The LV half is $0.045$ pu on the LV base, $Z_{b,LV}=13{800}^2/50\times {10}^6=3.809\mathrm{\Omega }$, giving $X_{LV}=0.171\mathrm{\Omega }$ and $L_{LV}=0.455\text{mH}$ — a hundredth of the HV value, as the square of the 10:1 turns ratio requires.

Problem 13.2. A short-circuit test on a 100 MVA unit draws rated current when 11.5 kV is applied to the 230 kV winding, and the wattmeter reads 600 kW. Find $r_{pu}$, $|Z{|}_{pu}$, $x_{pu}$, and $X/R$.

Solution 13.2

$|Z{|}_{pu}=11.5/230=0.05$ pu. $r_{pu}=P_{sc}/S_{\text{rated}}=0.6/100=0.006$ pu. $x_{pu}=\sqrt{{0.05}^2-{0.006}^2}=\sqrt{0.0025-0.000036}=0.0496$ pu. $X/R=0.0496/0.006=8.3$. Note how little the resistance changes the magnitude — 0.0496 against 0.05, a 0.7% correction — while dominating the DC-offset decay, whose time constant is $X/(\omega R)=22$ ms.

Problem 13.3. For the running example ($x=0.1$ pu, 1% magnetizing current), write the per-unit inductance matrix, then repeat it for a unit with 0.2% magnetizing current. Comment on what happens to the small eigenvalue.

Solution 13.3

At 1%: $X_{11}=1/0.01=100$, $L_1=0.05$, so $L_m=99.95$ and the matrix is $[[100,99.95],[99.95,100]]$. Eigenvalues $199.95$ and $0.05$.

At 0.2%: $X_{11}=1/0.002=500$, $L_1$ is unchanged at $0.05$, so $L_m=499.95$ and the matrix is $[[500,499.95],[499.95,500]]$. Eigenvalues $999.95$ and $0.05$.

The small eigenvalue is pinned at the leakage, 0.05, in both cases — it is a physical quantity that has nothing to do with the core. Only the large eigenvalue grows, so the condition number rises from 4000 to 20 000 purely because the magnetizing branch got weaker. Improving the core degrades the matrix while leaving the physics of interest untouched: a clear sign the matrix is the wrong set of coordinates.

Problem 13.4. A leakage branch has $R=1.32\mathrm{\Omega }$ and $L=70.2\text{mH}$. Compute $G_{RL}$ and $\alpha$ at $\mathrm{\Delta }t=50\mu \text{s}$ and again at $\mathrm{\Delta }t=10\mu \text{s}$. Which one changes more, and why?

Solution 13.4

At $\mathrm{\Delta }t=50\mu \text{s}$: $2L/\mathrm{\Delta }t=2808.0\mathrm{\Omega }$, so $G_{RL}=1/(2809.3)=3.560\times {10}^{-4}$ S and $\alpha =(2808.0-1.32)/(2808.0+1.32)=0.99906$.

At $\mathrm{\Delta }t=10\mu \text{s}$: $2L/\mathrm{\Delta }t=14040\mathrm{\Omega }$, so $G_{RL}=1/(14041)=7.122\times {10}^{-5}$ S and $\alpha =0.999812$.

$G_{RL}$ changes by a factor of five, essentially proportional to $\mathrm{\Delta }t$, because the inductive term dominates the denominator. $\alpha$ barely moves, from 0.9991 to 0.9998, because it measures how much current is lost to resistance in one step and a shorter step loses less. The physical decay is unchanged; only its per-step share differs.

Problem 13.5. A generator step-up transformer is specified Dy11 with the delta on the HV side. State the phase relationship between the two sides, and explain what happens to a zero-sequence current injected into the LV terminals from a downstream ground fault.

Solution 13.5

Clock 11 means the LV side leads the HV side by $11\times (-30°)=-330°$, i.e. $+30°$. The 30° comes from the delta winding spanning a line-to-line voltage while the wye winding spans line-to-neutral.

Zero-sequence current injected at the LV terminals flows in all three LV windings in phase. Their ampere-turns drive an in-phase current in the three HV delta windings, which — being in a closed triangle — circulates around the delta and never reaches the HV lines. The delta therefore acts as a zero-sequence sink: it accepts the current (so the LV system has a low zero-sequence impedance and a ground fault there draws real current) while completely isolating the HV system from it. This is why generator step-up transformers are wound this way.

Problem 13.6. Explain why NumaSim's even split of the leakage between the two windings has no effect on the short-circuit test result, but does affect an inrush simulation.

Solution 13.6

In the short-circuit test the magnetizing branch is negligible, so the circuit is the two leakages in series and only their sum matters. Any split with $X_{HV}+X_{LV}=x$ gives the same terminal impedance and the same current.

With the secondary open, the picture inverts. Almost no current flows, so almost no voltage is dropped on the HV leakage and the magnetizing branch sees nearly the full terminal voltage — and the LV leakage carries no current at all, so it is invisible. The split therefore has almost no effect in the open-circuit case either.

It matters in between, and inrush is exactly that case: the current is large, so the HV leakage drop is large, and the flux is what integrates the remaining voltage on the core. Moving leakage from the HV side to the LV side reduces the drop between the terminal and the core, admits more volt-seconds, and drives the core further over the knee. The split is a modelling choice with real consequences the moment saturation is active.

13.9 References

• P. Kundur, Power System Stability and Control, McGraw-Hill — Section 6.2 for transformer representation, per-unit conventions across a turns ratio, and the vector-group / phase-shift treatment.
• J. Arrillaga and N. R. Watson, Power Systems Electromagnetic Transients Simulation, IET Power and Energy Series 39 — Chapter 7 for the coupled-inductance model, its conditioning, and the trapezoidal companion of a multi-winding transformer.
• H. W. Dommel, Electromagnetic Transients Program (EMTP) Theory Book, Bonneville Power Administration — transformer branch formulation and the treatment of ideal ratio constraints.

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