Transformers

The four assumptions, each removing one real effect:

The voltage ratio, from Faraday's law applied to a common flux \(\phi\):

With \(k = 1\) the flux is genuinely common — every line links both windings — so the two voltages differ only by the turns count.

The current ratio, from the magnetomotive force. With \(L \to \infty\) the core needs no net mmf to carry the flux:

Note the two assumptions do different work: \(k = 1\) gives the voltage ratio, infinite inductance gives the current ratio.

Power is conserved exactly:

Both \(P\) and \(Q\) pass through unchanged — an ideal transformer stores no energy and dissipates none.

The impedance ratio follows by dividing:

Verify on the numbers. With \(a = 2400/240 = 10\) and \(\mathbf{Z}_L = 4\ \Omega\) on the secondary:

Note what the transformer does not change. The power factor is identical on both sides, since \(\mathbf{Z}_{in}\) is \(\mathbf{Z}_L\) multiplied by a real number. A transformer scales magnitudes; it never shifts phase.

The three rules, all following from Problem 1:

The last row is the check: whatever else is done, referring must not alter any power.

Applying at \(a = 10\), so \(a^2 = 100\):

The angle never changes. Multiplying by the real number \(a^2\) scales the magnitude and leaves the phase alone — so an inductive load stays inductive. Contrast Set 26, Problem 13, where reflection through a mutual coupling inverted the sign of the reactance.

Why referring is worth doing. Once every secondary element is referred, the ideal transformer can be deleted from the diagram and what remains is an ordinary single-loop circuit:

Every technique of Parts 1 and 2 then applies without restriction — the same benefit the T-equivalent gave in Set 26, Problem 19.

A worked check on power. Take 60 A through the 4 Ω secondary load:

The current is 10 times smaller and the resistance 100 times larger, so \(I^2R\) is unchanged. This is exactly why the exponent on \(a\) must be 2 and not 1.

Which side to refer to is a free choice, and worth making deliberately:

Mixing the two within one calculation is the commonest error in transformer work, and it produces answers wrong by factors of \(a\) or \(a^2\) — large enough to notice, but only if the magnitude is sanity-checked.

A linear transformer is simply the coupled pair of Set 26 with an iron-free (or at least unsaturated) core, so \(L_1\), \(L_2\) and \(M\) are constants. No idealisation is applied at all.

Loose coupling by transformer standards — an iron core would give 0.99 or better.

The two loop impedances:

The reflected impedance, from Set 26, Problem 13:

The currents:

Power balance — the essential check:

The efficiency is poor:

More power is lost in the primary resistance than reaches the load. This is what loose coupling costs: at \(k = 0.671\) a large primary current is needed to induce a modest secondary one, and that current dissipates in \(R_1\) whether or not it does useful work.

Contrast with a power transformer, where \(k \approx 0.998\) and the winding resistances are a fraction of a percent of the reactances. Problem 9 shows the same calculation there giving 98% efficiency. The difference is entirely in the coupling and the resistance-to-reactance ratio, not in any change of principle.

Start from the exact result, with the winding resistances set to zero to isolate the effect of the two limits:

aApply \(k = 1\) first, so \(M = \sqrt{L_1L_2}\) and \(\omega M = \sqrt{\omega L_1 \cdot \omega L_2}\):

Combining over a common denominator:

The two \(\omega^2L_1L_2\) terms cancel exactly — this is what perfect coupling buys.

bNow let \(L_2 \to \infty\). Divide numerator and denominator by \(j\omega L_2\):

and since \(L \propto N^2\):

Which limit does which job:

They are independent. A transformer can have excellent coupling and still draw large magnetising current if its core is small, or huge inductance with poor coupling if the windings are separated.

How good is the approximation in practice? The neglected term is \(\mathbf{Z}_L/(j\omega L_2)\). For the running example's 50 kVA unit with \(X_m = 4482\ \Omega\) referred to HV and a full-load referred impedance of about \(2400/20.83 = 115\ \Omega\):

A 2.6% correction — which is exactly the magnetising current as a fraction of load current, and Problem 6 confirms the figure independently.

The frequency dependence matters. \(\omega L_2 \gg |\mathbf{Z}_L|\) fails at low frequency, so the ideal model breaks down there — which is precisely the low-frequency roll-off of Problem 17, and why a 50 Hz transformer cannot be used at 5 Hz.

Restore the four assumptions one at a time, each adding one element:

The series/shunt distinction is the important one, and it follows from what each effect depends on: load current or applied voltage.

Why leakage reactance is a series element. Leakage flux links one winding only, so it behaves as ordinary self-inductance carrying that winding's current:

As \(k \to 1\) the leakage reactances vanish, recovering Problem 4's first limit.

Why core loss is a shunt resistance. Hysteresis and eddy-current losses depend on the peak flux, and the flux is fixed by the applied voltage through Faraday's law:

So core loss is essentially constant at constant voltage and frequency, whatever the load — which is why it is modelled as a resistance across the supply and why Problem 12 counts it for all 24 hours.

The full circuit, referred to the primary:

The shunt branch sits between the two series pairs, at the point where the induced emf \(\mathbf{E}_1\) appears.

Working the running example at full load, 0.8 pf lagging, with values from Problems 7 and 8:

The no-load current \(\mathbf{I}_0\) is only 2.6% of the load current and lags by more than 80° — almost purely magnetising, with a small in-phase component supplying the core loss.

Note the asymmetry of the shunt branch. \(R_c\) is very large (31 kΩ referred to HV) while \(X_m\) is much smaller (4.5 kΩ), so \(\mathbf{I}_0\) is dominated by its magnetising component. A no-load power factor of 0.14 is typical.

The simplification. Moving the shunt branch to the terminals means it sees \(\mathbf{V}_1\) rather than \(\mathbf{E}_1\), which allows the two series pairs to be combined:

The circuit becomes a single series impedance plus a shunt branch that no longer interacts with it — and the load calculation reduces to one loop.

The error introduced is the drop across \(R_1+jX_1\) caused by \(\mathbf{I}_0\) alone. Since \(\mathbf{I}_0\) is small, so is the error.

Quantify it on the running example. Comparing the two calculations at full load, 0.8 pf lagging:

A discrepancy of 0.023 percentage points in a regulation of about 2% — around 1% relative error, far inside the tolerance of the test data itself.

Why the error is second-order. The approximation misplaces a current of 2.6% of full load, and that current then flows through only half the series impedance. Two small factors multiply:

and the drop is itself only 2% of the voltage, so the effect on \(V_1\) is a few hundredths of a percent.

When it fails:

For large power transformers at or near rated load — the case that matters commercially — the approximation is excellent, which is why it is universal in practice.

A further simplification for regulation. Since the shunt branch is now across the supply, it affects neither the load current nor the series drop, so it can be ignored entirely for regulation calculations:

This is the form used in Problem 11. The shunt branch is still needed for efficiency, since it carries the core loss.

What the test does. With the secondary open, no load current flows, so the only current is \(\mathbf{I}_0\) through the shunt branch. The series elements carry so little current that their drop is negligible:

The copper loss at 5.41 A is around \((5.41)^2(0.0143) = 0.4\ \text{W}\) against 186 W measured — 0.2%, entirely negligible.

Why the test is done on the LV side. Full rated voltage must be applied so the flux — and hence the core loss — is normal. Applying 240 V is safer and needs no high-voltage supply, while the current drawn is modest.

The no-load power factor:

Very poor, as expected — the no-load current is almost entirely magnetising, and a transformer sitting energised with no load is a nearly pure inductive burden on the supply.

Resolve the current into components:

The shunt parameters, on the LV side:

Check \(R_c\) independently:

The second route is quicker and needs no power factor — worth preferring.

Refer to the HV side for use with the short-circuit data, multiplying by \(a^2 = 100\):

Both very large compared with the series impedance of about 2.3 Ω found in Problem 8 — a ratio of over 1000:1, which is the quantitative justification for Problem 6's approximation.

What the test does. With the secondary shorted, rated current flows at a small applied voltage — here 48 V, only 2% of rated. The flux is therefore about 2% of normal and the core loss, which goes roughly as flux squared, is negligible:

Core loss at 2% flux would be around \((0.02)^2(186) = 0.07\ \text{W}\) against 617 W measured.

Why the test is done on the HV side. Rated current must flow, and the HV winding's rated current is the smaller one:

Supplying 20.8 A at 48 V is easy; 208 A at 4.8 V needs heavy cabling. The two tests are done on opposite sides for complementary reasons — one needs full voltage, the other full current.

The equivalent impedance:

All referred to the HV side, because that is where the test was made. Dividing by 100 gives 0.01426 and 0.01814 Ω on the LV side.

The full-load copper loss. The test was made at essentially rated current (20.8 against 20.83 A), so

Had the test been at some other current, scaling by the square would be needed: \(P_{cu} = P_{sc}(I_{FL}/I_{sc})^2\).

What the test cannot give. It measures \(R_1+R_2'\) and \(X_1+X_2'\) as sums, and cannot separate them. For the approximate circuit this does not matter; where a split is needed, the usual assumption is

reasonable for a well-designed transformer with balanced windings, and the basis of the exact calculation in Problem 5.

The two tests together give a complete model from two measurements:

Neither test delivers appreciable power — the open-circuit test draws 186 W and the short-circuit 617 W from a 50 kVA machine. A complete characterisation costs about 1.6% of rated power.

The general formula. At a fraction \(x\) of rated kVA:

Note the two losses scale differently: core loss is constant, copper loss goes as \(x^2\). That difference is the whole content of Problem 10.

aFull load (\(x = 1\)):

bHalf load (\(x = 0.5\)):

Higher at half load than at full load. This is not a paradox — halving the load quarters the copper loss while the output only halves, so the loss-to-output ratio improves.

The efficiency curve:

A flat maximum near 55% of full load — Problem 10 locates it exactly.

Power factor matters too. At unity power factor and full load:

Better, because the same current delivers more real power while producing identical losses. Efficiency depends on the power factor even though the losses do not — the losses depend on current, the output on current times power factor.

Why efficiency is not measured directly. Measuring 98% efficiency by weighing input against output would require both wattmeters accurate to better than 0.1% — difficult and expensive. Measuring the losses instead needs only ordinary accuracy on two small readings, because a 1% error in 803 W is 8 W, or 0.02% of the output.

Write the efficiency and differentiate. Dividing through by \(x\):

The numerator is now constant, so \(\eta\) is maximised by minimising the bracketed term in the denominator.

Minimise:

An alternative proof by AM–GM, requiring no calculus. For positive quantities,

with equality exactly when the two terms are equal. The minimum total loss is therefore \(2\sqrt{P_cP_{cu}}\), which for the running example is \(2\sqrt{(186)(617)} = 678\ \text{W}\) per unit of \(x\) — a neat closed form.

For the running example:

so maximum efficiency occurs at 54.9% of full load:

The maximum is very flat. Comparing with Problem 9's table:

Missing the optimum by 20 percentage points of load costs less than a tenth of a percentage point of efficiency — so the design target need not be precise.

How designers use it. The ratio \(P_c/P_{cu}\) is a design choice, set by the relative amounts of iron and copper:

A distribution transformer is deliberately given low core loss relative to copper loss, so its efficiency peak sits at the light load it actually spends most of its life carrying — which Problem 12 evaluates properly.

Definition. Regulation is the rise in secondary voltage when the load is removed, as a fraction of the full-load value:

Working referred to the primary, the no-load secondary voltage is just \(|\mathbf{V}_1|\), since with no current there is no series drop.

a0.8 lagging. Take \(\mathbf{V}_2' = 2400\angle0°\) as reference:

The approximate formula avoids the complex arithmetic:

Agreeing to three figures. The approximation drops the quadrature component of the drop, which contributes only at second order.

b0.8 leading. The current now leads, so \(\sin\theta\) changes sign:

and exactly:

Essentially zero. Here the approximate formula is relatively poorer — 0.045% against 0.066% — because the two terms nearly cancel and the neglected quadrature part is no longer small by comparison. Cancellation amplifies relative error, exactly as in Set 26, Challenge C2.

Why leading loads improve regulation:

A sufficiently capacitive load gives negative regulation — the secondary voltage is higher on load than off it. The load's leading current produces a drop across \(X_{eq}\) that partly cancels rather than adds.

Why regulation matters. Consumers expect voltage within a few percent of nominal regardless of demand. A regulation of 2% means the voltage sags by 2% between no load and full load, which is tolerable; 10% would not be. Tap changers exist to correct the residual.

Why a different measure is needed. A distribution transformer is energised continuously but loaded only intermittently. Ordinary efficiency answers "how well does it work now"; the utility needs "how much energy did it waste today":

The key asymmetry:

Tabulate the cycle:

Copper energy for each period is \(x^2P_{cu} \times \text{hours}\) — for instance \((0.25)^2(617)(6)/1000 = 0.231\ \text{kWh}\).

The core loss runs all day:

Comparable with the entire day's copper loss of 5.27 kWh, despite the copper loss being 3.3 times larger at full load.

The all-day efficiency:

Slightly below the full-load figure of 98.03%, and well below the 98.33% peak — because six hours of energised idling produce loss with no output at all.

The design consequence. To maximise all-day efficiency, core loss must be reduced even at the cost of higher copper loss:

Amorphous-core distribution transformers cut core loss by 70% or more and are justified purely by this calculation over a 30-year life, despite costing more to build.

A useful check. If the transformer were de-energised during the idle six hours, the core energy would fall to 3.35 kWh and the all-day efficiency rise to 98.19%. Utilities do not do this — the switching cost and supply-continuity requirement outweigh 1.1 kWh — but the comparison shows exactly what idling costs.

Define the bases. Choose the rating as base power and the nominal voltage as base voltage; everything else follows:

Convert the series parameters:

Often quoted as a percentage: this is a 2% impedance transformer, a figure that appears on every nameplate.

Now compute on the LV side. The base changes:

and the referred impedance changes by the same factor of \(a^2\):

Identical. Both the impedance and the base scale by \(a^2\), so their ratio is invariant — which is the whole point of the system.

What per unit buys:

The first is decisive for power-system analysis: a network with a dozen transformers becomes a single per-unit circuit with no ratio boxes at all.

Reading physical quantities from per unit. The short-circuit test voltage was \(48/2400 = 0.02\) pu, which equals \(Z_{pu}\) — as it must, since the test drove 1.0 pu current through \(Z_{pu}\). Similarly:

50 times rated current on a bolted fault at the terminals — which is what protective devices must interrupt, and why the per-unit impedance is the single most important number on the nameplate.

The losses in per unit read directly as fractions:

And note \(P_{cu,pu} = R_{pu}\) exactly — because at 1.0 pu current, \(I^2R\) in per unit is just \(R_{pu}\). A satisfying consistency check.

The connection. The two windings are joined in series so the 240 V winding adds to the 2400 V winding:

The 240 V winding is now the series winding and the 2400 V winding the common winding, shared between input and output.

The windings keep their original ratings — the same copper, the same insulation, the same permissible current:

Trace the currents. The input current flows through the series winding, so it is limited to 208.3 A. At the output node the common winding's current adds:

The new rating:

An elevenfold increase from the identical windings, with no change to the copper or the core.

The general formula. With \(a = V_{\text{in}}/V_{\text{out}}\):

The advantage grows without limit as \(a \to 1\). Autotransformers are therefore used for small voltage changes — 11 kV to 10 kV, or 400 kV to 275 kV — and are pointless for large ratios, where \(a/(a-1) \to 1\).

Efficiency improves dramatically too. The losses are unchanged at 803 W, since the windings carry the same currents:

The same losses spread over eleven times the throughput. Losses fall from 2.0% to 0.18% of the rating.

The two paths. An autotransformer moves power in two quite different ways:

Only the first is limited by the core and windings; the second is limited only by the conductor's current rating.

The inductive share is what the windings actually transform — the original two-winding rating:

The conductive share is the remainder:

Over 90% of the power never enters the magnetic circuit at all. It simply flows through the shared winding from input to output, and that is precisely why so little iron and copper are needed.

The general split:

Confirming that the closer \(a\) is to 1, the more power is conducted and the greater the advantage — the same conclusion as Problem 14 from the other direction.

The safety consequence. Because the windings share a connection, there is no isolation:

The first row is the serious one. In the example, an open common winding would put 2640 V onto a circuit expecting 2400 V — and in a 240/120 V autotransformer, it puts the full 240 V onto a 120 V appliance.

Where autotransformers are used and avoided:

The requirement. Maximum power transfer needs the load, as seen by the source, to equal the source resistance — Set 22, Problem 14. The transformer supplies the \(a^2\) of Problem 1:

Without the transformer. With a source emf of 100 V rms behind 5000 Ω:

Almost all the power is lost inside the amplifier: the 8 Ω speaker is a near-short across a 5000 Ω source.

With the matching transformer:

Over 20 dB — the difference between an inaudible signal and a usable one.

Check the secondary conditions. At the match the speaker sees:

Taking \(V_1 = 50\) V because half the source emf drops internally at the match.

Other common matches:

Comparison with the L-network of Set 22, Challenge C1:

The transformer's wide bandwidth is decisive for audio, where a signal spans ten octaves. The L-network's frequency dependence would make it useless there.

The efficiency caveat. Matching gives maximum power to the load, not maximum efficiency — half the power still goes in the source resistance, exactly as Set 22 established. That is acceptable in a signal circuit and unacceptable in a power system, which is why matching transformers appear in amplifiers and never in distribution networks.

Two different elements limit the two ends, and both are ignored by the ideal model:

The low-frequency cut-off. The response falls 3 dB where the shunt reactance equals the circuit resistance:

Below this the magnetising branch draws more current than the load, and the transformer stops transforming — the failure of the \(L \to \infty\) limit identified in Problem 4.

The high-frequency cut-off:

Above this the leakage reactance dominates the series path — the failure of the \(k \to 1\) limit.

The bandwidth ratio has a memorable form:

The resistance cancels entirely. The bandwidth is set purely by the ratio of magnetising to leakage inductance — that is, by \(k\), since \(L_l \approx L_m(1-k^2)\). Here 5000:1 is about 12 octaves, comfortably covering the audio range of 20 Hz to 20 kHz.

What each end costs to improve:

The two are in tension, which is why a wide-band audio transformer is a difficult and expensive component while a 50 Hz power transformer — needing no bandwidth at all — is cheap.

Winding capacitance, ignored above, eventually resonates with the leakage inductance:

With \(C = 500\ \text{pF}\) this gives 159 kHz, well above \(f_H\) — so it produces a peak beyond the passband rather than inside it. In a poorly designed transformer the resonance can fall within the band and cause an audible peak.

The turns ratio is set by phase voltages, not line voltages — which is where the \(\sqrt3\) factors of Set 24 enter:

Yy and Dd give the same ratio, because both sides are treated alike. The mixed connections differ by \(\sqrt3\) in opposite directions — Dy needs a higher ratio, Yd a lower one.

Confirm one of them. For Dy, the primary winding sees the full line voltage while the secondary winding sees only the phase voltage:

A common error is to use \(11\,000/400 = 27.5\) for every connection. That is correct only for Yy and Dd.

The 30° phase shift. Mixed connections introduce a displacement between primary and secondary line voltages, because one side's line voltage leads its phase voltage by 30° and the other's does not — Set 24, Problem 3:

The clock-number notation gives the shift in units of 30°: "11" means the secondary lags by 330°, equivalently leads by 30°. Transformers of different vector groups cannot be paralleled — the 30° difference would drive a large circulating current.

Choosing a connection:

Dy is the standard distribution connection — HV delta, LV star with a neutral brought out to serve single-phase customers. Set 24, Problem 16's triplen harmonics circulate in the delta winding and never reach the supply.

A rating note. Whichever connection is used, the three-phase rating is three times the per-phase rating, and the line current follows from

so an 11 kV/400 V 500 kVA unit passes \(500\,000/(\sqrt3 \times 400) = 722\ \text{A}\) on the LV side regardless of connection.

Start from Faraday's law. The flux is the integral of the applied voltage:

With \(v = V_m\sin\omega t\) applied from \(t = 0\):

Compare with the steady state, where the constant of integration has settled to zero:

The steady-state flux swings symmetrically between \(\pm\phi_m\). The switched-on flux does not — it starts at zero and must swing a full \(2\phi_m\) upward.

The worst case is switching at a voltage zero with residual flux of the opposing sign:

With typical residual flux of 0.1 to 0.2 of \(\phi_m\):

Over twice the normal peak flux. The best case — switching at a voltage peak — gives no offset at all, because the required flux at that instant is zero and matches the residual.

Why this produces enormous current. A transformer core is deliberately operated close to the knee of its magnetisation curve, for economy of iron. Doubling the flux drives it deep into saturation:

In saturation the effective permeability collapses towards that of air, so \(L_m\) falls by orders of magnitude and the current is limited only by the winding resistance and leakage reactance.

How it decays. The offset is a transient, damped by the winding resistance with time constant \(L/R\). Because \(L\) is large and \(R\) small, the decay is slow:

Long enough to trip protection that cannot distinguish inrush from a genuine fault.

The mitigations:

The third is the standard protection solution, and it is a neat one: rather than avoiding the inrush, the relay identifies it by its harmonic signature and declines to trip.

The structure of the subject. Everything followed from Set 26's coupled equations plus two limits, then the systematic removal of those limits:

The running example in one table:

Nine derived quantities from two measurements — the economy of the loss-separation method.

The recurring theme is that each real effect appears where its cause dictates:

That one rule places all four elements, explains why the two tests separate cleanly, and explains why core loss must be counted for 24 hours while copper loss is counted only when the transformer is working.

The standing limitation. Every calculation assumed a single frequency. Problem 17 was the first crack: a transformer is a band-pass device, its response falling away at both ends for reasons the single-frequency model cannot express. Problem 19 was the second: saturation makes the core non-linear, so a sinusoidal voltage produces a non-sinusoidal current.

What comes next. Both cracks point the same way — treat \(\omega\) as a variable:

The last is closest to this set: Problem 5's equivalent circuit is a two-port, and Set 35 will show that its four parameters are one instance of a general description.