By the end of this chapter you should be able to:
Give the reasons for using instrument transformers.
Describe the construction and connection of a current transformer.
Compute the ratio and phase-angle errors from the exciting current.
Explain why those errors grow as the primary current falls.
Explain why a CT must never be open-circuited, and estimate the voltage produced.
Define burden and compute it for a given secondary circuit.
Compute the errors of a potential transformer.
Interpret accuracy classes and describe turns compensation.
Why They Are Needed
A power system carries thousands of amperes at hundreds of kilovolts. No instrument can be connected directly to it.
Scaling. They reduce large currents and voltages to standard low values — commonly 5 A or 1 A for current, and 110 V for voltage — so that one design of instrument serves every rating.
Insulation and safety. The instrument and the operator are isolated from the high-voltage circuit.
Economy. Instruments, relays and their wiring are built for low voltage and modest current, and can be sited remotely in a control room.
Standardisation. Protective relays are made for one input rating and applied to any system through the appropriate transformer.
The Current Transformer
Primary in series with the line, carrying the full line current.
Primary often a single turn — sometimes just the busbar passing through a toroidal core.
Secondary wound with many turns and closed through an ammeter or relay of very low impedance.
The secondary is effectively short-circuited in normal service.
The primary current is imposed by the line, not by the CT.
The core flux is therefore very small — set only by the small secondary burden voltage.
The core operates at a tiny fraction of saturation, which is what makes the ratio accurate.
Dividing by \(N_p\) and writing the turns ratio \(n = N_s/N_p\):
The exciting current \(I_0\) is stolen from the primary before anything reaches the secondary. That single term is the source of every CT error.
In an ideal CT \(I_0 = 0\) and the ratio is exactly \(n\). In a real one the exciting current of Chapter 42 subtracts a small phasor, so the actual ratio differs slightly from the nominal and the secondary current is not quite in phase opposition with the primary.
Ratio and Phase-Angle Errors
Resolving \(I_0\) into its magnetising component \(I_m\) and its loss component \(I_w\), and letting \(\delta\) be the angle of the secondary burden, the standard results are:
where \(R\) is the actual ratio and \(\theta\) the phase angle between the reversed secondary current and the primary current.
| Error | Affects | Matters most for |
|---|---|---|
| Ratio error | The magnitude read by the instrument | Ammeters, energy metering |
| Phase-angle error | The apparent power factor | Wattmeters, energy meters, directional relays |
The phase-angle error is the subtler one. An ammeter cannot detect it at all, but a wattmeter measures \(VI\cos\phi\), so a few minutes of phase error produces a proportional error in recorded energy — and at low power factor the effect is magnified, since \(\cos\phi\) is then changing rapidly with angle.
Problem. A 1000/5 A current transformer has a turns ratio of 200. The exciting current, referred to the primary, has a magnetising component of 1.50 A and a loss component of 0.40 A. The burden power factor is 0.8 lagging. Find the actual ratio, the ratio error, the phase angle, and the actual secondary current at rated primary current.
Burden angle.
Actual ratio.
Ratio error.
Phase angle.
Actual secondary current.
Comment. The ammeter reads 4.993 A where it should read 5.000 A, a shortfall of 0.14 %. The reading is always low, because the exciting current is subtracted from the primary before transformation. A CT under-reads by its nature.
Both errors depend on \(I_m\) and \(I_w\), so the way to a more accurate CT is a better core — high-permeability nickel-iron rather than ordinary silicon steel, which can cut the exciting current by a factor of ten or more.
Why the Errors Grow at Light Load
Both error expressions carry \(I_s\) in the denominator, and the numerators do not depend on the load at all. The consequence is important and often overlooked.
Problem. For the CT of Example 56.1, tabulate the ratio error and phase angle at 100 %, 50 %, 20 % and 10 % of rated primary current, and check against a class 0.5 requirement.
| % rated | \(I_p\) | \(I_s\) | \(R\) | Ratio error | Phase angle |
|---|---|---|---|---|---|
| 100 % | 1000 A | 5.00 A | 200.288 | −0.144 % | 3.30′ |
| 50 % | 500 A | 2.50 A | 200.576 | −0.287 % | 6.60′ |
| 20 % | 200 A | 1.00 A | 201.440 | −0.715 % | 16.50′ |
| 10 % | 100 A | 0.50 A | 202.880 | −1.420 % | 33.00′ |
Sample working, at 20 % of rated.
Comment. Every error is exactly inversely proportional to the secondary current — halve the load and both errors double, precisely. The table is a sequence of doublings.
Against a class 0.5 limit of \(\pm0.5\,\%\), this CT passes at 100 % and 50 % of rated but fails at 20 % and 10 %. It could be sold as class 0.5 only if its accuracy were declared over a restricted range.
This is why metering CTs are specified with their accuracy stated at several load points, typically 5 %, 20 %, 100 % and 120 % of rated, and why a CT should be chosen so that the expected load falls in the upper part of its range rather than the bottom.
Never Open-Circuit a CT
This is the one rule about instrument transformers that every engineer is expected to know, and it follows directly from the governing equation.
The primary current is fixed by the line and cannot change. If the secondary is opened, \(I_s = 0\), and the equation gives
Every ampere that was being transformed now goes into magnetising the core instead. The exciting ampere-turns rise by a factor of several hundred, and the core is driven violently into saturation on both half cycles.
The flux wave becomes almost square, so its rate of change is enormous at each transition. Since \(e = -N_s\,\mathrm{d}\Phi/\mathrm{d}t\), the secondary EMF becomes a train of tall narrow spikes.
Danger to life. Peak secondary voltages of several kilovolts appear at terminals designed for a few volts.
Insulation failure of the secondary winding and of the connected wiring.
Overheating of the core, since the iron loss rises with the greatly increased flux.
Permanent magnetisation of the core, leaving residual flux that spoils the accuracy even after normal service is restored.
The rule: before disconnecting any instrument from a CT secondary, short-circuit the secondary terminals first. A shorting link or shorting-type test block is fitted for exactly this purpose. A short circuit is the CT's normal condition and does it no harm whatever.
Problem. The CT of Example 56.1 has 200 secondary turns and a core of 8 cm² cross-section saturating at 1.8 T. Its normal burden is 5 VA at 5 A. Estimate the normal flux density, and the peak secondary voltage if the secondary is opened while the primary carries 1000 A. Assume the saturated flux reverses in 5 % of a half cycle at 50 Hz.
Normal condition. A 5 VA burden at 5 A means a secondary EMF of about 1.0 V:
The core normally works at 1.56 % of its saturation density — a factor of 64 below the knee. That is why the ratio is so accurate.
Rise in exciting current.
Saturated flux and the peak EMF.
Comment. The secondary terminals, which normally carry 1 V, now produce peaks of about 1152 V — more than a thousand times as much, on a circuit an operator may be about to touch.
Note that the danger comes from the waveform, not from the RMS value. A near-square flux wave has near-vertical edges, and it is the steepness of those edges that generates the spike. Averaged over a cycle the energy is modest; the peak is lethal.
The estimate is deliberately rough — the transition fraction is not precisely known — but the conclusion is robust across any reasonable assumption. Whatever the detail, the answer is kilovolts.
Burden
The burden of an instrument transformer is the total impedance connected to its secondary, quoted either in ohms or, more usually, as the volt-amperes absorbed at rated secondary current.
A larger burden needs a larger secondary EMF to drive the current through it, which needs more flux, which needs more exciting current — and more exciting current means larger errors.
The burden angle \(\delta\) also appears in both error formulas, so the power factor of the burden matters as well as its magnitude.
This is why a CT's accuracy class is always quoted with a stated burden — "class 0.5, 15 VA" — and why the connecting leads must be counted. On a long run from switchyard to control room the leads can dominate the burden entirely, which is one reason 1 A secondaries are preferred to 5 A for long cable runs: the same lead resistance then absorbs \(5^{2} = 25\) times fewer volt-amperes.
Problem. A 5 A CT secondary supplies an ammeter of 0.10 Ω, a relay of 0.35 Ω, and connecting leads totalling 0.05 Ω. Find the total burden in ohms and in VA. Then find the effect of doubling the lead length, and compare with the same installation using a 1 A CT.
Total impedance and burden.
Doubling the lead length to 0.10 Ω:
The same leads with a 1 A CT. The lead resistance is unchanged at 0.10 Ω, but the current is a fifth:
Comment. Doubling the leads adds 1.25 VA — 10 % more burden for no change to the instruments at all, which can be enough to push a marginal CT out of its accuracy class.
The 1 A comparison is the decisive one for long runs. The same cable imposes 25 times less burden, because burden goes as the square of the current. This is why 1 A secondaries are standard in large substations where the control room may be hundreds of metres from the switchyard, while 5 A remains usual for short, self-contained installations.
The Potential Transformer
Primary in parallel across the circuit whose voltage is to be measured.
Secondary standardised at 110 V, feeding voltmeters, wattmeter voltage coils and relays.
The secondary is lightly loaded — high-impedance instruments draw very little current.
It behaves like an ordinary small power transformer running almost on no load.
It may safely be open-circuited, and that is its normal light condition.
It must never be short-circuited — the reverse of the CT rule.
PT Errors
Since a PT is an ordinary transformer on light load, its errors come from the impedance drop of Chapter 45 rather than from the exciting current — this is the voltage-regulation calculation applied to a measuring instrument.
where \(R_e\) and \(X_e\) are the equivalent resistance and reactance referred to the secondary. These are exactly Chapter 45's regulation expressions, now read as measurement errors.
The secondary voltage is therefore always slightly low, so a PT under-reads — the same direction as a CT, and for an analogous reason.
Problem. An 11000/110 V potential transformer supplies a 100 VA burden at 0.8 power factor lagging. Its equivalent resistance and reactance referred to the secondary are 0.30 Ω and 0.40 Ω. Find the actual ratio, the ratio error and the phase angle, and check against class 0.5.
Secondary current.
Impedance drop.
Actual secondary voltage and ratio.
Ratio error.
Phase angle.
Comment. The voltmeter reads 109.56 V where it should read 110.00 V. At \(-0.397\,\%\) the PT scrapes inside class 0.5, but with almost no margin — and a heavier burden would put it outside.
The remedy is the same as for any regulation problem: reduce the impedance by using more copper, or compensate the turns. Reducing the primary turns by about 0.4 % raises the secondary voltage by the same fraction and cancels the error at this burden — which is what manufacturers actually do, adjusting the turns so that the error is zero at the middle of the declared burden range.
Note also that the error depends on the burden's power factor through \(\delta\), so compensation can be exact at only one operating point. Accuracy classes are always declared over a range, never at a point.
Accuracy Classes and Compensation
| Class | Ratio error at rated | Typical use |
|---|---|---|
| 0.1 | ±0.1 % | Laboratory standards, calibration |
| 0.2 | ±0.2 % | Precision revenue metering |
| 0.5 | ±0.5 % | Ordinary revenue metering |
| 1.0 | ±1.0 % | Industrial metering, indication |
| 3.0 | ±3.0 % | Indicating instruments only |
| 5P, 10P | ±1 %, ±3 % ratio | Protection — accuracy up to 5 or 10 % composite error at the accuracy limit factor |
A metering CT must be accurate at and below rated current, and should saturate early on fault current so as to protect the instruments.
A protection CT must stay accurate at many times rated current, so it must not saturate until well beyond the fault level.
The two requirements are incompatible, so large installations use separate cores — often several cores on the same primary conductor, each with its own secondary and its own class.
The protection requirement is expressed by the accuracy limit factor (the multiple of rated current up to which the class is held, typically 10 or 20) and by the knee-point voltage — the secondary EMF at which a 10 % increase in voltage causes a 50 % increase in exciting current, marking the onset of saturation.
Because a CT always under-reads, the error can be cancelled by giving the secondary slightly fewer turns than the nominal ratio requires. A 1000/5 A CT with 200 turns might be wound with 199, so that the deficit introduced by the exciting current is offset by a surplus in the turns ratio.
The compensation is exact at only one current and one burden, so it is chosen to centre the error band over the declared operating range rather than to null it at a single point.
Summary and Key Formulas
Instrument transformers provide scaling, isolation, economy and standardisation, giving standard 5 A or 1 A and 110 V outputs.
Their design objective is accuracy of ratio and phase angle, not efficiency or regulation.
A CT has its primary in series with the line and its secondary effectively short-circuited. Its governing equation is \(I_p = nI_s + I_0\).
The exciting current is subtracted from the primary, so a CT always under-reads.
Both CT errors carry \(I_s\) in the denominator, so they are inversely proportional to load — halve the current and both errors double exactly.
Never open-circuit a CT. With \(I_s = 0\) the whole primary current becomes exciting current, the core saturates, and the spiked secondary EMF reaches kilovolts. Short the secondary before disconnecting anything.
Burden is the secondary impedance, quoted in VA. A larger burden means more flux, more exciting current and larger errors. Leads count, and 1 A secondaries impose \(25\times\) less lead burden than 5 A.
A PT has its primary across the circuit and behaves as a lightly loaded power transformer. It may be open-circuited but never short-circuited.
PT errors come from the impedance drop, so they are Chapter 45's regulation expressions read as measurement errors.
Metering and protection classes conflict: metering cores should saturate early, protection cores late. Large installations use separate cores.
| Quantity | Formula | Notes |
|---|---|---|
| CT ampere-turn balance | \(I_p = n\,I_s + I_0\) | source of every error |
| CT actual ratio | \(R = n + \dfrac{I_w\sin\delta + I_m\cos\delta}{I_s}\) | always \(\gt n\) |
| CT phase angle | \(\theta = \dfrac{I_m\cos\delta - I_w\sin\delta}{n\,I_s}\) rad | — |
| Ratio error | \(\dfrac{n - R}{R}\times100\,\%\) | negative: under-reads |
| Error scaling | \(\propto 1/I_s\) | doubles at half load |
| CT open circuit | \(I_0 = I_p\) | core saturates |
| Peak EMF estimate | \(\hat{e} = N_s\dfrac{2\Phi_{sat}}{t_{trans}}\) | kilovolts |
| CT burden | \(S = I_s^{2}Z\) | leads included |
| PT burden | \(S = V_s^{2}/Z\) | — |
| PT drop | \(I_s\left(R_e\cos\delta + X_e\sin\delta\right)\) | Chapter 45 again |
| PT phase angle | \(\dfrac{I_s\left(X_e\cos\delta - R_e\sin\delta\right)}{V_s}\) rad | quadrature term |
Common Mistakes
Opening a CT secondary to change an instrument. Short the terminals first, every time.
Short-circuiting a PT secondary. The opposite rule applies — a PT is a small power transformer.
Assuming CT error is a fixed percentage. The exciting current is a fixed subtraction, so the percentage error doubles when the load halves.
Expecting a CT to over-read. It always under-reads, since \(I_0\) is taken from the primary first.
Ignoring the leads in the burden. On a long run they can exceed the instruments themselves.
Quoting an accuracy class without a burden. The class is meaningless on its own.
Using a metering CT for protection. It saturates early by design, and will not reproduce fault current.
Using a protection CT for revenue metering. It is accurate at fault levels, not at a fraction of rated.
Neglecting phase-angle error for wattmeters. An ammeter cannot see it, but a wattmeter's reading depends on it directly.
Believing turns compensation removes the error entirely. It nulls it at one current and one burden only.
Chapter Review
For CT problems, compute the two numerators first — they do not change with load — then divide by the secondary current at each operating point.
P56.1 A 500/5 A CT has a turns ratio of 100. Referred to the primary, \(I_m = 2.0\) A and \(I_w = 0.5\) A. The burden power factor is 0.866 lagging. Find the actual ratio and the ratio error at rated current.
Show answer
\[\cos\delta = 0.866, \quad \sin\delta = 0.5\]\[I_w\sin\delta + I_m\cos\delta = (0.5)(0.5) + (2.0)(0.866) = 0.25 + 1.732 = 1.982~\mathrm{A}\]\[R = 100 + \frac{1.982}{5} = 100.3964\]\[\text{error} = \frac{100 - 100.3964}{100.3964}\times100 = -0.3948\,\%\]P56.2 For P56.1, find the phase angle in minutes.
Show answer
\[I_m\cos\delta - I_w\sin\delta = (2.0)(0.866) - (0.5)(0.5) = 1.732 - 0.25 = 1.482~\mathrm{A}\]\[\theta = \frac{1.482}{(100)(5)} = 2.964\times10^{-3}~\mathrm{rad} = 0.1698^{\circ} = 10.19\ \text{minutes}\]P56.3 For P56.1, find the ratio error at 25 % of rated current, and say whether the CT then meets class 1.0.
Show answer
At 25 %, \(I_s = 1.25\) A:\[R = 100 + \frac{1.982}{1.25} = 101.5856\]Outside \(\pm1.0\,\%\), so it fails class 1.0 at quarter load, though it passed comfortably at rated.\[\text{error} = \frac{100 - 101.5856}{101.5856}\times100 = -1.5608\,\%\]P56.4 A 5 A CT secondary carries an ammeter of 0.2 Ω and leads of 0.15 Ω. Find the burden in VA, and the burden if a relay of 0.4 Ω is added.
Show answer
\[Z = 0.20 + 0.15 = 0.35~\Omega \quad\Longrightarrow\quad S = (25)(0.35) = 8.75~\mathrm{VA}\]More than doubling the burden, which will roughly double the errors.\[Z = 0.75~\Omega \quad\Longrightarrow\quad S = (25)(0.75) = 18.75~\mathrm{VA}\]P56.5 A 6600/110 V PT supplies a 50 VA burden at unity power factor. With \(R_e = 0.5~\Omega\) and \(X_e = 0.7~\Omega\) referred to the secondary, find the ratio error.
Show answer
\[I_s = \frac{50}{110} = 0.4545~\mathrm{A}, \qquad \cos\delta = 1, \ \sin\delta = 0\]\[\text{drop} = (0.4545)(0.5) = 0.2273~\mathrm{V}\]\[V_s = 109.7727~\mathrm{V}, \qquad R = \frac{6600}{109.7727} = 60.1243\]At unity power factor only \(R_e\) contributes to the drop.\[\text{error} = \frac{60 - 60.1243}{60.1243}\times100 = -0.2068\,\%\]P56.6 For P56.5, find the phase angle.
Show answer
\[\theta = \frac{I_s\left(X_e\cos\delta - R_e\sin\delta\right)}{V_s} = \frac{(0.4545)(0.7 - 0)}{110}\]At unity power factor the reactance alone produces the phase error, as the resistance alone produced the ratio error.\[= \frac{0.3182}{110} = 2.893\times10^{-3}~\mathrm{rad} = 9.94\ \text{minutes}\]P56.7 Explain why a current transformer must never be open-circuited, and estimate the consequences.
Show answer
The governing equation is \(I_p = nI_s + I_0\), and the primary current is imposed by the line — the CT cannot change it.If the secondary is opened, \(I_s = 0\) and therefore
\[I_0 = I_p\]The entire primary current becomes exciting current, several hundred times the normal value. The core is driven hard into saturation on both half cycles, so the flux wave becomes almost square.
Since \(e = -N_s\,\mathrm{d}\Phi/\mathrm{d}t\) and a square wave has near-vertical edges, the secondary EMF becomes a train of tall narrow spikes, reaching kilovolts on terminals rated for a few volts.
Consequences: danger to life, insulation failure, core overheating from the greatly increased iron loss, and permanent magnetisation that spoils the accuracy afterwards.
The rule: short-circuit the secondary before disconnecting anything. A short circuit is the CT's normal condition and harms it not at all.
P56.8 Explain why CT errors grow as the primary current falls.
Show answer
Both error expressions have \(I_s\) in the denominator while their numerators depend only on the exciting current:The exciting current is set by the core and the burden, not by the load, so it is a fixed subtraction rather than a fixed fraction.\[R - n = \frac{I_w\sin\delta + I_m\cos\delta}{I_s}, \qquad \theta = \frac{I_m\cos\delta - I_w\sin\delta}{n\,I_s}\]The consequence: halving the primary current halves \(I_s\) and therefore doubles both errors, exactly. In Example 56.2 the ratio error ran −0.144 %, −0.287 %, −0.715 % and −1.420 % at 100 %, 50 %, 20 % and 10 % of rated — a sequence of doublings.
Practical significance: a CT comfortably inside class at rated current may be well outside it at a fifth of rated, which is where a metering CT on a lightly loaded feeder often operates. Accuracy is therefore declared at several load points, and the CT should be chosen so the expected load sits in the upper part of its range.
P56.9 Compare the CT and the PT: connection, normal secondary condition, source of error, and the safety rule for each.
Show answer
CT PT Primary In series with the line Across the circuit Primary quantity Current, imposed by the line Voltage, imposed by the system Secondary Effectively short-circuited Nearly open-circuited Normal flux Very small Full, as a power transformer Source of error Exciting current \(I_0\) Impedance drop Safety rule Never open-circuit Never short-circuit The unifying principle: each must be left in the condition its primary quantity imposes. A CT's current is fixed, so the secondary must offer a path; a PT's voltage is fixed, so the secondary must not draw heavy current. Both rules are the same rule seen from the two sides.
P56.10 Why do metering and protection CTs need different cores?
Show answer
Because the two duties want opposite saturation behaviour.Metering. Accuracy is needed at and below rated current, where the meter actually operates. On a fault the CT should saturate early, limiting the secondary current and protecting the delicate instruments from damage.
Protection. Accuracy is needed at many times rated current, since that is when the relay must operate and must see the true fault current. The core must not saturate until well beyond the fault level.
A core that saturates early cannot reproduce fault current; a core that saturates late offers the instruments no protection. The requirements are incompatible in one core.
The solution is several cores on the same primary conductor, each with its own secondary and its own class — a metering core of class 0.5 and one or more protection cores of class 5P or 10P, specified by their accuracy limit factor and knee-point voltage.
MCQ 1. A current transformer primary is connected:
(a) across the line (b) in series with the line (c) to earth (d) to the neutralShow answer
(b) in series with the line, carrying the full line current.MCQ 2. The standard CT secondary rating is:
(a) 110 V (b) 5 A or 1 A (c) 100 A (d) 240 VShow answer
(b) 5 A or 1 A.MCQ 3. CT ratio error arises from:
(a) copper loss (b) the exciting current (c) leakage flux only (d) the burden aloneShow answer
(b) the exciting current, subtracted from the primary before transformation.MCQ 4. If the primary current halves, the CT ratio error:
(a) halves (b) is unchanged (c) doubles (d) quadruplesShow answer
(c) doubles — the exciting current is a fixed subtraction.MCQ 5. A CT secondary must never be:
(a) short-circuited (b) open-circuited (c) earthed (d) loadedShow answer
(b) open-circuited, or the whole primary current becomes exciting current.MCQ 6. A PT secondary must never be:
(a) short-circuited (b) open-circuited (c) earthed (d) unloadedShow answer
(a) short-circuited — the opposite rule to the CT.MCQ 7. Burden of a CT is measured in:
(a) amperes (b) volt-amperes (c) watts only (d) henriesShow answer
(b) volt-amperes, at rated secondary current — or equivalently in ohms.MCQ 8. Changing from a 5 A to a 1 A secondary reduces the lead burden by a factor of:
(a) 5 (b) 10 (c) 25 (d) 125Show answer
(c) 25, since burden goes as \(I^{2}\).MCQ 9. PT errors arise chiefly from:
(a) the exciting current (b) the impedance drop (c) saturation (d) harmonicsShow answer
(b) the impedance drop — Chapter 45's regulation, read as an error.MCQ 10. A metering CT should, on a heavy fault:
(a) stay accurate (b) saturate early (c) open-circuit (d) reverse polarityShow answer
(b) saturate early, to protect the instruments — the opposite of a protection CT.
Give the reasons for using instrument transformers.
Derive the CT ratio and phase-angle errors from the ampere-turn equation.
Explain why CT errors grow as the primary current falls.
Explain why a CT must never be open-circuited, and estimate the voltage produced.
Define burden and explain why the connecting leads matter.
Derive the PT errors and relate them to voltage regulation.
Compare the CT and the PT under the headings of connection, secondary condition, error source and safety rule.
Explain why metering and protection CTs need different cores.
This chapter closes Part 3. The transformer has been followed from the ideal model of Chapter 41 through no-load and on-load behaviour, losses, regulation, testing, efficiency and autotransformers, parallel operation, the practical matters of inrush, tapping and cooling, the per-unit system, and the whole of three-phase working — connections, open delta, vector groups, harmonics, the Scott connection, and finally the instrument transformers on which measurement and protection rest.
Two ideas have run through all of it. The first is that flux is the integral of voltage, which gave the EMF equation, the shape of the magnetising current, the inrush transient, and the saturation of an open-circuited CT. The second is ampere-turn balance, which gave the current ratio, the equivalent circuit, the autotransformer's uprating, and every result in this chapter.
Part 4 opens on the three-phase induction motor, the most widely used machine in the world. Its theory is a direct continuation of what has just been completed: an induction motor is a transformer whose secondary is free to rotate.
The stator winding sets up the rotating magnetic field of Chapter 22. The rotor, acting as a short-circuited secondary, has EMF induced in it exactly as in a transformer — but because the rotor turns, the frequency of that induced EMF is not the supply frequency but \(sf\), where \(s\) is the slip.
Everything follows from that one modification. The equivalent circuit is the transformer circuit of Chapter 44 with the rotor resistance divided by the slip; the air-gap power divides in the ratio \(s : (1-s)\) between rotor copper loss and mechanical output; and the torque-slip curve, with its maximum torque independent of rotor resistance, follows from the same maximum-power-transfer argument met in Chapter 4.