AEE Electrical Engineering Core · Chapter 3

Transformers: Single-Phase, Three-Phase and Auto-Transformer

What to remember

A transformer changes voltage and current levels at the same frequency, with no moving parts. Its rating is in kVA because copper loss depends on current and iron loss depends on voltage.
EMF equation: E = 4.44 f N Φm (rms). Maximum efficiency occurs when copper loss equals iron loss. Fraction of full load at maximum efficiency = √(Pi / Pcu,fl).
An auto-transformer has one winding shared by primary and secondary. It saves copper in proportion to how close the voltage ratio is to 1, but it gives no electrical isolation.

1. Construction and working

A transformer has two or more windings on a laminated core. The core is made of CRGO (cold-rolled grain-oriented) silicon steel. Silicon raises resistivity and so cuts eddy-current loss. Grain orientation cuts hysteresis loss. Laminations are thin and insulated from each other.

Two types of construction are common. In the core type, the windings surround the core limbs. In the shell type, the core surrounds the windings. Power transformers are mostly core type with three limbs.

An alternating voltage on the primary drives an alternating flux in the core. This flux links the secondary and induces an EMF (Faraday's law). Only the changing flux induces an EMF, so a transformer does not work on DC.

EMF equation: E1 = 4.44 f N1 Φm and E2 = 4.44 f N2 Φm, where Φm is the peak flux in weber. Also Φm = Bm × A.

Transformation ratio: K = N2/N1 = E2/E1 = V2/V1 = I1/I2 for an ideal transformer.

For a fixed supply voltage, flux is proportional to V/f. If frequency falls at the same voltage, flux rises and the core saturates. Hence a transformer rated for 50 Hz must not be run on a lower frequency at full voltage.

2. Equivalent circuit and referred values

A real transformer has winding resistances R1, R2, leakage reactances X1, X2, and a no-load branch. The no-load branch has a core-loss resistance R0 (carries Iw) and a magnetising reactance X0 (carries Im). The no-load current I0 is about 2 to 5 percent of rated current. It is non-sinusoidal because of saturation and lags the voltage by a large angle, so the no-load power factor is low (about 0.1 to 0.2).

Referring values from secondary to primary: R2' = R2/K² and X2' = X2/K², where K = N2/N1. Referred to the primary: Re = R1 + R2', Xe = X1 + X2', Ze = √(Re² + Xe²). In general, impedance referred to a side is multiplied by the square of the turns ratio towards that side.

Voltage regulation = (V0 − VL)/V0. For small loads, the approximation is:

Regulation ≈ (I Re cosφ ± I Xe sinφ)/V × 100 percent. Use plus for lagging power factor and minus for leading power factor.

Regulation is maximum when tanφ = Xe/Re (lagging); its value then equals the per-unit impedance Z.
Regulation is zero when the load is leading and tanφ = −Re/Xe.

3. Tests on a transformer

Test	Applied to	Side usually used	Gives
Open-circuit (no-load)	Rated voltage, other side open	LV side	Core (iron) loss, R0, X0
Short-circuit	Reduced voltage so rated current flows, other side shorted	HV side	Full-load copper loss, Re, Xe
Sumpner's (back-to-back)	Two identical units, one supply and one low-voltage injection	Both	Full-load temperature rise, both losses
Polarity test	Small voltage on one winding	Either	Subtractive or additive polarity

In the OC test, the wattmeter reading equals the iron loss, because the copper loss at no-load current is negligible. In the SC test the voltage is only about 5 to 10 percent of rated, so the core loss is negligible and the wattmeter reading is the full-load copper loss.

4. Losses and efficiency

Iron loss = hysteresis loss + eddy-current loss. It is constant from no load to full load at fixed voltage and frequency.
Hysteresis loss ∝ f × Bm^n (Steinmetz index n is about 1.6 for modern steel). Eddy loss ∝ f² × Bm² × t², where t is lamination thickness.
Copper loss ∝ I², so it varies with the square of the load.

Efficiency = Output / (Output + Pi + Pcu).

For a load fraction x of full load: η = x S cosφ / (x S cosφ + Pi + x² Pcu,fl).

Condition for maximum efficiency: x² Pcu,fl = Pi, so x = √(Pi/Pcu,fl). Maximum efficiency occurs at a load where variable loss equals constant loss. It does not depend on the power factor, but its value does.

All-day (energy) efficiency = energy output in 24 hours / energy input in 24 hours. It is used for distribution transformers, which are on all day but loaded only part of the time. Such transformers are designed with low iron loss, because the core is energised for the whole day.

5. Three-phase transformers

A three-phase bank can be three single-phase units or one three-phase unit on a common core. A single three-phase unit is cheaper, lighter and smaller. A bank of three single-phase units is easier to repair, because one faulty unit can be replaced.

Common connections: Star-star (Yy), Delta-delta (Dd), Star-delta (Yd), Delta-star (Dy).

Connection	Line voltage ratio	Remarks
Y-Y	Same as turns ratio	Needs neutral; third-harmonic problems unless a tertiary delta is added
Δ-Δ	Same as turns ratio	No third-harmonic problem; one unit can be removed to give open-delta
Y-Δ	Turns ratio × (1/√3)	Used at the receiving-end substation to step down; 30° phase shift
Δ-Y	Turns ratio × √3	Used at the distribution end to step down, with neutral available for four-wire supply

In star, line voltage = √3 × phase voltage and line current = phase current. In delta, line voltage = phase voltage and line current = √3 × phase current.

Open-delta (V-V) connection: if one transformer of a delta bank is removed, the remaining two supply three-phase power at 57.7 percent (1/√3) of the original bank capacity. Utilisation of each of the two units is 86.6 percent.

Scott connection: two single-phase transformers (main and teaser) convert three-phase to two-phase. The teaser has 86.6 percent of the main turns and is tapped at the centre of the main.

Vector group: Written as, for example, Dyn11. Capital letter is HV winding connection, small letter is LV winding connection, n means neutral brought out, and the number is the LV phase displacement in units of 30° (clock method). Dyn11 means the LV leads the HV by 30°. Dyn11 is the common distribution transformer group. Yy0 has zero displacement. In Yd1, LV lags HV by 30°.

6. Parallel operation

Conditions for transformers to work in parallel:

1. Same voltage ratio (otherwise circulating current flows at no load).
2. Same polarity.
3. Same phase sequence and zero relative phase displacement (three-phase).
4. For proper load sharing: per-unit (percentage) impedances should be equal, with the same R/X ratio.

Load shared is proportional to kVA rating if per-unit impedances are equal. If they are unequal, the unit with the lower per-unit impedance takes a higher share and may overload first.

7. Auto-transformer

An auto-transformer has a single winding with a tap. Part of the winding is common to input and output. Let K = V_low/V_high (K < 1).

Power transferred by transformer action = (1 − K) × input kVA.
Power transferred directly by conduction = K × input kVA.
Copper required = (1 − K) × copper required for a two-winding transformer of the same rating. Copper saved = K (as a fraction).
Advantages: lower cost, lower losses, better regulation, higher efficiency, smaller size. All are best when K is near 1.
Disadvantages: no electrical isolation, so a fault on the HV side can reach the LV side; high short-circuit current because of low impedance.
Uses: starting of induction and synchronous motors, voltage boosting or bucking in distribution, variable voltage supply (variac).

8. Accessories and cooling

Conservator: takes oil expansion; breather with silica gel removes moisture from air entering.
Buchholz relay: gas-operated relay placed between the tank and the conservator. It detects incipient internal faults through gas collection (alarm) and severe faults through oil surge (trip). It works only on oil-filled transformers with a conservator.
Tap changer: changes turns to adjust voltage. It is usually on the HV side, where current is lower. OLTC means on-load tap changer.
Oil gives insulation and cooling. Bushings bring the terminals out. Explosion vent and temperature indicators are protective accessories.
Cooling methods: ONAN (oil natural, air natural), ONAF, OFAF, and AN for dry types.
Instrument transformers: CT secondary must never be open-circuited (very high voltage). PT secondary must never be short-circuited (very high current).
Inrush current: switching on can draw several times the rated current for a few cycles because of core saturation. It is rich in the second harmonic, which is used by differential protection to restrain tripping.

Exam traps

1. Transformer rating is in kVA, not kW, because losses do not depend on power factor.
2. OC test is done on the LV side and gives iron loss; SC test is done on the HV side and gives copper loss.
3. Maximum efficiency is at Pcu = Pi, not at full load, unless the design makes them meet there.
4. Maximum regulation is at tanφ = X/R, not at unity power factor.
5. Referring from secondary to primary divides impedance by K² (K = N2/N1); a careless step multiplies it.
6. Copper saved in an auto-transformer is K, but copper used is (1 − K) of the two-winding value.
7. Open-delta capacity is 57.7 percent of the bank, not 66.7 percent. Each unit is 86.6 percent used.
8. A CT is never open-circuited; a PT is never short-circuited.

One-liners

1. EMF per turn = 4.44 f Φm.
2. Core material: CRGO silicon steel, laminated.
3. Hysteresis loss varies with frequency; eddy loss varies with the square of frequency.
4. No-load current is small, rich in harmonics and lags the voltage by nearly 90°.
5. Efficiency is maximum when copper loss equals iron loss.
6. Distribution transformers are designed for good all-day efficiency.
7. Dyn11: LV leads HV by 30°.
8. Open-delta delivers 57.7 percent of the delta bank rating.
9. Scott connection gives two-phase from three-phase.
10. Buchholz relay sits between tank and conservator.
11. Sumpner's test needs two identical transformers and gives temperature rise.
12. Per-unit impedance is the same when seen from either side.

Practice questions

A transformer is rated in kVA rather than kW because
1. its output is always at unity power factor
2. copper loss depends on current and iron loss depends on voltage, not on power factor
3. it works only on DC supply
4. its efficiency is independent of load
Answer
B. copper loss depends on current and iron loss depends on voltage, not on power factor
Both losses are independent of load power factor, so kVA fixes the limit.
A 50 Hz transformer has 200 primary turns and a peak core flux of 0.02 Wb. The primary induced EMF (rms) is
1. 888 V
2. 444 V
3. 1256 V
4. 628 V
Answer
A. 888 V
E = 4.44 f N Φm = 4.44 × 50 × 200 × 0.02 = 888 V.
Which core material is used in power transformers to reduce core loss?
1. Cast iron
2. Copper-clad aluminium
3. Solid mild steel
4. CRGO silicon steel laminations
Answer
D. CRGO silicon steel laminations
Silicon raises resistivity and grain orientation lowers hysteresis loss; laminations cut eddy loss.
A 11 kV/400 V single-phase transformer has a voltage ratio (V1/V2) of
1. 0.0364
2. 2.75
3. 27.5
4. 36.4
Answer
C. 27.5
11000/400 = 27.5.
A 100 kVA, 2000 V/200 V single-phase transformer has a full-load secondary current of
1. 500 A
2. 250 A
3. 5000 A
4. 50 A
Answer
A. 500 A
I2 = 100000/200 = 500 A.
Maximum efficiency of a transformer occurs when
1. copper loss is zero
2. copper loss equals iron loss
3. the load is at full rated value always
4. iron loss is zero
Answer
B. copper loss equals iron loss
Variable loss equals constant loss at maximum efficiency.
A transformer has iron loss 1 kW and full-load copper loss 4 kW. At what fraction of full load is its efficiency maximum?
1. 0.25
2. 2.0
3. 0.5
4. 0.75
Answer
C. 0.5
x = √(Pi/Pcu) = √(1/4) = 0.5.
A transformer delivers 49 kW and has total losses of 1 kW. Its efficiency is
1. 99%
2. 96%
3. 97%
4. 98%
Answer
D. 98%
η = 49/(49+1) = 0.98.
Statements about the open-circuit test: 1. It is usually done on the LV side. 2. The wattmeter reading gives the full-load copper loss.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
OC test gives iron loss, not copper loss, so only statement 1 is correct.
The short-circuit test on a transformer gives
1. iron loss and magnetising current
2. only the turns ratio
3. only the polarity
4. full-load copper loss and equivalent impedance
Answer
D. full-load copper loss and equivalent impedance
At rated current with low voltage, core loss is negligible.
A transformer has R2 = 0.02 Ω on the secondary and N2/N1 = 0.1. R2 referred to the primary is
1. 0.0002 Ω
2. 2 Ω
3. 0.2 Ω
4. 20 Ω
Answer
B. 2 Ω
R2' = R2/K² = 0.02/0.01 = 2 Ω.
A transformer has per-unit resistance 2% and reactance 4%. Regulation at full load, 0.8 power factor lagging, is about
1. 2.4%
2. 3.2%
3. 4.0%
4. 5.6%
Answer
C. 4.0%
2×0.8 + 4×0.6 = 1.6 + 2.4 = 4.0%.
A transformer has 3% resistance and 4% reactance (per unit on its rating). Its maximum possible regulation is
1. 7%
2. 5%
3. 4%
4. 1%
Answer
B. 5%
Max regulation = Z = √(3²+4²) = 5%, at tanφ = 4/3 lagging.
Voltage regulation is zero for a transformer when the load power factor is
1. leading, with tanφ = Re/Xe
2. unity always
3. lagging at 0.8
4. zero lagging
Answer
A. leading, with tanφ = Re/Xe
The drop IRcosφ − IXsinφ cancels for a leading load at that angle.
Statements about iron loss: 1. Eddy-current loss varies with the square of frequency. 2. Hysteresis loss varies with the square of lamination thickness.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Eddy loss ∝ f²; hysteresis loss does not depend on lamination thickness (eddy loss does).
If the supply frequency is doubled with Bm constant, the eddy-current loss becomes
1. two times
2. four times
3. half
4. unchanged
Answer
B. four times
Eddy loss ∝ f², so 2² = 4.
A transformer designed for 50 Hz is run at 40 Hz at the same voltage. The peak flux becomes
1. 0.8 times
2. unchanged
3. 1.25 times
4. 1.5 times
Answer
C. 1.25 times
Φm ∝ V/f, so 50/40 = 1.25; the core may saturate.
The no-load current of a transformer is
1. large and in phase with voltage
2. purely resistive
3. equal to full-load current
4. small, non-sinusoidal and lagging the voltage by a large angle
Answer
D. small, non-sinusoidal and lagging the voltage by a large angle
Saturation makes it non-sinusoidal; no-load power factor is low.
The all-day efficiency is most important for
1. Instrument transformers
2. distribution transformers
3. Furnace transformers
4. Large generator transformers
Answer
B. distribution transformers
They are energised all day but lightly loaded much of the time.
A 400 V/300 V auto-transformer is compared with a two-winding transformer of the same rating. The fraction of copper saved is
1. 0.5
2. 0.33
3. 0.25
4. 0.75
Answer
D. 0.75
Saving = K = V_low/V_high = 300/400 = 0.75.
In an auto-transformer with input 10 kVA, 400 V to 200 V, the power transferred by transformer action is
1. 5 kVA
2. 10 kVA
3. 7.5 kVA
4. 2.5 kVA
Answer
A. 5 kVA
Transformed = (1 − K) × input = 0.5 × 10 = 5 kVA.
Statements about auto-transformers: 1. They give electrical isolation between the two sides. 2. They are economical when the voltage ratio is close to 1.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
They have one common winding, so no isolation; saving is large when K is near 1.
An auto-transformer is NOT suitable when
1. a variable AC voltage is needed
2. a small boost in voltage is required
3. isolation between primary and secondary is required
4. a motor needs reduced starting voltage
Answer
C. isolation between primary and secondary is required
It has a direct electrical connection between the sides.
Two transformers of 100 kVA and 200 kVA have equal per-unit impedance and share a 150 kVA load. The 200 kVA unit takes
1. 100 kVA
2. 75 kVA
3. 150 kVA
4. 50 kVA
Answer
A. 100 kVA
Load shares in the ratio of ratings, 1:2, so 50 and 100 kVA.
Which is NOT a condition for parallel operation of transformers?
1. Same polarity
2. Same phase sequence
3. Same voltage ratio
4. Same kVA rating
Answer
D. Same kVA rating
Ratings may differ; per-unit impedances should match for proper sharing.
In a Y-Y transformer with a turns ratio of 20:1 and a 11 kV line voltage on the primary, the secondary line voltage is
1. 190 V
2. 220 V
3. 550 V
4. 1100 V
Answer
C. 550 V
Line ratio equals turns ratio in Y-Y: 11000/20 = 550 V.
Three 10 kVA single-phase transformers in delta are changed to open delta. The three-phase capacity becomes about
1. 20 kVA
2. 17.3 kVA
3. 30 kVA
4. 15 kVA
Answer
B. 17.3 kVA
Open-delta = 0.577 × 30 = 17.3 kVA.
The Scott connection is used to convert
1. single-phase to three-phase
2. AC to DC
3. three-phase to two-phase
4. two-phase to single-phase only
Answer
C. three-phase to two-phase
Main and teaser transformers give two phases 90° apart.
In a Scott connection the main winding has 1000 turns. The teaser winding turns are
1. 500
2. 577
3. 1732
4. 866
Answer
D. 866
Teaser = 0.866 × main = 866 turns.
A Dyn11 transformer has the
1. LV leading the HV by 30°
2. LV lagging the HV by 30°
3. LV in phase with HV
4. LV leading the HV by 60°
Answer
A. LV leading the HV by 30°
The number 11 × 30° = 330° lag, which equals 30° lead.
Match the connection with a feature: (a) Y-Y (b) Δ-Δ (c) Y-Δ 1. Open-delta operation possible 2. Problem with third harmonic without a tertiary 3. 30° phase shift
1. a-2, b-3, c-1
2. a-1, b-2, c-3
3. a-3, b-1, c-2
4. a-2, b-1, c-3
Answer
D. a-2, b-1, c-3
Y-Y has third-harmonic trouble; Δ-Δ allows open-delta; Y-Δ gives 30° shift.
Buchholz relay is located
1. on the breather
2. inside the winding
3. in the pipe between the main tank and the conservator
4. in the secondary circuit
Answer
C. in the pipe between the main tank and the conservator
It collects gas rising from faults toward the conservator.
A Buchholz relay operates on
1. overvoltage on the line
2. gas or oil surge from internal faults
3. external short circuits only
4. low oil temperature
Answer
B. gas or oil surge from internal faults
Gas triggers alarm; oil surge triggers trip.
The tap changer is generally on the HV side because
1. the current is lower there
2. the HV side has fewer turns
3. the voltage is lower there
4. the winding is thicker
Answer
A. the current is lower there
Lower current makes switching easier and cheaper.
The oil in a transformer serves as
1. lubricant and insulation
2. magnetic core
3. insulation and coolant
4. fuel and coolant
Answer
C. insulation and coolant
It insulates the windings and carries heat to the tank.
ONAN cooling stands for
1. oil natural, air natural
2. oil neutral, air natural
3. oil normal, air normal
4. oil natural, air forced
Answer
A. oil natural, air natural
ONAN: natural circulation of oil and natural air cooling.
Statements about instrument transformers: 1. A CT secondary should never be open-circuited. 2. A PT secondary should never be short-circuited.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Open CT gives dangerous voltage; short PT draws very high current.
A transformer with 5% per-unit impedance is short-circuited at rated voltage. The steady fault current is
1. 2 times rated current
2. 20 times rated current
3. 0.05 times rated current
4. 5 times rated current
Answer
B. 20 times rated current
I_sc = 1/Z(pu) = 1/0.05 = 20 pu.
The inrush current when a transformer is switched on is rich in
1. fifth harmonic only
2. DC only
3. no harmonic
4. second harmonic
Answer
D. second harmonic
Differential relays use the second harmonic to restrain tripping.
The silica gel breather of a transformer is used to
1. remove moisture from air entering the conservator
2. prevent gas formation
3. cool the oil
4. measure oil level
Answer
A. remove moisture from air entering the conservator
Silica gel absorbs moisture.
Which test uses two identical transformers and determines temperature rise?
1. Open-circuit test
2. Sumpner's (back-to-back) test
3. Short-circuit test
4. Polarity test
Answer
B. Sumpner's (back-to-back) test
Both are connected and loaded by circulating power, so only losses are supplied.
Impedance referred from secondary to primary is multiplied by
1. N2/N1
2. N1/N2
3. (N2/N1)²
4. (N1/N2)²
Answer
D. (N1/N2)²
Z2' = Z2 (N1/N2)².
The per-unit impedance of a transformer is
1. different by the turns ratio
2. the same whether viewed from the primary or secondary
3. double on the HV side
4. four times on the LV side
Answer
B. the same whether viewed from the primary or secondary
Per-unit values use matching bases on each side, so they are equal.
Statements about iron and copper loss: 1. Iron loss is nearly constant from no load to full load. 2. Copper loss varies with the square of the load current.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are correct standard facts.
A three-phase unit compared with a bank of three single-phase units is
1. more expensive and heavier
2. easier to repair by replacing one unit
3. cheaper, lighter and smaller
4. used only for very small ratings
Answer
C. cheaper, lighter and smaller
A common core saves material; a bank is better for spare-unit repair.

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