Single Phase Transformer Notes FE Sem 1 BEE | MU

Module 3 explains the single-phase transformer, the static machine that makes long-distance AC power transmission possible by stepping voltages up and down. It covers the working principle based on mutual induction, the core-type and shell-type constructions, the transformation ratio, and the difference between an ideal and a practical transformer. It then explains the iron and copper losses, efficiency and the main applications. The syllabus states that numericals are not expected, so the focus is on clear explanations, diagrams and comparisons.

Summary

Working principle: a transformer works on mutual induction between two windings placed on a common laminated magnetic core. When AC is applied to the primary winding, it sets up an alternating flux in the core. This flux links the secondary winding and induces an EMF in it by Faraday's law; it also induces a back EMF in the primary. Since the flux is common, the EMF per turn is the same in both windings. The transformer changes voltage and current levels at the same frequency with no moving parts. It cannot work on DC because a steady current produces no change of flux.

Construction and types: the core is built of thin silicon steel laminations (about 0.35 to 0.5 mm) insulated from each other to reduce eddy current loss; windings are of insulated copper. In a core-type transformer the windings surround the core, the core has two limbs and a single magnetic circuit, and the windings are split on both limbs. In a shell-type transformer the core surrounds the windings, it has three limbs with the windings on the central limb and a double magnetic circuit, giving lower leakage. Based on voltage, a transformer is step-up (N2 greater than N1) or step-down (N2 less than N1). Other single-phase types include the auto-transformer (single winding with a tapping), and instrument transformers (current and potential transformers).

Transformation ratio: the EMF induced in each winding is E = 4.44 f N phi_m. Hence E2/E1 = N2/N1 = K, the voltage (transformation) ratio. For an ideal transformer V2/V1 = N2/N1 = I1/I2 = K, because input VA equals output VA. If K is greater than 1 it is step-up, if less than 1 it is step-down, and if equal to 1 it is an isolation (one-to-one) transformer.

An ideal transformer has no losses (no winding resistance, no hysteresis or eddy current loss), no leakage flux, an infinitely permeable core so that negligible magnetising current is needed, and 100 percent efficiency. A practical (actual) transformer has winding resistances R1 and R2, leakage reactances X1 and X2 due to leakage flux, and core losses; it draws a no-load current I0 made up of a magnetising component Im (lagging V1 by 90 degrees, producing the flux) and a working component Iw (in phase with V1, supplying core loss). Hence its secondary terminal voltage falls with load and its efficiency is below 100 percent, typically 95 to 99 percent.

Transformer losses: (1) Iron or core losses occur in the core. Hysteresis loss Ph = Kh Bm^1.6 f V is due to repeated magnetisation reversal and is reduced by using silicon steel; eddy current loss Pe = Ke Bm^2 f^2 t^2 V is due to circulating currents in the core and is reduced by thin insulated laminations. Iron losses depend on voltage and frequency and are practically constant at all loads. (2) Copper losses I1^2 R1 + I2^2 R2 occur in the windings and vary with the square of the load current (variable loss). There are no friction or windage losses because there are no moving parts.

Efficiency is the ratio of output power to input power: eta = output/(output + iron loss + copper loss). Because the output depends on load and power factor while copper loss varies with the square of load, efficiency varies with load. Efficiency is maximum when copper loss equals iron loss (variable loss = constant loss). Transformers have the highest efficiency among electrical machines because they have no rotating parts. All-day efficiency (energy output over 24 hours/energy input) is used for distribution transformers which are energised all the time.

Applications: step-up transformers at generating stations raise the voltage for transmission, reducing line current and I^2 R losses; step-down transformers at substations and distribution points reduce the voltage for consumers; small transformers in chargers, adapters and electronic power supplies; isolation transformers for safety; auto-transformers for motor starting and variable AC supplies (variac); current and potential transformers for measurement and protection; impedance matching in audio circuits; welding transformers.

Mind map

How a transformer transfers energy

Important definitions

Formulas

Key points to remember

• Numericals are not expected in this module; answers should be explanatory with neat diagrams.
• A transformer works only on AC; on DC there is no change of flux, no back EMF, and the primary draws a very large current that can burn the winding.
• Frequency remains unchanged; power (VA) remains nearly the same; only voltage and current levels change.
• Laminations reduce eddy current loss; silicon steel reduces hysteresis loss.
• Iron loss is constant (depends on V and f); copper loss is variable (depends on load current squared).
• Efficiency is maximum when copper loss equals iron loss.
• Transformer ratings are in kVA, not kW, because losses depend on voltage and current and not on load power factor.
• Core type has windings around the core (good for high voltage, easier repair); shell type has the core around the windings (lower leakage, better mechanical strength, used for low voltage high current).
• In a practical transformer the no-load current is only about 2 to 5 percent of full-load current and lags V1 by nearly 90 degrees.
• Step-up at generating stations reduces transmission current and line losses.

Solved examples

Example 1. (Illustrative, numericals not asked in the MU paper.) A 2200/220 V single-phase transformer has 1000 primary turns. Find the transformation ratio and the number of secondary turns, and state whether it is step-up or step-down.

K = V2/V1 = 220/2200 = 0.1.

N2 = K N1 = 0.1 x 1000 = 100 turns.

Since K is less than 1, it is a step-down transformer.

Example 2. (Illustrative.) For the same 10 kVA, 2200/220 V ideal transformer, find the full-load primary and secondary currents and show that they are in the inverse ratio of the turns.

I1 = VA/V1 = 10000/2200 = 4.545 A.

I2 = VA/V2 = 10000/220 = 45.45 A.

I1/I2 = 4.545/45.45 = 0.1 = K = N2/N1. Hence current ratio is the inverse of the voltage ratio, because V1 I1 = V2 I2 in an ideal transformer.

Example 3. (Illustrative.) Show how the transformation ratio follows from the EMF equation for a transformer supplied at 50 Hz with a maximum core flux of 2 mWb, if N1 = 1000 and N2 = 100.

E1 = 4.44 f N1 phi_m = 4.44 x 50 x 1000 x 0.002 = 444 V.

E2 = 4.44 f N2 phi_m = 4.44 x 50 x 100 x 0.002 = 44.4 V.

E2/E1 = 44.4/444 = 0.1 = N2/N1 = K. Because the flux and frequency are common to both windings, the EMF per turn (0.444 V) is the same and the EMFs are directly proportional to the turns.

Example 4. (Illustrative.) A 10 kVA transformer has an iron loss of 100 W and a full-load copper loss of 200 W. Find its full-load efficiency at 0.8 power factor lagging, and comment on when maximum efficiency occurs.

Output = 10 x 0.8 = 8 kW = 8000 W.

Total loss = 100 + 200 = 300 W.

Efficiency = 8000/(8000 + 300) x 100 = 96.39 percent.

Maximum efficiency occurs when copper loss = iron loss = 100 W. Copper loss varies as (load fraction)^2, so the load fraction x = sqrt(100/200) = 0.707, i.e. at about 70.7 percent of full load.

Example 5. (Conceptual.) A transformer designed for 230 V, 50 Hz is connected by mistake to a 230 V DC supply. Explain what happens.

On DC the flux is constant, so no EMF (back EMF) is induced in the primary (d phi/dt = 0).

The primary current is then limited only by the very small winding resistance: I = V/R1, which is very large.

This causes heavy I^2 R heating and the core saturates; the insulation and winding may burn.

The secondary gets no induced EMF because there is no change of flux. Hence a transformer must never be connected to DC.

Important questions and answers

Q1. Explain the working principle of a single-phase transformer with a neat diagram. (5 marks)

A transformer consists of two windings, primary and secondary, wound on a common laminated silicon-steel core and magnetically coupled but electrically isolated.

When an alternating voltage V1 is applied to the primary, a small magnetising current flows and produces an alternating flux phi in the core. This flux links both windings. By Faraday's law of electromagnetic induction, it induces a self-induced EMF E1 = 4.44 f N1 phi_m in the primary (opposing V1) and a mutually induced EMF E2 = 4.44 f N2 phi_m in the secondary.

If a load is connected, current I2 flows in the secondary and power is delivered. Since the EMF per turn is the same, E2/E1 = N2/N1, so the voltage can be stepped up or down by choosing the turns ratio. The frequency remains unchanged. Diagram: core with primary on one limb connected to AC supply, secondary on the other limb connected to load, flux path shown in the core.

Q2. Compare core-type and shell-type transformers. (4 marks)

Core type: windings surround a considerable part of the core; single magnetic circuit; two limbs; windings split between both limbs; more leakage flux; easier to inspect and repair; natural cooling is better; preferred for high voltage.

Shell type: core surrounds the windings; double magnetic circuit; three limbs with both windings on the central limb (sandwich/interleaved coils); less leakage; better mechanical support against short-circuit forces; repair is difficult; preferred for low voltage and high current applications.

Diagrams of both constructions should be drawn.

Q3. What is transformation ratio? Show that V2/V1 = N2/N1 = I1/I2 for an ideal transformer. (4 marks)

Transformation ratio K is the ratio of secondary to primary turns, K = N2/N1.

EMF equation: E1 = 4.44 f N1 phi_m and E2 = 4.44 f N2 phi_m. Dividing, E2/E1 = N2/N1 = K.

In an ideal transformer there are no resistive or leakage drops, so V1 = E1 and V2 = E2. Hence V2/V1 = N2/N1 = K.

Also, with no losses, input VA = output VA: V1 I1 = V2 I2, so I1/I2 = V2/V1 = K.

Thus the voltages are in direct ratio and the currents in inverse ratio of the turns. K greater than 1 gives step-up, K less than 1 gives step-down.

Q4. Distinguish between an ideal and a practical transformer. (4 marks)

Ideal: zero winding resistance; no leakage flux (all flux links both windings); no core losses; core of infinite permeability so no magnetising current; efficiency 100 percent; V2/V1 exactly equal to N2/N1 at all loads.

Practical: windings have resistances R1 and R2 causing copper loss and voltage drop; leakage flux gives leakage reactances X1 and X2; hysteresis and eddy current losses occur in the core; a no-load current I0 (2 to 5 percent of rated) is drawn, with magnetising component Im and core-loss component Iw; efficiency is 95 to 99 percent; secondary voltage falls with load (regulation).

Q5. Explain the losses in a transformer and how they are minimised. (6 marks)

1. Core (iron) losses: occur in the magnetic core because of alternating flux.

(a) Hysteresis loss, Ph = Kh Bm^1.6 f V: energy lost in repeatedly magnetising and demagnetising the core; minimised by using high-grade silicon steel with a narrow hysteresis loop.

(b) Eddy current loss, Pe = Ke Bm^2 f^2 t^2 V: alternating flux induces circulating currents in the core which produce heat; minimised by making the core of thin laminations insulated with varnish, since loss is proportional to t^2.

Iron losses depend on supply voltage and frequency and remain practically constant from no load to full load (constant losses).

2. Copper losses: I1^2 R1 + I2^2 R2 in the winding resistances; they vary with the square of load current (variable losses); minimised by using copper conductors of adequate cross-section.

There are no mechanical losses in a transformer as it is a static device.

Q6. Define efficiency of a transformer. State and explain the condition for maximum efficiency. (5 marks)

Efficiency eta = output power/input power = V2 I2 cos phi/(V2 I2 cos phi + Pi + Pcu), where Pi is iron loss and Pcu is copper loss.

For a given secondary voltage and power factor, writing Pcu = I2^2 R02 (R02 = equivalent resistance referred to the secondary), eta is maximum when the denominator divided by I2 is minimum: d/dI2 [V2 cos phi + Pi/I2 + I2 R02] = 0, giving -Pi/I2^2 + R02 = 0, i.e. I2^2 R02 = Pi.

Thus efficiency is maximum when copper loss (variable loss) equals iron loss (constant loss). The load at which this occurs is x = sqrt(Pi/Pcu,fl) times full load. Transformers are designed so that maximum efficiency occurs near the load at which they usually operate.

Q7. Why is a transformer rated in kVA and not in kW? (3 marks)

Copper loss depends on current and iron loss depends on voltage, so the total loss (which decides the temperature rise and hence the rating) depends on volt-amperes, not on the load power factor. The manufacturer does not know the power factor of the load that will be connected. Therefore the rating is given in kVA, the product of voltage and current, which the transformer can carry safely irrespective of power factor.

Q8. Why is the transformer core laminated? Why does a transformer not work on DC? (3 marks)

The core is laminated to reduce eddy current loss: thin insulated laminations increase the resistance of the eddy current paths, and since eddy loss is proportional to the square of lamination thickness, it reduces sharply.

On DC the flux is constant, so no EMF is induced in either winding. Without back EMF, the primary current is limited only by its small resistance, so a very large current flows, overheating and possibly burning the winding. Hence a transformer works only on AC.

Q9. State the applications of single-phase transformers. (3 marks)

Step-down transformers in distribution networks to supply domestic consumers; small step-down transformers in mobile chargers, adapters, battery chargers and electronic power supplies; isolation transformers for safety in medical and test equipment; auto-transformers (variac) for variable AC supply and motor starting; current and potential transformers for measurement and protection; audio transformers for impedance matching; welding transformers; step-up transformers in microwave ovens and for high-voltage testing.

Q10. Explain the behaviour of a practical transformer on no load. (4 marks)

On no load the secondary is open, so I2 = 0. The primary draws a small no-load current I0 (2 to 5 percent of full-load current) which lags V1 by an angle phi0 close to 90 degrees.

I0 has two components: magnetising component Im = I0 sin phi0, lagging V1 by 90 degrees, which produces the core flux; and working (active) component Iw = I0 cos phi0, in phase with V1, which supplies the core losses (and a negligible copper loss).

No-load input power = V1 I0 cos phi0, which is practically equal to the iron loss. The no-load power factor is very low, typically 0.1 to 0.2. Phasor diagram: V1 reference, phi lagging V1 by 90 degrees, I0 between them close to phi, E1 and E2 in antiphase to V1.

Exam tips

• Draw a labelled diagram (core, primary, secondary, flux, supply and load) for the working principle; it carries marks.
• Learn the comparisons as tables: core versus shell type, ideal versus practical transformer.
• Remember the two types of iron loss, their formulas and how each is reduced.
• Memorise the maximum efficiency condition: copper loss = iron loss.
• Short reasoning questions (why kVA rating, why laminations, why not DC) are common 2-3 mark items.
• Since numericals are not expected, write answers in point form with diagrams rather than lengthy paragraphs.

FAQs

Are numericals asked on the transformer in FE BEE?

No. The MU syllabus states numericals are not expected for this module. Questions are on working principle, types, transformation ratio, ideal versus practical, losses, efficiency and applications.

Does a transformer change frequency or power?

No. Frequency stays the same, and power is transferred almost unchanged (minus small losses). Only voltage and current levels change, in inverse proportion.

Why is transformer efficiency so high?

A transformer has no moving parts, so there are no friction or windage losses; only iron and copper losses exist, giving efficiencies of 95 to 99 percent.

What is the difference between an auto-transformer and a two-winding transformer?

An auto-transformer has a single winding with a tapping, part of which is common to primary and secondary, so there is no electrical isolation. A two-winding transformer has separate, electrically isolated primary and secondary windings.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Basic Electrical & Electronics Engineering) and were prepared with AI help and checked.