Transformers

In short

A transformer is a static device that transfers electrical energy between two or more windings by mutual induction at constant frequency, changing voltage and current in inverse proportion while keeping power (almost) unchanged. It makes high-voltage transmission possible, so it sits at the heart of every Indian substation, from 765 kV POWERGRID lines down to the 11 kV/433 V pole-mounted distribution transformer on a street corner. It is the highest-scoring and most frequently asked unit in Electrical Machines - GATE EE asks regulation, efficiency, equivalent circuit and test-based numericals almost every year. Interviewers at NTPC, POWERGRID, state discoms, BHEL, Siemens, ABB, Schneider, CG Power, L&T and TRANSCO ask about losses, cooling, vector groups, CT/PT selection and Buchholz relays. Mastering it also underpins induction motor theory, since the motor is a rotating transformer.

Explained

Construction, principle and classification

• Principle: an alternating current in the primary creates an alternating flux in the core; the flux links the secondary and induces an EMF (Faraday's law). There is no electrical connection and no change of frequency - only mutual induction.
• Core material: cold-rolled grain-oriented (CRGO) silicon steel, 0.23-0.35 mm laminations, insulated and stacked; silicon raises resistivity (lower eddy loss) and lamination breaks eddy current paths. Amorphous-core distribution transformers give far lower no-load loss and are used by several Indian discoms.
• Core type: windings surround the core; two limbs, one yoke pair; less iron, more copper; easier insulation and repair - preferred for high-voltage, high-capacity units.
• Shell type: the core surrounds the windings; central limb carries full flux; mechanically stronger against short-circuit forces; used in some HV and furnace transformers.
• Step-up / step-down: decided by turns ratio, not by construction - the same transformer can work either way.
• By use: power (transmission, 100% loaded continuously), distribution (11 kV/433 V, lightly and variably loaded), instrument (CT, PT), autotransformer, welding, furnace, isolation, pulse, converter/traction transformers (25 kV AC traction in Indian Railways).
• Parts of an oil-filled transformer: tank, mineral insulating oil (IS 335 / IEC 60296), HV and LV windings (concentric; LV inside, nearer the core, for cheaper insulation), conservator, breather with silica gel, Buchholz relay, radiators/fans, bushings, tap changer (off-circuit OCTC or on-load OLTC), explosion vent, oil and winding temperature indicators, Marshalling box.
• Why windings are concentric and interleaved: reduces leakage reactance; interleaved/sandwiched discs in shell type further cut leakage.
• Dry-type (cast resin) transformers: no oil, so fire-safe - used inside buildings, malls, metro stations, hospitals (Class F/H insulation).

EMF equation, ideal transformer and transformation ratio

• Flux φ = φm sin ωt. Induced EMF per turn is the rate of change of flux; the RMS value works out to 4.44 f φm, giving E = 4.44 N f φm.
• Turns ratio: E1/E2 = N1/N2 = a. In an ideal transformer (no losses, no leakage, infinite permeability) V1/V2 = N1/N2 = I2/I1, so VA in = VA out.
• Impedance transformation: an impedance Z2 on the secondary appears as a²Z2 on the primary - the basis of impedance matching and of referring the whole machine to one side.
• Flux is load-independent: V1 ≈ E1 = 4.44 N1 f φm, so with supply voltage and frequency fixed, the core flux (and hence iron loss) stays nearly constant from no load to full load. The secondary's demagnetising MMF is instantly balanced by extra primary current.
• Volts per turn is the same for both windings - a quick design check.
• No-load condition: the primary draws a small current I0 (2-6% of full load) with two components - Iμ (magnetising, lags V1 by 90°, sets up the flux) and Iw (core-loss/working component, in phase with V1). I0 = √(Iμ² + Iw²); no-load power factor is very low (0.1-0.3).
• Magnetising current is non-sinusoidal (peaky) because of the non-linear B-H curve and saturation; it is rich in third harmonic - the reason delta windings or a tertiary winding are needed in three-phase banks.

Equivalent circuit, phasor diagram and tests

• Parameters: R1, X1 (primary resistance and leakage reactance), R2, X2 (secondary), R0 (core loss) and X0 or Xm (magnetising) in the shunt branch.
• Referred to primary: R2' = a²R2, X2' = a²X2; equivalent resistance Req1 = R1 + a²R2, equivalent reactance Xeq1 = X1 + a²X2, Zeq1 = √(Req1² + Xeq1²). The approximate equivalent circuit shifts the shunt branch to the input terminals.
• Open-circuit (no-load) test - done on the LV side with HV open, at rated voltage: gives iron (core) loss Wi, R0 = V1²/Wi, Iw = Wi/V1, Iμ = √(I0² − Iw²), X0 = V1/Iμ. Measures the shunt branch; low power factor, so a LPF wattmeter is needed.
• Short-circuit (impedance) test - done on the HV side with LV shorted, at rated current and reduced voltage (5-10%): gives full-load copper loss Wcu, Zeq = Vsc/Isc, Req = Wsc/Isc², Xeq = √(Zeq² − Req²). Measures the series branch; flux and hence iron loss are negligible.
• Why these two sides: OC on LV so the applied voltage and instruments are manageable; SC on HV so the reduced voltage is easy to set and rated current is small.
• Sumpner's (back-to-back) test: two identical transformers, primaries in parallel across rated supply, secondaries in phase opposition; gives both iron and copper loss simultaneously under full-load heating - a true heat-run test without dissipating full output power.
• Polarity test (additive/subtractive) before parallel operation; ratio test, winding resistance test, insulation resistance and tan-delta test, induced and applied over-potential test, impulse test, magnetic balance test, oil BDV and DGA (dissolved gas analysis) are the routine and type tests per IS 2026 / IEC 60076.

Losses, efficiency and voltage regulation

• Iron (core) loss = hysteresis + eddy current, roughly constant: Ph = Kh·f·Bm^1.6 (Steinmetz exponent 1.6-2), Pe = Ke·f²·Bm²·t². Reduced by CRGO steel, thin laminations, lower flux density.
• Copper (I²R) loss varies with the square of the load current; stray loss in tank and clamps and dielectric loss in insulation are small.
• Efficiency η = output/(output + Wi + Wcu). Maximum efficiency occurs when variable loss = constant loss, i.e. copper loss = iron loss, at a load fraction x = √(Wi/Wcu,FL).
• Power transformers are designed with Wi/Wcu close to 1 (max efficiency near full load); distribution transformers with a much lower ratio (max efficiency at 50-75% load) because they are lightly loaded for most of the day. Typical efficiencies: 97-99% for distribution, above 99.5% for large power transformers.
• All-day (energy) efficiency = kWh output in 24 h / kWh input in 24 h - the correct measure for a distribution transformer, where iron loss runs 24 hours but copper loss follows the load curve. It is normally lower than the commercial (full-load) efficiency.
• Voltage regulation = (V2,no-load − V2,full-load)/V2,full-load, expressed as a percentage - a measure of how much the terminal voltage droops with load. Approximate formula: %R ≈ (I2(R cos φ ± X sin φ)/V2) × 100, '+' for lagging and '−' for leading loads.
• Zero regulation occurs at a leading power factor where tan φ = R/X; at a capacitive load the secondary voltage can rise above no-load value (negative regulation).
• Indian loss norms: IS 1180 (Part 1) specifies maximum no-load and load losses for distribution transformers up to 2500 kVA, 33 kV, with star-rating / efficiency levels (Level 1, 2, 3 ...); BEE star labelling for distribution transformers ties procurement by discoms to these loss figures. Exact loss tables and level names have been revised more than once - cite the current amendment rather than memorised numbers.
• Per-unit impedance: Zpu = Z·Ibase/Vbase = Z·Sbase/Vbase²; % impedance = SC test voltage as % of rated voltage. Typically 4-6% for distribution, 8-15% for large power transformers. Low %Z means better regulation but higher fault current.

Autotransformer, instrument transformers and special points

• Autotransformer: a single tapped winding; part is common to both sides, so energy is partly transferred conductively and partly inductively. Saving of copper = (1 − 1/a) × that of a two-winding transformer; the common winding carries (I2 − I1).
• Advantages: smaller, cheaper, lower losses, better regulation and efficiency, compact - used as 400/220 kV interconnecting transformers (ICT), motor starters, variacs (dimmerstats) and railway traction feeds.
• Disadvantage: no electrical isolation; a break in the common winding puts full HV on the LV side; higher short-circuit current. Only economical when the ratio is near unity (a < 2 roughly).
• Current transformer (CT): secondary nearly short-circuited through a 1 A or 5 A burden; never open-circuit a live CT - the loss of the secondary's demagnetising MMF saturates the core and induces dangerously high voltage. Classes: metering (0.2S, 0.5S) and protection (5P, 10P, PS class for differential), with accuracy limit factor (ALF) and knee-point voltage. Standard: IS 2705 / IEC 61869-2.
• Potential/voltage transformer (PT/VT): secondary 110 V (line) typically, high-impedance burden, never short-circuited; electromagnetic or capacitive (CVT) at EHV. Standard: IS 3156 / IEC 61869-3.
• Burden, ratio error and phase angle error are the key accuracy terms for both.
• Inrush current: switching on at an instant of unfavourable residual flux drives the core deep into saturation; the first-cycle peak can be 6-10 times rated current, is rich in second harmonic and DC, and decays in cycles to seconds. Differential relays use second-harmonic restraint to avoid tripping on it.
• Parallel operation - conditions: (1) same voltage ratio and tapping, (2) same polarity (mandatory), (3) same phase sequence and phase displacement/vector group for three-phase, (4) same per-unit impedance, and (5) impedances with the same X/R ratio. Unequal ratios cause circulating current at no load; unequal pu impedance causes load sharing in inverse proportion to impedance, so one unit may overload before the other is full.
• Effect of frequency and voltage: operating a 50 Hz transformer at 60 Hz with the same voltage lowers flux (safe, lower iron loss, higher kVA possible); a 60 Hz unit on 50 Hz at the same voltage raises flux by 20% and may saturate - V/f must be respected.
• Cooling classes (first letter internal medium, then its circulation, then external medium and its circulation): ONAN (oil natural, air natural), ONAF, OFAF (oil forced, air forced), ODAF (directed oil), OFWF; small units up to about 1 MVA use AN or ONAN, large units ONAN/ONAF/OFAF dual or triple rated.
• Protection: Buchholz relay (gas-actuated, only for conservator-type oil transformers - alarm on gas accumulation, trip on surge), differential protection (87T), REF, overcurrent and earth fault, oil and winding temperature trips, pressure relief device, and neutral earthing per IS 3043. Tap changing, earthing and clearances are governed by CEA (Measures relating to Safety and Electric Supply) Regulations, 2023 and the Electricity Act, 2003.
• Tertiary (stabilising) winding provides a path for third-harmonic and zero-sequence current and can supply station auxiliaries.

Key points

• Transformer - static machine, transfers energy by mutual induction at constant frequency; changes V and I, not power or frequency.
• EMF equation - E = 4.44 f N φm; volts per turn is identical on both windings.
• Flux ≈ constant from no load to full load because V1 ≈ E1 with fixed supply voltage and frequency; so iron loss is constant.
• No-load current - I0 = 2-6% of rated, at 0.1-0.3 pf lagging; two components, magnetising Iμ and core loss Iw.
• Magnetising current is peaky and third-harmonic rich due to core saturation.
• Laminations and CRGO cut eddy current loss; eddy loss ∝ f²Bm²t², hysteresis loss ∝ f·Bm^1.6.
• OC test gives iron loss and the shunt branch (R0, X0); SC test gives copper loss and the series branch (Req, Xeq).
• Sumpner's test gives both losses together at full-load temperature rise on two identical units.
• Maximum efficiency when copper loss = iron loss, at load fraction √(Wi/Wcu,FL).
• All-day efficiency uses 24-hour kWh and is the right metric for distribution transformers.
• Voltage regulation = (V2,NL − V2,FL)/V2,FL; worse at lagging pf, can be negative at leading pf.
• Per-unit impedance 4-6% (distribution) to 8-15% (power); lower %Z → better regulation but larger fault current.
• Autotransformer saves copper by (1 − 1/a) but gives no isolation; ideal for ratios near unity (ICTs, variacs).
• CT secondary must never be opened; PT secondary must never be shorted.
• Inrush current 6-10 × rated, second-harmonic rich; blocked by harmonic restraint in differential relays.
• Parallel operation needs same ratio, polarity, phase sequence/displacement and preferably same pu impedance and X/R.
• Cooling codes - ONAN, ONAF, OFAF, ODAF, OFWF; Buchholz relay works only with a conservator.
• Indian codes - IS 2026 (power transformers), IS 1180 (distribution transformer losses, BEE star labels), IS 335 (oil), IS 2705 (CT), IS 3156 (PT), CEA Safety Regulations 2023.

Formulas and examples

Example 1 - EMF equation and design check. A 50 Hz single-phase transformer has a net core area of 0.02 m², peak flux density 1.5 T, and must give 11,000 V / 415 V.

• φm = 1.5 × 0.02 = 0.03 Wb.
• Volts per turn = 4.44 × 50 × 0.03 = 6.66 V/turn.
• N1 = 11,000/6.66 ≈ 1,652 turns; N2 = 415/6.66 ≈ 62 turns (nearest integers; a ≈ 26.5).
• Check: 1652/62 = 26.6, close to 11,000/415 = 26.5 - acceptable rounding.

Example 2 - OC and SC tests, then regulation and efficiency. A 25 kVA, 2,200 V/220 V, 50 Hz transformer gives: OC test on LV - 220 V, 4.0 A, 180 W; SC test on HV - 95 V, 11.36 A, 350 W.

• Rated currents: HV = 25,000/2,200 = 11.36 A; LV = 25,000/220 = 113.6 A. SC was done at rated HV current, so Wcu,FL = 350 W; Wi = 180 W.
• From OC (LV side): Iw = 180/220 = 0.818 A; Iμ = √(4.0² − 0.818²) = √(16 − 0.669) = 3.915 A; R0 = 220²/180 = 268.9 Ω; X0 = 220/3.915 = 56.2 Ω; cos φ0 = 180/(220 × 4) = 0.205.
• From SC (HV side): Zeq1 = 95/11.36 = 8.36 Ω; Req1 = 350/11.36² = 350/129.0 = 2.71 Ω; Xeq1 = √(8.36² − 2.71²) = √(69.9 − 7.35) = 7.91 Ω.
• Regulation at full load, 0.8 pf lagging (work on HV side, V1 = 2,200 V): drop = 11.36(2.71 × 0.8 + 7.91 × 0.6) = 11.36(2.168 + 4.746) = 11.36 × 6.914 = 78.5 V → %R = 78.5/2,200 = 3.57%.
• At 0.8 pf leading: drop = 11.36(2.168 − 4.746) = −29.3 V → %R = −1.33% (voltage rises).
• Full-load efficiency at 0.8 pf: output = 25,000 × 0.8 = 20 kW; losses = 180 + 350 = 530 W → η = 20,000/20,530 = 97.42%.
• Maximum efficiency: x = √(180/350) = √0.514 = 0.717 → at 17.93 kVA. Output at 0.8 pf = 14.34 kW, losses = 2 × 180 = 360 W → ηmax = 14,340/14,700 = 97.55%.
• % impedance = 95/2,200 × 100 = 4.32%; so symmetrical fault current ≈ 11.36 × 100/4.32 ≈ 263 A on the HV side.

Example 3 - All-day efficiency. A 20 kVA distribution transformer has Wi = 200 W and full-load Wcu = 400 W. Daily load: 4 kW at unity pf for 8 h, 10 kW at unity pf for 8 h, no load for 8 h.

• Energy output = 4 × 8 + 10 × 8 = 32 + 80 = 112 kWh.
• Iron loss energy = 0.2 kW × 24 = 4.8 kWh.
• Copper loss: at 4 kVA, x = 0.2 → 0.4 × 0.04 = 0.016 kW for 8 h = 0.128 kWh; at 10 kVA, x = 0.5 → 0.4 × 0.25 = 0.1 kW for 8 h = 0.8 kWh; at no load = 0.
• Total copper energy = 0.928 kWh; total loss = 4.8 + 0.928 = 5.728 kWh.
• All-day efficiency = 112/(112 + 5.728) = 112/117.728 = 95.14% - well below the commercial efficiency, because iron loss runs all 24 hours.

Example 4 - Autotransformer saving. A 230 V/115 V, 10 kVA autotransformer.

• a = 2, so I1 = 10,000/230 = 43.5 A, I2 = 10,000/115 = 87.0 A; the common winding carries I2 − I1 = 43.5 A.
• Copper saving = (1 − 1/2) = 50% of the copper of an equivalent two-winding transformer.
• For a 400/220 kV ICT, a = 1.82 → saving = 1 − 1/1.82 = 45%; for a 11 kV/415 V ratio, a = 26.5 → saving = 96.2% but isolation is lost and insulation must be rated for 11 kV throughout, so an autotransformer is not used there.

Example 5 - Parallel operation and load sharing. Transformer A: 500 kVA, 5% impedance. Transformer B: 250 kVA, 4% impedance. Combined load 700 kVA, same ratio and vector group.

• On a common 500 kVA base: ZA = 5%, ZB = 4% × (500/250) = 8%.
• Share ∝ 1/Z: A : B = 1/5 : 1/8 = 8 : 5 → A = 700 × 8/13 = 430.8 kVA, B = 700 × 5/13 = 269.2 kVA.
• B is overloaded (269.2 against 250 kVA) while A is only 86% loaded - the classic penalty of unequal per-unit impedance. Safe total = limited by B: 250 × 13/5 = 650 kVA.

Where it is used

• Transmission and substations: POWERGRID and state TRANSCOs operate 765/400/220/132 kV transformers and 400/220 kV autotransformers; substation engineers handle commissioning, DGA and tap-changer maintenance.
• Distribution: 11 kV/433 V pole-mounted and compact substation units in every discom (BESCOM, MSEDCL, Tata Power, Adani Electricity); IS 1180 / BEE star-rated loss levels drive tendering, and loss reduction is central to RDSS and smart-metering programmes.
• Manufacturing: BHEL (Bhopal, Jhansi), CG Power, TRANSCO vendors, Siemens Energy, Hitachi Energy, Schneider, Vijai Electricals, Voltamp, Kirloskar - design, testing (CPRI Bengaluru short-circuit and type testing), quality and field service roles.
• Industry and buildings: dry-type cast-resin transformers in metros, malls, IT parks and hospitals; furnace and rectifier transformers in steel and aluminium plants; welding transformers; isolation transformers for medical and sensitive loads.
• Railways: 25 kV AC traction substations with single-phase 132/25 kV transformers and onboard traction transformers on electric locos and EMUs.
• Protection and metering: CTs and PTs for every energy meter, relay and SCADA input; instrument transformer accuracy class selection is part of discom metering and protection-coordination jobs.
• Renewables: inverter-duty step-up transformers in solar parks and wind farms (0.4/0.69 kV to 33 kV), where harmonics and frequent cycling influence design.

Interview questions

• Q: Why can a transformer not work on DC? A: DC gives constant flux, so dφ/dt is zero and no EMF is induced; worse, the winding draws only V/R, a very large current that burns it.
• Q: Why does iron loss stay constant while copper loss varies with load? A: Iron loss depends on flux, which is fixed by the applied voltage and frequency (V ≈ 4.44 fNφm); copper loss is I²R and so follows the square of load current.
• Q: At what load is efficiency maximum, and why? A: Where variable (copper) loss equals constant (iron) loss, i.e. at load fraction √(Wi/Wcu,FL) - differentiating the efficiency expression gives this condition.
• Q: Why is the OC test done on the LV side and the SC test on the HV side? A: The OC test needs rated voltage, which is easier and safer on LV; the SC test needs rated current at a small voltage, which is easier on HV where rated current is low.
• Q: What is all-day efficiency and when do you use it? A: The ratio of 24-hour energy output to energy input; it is used for distribution transformers because they are lightly loaded while iron loss runs continuously.
• Q: Why must a CT secondary never be left open? A: With no secondary current there is no counter-MMF, the core saturates and a very high voltage appears across the open terminals - dangerous to personnel and insulation.
• Q: What can happen if a 50 Hz transformer is run on 50 Hz but at 10% over-voltage? A: Flux rises about 10%, pushing the core towards saturation; magnetising current and iron loss increase sharply and the unit overheats and gets noisier.
• Q: Why do two transformers in parallel need the same per-unit impedance? A: Load divides inversely with per-unit impedance, so with unequal values the lower-impedance unit takes a disproportionate share and overloads before the other reaches rating.
• Q: What causes inrush current and how is nuisance tripping avoided? A: Switching at an unfavourable point on the voltage wave with residual core flux causes deep saturation and a 6-10 pu first peak; differential relays use second-harmonic restraint to block tripping.
• Q: Why is a Buchholz relay absent on sealed transformers? A: It is fitted in the pipe between tank and conservator to catch gas and oil surges; a sealed or conservator-less unit has no such pipe, so pressure relief devices are used instead.
• Q: When is an autotransformer preferred, and why not for 11 kV/415 V? A: When the ratio is close to unity, as in 400/220 kV ICTs, since copper saving is large; at a high ratio the saving is offset by loss of isolation and the need to insulate the whole winding for HV.
• Q: Can voltage regulation be negative? A: Yes - at a sufficiently leading power factor the reactive drop opposes the resistive drop and the secondary terminal voltage at load exceeds its no-load value.
• Q: Why is the LV winding placed next to the core? A: It needs less insulation to the earthed core, reducing insulation cost and simplifying HV bushing and clearance design.

Common mistakes

• Writing E = 4.44 N φm and forgetting f, or using average instead of RMS (4.44 = 4 × 1.11 form factor).
• Using φm in Wb when only flux density is given - φm = Bm × net iron area must be computed first.
• Mixing referred quantities: adding R1 to R2 without multiplying by a², or applying a primary-side Xeq in a secondary-side regulation formula.
• Assuming SC test power is the full-load copper loss when the test was not done at rated current - it must be scaled by (Irated/Itest)².
• Treating iron loss as load-dependent, or copper loss as constant, in efficiency and all-day efficiency sums.
• Using the kVA rating directly as output power in the efficiency formula instead of kVA × cos φ.
• Confusing regulation sign conventions - taking '+' for leading loads, or quoting regulation with respect to the full-load instead of the no-load voltage definition in use.
• Shorting a PT secondary or opening a CT secondary in the lab; also forgetting a low-power-factor wattmeter for the OC test, which makes the iron-loss reading badly inaccurate.

Related topics

Three-phase transformers and connections · Induction motors · Power system analysis and per-unit systems · Switchgear and protection · Electrical machine design basics · Electrical measurements and instrumentation