Parallel operation of Yy0 and Dd0 transformers is

not possible due to phase shift mismatch

Three-phase transformer vector group Dy11 means

delta primary, star secondary, 11 o'clock shift (−30°)

Pulse transformer design emphasizes

fast rise time and low leakage inductance

A welding transformer is designed for

high drooping volt-ampere characteristic

Transformers MCQs 2026

88 questions with detailed answers · 31 from past papers · 9 quiz batches available

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Page 1 of 1Questions 1–10 of 88

Q1medium
Parallel operation of Yy0 and Dd0 transformers is
1. Anot possible due to phase shift mismatch
2. Balways ideal
3. Crequired for Scott
4. Dsame as Yd11 parallel
💡 Explanation:
Different phase shifts cause large circulating currents even if voltage ratios match.
Q2hard
Vector group Yd1 indicates phase displacement of
1. A−30°
2. B0°
3. C+30° (1 o'clock)
4. D180°
💡 Explanation:
Clock number 1 means secondary leads primary by 30° in positive sequence.
Q3hard
Three-phase transformer vector group Dy11 means
1. Astar primary delta secondary 0°
2. Bdelta-delta 30° lag
3. Cstar-star 11°
4. Ddelta primary, star secondary, 11 o'clock shift (−30°)
💡 Explanation:
Dy11: secondary line voltage leads primary by 30° (clock position 11 = −30° from 12).
Q4hard
Pulse transformer design emphasizes
1. Ahigh iron loss
2. Bfast rise time and low leakage inductance
3. Clarge magnetizing current
4. DONAN cooling only
💡 Explanation:
Signal integrity requires minimal leakage L and distributed capacitance for sharp pulses.
Q5medium
A welding transformer is designed for
1. Aconstant voltage like PT
2. Bhigh drooping volt-ampere characteristic
3. C1:1 isolation only
4. Dzero leakage reactance
💡 Explanation:
High leakage reactance limits short-circuit current and provides stable arc voltage.
Q6Past Paper · PPSC/FPSC/CSShard
Equivalent resistance referred to secondary equals
1. AR1 only
2. BR2 only
3. CR2 + R1 × (Ns/Np)²
4. DR1 × Np/Ns
💡 Explanation:
Primary resistance referred to secondary is scaled by inverse turns ratio squared.
Q7easy
Copper losses in a transformer vary as
1. Aload current linearly
2. Binversely with load
3. Cindependent of load
4. Dsquare of load current
💡 Explanation:
Pc = I²R; at x per unit load, Pc = x² × full-load copper loss.
Q8Past Paper · PPSC/FPSC/CSSeasy
Iron losses in a transformer are
1. Aconstant from no load to full load
2. Bproportional to square of load
3. Czero at no load
4. Donly at short circuit
💡 Explanation:
Core flux remains near rated value at normal operation, so Pi stays essentially constant.
Q9easy
Transformers are rated in
1. AkW only
2. BkVA or MVA
3. CkVAR only
4. Dhp only
💡 Explanation:
Rating is apparent power because load power factor is unknown; losses depend on current not PF alone.
Q10Past Paper · PPSC/FPSC/CSSmedium
Conservator tank on a transformer maintains
1. Asecondary voltage constant
2. Bthree-phase balance
3. Czero iron loss
4. Dpositive oil pressure and accommodates expansion
💡 Explanation:
Oil expands/contracts with temperature; conservator allows breathing without exposing oil to air.
Q11easy
Silica gel breather in a transformer prevents
1. Amoisture entering transformer oil
2. Boil overheating
3. Cshort circuit
4. Dphase unbalance
💡 Explanation:
Silica gel absorbs moisture from air drawn into conservator during oil contraction.
Q12Past Paper · PPSC/FPSC/CSSeasy
The no-load current of a power transformer is typically
1. A50 to 60 percent
2. B2 to 5 percent of rated current
3. Cequal to full-load current
4. Dzero always
💡 Explanation:
Mostly magnetizing component; small compared to rated current in large power transformers.
Q13medium
Stacking laminations reduces eddy current loss because
1. Athey increase flux density
2. Bthey eliminate hysteresis
3. Cthin sheets increase resistance to eddy paths
4. Dthey increase core volume
💡 Explanation:
Thinner laminations confine eddy currents to small loops, reducing I²R eddy loss.
Q14Past Paper · PPSC/FPSC/CSShard
Eddy current loss in transformer laminations varies approximately with
1. Aflux density only
2. Bfrequency only
3. Cload power factor
4. Dflux density squared and frequency squared
💡 Explanation:
Pe ∝ f² × B² × t² for thin laminations, rising sharply with flux and frequency.
Q15Past Paper · PPSC/FPSC/CSSeasy
The EMF induced in a transformer primary winding is given by the equation
1. AE = 2π f N Φm
2. BE = 4.44 f N Φm
3. CE = 4.44 f N I
4. DE = f N Φm
💡 Explanation:
The RMS induced EMF per phase equals 4.44 times frequency, number of turns, and maximum flux in webers.
Q16Past Paper · PPSC/FPSC/CSSmedium
Maximum flux Φm in a transformer core is directly proportional to
1. Acurrent and turns
2. Bfrequency squared
3. Capplied voltage and inversely proportional to frequency
4. Dcore area only
💡 Explanation:
From E = 4.44 f N Φm, for fixed N, Φm ∝ V/f, which is why flux must be limited at low frequency.
Q17easy
The turns ratio of an ideal transformer is defined as
1. ANs/Np
2. BVp/Is
3. CEp × Is
4. DNp/Ns
💡 Explanation:
Turns ratio a = Np/Ns relates primary and secondary voltages and currents inversely.
Q18Past Paper · PPSC/FPSC/CSSeasy
For an ideal transformer, if primary has 1000 turns and secondary has 100 turns, the turns ratio is
1. A0.1
2. B10
3. C100
4. D100000
💡 Explanation:
a = Np/Ns = 1000/100 = 10, meaning primary voltage is ten times secondary voltage.
Q19easy
In an ideal step-down transformer, the secondary current compared to primary current is
1. Ahigher
2. Blower
3. Cequal
4. Dzero
💡 Explanation:
Power conservation gives VpIp = VsIs, so lower secondary voltage means higher secondary current.
Q20Past Paper · PPSC/FPSC/CSSeasy
The voltage ratio of an ideal transformer equals
1. Ainverse of turns ratio squared
2. Bturns ratio
3. Csquare of turns ratio
4. Dindependent of turns
💡 Explanation:
Vp/Vs = Np/Ns for an ideal transformer with same flux linking both windings.
Q21Past Paper · PPSC/FPSC/CSSmedium
Open-circuit (OC) test on a transformer is performed on the
1. Alow-voltage winding
2. Bhigh-voltage winding
3. Cboth windings simultaneously
4. Dtertiary winding only
💡 Explanation:
OC test uses rated voltage on LV side while HV is open, keeping test voltage safely accessible.
Q22medium
The open-circuit test of a transformer primarily determines
1. Acopper losses
2. Bfull-load efficiency only
3. Cshort-circuit impedance
4. Dcore (iron) losses and no-load current
💡 Explanation:
At no load, copper loss is negligible; measured input gives iron loss and magnetizing parameters.
Q23Past Paper · PPSC/FPSC/CSSmedium
Short-circuit (SC) test on a transformer is normally conducted on the
1. Alow-voltage winding
2. Bneutral winding
3. Chigh-voltage winding
4. Dopen secondary
💡 Explanation:
Reduced voltage (5-10% rated) is applied to HV with LV shorted to measure impedance and copper loss.
Q24medium
The short-circuit test of a transformer is used to determine
1. Airon losses
2. Bcopper losses and equivalent impedance
3. Cturns ratio
4. Dno-load power factor
💡 Explanation:
Rated current flows at reduced voltage; iron loss is small and copper loss dominates.
Q25Past Paper · PPSC/FPSC/CSSeasy
During OC test, wattmeter reading represents mainly
1. AI²R copper loss at full load
2. Bdielectric loss only
3. Cstray load loss
4. Dhysteresis and eddy current losses
💡 Explanation:
No-load current is small, so copper loss is negligible compared to constant core losses.
Q26medium
During SC test, the applied voltage is approximately
1. A5 to 10 percent of rated voltage
2. B50 percent of rated voltage
3. C100 percent of rated voltage
4. D0.1 percent of rated voltage
💡 Explanation:
Short-circuit impedance limits current to rated value at a small fraction of rated voltage.
Q27Past Paper · PPSC/FPSC/CSSmedium
All-day efficiency of a distribution transformer is also called
1. Acommercial efficiency
2. Bmaximum efficiency
3. Cenergy efficiency
4. Dregulation efficiency
💡 Explanation:
It considers energy losses over 24 hours at varying load cycles, not just one load point.
Q28easy
An auto-transformer differs from a two-winding transformer because
1. Ait has no core
2. Bit cannot step up voltage
3. Cit has no losses
4. Dprimary and secondary share a common winding
💡 Explanation:
Part of the winding is common, reducing copper and size for same voltage ratio.
Q29Past Paper · PPSC/FPSC/CSShard
Savings in copper of an auto-transformer compared to two-winding type is approximately
1. A(1 − Vlow/Vhigh) × 100%
2. Balways 50%
3. Czero always
4. Dtwice the rating
💡 Explanation:
Common winding carries only the difference current, saving conductor material.
Q30medium
Auto-transformers are NOT used when
1. Aeconomy is important
2. Bsmall voltage change is needed
3. Cgalvanic isolation is required
4. Dstarting large motors
💡 Explanation:
Shared winding provides no electrical isolation between primary and secondary.
Q31Past Paper · PPSC/FPSC/CSSeasy
An isolation transformer has turns ratio of
1. A10:1
2. B1:1
3. C1:10
4. Dvariable only
💡 Explanation:
Isolation transformers typically have equal turns to transfer voltage while isolating circuits.
Q32easy
Primary purpose of an isolation transformer is to
1. Amaximize efficiency at 1:10 ratio
2. Bprovide galvanic isolation and safety
3. Celiminate all losses
4. Dincrease frequency
💡 Explanation:
It separates equipment from supply ground, reducing shock hazard and noise coupling.
Q33Past Paper · PPSC/FPSC/CSSeasy
A current transformer (CT) is connected in
1. Aparallel with the line
2. Bacross the bus bar in shunt
3. Cseries with the line
4. Donly in neutral ground
💡 Explanation:
CT primary carries line current; secondary feeds measuring instruments with stepped-down current.
Q34easy
Standard secondary current of most instrument CTs is
1. A5 A or 1 A
2. B110 V
3. C415 V
4. D0.1 A only
💡 Explanation:
Industry standardizes on 5 A (or 1 A) secondary for ammeters, relays, and energy meters.
Q35Past Paper · PPSC/FPSC/CSSeasy
A potential transformer (PT) is connected in
1. Aseries with load current
2. Binside transformer tank
3. Conly on secondary of CT
4. Dparallel (shunt) across the line
💡 Explanation:
PT steps down high voltage to standard 110 V or 100 V for metering and protection.
Q36Past Paper · PPSC/FPSC/CSSmedium
The secondary of a CT must never be
1. Aopen-circuited on load
2. Bshorted
3. Cgrounded
4. Dconnected to relay
💡 Explanation:
Open secondary with primary energized causes dangerous overvoltage due to magnetizing flux buildup.
Q37easy
CT ratio 200/5 means secondary current of 5 A when primary carries
1. A5 A
2. B2000 A
3. C200 A
4. D20 A
💡 Explanation:
Ratio expresses primary to secondary current transformation for measurement scaling.
Q38medium
Per-unit system for transformers uses base quantities such that
1. Aall values are in amperes only
2. Bper-unit values are dimensionless ratios
3. Cper-unit has physical units
4. Donly voltage has per-unit
💡 Explanation:
pu = actual value / base value; bases chosen for convenient calculation on common MVA and kV.
Q39Past Paper · PPSC/FPSC/CSShard
Base impedance in per-unit system is given by
1. ASbase / Vbase
2. BVbase × Ibase²
3. CIbase / Vbase
4. DVbase² / Sbase
💡 Explanation:
Zbase = V²/S in consistent units (e.g., kV and MVA) for per-unit impedance conversion.
Q40hard
Transformer per-unit impedance is independent of
1. Athe base MVA
2. Bthe turns ratio
3. Cwhich side it is referred to
4. Dthe frequency
💡 Explanation:
When properly referred, Zpu is the same whether calculated on HV or LV side.
Q41medium
Hysteresis loss in transformer core depends on
1. Aload current only
2. Bmaximum flux density and frequency
3. Csecondary voltage squared only
4. Dtap position only
💡 Explanation:
Ph ∝ f × Bmax^n (Steinmetz); hysteresis varies with flux density and frequency.
Q42easy
Commercial efficiency of a transformer is defined at
1. Ano load
2. Bfull load
3. Chalf load only
4. Dquarter load only
💡 Explanation:
Commercial efficiency = output at full load / (output + losses at full load), used for rating comparisons.
Q43Past Paper · PPSC/FPSC/CSSmedium
Maximum efficiency of a transformer occurs when
1. Acopper loss is zero
2. Biron loss is twice copper loss
3. Cload is zero
4. Dcopper loss equals iron loss
💡 Explanation:
Total loss Pt = Pi + Pc×(x)²; minimum when dPt/dx = 0 gives x = √(Pi/Pc), i.e., Cu loss = iron loss.
Q44hard
If iron loss is 2 kW and full-load copper loss is 8 kW, load fraction for maximum efficiency is
1. A0.25
2. B1.0
3. C0.5
4. D0.707
💡 Explanation:
x = √(Pi/Pc) = √(2/8) = √(0.25) = 0.5 or 50% load.
Q45Past Paper · PPSC/FPSC/CSSeasy
At maximum efficiency, the variable copper loss equals
1. Aconstant iron loss
2. Bhalf the iron loss
3. Ctwice the iron loss
4. Dzero
💡 Explanation:
This is the fundamental condition for minimum total loss at a given load fraction.
Q46medium
A transformer has Pi = 1 kW and Pc = 4 kW. Maximum efficiency occurs at
1. A50% load
2. B25% load
3. C100% load
4. D75% load
💡 Explanation:
x = √(1/4) = 0.5, so maximum efficiency is at half rated load.
Q47Past Paper · PPSC/FPSC/CSSmedium
Star-star (Y-Y) connection of three-phase transformers gives
1. A30° phase shift
2. B180° phase shift
3. C60° phase shift
4. Dzero phase shift between primary and secondary line voltages
💡 Explanation:
Y-Y connection preserves phase sequence with no inherent angular displacement.
Q48medium
Delta-delta (Δ-Δ) connection is preferred when
1. Aminimum copper is needed
2. Bphase shift of 30° is required
3. Cone transformer can be removed for maintenance (open delta)
4. Dneutral is mandatory
💡 Explanation:
Open-delta operation allows continued two-phase service at reduced capacity using two transformers.
Q49Past Paper · PPSC/FPSC/CSSmedium
Star-delta (Y-Δ) connection introduces a phase shift of
1. A0°
2. B30° lagging of secondary
3. C60° leading
4. D180°
💡 Explanation:
Y-Δ shifts secondary voltage by 30° relative to primary, important for parallel operation.
Q50hard
Delta-star (Δ-Y) connection produces a phase shift of
1. A30° lagging
2. B0°
3. C30° leading on secondary
4. D45°
💡 Explanation:
Δ-Y is the mirror of Y-Δ; secondary leads primary by 30 electrical degrees.
Q51Past Paper · PPSC/FPSC/CSSmedium
Scott connection uses two transformers to convert
1. Athree-phase to two-phase
2. Bsingle-phase to three-phase
3. CDC to AC
4. Dtwo-phase to single-phase only
💡 Explanation:
Scott T-connection provides balanced two-phase supply from a three-phase source.
Q52hard
In Scott connection, one transformer is called teaser and has
1. A100% same rating
2. B50% rating
3. C173% rating
4. D86.6% of main transformer voltage rating
💡 Explanation:
Teaser transformer winding is tapped at 86.6% (√3/2 × 50%) to obtain 90° displacement.
Q53Past Paper · PPSC/FPSC/CSSeasy
ONAN cooling classification means
1. AOil Forced Air Forced
2. BOil Natural Air Natural
3. COil Natural Air Forced
4. DOil Forced Air Natural
💡 Explanation:
ONAN relies on natural oil circulation and natural air cooling without pumps or fans.
Q54easy
OFAF cooling in transformers stands for
1. AOil Natural Air Forced
2. BOil Forced Air Natural
3. COil Forced Air Forced
4. DOil Free Air Forced
💡 Explanation:
OFAF uses oil pumps and forced air fans for higher capacity heat removal.
Q55Past Paper · PPSC/FPSC/CSSeasy
As transformer load increases, voltage regulation generally
1. Adecreases
2. Bremains constant
3. Cbecomes negative always
4. Dincreases
💡 Explanation:
Greater load current causes larger resistive and reactive drops, worsening regulation.
Q56medium
Voltage regulation of a transformer is defined as
1. A(V2 no-load − V2 full-load) / V2 rated × 100%
2. B(V2 full-load − V2 no-load) / V2 rated
3. CV2 full-load / V1
4. DI2 / I1
💡 Explanation:
Regulation measures percentage drop in secondary terminal voltage from no-load to full-load.
Q57Past Paper · PPSC/FPSC/CSShard
Negative voltage regulation in a transformer can occur when
1. Aload is purely resistive
2. Bload power factor is leading
3. Cload power factor is lagging unity
4. Dtransformer is open-circuited
💡 Explanation:
Leading PF can cause voltage rise at terminals, giving negative regulation.
Q58medium
The approximate equivalent circuit of a transformer is referred to
1. Aboth sides simultaneously without referral
2. Bthe core only
3. Cthe tank
4. Done side (primary or secondary)
💡 Explanation:
Parameters are referred to one winding for simplified analysis using turns ratio.
Q59Past Paper · PPSC/FPSC/CSSmedium
In the exact equivalent circuit, R0 and X0 represent
1. Aseries leakage impedance
2. Bload resistance
3. Cno-load (core loss and magnetizing) branch
4. Dtap changer resistance
💡 Explanation:
Parallel branch R0-X0 accounts for iron loss and magnetizing reactance.
Q60medium
Series components R1 and X1 in transformer equivalent circuit represent
1. Acore loss
2. Bwinding resistance and leakage reactance
3. Cmagnetizing current
4. Dcooling system
💡 Explanation:
Impedance drop in windings is modeled by series R and X referred to one side.
Q61Past Paper · PPSC/FPSC/CSSeasy
Buchholz relay is installed in
1. Atransformer conservator pipe / main tank pipe
2. Bcircuit breaker panel
3. Csecondary bus bar
4. Dlightning arrester
💡 Explanation:
It detects gas accumulation and oil surge from internal faults such as arcing or overheating.
Q62medium
Buchholz relay operates on the principle of
1. Agas accumulation and oil flow displacement
2. Bovercurrent in secondary
3. Covervoltage on primary
4. Dexternal temperature only
💡 Explanation:
Decomposing oil releases gas; severe faults displace oil, tripping the relay.
Q63Past Paper · PPSC/FPSC/CSSeasy
A Buchholz relay provides protection against
1. Aexternal line faults only
2. Blightning on transmission towers
3. Cincipient internal faults
4. Dmetering errors
💡 Explanation:
It gives early warning for winding faults, core overheating, and loss of oil before major damage.
Q64medium
On-load tap changer (OLTC) allows
1. Aonly off-load tapping
2. Bfrequency change
3. Cphase reversal
4. Dvoltage adjustment without interrupting load
💡 Explanation:
OLTC switches taps under load using resistors or reactor transition to maintain voltage.
Q65Past Paper · PPSC/FPSC/CSSeasy
Off-load tap changer must be operated when
1. Aat full rated load
2. Btransformer is de-energized or unloaded
3. Cduring short circuit
4. Donly at maximum tap
💡 Explanation:
De-energizing prevents arcing and ensures safe tap contact change.
Q66medium
Leakage flux in a transformer links
1. Aboth windings equally
2. Bneither winding
3. Conly one winding
4. Donly the core
💡 Explanation:
Leakage flux does not link both windings and causes voltage drop modeled as leakage reactance.
Q67medium
Percentage impedance of a transformer indicates
1. Airon loss in watts
2. Bno-load current
3. Cturns ratio
4. Dvoltage drop at rated current
💡 Explanation:
Z% = (short-circuit voltage / rated voltage) × 100 at rated current.
Q68medium
Two transformers can operate in parallel if they have
1. Asame voltage ratio, phase shift, percent impedance, and polarity
2. Bdifferent voltage ratios
3. Copposite polarity only
4. Dany impedance
💡 Explanation:
Mismatch causes circulating current and unequal load sharing.
Q69medium
Polarity test of a transformer determines
1. Acore material type
2. Brelative direction of induced EMF in windings
3. Coil dielectric strength
4. Dcooling class
💡 Explanation:
Additive or subtractive polarity affects parallel connection and relay CT connections.
Q70hard
Subtractive polarity means
1. Aopposite side terminals same polarity
2. Bno polarity exists
3. CHV and LV terminals on same side have same instantaneous polarity
4. Donly delta connection possible
💡 Explanation:
Terminals 1-H1 and 2-H2 are on same physical side for subtractive marking.
Q71easy
Three single-phase transformers in bank can replace
1. Aonly auto-transformer
2. Bone three-phase transformer
3. Conly Scott connection
4. Dno other configuration
💡 Explanation:
Banking provides flexibility; one unit can be spare or replaced individually.
Q72hard
Open-delta connection delivers
1. A57.7% of full delta-delta capacity
2. B100% capacity
3. C33% capacity
4. D86.6% capacity
💡 Explanation:
With one transformer removed, capacity = 1/√3 ≈ 57.7% of complete delta bank rating.
Q73hard
Tertiary winding in Y-Y-Δ transformer bank is often
1. Aopen circuited always
2. Bused only for metering
3. Cconnected to primary only
4. Ddelta connected unbalanced load carrier
💡 Explanation:
Delta tertiary provides path for zero-sequence current and stabilizes tertiary flux unbalance.
Q74hard
Zero-sequence impedance of Y-Δ transformer is
1. Azero on all sides
2. Bequal to positive sequence
3. Cvery high on Y side
4. Dnegative value
💡 Explanation:
Delta blocks zero-sequence current from passing to Y primary, giving high Z0.
Q75medium
Instrument PT secondary is normally grounded for
1. Aincreasing voltage
2. Breducing iron loss
3. Cchanging turns ratio
4. Dsafety and relay reference
💡 Explanation:
Grounding prevents dangerous floating potential on metering circuits and provides reference for relays.
Q76hard
CT accuracy class 5P20 means
1. A20% error at 5 A
2. B5% error up to 20 times rated current
3. C5 VA burden only
4. D20 kV insulation
💡 Explanation:
Protection class CT maintains specified accuracy up to 20× rated primary for fault currents.
Q77medium
Magnetizing inrush current occurs when transformer is
1. Aswitched on at voltage zero crossing unfavorably
2. Bat full load steady state
3. Cduring tap change off load
4. Dwith secondary shorted
💡 Explanation:
Residual flux plus applied flux can saturate core momentarily, drawing large inrush.
Q78hard
To reduce switching inrush, transformers may use
1. Aopen secondary permanently
2. Bremove core laminations
3. Cincrease frequency only
4. Dcontrolled switching or series resistance
💡 Explanation:
Point-on-wave switching and pre-insertion resistors limit peak inrush current magnitude.
Q79medium
Core saturation during inrush causes
1. Alarge magnetizing current with harmonics
2. Bzero current
3. Conly fundamental at rated value
4. Ddecreased iron loss
💡 Explanation:
Saturated core needs high magnetizing MMF, producing harmonic-rich inrush lasting several cycles.
Q80easy
Distribution transformers use
1. Aonly step-up for generation
2. Bonly isolation 1:1
3. Cstep-down voltage for consumers
4. DDC conversion
💡 Explanation:
They reduce HV distribution to LV utilization voltage at poles or pads.
Q81medium
Power transformer impedance is designed mainly for
1. Amaximizing inrush
2. Bfault current limitation and system stability
3. Celiminating regulation
4. Dzero copper loss
💡 Explanation:
Higher Z% limits short-circuit current but increases voltage regulation trade-off.
Q82easy
Transformer oil serves as
1. Aonly structural support
2. Bonly flux path
3. Cinsulation and coolant
4. Donly current carrier
💡 Explanation:
Mineral oil insulates windings and transfers heat from core and coils to tank walls.
Q83medium
Dissolved gas analysis (DGA) in transformers detects
1. Aexternal lightning only
2. Bincipient insulation breakdown
3. Cmeter calibration
4. Dtap position
💡 Explanation:
Gases like H2, CH4, C2H2 indicate overheating, partial discharge, or arcing in oil/paper.
Q84hard
Winding hot-spot temperature limit for oil transformers is typically guided by
1. Ainsulation aging (IEEE/IEC standards)
2. Bonly ambient air
3. Csecondary current alone
4. Dtap number
💡 Explanation:
Hot-spot above 98-110°C accelerates cellulose aging; insulation life halves roughly every 6-7°C rise.
Q85medium
FR3 or ester oil in transformers offers
1. Alower insulation than mineral oil
2. Bno cooling function
3. Cincreased flammability
4. Dhigher fire point and biodegradability
💡 Explanation:
Natural ester fluids have fire points above 300°C and are environmentally friendlier.
Q86easy
Transformer tank is earthed to
1. Aprovide safety and fault current path
2. Bincrease secondary voltage
3. Cblock fault detection
4. Deliminate Buchholz
💡 Explanation:
Grounding the tank protects personnel and ensures proper operation of protective devices.
Q87easy
Lightning arrester on transformer protects against
1. Asteady state overload
2. Blow power factor
3. Cnegative regulation
4. Dsurge overvoltages
💡 Explanation:
Arresters clamp transient overvoltages from lightning or switching before they damage insulation.
Q88hard
Surge impedance loading (SIL) relates to
1. Atransformer iron loss
2. BCT ratio
3. Ctransmission line reactive power balance
4. Dtap changer steps
💡 Explanation:
While line-related, transformer-line interface design considers SIL for voltage profile on long lines.

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