Primary purpose of an isolation transformer is to

provide galvanic isolation and safety

Maximum flux Φm in a transformer core is directly proportional to

applied voltage and inversely proportional to frequency

Transformers MCQs 2026

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

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Page 1 of 1 Questions 1–10 of 88

Q1 easy

Primary purpose of an isolation transformer is to
1. A maximize efficiency at 1:10 ratio
2. B provide galvanic isolation and safety
3. C eliminate all losses
4. D increase frequency
💡 Explanation:
It separates equipment from supply ground, reducing shock hazard and noise coupling.
Q2 Past Paper · PPSC/FPSC/CSS easy

The EMF induced in a transformer primary winding is given by the equation
1. A E = 2π f N Φm
2. B E = 4.44 f N Φm
3. C E = 4.44 f N I
4. D E = f N Φm
💡 Explanation:
The RMS induced EMF per phase equals 4.44 times frequency, number of turns, and maximum flux in webers.
Q3 Past Paper · PPSC/FPSC/CSS medium

Maximum flux Φm in a transformer core is directly proportional to
1. A current and turns
2. B frequency squared
3. C applied voltage and inversely proportional to frequency
4. D core 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.
Q4 easy

The turns ratio of an ideal transformer is defined as
1. A Ns/Np
2. B Vp/Is
3. C Ep × Is
4. D Np/Ns
💡 Explanation:
Turns ratio a = Np/Ns relates primary and secondary voltages and currents inversely.
Q5 Past Paper · PPSC/FPSC/CSS easy

For an ideal transformer, if primary has 1000 turns and secondary has 100 turns, the turns ratio is
1. A 0.1
2. B 10
3. C 100
4. D 100000
💡 Explanation:
a = Np/Ns = 1000/100 = 10, meaning primary voltage is ten times secondary voltage.
Q6 easy

In an ideal step-down transformer, the secondary current compared to primary current is
1. A higher
2. B lower
3. C equal
4. D zero
💡 Explanation:
Power conservation gives VpIp = VsIs, so lower secondary voltage means higher secondary current.
Q7 Past Paper · PPSC/FPSC/CSS easy

The voltage ratio of an ideal transformer equals
1. A inverse of turns ratio squared
2. B turns ratio
3. C square of turns ratio
4. D independent of turns
💡 Explanation:
Vp/Vs = Np/Ns for an ideal transformer with same flux linking both windings.
Q8 Past Paper · PPSC/FPSC/CSS medium

Open-circuit (OC) test on a transformer is performed on the
1. A low-voltage winding
2. B high-voltage winding
3. C both windings simultaneously
4. D tertiary winding only
💡 Explanation:
OC test uses rated voltage on LV side while HV is open, keeping test voltage safely accessible.
Q9 medium

The open-circuit test of a transformer primarily determines
1. A copper losses
2. B full-load efficiency only
3. C short-circuit impedance
4. D core (iron) losses and no-load current
💡 Explanation:
At no load, copper loss is negligible; measured input gives iron loss and magnetizing parameters.
Q10 Past Paper · PPSC/FPSC/CSS medium

Short-circuit (SC) test on a transformer is normally conducted on the
1. A low-voltage winding
2. B neutral winding
3. C high-voltage winding
4. D open secondary
💡 Explanation:
Reduced voltage (5-10% rated) is applied to HV with LV shorted to measure impedance and copper loss.
Q11 medium

The short-circuit test of a transformer is used to determine
1. A iron losses
2. B copper losses and equivalent impedance
3. C turns ratio
4. D no-load power factor
💡 Explanation:
Rated current flows at reduced voltage; iron loss is small and copper loss dominates.
Q12 Past Paper · PPSC/FPSC/CSS easy

During OC test, wattmeter reading represents mainly
1. A I²R copper loss at full load
2. B dielectric loss only
3. C stray load loss
4. D hysteresis and eddy current losses
💡 Explanation:
No-load current is small, so copper loss is negligible compared to constant core losses.
Q13 medium

During SC test, the applied voltage is approximately
1. A 5 to 10 percent of rated voltage
2. B 50 percent of rated voltage
3. C 100 percent of rated voltage
4. D 0.1 percent of rated voltage
💡 Explanation:
Short-circuit impedance limits current to rated value at a small fraction of rated voltage.
Q14 Past Paper · PPSC/FPSC/CSS medium

All-day efficiency of a distribution transformer is also called
1. A commercial efficiency
2. B maximum efficiency
3. C energy efficiency
4. D regulation efficiency
💡 Explanation:
It considers energy losses over 24 hours at varying load cycles, not just one load point.
Q15 easy

An auto-transformer differs from a two-winding transformer because
1. A it has no core
2. B it cannot step up voltage
3. C it has no losses
4. D primary and secondary share a common winding
💡 Explanation:
Part of the winding is common, reducing copper and size for same voltage ratio.
Q16 Past Paper · PPSC/FPSC/CSS hard

Savings in copper of an auto-transformer compared to two-winding type is approximately
1. A (1 − Vlow/Vhigh) × 100%
2. B always 50%
3. C zero always
4. D twice the rating
💡 Explanation:
Common winding carries only the difference current, saving conductor material.
Q17 medium

Auto-transformers are NOT used when
1. A economy is important
2. B small voltage change is needed
3. C galvanic isolation is required
4. D starting large motors
💡 Explanation:
Shared winding provides no electrical isolation between primary and secondary.
Q18 Past Paper · PPSC/FPSC/CSS easy

An isolation transformer has turns ratio of
1. A 10:1
2. B 1:1
3. C 1:10
4. D variable only
💡 Explanation:
Isolation transformers typically have equal turns to transfer voltage while isolating circuits.
Q19 Past Paper · PPSC/FPSC/CSS easy

A current transformer (CT) is connected in
1. A parallel with the line
2. B across the bus bar in shunt
3. C series with the line
4. D only in neutral ground
💡 Explanation:
CT primary carries line current; secondary feeds measuring instruments with stepped-down current.
Q20 easy

Standard secondary current of most instrument CTs is
1. A 5 A or 1 A
2. B 110 V
3. C 415 V
4. D 0.1 A only
💡 Explanation:
Industry standardizes on 5 A (or 1 A) secondary for ammeters, relays, and energy meters.
Q21 Past Paper · PPSC/FPSC/CSS easy

A potential transformer (PT) is connected in
1. A series with load current
2. B inside transformer tank
3. C only on secondary of CT
4. D parallel (shunt) across the line
💡 Explanation:
PT steps down high voltage to standard 110 V or 100 V for metering and protection.
Q22 Past Paper · PPSC/FPSC/CSS medium

The secondary of a CT must never be
1. A open-circuited on load
2. B shorted
3. C grounded
4. D connected to relay
💡 Explanation:
Open secondary with primary energized causes dangerous overvoltage due to magnetizing flux buildup.
Q23 easy

CT ratio 200/5 means secondary current of 5 A when primary carries
1. A 5 A
2. B 2000 A
3. C 200 A
4. D 20 A
💡 Explanation:
Ratio expresses primary to secondary current transformation for measurement scaling.
Q24 medium

Per-unit system for transformers uses base quantities such that
1. A all values are in amperes only
2. B per-unit values are dimensionless ratios
3. C per-unit has physical units
4. D only voltage has per-unit
💡 Explanation:
pu = actual value / base value; bases chosen for convenient calculation on common MVA and kV.
Q25 Past Paper · PPSC/FPSC/CSS hard

Base impedance in per-unit system is given by
1. A Sbase / Vbase
2. B Vbase × Ibase²
3. C Ibase / Vbase
4. D Vbase² / Sbase
💡 Explanation:
Zbase = V²/S in consistent units (e.g., kV and MVA) for per-unit impedance conversion.
Q26 hard

Transformer per-unit impedance is independent of
1. A the base MVA
2. B the turns ratio
3. C which side it is referred to
4. D the frequency
💡 Explanation:
When properly referred, Zpu is the same whether calculated on HV or LV side.
Q27 medium

Hysteresis loss in transformer core depends on
1. A load current only
2. B maximum flux density and frequency
3. C secondary voltage squared only
4. D tap position only
💡 Explanation:
Ph ∝ f × Bmax^n (Steinmetz); hysteresis varies with flux density and frequency.
Q28 Past Paper · PPSC/FPSC/CSS hard

Eddy current loss in transformer laminations varies approximately with
1. A flux density only
2. B frequency only
3. C load power factor
4. D flux density squared and frequency squared
💡 Explanation:
Pe ∝ f² × B² × t² for thin laminations, rising sharply with flux and frequency.
Q29 medium

Stacking laminations reduces eddy current loss because
1. A they increase flux density
2. B they eliminate hysteresis
3. C thin sheets increase resistance to eddy paths
4. D they increase core volume
💡 Explanation:
Thinner laminations confine eddy currents to small loops, reducing I²R eddy loss.
Q30 Past Paper · PPSC/FPSC/CSS easy

The no-load current of a power transformer is typically
1. A 50 to 60 percent
2. B 2 to 5 percent of rated current
3. C equal to full-load current
4. D zero always
💡 Explanation:
Mostly magnetizing component; small compared to rated current in large power transformers.
Q31 easy

Silica gel breather in a transformer prevents
1. A moisture entering transformer oil
2. B oil overheating
3. C short circuit
4. D phase unbalance
💡 Explanation:
Silica gel absorbs moisture from air drawn into conservator during oil contraction.
Q32 Past Paper · PPSC/FPSC/CSS medium

Conservator tank on a transformer maintains
1. A secondary voltage constant
2. B three-phase balance
3. C zero iron loss
4. D positive oil pressure and accommodates expansion
💡 Explanation:
Oil expands/contracts with temperature; conservator allows breathing without exposing oil to air.
Q33 easy

Transformers are rated in
1. A kW only
2. B kVA or MVA
3. C kVAR only
4. D hp only
💡 Explanation:
Rating is apparent power because load power factor is unknown; losses depend on current not PF alone.
Q34 Past Paper · PPSC/FPSC/CSS easy

Iron losses in a transformer are
1. A constant from no load to full load
2. B proportional to square of load
3. C zero at no load
4. D only at short circuit
💡 Explanation:
Core flux remains near rated value at normal operation, so Pi stays essentially constant.
Q35 easy

Copper losses in a transformer vary as
1. A load current linearly
2. B inversely with load
3. C independent of load
4. D square of load current
💡 Explanation:
Pc = I²R; at x per unit load, Pc = x² × full-load copper loss.
Q36 Past Paper · PPSC/FPSC/CSS hard

Equivalent resistance referred to secondary equals
1. A R1 only
2. B R2 only
3. C R2 + R1 × (Ns/Np)²
4. D R1 × Np/Ns
💡 Explanation:
Primary resistance referred to secondary is scaled by inverse turns ratio squared.
Q37 medium

A welding transformer is designed for
1. A constant voltage like PT
2. B high drooping volt-ampere characteristic
3. C 1:1 isolation only
4. D zero leakage reactance
💡 Explanation:
High leakage reactance limits short-circuit current and provides stable arc voltage.
Q38 hard

Pulse transformer design emphasizes
1. A high iron loss
2. B fast rise time and low leakage inductance
3. C large magnetizing current
4. D ONAN cooling only
💡 Explanation:
Signal integrity requires minimal leakage L and distributed capacitance for sharp pulses.
Q39 hard

Three-phase transformer vector group Dy11 means
1. A star primary delta secondary 0°
2. B delta-delta 30° lag
3. C star-star 11°
4. D delta primary, star secondary, 11 o'clock shift (−30°)
💡 Explanation:
Dy11: secondary line voltage leads primary by 30° (clock position 11 = −30° from 12).
Q40 hard

Vector group Yd1 indicates phase displacement of
1. A −30°
2. B 0°
3. C +30° (1 o'clock)
4. D 180°
💡 Explanation:
Clock number 1 means secondary leads primary by 30° in positive sequence.
Q41 medium

Parallel operation of Yy0 and Dd0 transformers is
1. A not possible due to phase shift mismatch
2. B always ideal
3. C required for Scott
4. D same as Yd11 parallel
💡 Explanation:
Different phase shifts cause large circulating currents even if voltage ratios match.
Q42 easy

Commercial efficiency of a transformer is defined at
1. A no load
2. B full load
3. C half load only
4. D quarter load only
💡 Explanation:
Commercial efficiency = output at full load / (output + losses at full load), used for rating comparisons.
Q43 Past Paper · PPSC/FPSC/CSS medium

Maximum efficiency of a transformer occurs when
1. A copper loss is zero
2. B iron loss is twice copper loss
3. C load is zero
4. D copper 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.
Q44 hard

If iron loss is 2 kW and full-load copper loss is 8 kW, load fraction for maximum efficiency is
1. A 0.25
2. B 1.0
3. C 0.5
4. D 0.707
💡 Explanation:
x = √(Pi/Pc) = √(2/8) = √(0.25) = 0.5 or 50% load.
Q45 Past Paper · PPSC/FPSC/CSS easy

At maximum efficiency, the variable copper loss equals
1. A constant iron loss
2. B half the iron loss
3. C twice the iron loss
4. D zero
💡 Explanation:
This is the fundamental condition for minimum total loss at a given load fraction.
Q46 medium

A transformer has Pi = 1 kW and Pc = 4 kW. Maximum efficiency occurs at
1. A 50% load
2. B 25% load
3. C 100% load
4. D 75% load
💡 Explanation:
x = √(1/4) = 0.5, so maximum efficiency is at half rated load.
Q47 Past Paper · PPSC/FPSC/CSS medium

Star-star (Y-Y) connection of three-phase transformers gives
1. A 30° phase shift
2. B 180° phase shift
3. C 60° phase shift
4. D zero phase shift between primary and secondary line voltages
💡 Explanation:
Y-Y connection preserves phase sequence with no inherent angular displacement.
Q48 medium

Delta-delta (Δ-Δ) connection is preferred when
1. A minimum copper is needed
2. B phase shift of 30° is required
3. C one transformer can be removed for maintenance (open delta)
4. D neutral is mandatory
💡 Explanation:
Open-delta operation allows continued two-phase service at reduced capacity using two transformers.
Q49 Past Paper · PPSC/FPSC/CSS medium

Star-delta (Y-Δ) connection introduces a phase shift of
1. A 0°
2. B 30° lagging of secondary
3. C 60° leading
4. D 180°
💡 Explanation:
Y-Δ shifts secondary voltage by 30° relative to primary, important for parallel operation.
Q50 hard

Delta-star (Δ-Y) connection produces a phase shift of
1. A 30° lagging
2. B 0°
3. C 30° leading on secondary
4. D 45°
💡 Explanation:
Δ-Y is the mirror of Y-Δ; secondary leads primary by 30 electrical degrees.
Q51 Past Paper · PPSC/FPSC/CSS medium

Scott connection uses two transformers to convert
1. A three-phase to two-phase
2. B single-phase to three-phase
3. C DC to AC
4. D two-phase to single-phase only
💡 Explanation:
Scott T-connection provides balanced two-phase supply from a three-phase source.
Q52 hard

In Scott connection, one transformer is called teaser and has
1. A 100% same rating
2. B 50% rating
3. C 173% rating
4. D 86.6% of main transformer voltage rating
💡 Explanation:
Teaser transformer winding is tapped at 86.6% (√3/2 × 50%) to obtain 90° displacement.
Q53 Past Paper · PPSC/FPSC/CSS easy

ONAN cooling classification means
1. A Oil Forced Air Forced
2. B Oil Natural Air Natural
3. C Oil Natural Air Forced
4. D Oil Forced Air Natural
💡 Explanation:
ONAN relies on natural oil circulation and natural air cooling without pumps or fans.
Q54 easy

OFAF cooling in transformers stands for
1. A Oil Natural Air Forced
2. B Oil Forced Air Natural
3. C Oil Forced Air Forced
4. D Oil Free Air Forced
💡 Explanation:
OFAF uses oil pumps and forced air fans for higher capacity heat removal.
Q55 Past Paper · PPSC/FPSC/CSS easy

As transformer load increases, voltage regulation generally
1. A decreases
2. B remains constant
3. C becomes negative always
4. D increases
💡 Explanation:
Greater load current causes larger resistive and reactive drops, worsening regulation.
Q56 medium

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. C V2 full-load / V1
4. D I2 / I1
💡 Explanation:
Regulation measures percentage drop in secondary terminal voltage from no-load to full-load.
Q57 Past Paper · PPSC/FPSC/CSS hard

Negative voltage regulation in a transformer can occur when
1. A load is purely resistive
2. B load power factor is leading
3. C load power factor is lagging unity
4. D transformer is open-circuited
💡 Explanation:
Leading PF can cause voltage rise at terminals, giving negative regulation.
Q58 medium

The approximate equivalent circuit of a transformer is referred to
1. A both sides simultaneously without referral
2. B the core only
3. C the tank
4. D one side (primary or secondary)
💡 Explanation:
Parameters are referred to one winding for simplified analysis using turns ratio.
Q59 Past Paper · PPSC/FPSC/CSS medium

In the exact equivalent circuit, R0 and X0 represent
1. A series leakage impedance
2. B load resistance
3. C no-load (core loss and magnetizing) branch
4. D tap changer resistance
💡 Explanation:
Parallel branch R0-X0 accounts for iron loss and magnetizing reactance.
Q60 medium

Series components R1 and X1 in transformer equivalent circuit represent
1. A core loss
2. B winding resistance and leakage reactance
3. C magnetizing current
4. D cooling system
💡 Explanation:
Impedance drop in windings is modeled by series R and X referred to one side.
Q61 Past Paper · PPSC/FPSC/CSS easy

Buchholz relay is installed in
1. A transformer conservator pipe / main tank pipe
2. B circuit breaker panel
3. C secondary bus bar
4. D lightning arrester
💡 Explanation:
It detects gas accumulation and oil surge from internal faults such as arcing or overheating.
Q62 medium

Buchholz relay operates on the principle of
1. A gas accumulation and oil flow displacement
2. B overcurrent in secondary
3. C overvoltage on primary
4. D external temperature only
💡 Explanation:
Decomposing oil releases gas; severe faults displace oil, tripping the relay.
Q63 Past Paper · PPSC/FPSC/CSS easy

A Buchholz relay provides protection against
1. A external line faults only
2. B lightning on transmission towers
3. C incipient internal faults
4. D metering errors
💡 Explanation:
It gives early warning for winding faults, core overheating, and loss of oil before major damage.
Q64 medium

On-load tap changer (OLTC) allows
1. A only off-load tapping
2. B frequency change
3. C phase reversal
4. D voltage adjustment without interrupting load
💡 Explanation:
OLTC switches taps under load using resistors or reactor transition to maintain voltage.
Q65 Past Paper · PPSC/FPSC/CSS easy

Off-load tap changer must be operated when
1. A at full rated load
2. B transformer is de-energized or unloaded
3. C during short circuit
4. D only at maximum tap
💡 Explanation:
De-energizing prevents arcing and ensures safe tap contact change.
Q66 medium

Leakage flux in a transformer links
1. A both windings equally
2. B neither winding
3. C only one winding
4. D only the core
💡 Explanation:
Leakage flux does not link both windings and causes voltage drop modeled as leakage reactance.
Q67 medium

Percentage impedance of a transformer indicates
1. A iron loss in watts
2. B no-load current
3. C turns ratio
4. D voltage drop at rated current
💡 Explanation:
Z% = (short-circuit voltage / rated voltage) × 100 at rated current.
Q68 medium

Two transformers can operate in parallel if they have
1. A same voltage ratio, phase shift, percent impedance, and polarity
2. B different voltage ratios
3. C opposite polarity only
4. D any impedance
💡 Explanation:
Mismatch causes circulating current and unequal load sharing.
Q69 medium

Polarity test of a transformer determines
1. A core material type
2. B relative direction of induced EMF in windings
3. C oil dielectric strength
4. D cooling class
💡 Explanation:
Additive or subtractive polarity affects parallel connection and relay CT connections.
Q70 hard

Subtractive polarity means
1. A opposite side terminals same polarity
2. B no polarity exists
3. C HV and LV terminals on same side have same instantaneous polarity
4. D only delta connection possible
💡 Explanation:
Terminals 1-H1 and 2-H2 are on same physical side for subtractive marking.
Q71 easy

Three single-phase transformers in bank can replace
1. A only auto-transformer
2. B one three-phase transformer
3. C only Scott connection
4. D no other configuration
💡 Explanation:
Banking provides flexibility; one unit can be spare or replaced individually.
Q72 hard

Open-delta connection delivers
1. A 57.7% of full delta-delta capacity
2. B 100% capacity
3. C 33% capacity
4. D 86.6% capacity
💡 Explanation:
With one transformer removed, capacity = 1/√3 ≈ 57.7% of complete delta bank rating.
Q73 hard

Tertiary winding in Y-Y-Δ transformer bank is often
1. A open circuited always
2. B used only for metering
3. C connected to primary only
4. D delta connected unbalanced load carrier
💡 Explanation:
Delta tertiary provides path for zero-sequence current and stabilizes tertiary flux unbalance.
Q74 hard

Zero-sequence impedance of Y-Δ transformer is
1. A zero on all sides
2. B equal to positive sequence
3. C very high on Y side
4. D negative value
💡 Explanation:
Delta blocks zero-sequence current from passing to Y primary, giving high Z0.
Q75 medium

Instrument PT secondary is normally grounded for
1. A increasing voltage
2. B reducing iron loss
3. C changing turns ratio
4. D safety and relay reference
💡 Explanation:
Grounding prevents dangerous floating potential on metering circuits and provides reference for relays.
Q76 hard

CT accuracy class 5P20 means
1. A 20% error at 5 A
2. B 5% error up to 20 times rated current
3. C 5 VA burden only
4. D 20 kV insulation
💡 Explanation:
Protection class CT maintains specified accuracy up to 20× rated primary for fault currents.
Q77 medium

Magnetizing inrush current occurs when transformer is
1. A switched on at voltage zero crossing unfavorably
2. B at full load steady state
3. C during tap change off load
4. D with secondary shorted
💡 Explanation:
Residual flux plus applied flux can saturate core momentarily, drawing large inrush.
Q78 hard

To reduce switching inrush, transformers may use
1. A open secondary permanently
2. B remove core laminations
3. C increase frequency only
4. D controlled switching or series resistance
💡 Explanation:
Point-on-wave switching and pre-insertion resistors limit peak inrush current magnitude.
Q79 medium

Core saturation during inrush causes
1. A large magnetizing current with harmonics
2. B zero current
3. C only fundamental at rated value
4. D decreased iron loss
💡 Explanation:
Saturated core needs high magnetizing MMF, producing harmonic-rich inrush lasting several cycles.
Q80 easy

Distribution transformers use
1. A only step-up for generation
2. B only isolation 1:1
3. C step-down voltage for consumers
4. D DC conversion
💡 Explanation:
They reduce HV distribution to LV utilization voltage at poles or pads.
Q81 medium

Power transformer impedance is designed mainly for
1. A maximizing inrush
2. B fault current limitation and system stability
3. C eliminating regulation
4. D zero copper loss
💡 Explanation:
Higher Z% limits short-circuit current but increases voltage regulation trade-off.
Q82 easy

Transformer oil serves as
1. A only structural support
2. B only flux path
3. C insulation and coolant
4. D only current carrier
💡 Explanation:
Mineral oil insulates windings and transfers heat from core and coils to tank walls.
Q83 medium

Dissolved gas analysis (DGA) in transformers detects
1. A external lightning only
2. B incipient insulation breakdown
3. C meter calibration
4. D tap position
💡 Explanation:
Gases like H2, CH4, C2H2 indicate overheating, partial discharge, or arcing in oil/paper.
Q84 hard

Winding hot-spot temperature limit for oil transformers is typically guided by
1. A insulation aging (IEEE/IEC standards)
2. B only ambient air
3. C secondary current alone
4. D tap number
💡 Explanation:
Hot-spot above 98-110°C accelerates cellulose aging; insulation life halves roughly every 6-7°C rise.
Q85 medium

FR3 or ester oil in transformers offers
1. A lower insulation than mineral oil
2. B no cooling function
3. C increased flammability
4. D higher fire point and biodegradability
💡 Explanation:
Natural ester fluids have fire points above 300°C and are environmentally friendlier.
Q86 easy

Transformer tank is earthed to
1. A provide safety and fault current path
2. B increase secondary voltage
3. C block fault detection
4. D eliminate Buchholz
💡 Explanation:
Grounding the tank protects personnel and ensures proper operation of protective devices.
Q87 easy

Lightning arrester on transformer protects against
1. A steady state overload
2. B low power factor
3. C negative regulation
4. D surge overvoltages
💡 Explanation:
Arresters clamp transient overvoltages from lightning or switching before they damage insulation.
Q88 hard

Surge impedance loading (SIL) relates to
1. A transformer iron loss
2. B CT ratio
3. C transmission line reactive power balance
4. D tap changer steps
💡 Explanation:
While line-related, transformer-line interface design considers SIL for voltage profile on long lines.

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