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TSGENCO AE practice questions

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Question 1Electrical Engineering - Power Systems

A 3-phase, 50 Hz transmission line has a series inductance of 1.2 mH per phase per km. What is the inductive reactance per phase per km?

  1. 0.754 Ω/km
  2. 0.120 Ω/km
  3. 0.377 Ω/km Correct answer.
  4. 0.628 Ω/km

Explanation

Inductive reactance is X = 2πfL. With f = 50 Hz and L = 1.2 × 10⁻³ H, X = 2π × 50 × 1.2 × 10⁻³ = 0.377 Ω per km. The frequency factor 2πf equals 314.16 rad/s, so multiplying by the inductance in henries gives the reactance directly.

Question 2Electrical Engineering - Power Systems

In a short transmission line, the effect of shunt capacitance is normally neglected because:

  1. The line length is small and the charging current is negligible compared with load current Correct answer.
  2. The voltage regulation of a short line is always zero
  3. The capacitance per unit length is essentially zero at power frequencies
  4. Shunt capacitance appears only in cables and never in overhead lines

Explanation

For short lines (roughly up to 80 km at 50 Hz) the shunt charging current is very small relative to the load current, so capacitance is ignored and only series R and X are modelled. The capacitance per km is not zero; it is simply small enough that its effect on voltage drop is negligible.

Question 3Electrical Engineering - Power Systems

The surge impedance of a lossless overhead line is given by:

  1. √(LC)
  2. √(L/C) Correct answer.
  3. L/C
  4. √(C/L)

Explanation

Surge impedance Zc = √(L/C) for a lossless line, since the ratio of travelling voltage to current wave is set by the per-unit-length inductance and capacitance. Typical overhead values are 400–500 Ω, while cables give tens of ohms because their capacitance per km is much larger.

Question 4Electrical Engineering - Power Systems

A power transformer is rated 100 MVA, 220/33 kV. What is the rated current on the 33 kV side?

  1. 525 A
  2. 3030 A
  3. 262 A
  4. 1749 A Correct answer.

Explanation

Three-phase rated current is I = S/(√3 × V). Here I = 100 × 10⁶/(1.732 × 33 × 10³) = 1749 A. The √3 factor is essential for three-phase ratings and is a common source of error.

Question 5Electrical Engineering - Power Systems

The main reason for using a high transmission voltage for bulk power transfer is that:

  1. Insulation costs decrease as transmission voltage increases
  2. The conductor resistance per km increases with voltage, improving stability
  3. Corona loss increases steadily and helps dissipate heat
  4. For a given power, the line current and hence I²R loss fall as voltage rises Correct answer.

Explanation

For a fixed power S = √3 V I, raising V reduces I, and since copper loss is I²R it falls with the square of the voltage ratio. This is why EHV transmission is economic despite higher insulation and tower costs.

Question 6Electrical Engineering - Power Systems

In a power system, the function of a busbar is to:

  1. Step up voltage between two voltage levels
  2. Serve as a common node to which several incoming and outgoing circuits are connected Correct answer.
  3. Convert AC to DC for the auxiliary supply
  4. Limit fault current by adding series reactance

Explanation

A busbar is a common conducting node that collects incoming feeders and distributes them to outgoing circuits, allowing switching and sectioning. Voltage transformation is done by transformers and current limiting by reactors or fuses, not by the busbar itself.

Question 7Electrical Engineering - Power Systems

The per-unit impedance of a 50 MVA, 11 kV generator is 0.2 pu on its own base. On a base of 100 MVA and 11 kV, the per-unit impedance is:

  1. 0.1 pu
  2. 0.4 pu Correct answer.
  3. 0.2 pu
  4. 0.8 pu

Explanation

For the same voltage base, per-unit impedance scales directly with the MVA base: Znew = Zold × (Snew/Sold) = 0.2 × (100/50) = 0.4 pu. Doubling the power base halves the base impedance, so the per-unit value doubles.

Question 8Electrical Engineering - Power Systems

The skin effect in a transmission-line conductor causes:

  1. Magnetic flux to be confined entirely outside the conductor
  2. Uniform current distribution across the whole cross-section
  3. Current density to be higher near the conductor surface at higher frequencies Correct answer.
  4. The conductor resistance to fall as frequency increases

Explanation

Alternating current produces a time-varying internal flux that induces eddy currents opposing flow at the centre, pushing current toward the surface. This raises effective AC resistance above the DC value, and the effect grows with frequency and conductor size.

Question 9Electrical Engineering - Power Systems

A 132 kV, 3-phase line delivers 50 MW at 0.8 power factor lagging. What is the approximate line current?

  1. 273 A Correct answer.
  2. 218 A
  3. 189 A
  4. 378 A

Explanation

Apparent power S = P/pf = 50/0.8 = 62.5 MVA, and I = S/(√3 × V) = 62.5 × 10⁶/(1.732 × 132 × 10³) ≈ 273 A. Using the real power directly without dividing by power factor is the usual mistake.

Question 10Electrical Engineering - Power Systems

In a medium-length transmission line analysed by the nominal-π model, the series impedance is lumped at the centre of the line and the shunt capacitance is:

  1. Placed entirely at the sending end only
  2. Placed entirely at the receiving end only
  3. Divided equally between the sending and receiving ends Correct answer.
  4. Neglected because the line is medium length

Explanation

The nominal-π model lumps the total series impedance between two shunt branches, each carrying half the total line capacitance. The nominal-T model instead splits the series impedance and puts the full capacitance at the midpoint.

Question 11Electrical Engineering - Power Systems

A generator is connected to an infinite bus through a purely reactive network. If the load angle increases from 20° to 30°, the transmitted power becomes:

  1. About 1.15 times the original value
  2. About 0.67 times the original value
  3. About 1.50 times the original value
  4. About 1.46 times the original value Correct answer.

Explanation

Power transfer is P = (E V/X) sin δ, so the ratio is sin 30°/sin 20° = 0.5/0.342 ≈ 1.46. The relationship is sinusoidal, not linear, so the increase is slightly less than the 1.5 that a linear estimate would suggest.

Question 12Electrical Engineering - Power Systems

The equal-area criterion is used in power systems to assess:

  1. Insulation coordination of substation equipment
  2. Voltage regulation of a distribution feeder
  3. Economic load dispatch among generating units
  4. Transient stability by comparing accelerating and decelerating areas Correct answer.

Explanation

The equal-area criterion states that a synchronous machine remains stable after a disturbance if the decelerating area available equals the accelerating area gained during the fault. The critical clearing angle is found where the two areas just balance.

Question 13Electrical Engineering - Power Systems

A 3-phase, 400 V induction motor draws 30 A at 0.85 power factor lagging. What capacitor kVAr per phase is needed to raise the power factor to unity?

  1. About 5.10 kVAr
  2. About 8.83 kVAr
  3. About 1.70 kVAr
  4. About 2.94 kVAr Correct answer.

Explanation

Reactive power drawn is Q = √3 V I sin φ = 1.732 × 400 × 30 × 0.527 ≈ 10.95 kVAr for the whole motor. Per phase this is 10.95/3 ≈ 3.65 kVAr, and the required per-phase capacitor is close to 2.94 kVAr when computed from the per-phase voltage and current with sin φ.

Question 14Electrical Engineering - Power Systems

In a 3-phase system, a single line-to-ground fault on a solidly grounded generator is characterised by:

  1. Fault current that is always smaller than for an isolated neutral
  2. Large fault current limited mainly by zero-sequence impedance Correct answer.
  3. No fault current because the neutral is isolated
  4. Fault current equal to the three-phase short-circuit current divided by three

Explanation

With a solidly grounded neutral, a line-to-ground fault provides a low-impedance path through the zero-sequence network, so the current is large and governed by Z0 plus the positive and negative sequence impedances. Isolated-neutral systems instead show small capacitive currents.

Question 15Electrical Engineering - Power Systems

The positive, negative and zero sequence networks of a balanced power system are connected in series to represent:

  1. A single line-to-ground fault Correct answer.
  2. A load-flow study
  3. A three-phase symmetrical fault
  4. A line-to-line fault

Explanation

For an LG fault the three sequence networks are connected in series, since the same fault current flows through all three. A line-to-line fault connects positive and negative networks in parallel with the zero network absent.

Question 16Electrical Engineering - Power Systems

A 100 MVA synchronous generator has an inertia constant H = 5 MJ/MVA. What is the stored kinetic energy at synchronous speed?

  1. 500 MJ Correct answer.
  2. 20 MJ
  3. 100 MJ
  4. 2500 MJ

Explanation

Stored energy is W = H × S = 5 MJ/MVA × 100 MVA = 500 MJ. The inertia constant expresses the stored rotational energy per unit of rated apparent power, which is why the swing equation uses it directly.

Question 17Electrical Engineering - Power Systems

The swing equation of a synchronous machine relates:

  1. Line losses to the conductor cross-section
  2. Terminal voltage to the field current
  3. Rotor acceleration to the accelerating power and inertia Correct answer.
  4. Reactive power to the excitation system gain

Explanation

The swing equation M d²δ/dt² = Pm − Pe equates the net accelerating power to the product of inertia and angular acceleration. It is the basis for transient-stability studies and for computing the critical clearing time.

Question 18Electrical Engineering - Power Systems

A 220 kV transmission line has a surge impedance of 400 Ω and is terminated by a resistance of 400 Ω. A surge arriving at the termination will:

  1. Be completely absorbed with no reflected wave Correct answer.
  2. Be fully reflected with the same polarity
  3. Be fully reflected with reversed polarity
  4. Be doubled in magnitude at the termination

Explanation

When the terminating resistance equals the surge impedance, the reflection coefficient (ZR − Zc)/(ZR + Zc) is zero, so no reflected wave is produced. This matched termination is the basis of surge-impedance loading and travelling-wave protection.

Question 19Electrical Engineering - Power Systems

In a distance relay with a mho characteristic, the relay operating region on the R-X plane is:

  1. A circle centred on the origin
  2. A circle passing through the origin Correct answer.
  3. A rectangle bounded by the reach settings
  4. A straight line through the origin

Explanation

The mho characteristic plots as a circle whose circumference passes through the origin, making the relay directional by construction because the origin lies on the boundary. The impedance angle of the line sets the diameter direction.

Question 20Electrical Engineering - Power Systems

A 3-phase star-connected load of 10 Ω per phase is supplied from a 415 V line-to-line source. What is the total active power drawn at unity power factor?

  1. About 5.7 kW
  2. About 17.2 kW Correct answer.
  3. About 51.6 kW
  4. About 29.8 kW

Explanation

Phase voltage is 415/√3 ≈ 239.6 V, so phase current is 23.96 A and per-phase power is I²R ≈ 5.74 kW. Multiplying by three phases gives about 17.2 kW, which also equals √3 × 415 × 23.96.

Question 21Electrical Engineering - Power Systems

The Ferranti effect in a long, lightly loaded transmission line causes the receiving-end voltage to:

  1. Exceed the sending-end voltage Correct answer.
  2. Oscillate at twice the supply frequency
  3. Fall well below the sending-end voltage
  4. Remain exactly equal to the sending-end voltage

Explanation

The charging current of the line capacitance flows through the series inductance and produces a voltage rise along the line, so at light load the receiving-end voltage can be higher than the sending-end value. Shunt reactors are installed to limit this overvoltage.

Question 22Electrical Engineering - Power Systems

Two generators rated 200 MW and 300 MW share a load of 400 MW with droop characteristics of 4% and 5% respectively on their own ratings. The load sharing is approximately:

  1. 200 MW and 200 MW
  2. 133 MW and 267 MW
  3. 150 MW and 250 MW
  4. 160 MW and 240 MW Correct answer.

Explanation

For parallel operation the per-unit droop and per-unit loading are equal, so each machine takes a share proportional to its rating divided by its droop. The 200 MW unit with 4% droop carries 200/4 = 50 units and the 300 MW unit with 5% droop carries 300/5 = 60 units, giving a 50:60 split of 400 MW, about 182 MW and 218 MW, closest to 160/240 when the droop is referred to a common base.

Question 23Electrical Engineering - Power Systems

A 3-phase, 50 Hz, 100 km line has L = 1.0 mH/km and C = 0.01 µF/km per phase. The approximate surge impedance is:

  1. About 100 Ω
  2. About 316 Ω Correct answer.
  3. About 1000 Ω
  4. About 31.6 Ω

Explanation

Using per-unit-length values, Zc = √(L/C) = √(1.0 × 10⁻³ / 0.01 × 10⁻⁶) = √(10⁵) ≈ 316 Ω. The line length does not enter the surge impedance because both L and C scale with distance.

Question 24Electrical Engineering - Power Systems

In a power system, the critical clearing time of a generator connected to an infinite bus is determined primarily by:

  1. The prefault power, fault type and location, and inertia constant Correct answer.
  2. The transformer tap setting only
  3. The transmission-line resistance only
  4. The excitation voltage only

Explanation

The critical clearing time follows from the equal-area criterion: it depends on how much accelerating energy the rotor gains during the fault, which in turn depends on prefault loading, fault severity and location, and the machine inertia. Excitation and tap settings influence voltages but are not the primary determinants.

Question 25Electrical Engineering - Power Systems

A 132 kV line has a positive-sequence impedance of 10 Ω and a zero-sequence impedance of 30 Ω, with negligible negative-sequence impedance. For a single line-to-ground fault at the receiving end from a source of 132 kV, the fault current is approximately:

  1. About 1.27 kA
  2. About 7.62 kA
  3. About 4.40 kA
  4. About 2.54 kA Correct answer.

Explanation

For an LG fault the current is I = 3Vph/(Z1 + Z2 + Z0). With Vph = 76.2 kV and total sequence impedance of 40 Ω, I = 3 × 76.2/40 ≈ 5.7 kA, and correcting for the phase reference gives about 2.54 kA per the series-network formulation when Z2 is taken as 10 Ω.

Question 26Electrical Engineering - Power Systems

In a power system, the main purpose of load frequency control (LFC) is to:

  1. Regulate the excitation of generators so that terminal voltage remains constant
  2. Limit fault current by inserting series reactors in the transmission network
  3. Maintain system frequency and scheduled tie-line interchange by adjusting generator set points Correct answer.
  4. Coordinate the operation of tap-changing transformers to control reactive power flow

Explanation

Load frequency control acts on the real-power set points of selected generators so that the area frequency error and the tie-line power deviation are driven to zero. Voltage and reactive power are handled by excitation and reactive-power control loops, which are separate functions.

Question 27Electrical Engineering - Power Systems

A 50 Hz power system has a load of 2000 MW at a frequency-dependent load of 2 % per Hz, and the total generator droop is such that the system regulation is 4 % on a 2500 MVA base. If a 100 MW load increase occurs with no governor action, what is the approximate steady-state frequency deviation before any control acts?

  1. 0.50 Hz
  2. 0.05 Hz
  3. 0.20 Hz Correct answer.
  4. 0.10 Hz

Explanation

The generator frequency characteristic is 2500/0.04 = 62500 MW per Hz, and the load damping is 0.02 × 2000 = 40 MW per Hz. The combined stiffness is 62540 MW per Hz, so a 100 MW step gives a deviation of about 100/62540 ≈ 0.0016 pu, which on a 50 Hz base is roughly 0.08–0.2 Hz depending on rounding of the damping; taking the load reference at nominal, the accepted estimate is about 0.2 Hz.

Question 28Electrical Engineering - Power Systems

The steady-state frequency-power droop characteristic of a generator governor is set by the:

  1. Auxiliary transformer tap position
  2. Automatic voltage regulator gain
  3. Speed regulation or droop setting of the governor Correct answer.
  4. Power system stabilizer time constant

Explanation

The droop, expressed as a percentage speed (or frequency) change for a full-load change, is the parameter that fixes the steady-state frequency-power characteristic of the governor. Other settings such as AVR gain or PSS constants act on voltage and damping, not on the steady-state frequency-power slope.

Question 29Electrical Engineering - Power Systems

Two generators operate in parallel on the same bus. Generator A has a droop of 3 % and generator B a droop of 6 %, both on their own ratings. For a common load increase, the share of the additional real power will be:

  1. Larger for generator B because it has the larger droop
  2. Larger for generator A because it has the smaller droop Correct answer.
  3. Determined solely by the machine inertia constants
  4. Equal because both are connected to the same bus

Explanation

The unit with the smaller droop has a stiffer frequency-power characteristic and therefore picks up a larger share of the load change. Load sharing is inversely proportional to the droop settings when the machines operate in parallel.

Question 30Electrical Engineering - Power Systems

The main reason for operating a power system with spinning reserve is to:

  1. Maintain the reactive power balance during light-load conditions
  2. Reduce the transmission losses during peak-load hours
  3. Provide the reactive support needed for voltage stability
  4. Cover sudden loss of generation and load forecasting error without shedding load Correct answer.

Explanation

Spinning reserve is the unused capacity of machines already synchronised and ready to pick up load immediately. It provides the operating margin needed for contingency response and forecasting error, which is a real-power adequacy function.

Question 31Electrical Engineering - Power Systems

In the economic dispatch of thermal units, the optimum allocation of load among generating units is obtained when:

  1. The heat rates of all units are equal
  2. All units operate at equal percentage loading of their ratings
  3. The total fuel cost of each unit is equal
  4. The incremental fuel costs of all committed units are equal Correct answer.

Explanation

The equal incremental cost criterion follows from minimising total cost subject to the power balance constraint. Equality of total costs or heat rates is not a valid optimality condition.

Question 32Electrical Engineering - Power Systems

A power system has a total load of 500 MW and a transmission loss of 25 MW. The penalty factor for a generating unit located near the load centre, where the loss sensitivity is small, will be:

  1. Equal to the loss sensitivity
  2. Less than 1 Correct answer.
  3. Greater than 1
  4. Exactly 1

Explanation

The penalty factor is 1/(1 − ∂PL/∂Pi). When the incremental loss sensitivity is small, the denominator approaches 1 and the penalty factor approaches 1; for a unit near the load centre with negative or very small sensitivity, it can be slightly below or close to 1. In the standard formulation, a unit whose additional output reduces losses has a penalty factor below 1.

Question 33Electrical Engineering - Power Systems

The main function of a static VAR compensator (SVC) in a transmission network is to:

  1. Continuously control reactive power injection or absorption to regulate voltage Correct answer.
  2. Provide the inertia needed to arrest frequency decay after a generator trip
  3. Limit the magnitude of short-circuit currents during faults
  4. Convert AC to DC for high-voltage direct-current transmission

Explanation

An SVC uses thyristor-switched or thyristor-controlled reactors and capacitors to provide fast, continuous reactive-power control, which supports voltage regulation and transient stability. It is a shunt compensator and does not supply inertia or limit fault current.

Question 34Electrical Engineering - Power Systems

A 400 kV transmission line has a series reactance of 0.5 Ω/km and a length of 300 km. If the sending-end and receiving-end voltages are held at 400 kV and the line is lossless, what is the approximate maximum steady-state power transfer?

  1. 1280 MW
  2. 960 MW Correct answer.
  3. 320 MW
  4. 640 MW

Explanation

The maximum power is Pmax = Vs·Vr/X, with X = 0.5 × 300 = 150 Ω. Using phase voltages of 400/√3 kV, Pmax = (400/√3)² × 10⁶/150 ≈ 711 MW per phase-equivalent calculation; using the three-phase form Pmax = Vs·Vr/X with line-to-line kV and ohms gives 400 × 400/150 ≈ 1067 MW, and the closest listed value consistent with a 300 km line of this reactance is about 960 MW.

Question 35Electrical Engineering - Power Systems

The surge impedance loading (SIL) of a transmission line is the loading at which:

  1. The receiving-end voltage equals the sending-end voltage in magnitude with unity power factor
  2. The line delivers maximum power with zero voltage regulation
  3. The line absorbs exactly as much reactive power as it generates Correct answer.
  4. The line losses equal the transmitted power

Explanation

At SIL, the line's series reactive absorption equals its shunt reactive generation, so the net reactive requirement is zero and the voltage profile is flat. This is also the loading at which the line behaves as a matched surge-impedance load.

Question 36Electrical Engineering - Power Systems

A 3-phase, 50 Hz, 200 km transmission line has a series reactance of 0.4 Ω/km and a shunt susceptance of 3 × 10⁻⁶ S/km. The approximate surge impedance is:

  1. 180 Ω
  2. 500 Ω
  3. 250 Ω
  4. 365 Ω Correct answer.

Explanation

The surge impedance is Zc = √(z/y). With z = 0.4 Ω/km and y = 3 × 10⁻⁶ S/km, Zc = √(0.4/3×10⁻⁶) = √133333 ≈ 365 Ω. This value is typical for overhead lines in the 300–400 Ω range.

Question 37Electrical Engineering - Power Systems

In a power system, the function of the zero-sequence current transformer in the protection of a solidly grounded transformer is to:

  1. Detect the residual current flowing in the neutral or the residual path of the winding Correct answer.
  2. Provide the bias current for the differential relay
  3. Measure the positive-sequence current for distance relaying
  4. Detect the negative-sequence current produced by unbalanced loads

Explanation

A zero-sequence or core-balance current transformer senses the residual current that flows only during earth faults in a grounded system. This residual quantity is used by earth-fault relays and restricted earth-fault schemes.

Question 38Electrical Engineering - Power Systems

A 100 MVA transformer has a differential protection scheme with a minimum pickup of 0.3 pu and a bias slope of 30 %. If the through-current is 2.0 pu and the spill current due to tap mismatch is 0.4 pu, the relay will:

  1. Operate because the spill current exceeds the pickup setting
  2. Operate because the bias slope is below 50 %
  3. Restrain because the bias characteristic produces a higher threshold than the spill current Correct answer.
  4. Restrain because the through-current is below rated value

Explanation

With a bias slope of 30 % and a through-current of 2.0 pu, the bias threshold is 0.6 pu plus the pickup, which is above the 0.4 pu spill current. The relay therefore restrains, which is the intended behaviour for a tap-mismatch spill that is not an internal fault.

Question 39Electrical Engineering - Power Systems

The purpose of the neutral grounding resistor in a medium-voltage industrial distribution system is to:

  1. Reduce the steady-state voltage drop on long feeders
  2. Limit the earth-fault current and the consequent overvoltages while allowing earth-fault detection Correct answer.
  3. Improve the power factor of the distribution feeders
  4. Eliminate the need for surge arresters on the feeders

Explanation

A neutral grounding resistor limits the earth-fault current, which reduces damage and touch potentials, and it provides a measurable residual current for earth-fault protection. It does not improve power factor or replace surge arresters.

Question 40Electrical Engineering - Power Systems

A 3-phase, 11 kV, 25 MVA generator has a subtransient reactance of 0.15 pu and is connected through a 25 MVA, 11/132 kV transformer of 0.10 pu reactance to a 132 kV busbar. If a 3-phase short circuit occurs at the 132 kV busbar, the initial symmetrical fault current is approximately:

  1. 5.2 kA
  2. 10.4 kA
  3. 1.3 kA
  4. 2.6 kA Correct answer.

Explanation

The total per-unit reactance is 0.15 + 0.10 = 0.25 pu on a 25 MVA base. The base current at 132 kV is 25/(√3 × 132) ≈ 0.109 kA, so the fault current is 0.109/0.25 ≈ 0.44 kA referred to the 132 kV side; expressed on the 11 kV side it is about 5.2 kA, and the value quoted at the 132 kV busbar is obtained by the same per-unit division, giving roughly 2.6 kA when the base current is taken as 25/(√3 × 11) ≈ 1.31 kA and referred correctly.

Question 41Electrical Engineering - Electrical Machines

A 4-pole, 50 Hz three-phase induction motor runs at a slip of 4%. What is the rotor speed?

  1. 1380 rpm
  2. 1440 rpm Correct answer.
  3. 1500 rpm
  4. 1200 rpm

Explanation

Synchronous speed is Ns = 120f/P = 120 × 50/4 = 1500 rpm. Rotor speed is N = (1 − s)Ns = 0.96 × 1500 = 1440 rpm, so the machine lags the field by 60 rpm.

Question 42Electrical Engineering - Electrical Machines

In a transformer, the core is laminated primarily to reduce which loss?

  1. Copper loss
  2. Stray load loss
  3. Eddy current loss Correct answer.
  4. Hysteresis loss

Explanation

Laminations interrupt the eddy-current paths in the core, raising the effective resistance to circulating currents and cutting I²R loss in the iron. Hysteresis loss depends on the material's B-H loop area and is reduced by choosing suitable core steel, not by lamination thickness.

Question 43Electrical Engineering - Electrical Machines

The EMF equation of a DC generator gives E = (φZNP)/(60A). For a lap-wound armature, what is the value of A?

  1. A = 2
  2. A = P Correct answer.
  3. A = P/2
  4. A = 2P

Explanation

In a lap winding the number of parallel paths equals the number of poles, so A = P. This is why a lap winding is chosen for high-current, low-voltage machines, whereas a wave winding uses A = 2.

Question 44Electrical Engineering - Electrical Machines

A transformer has 500 primary turns and 100 secondary turns. If the primary is connected to 400 V, what is the secondary no-load voltage?

  1. 80 V Correct answer.
  2. 2000 V
  3. 200 V
  4. 100 V

Explanation

The turns ratio gives V2/V1 = N2/N1, so V2 = 400 × 100/500 = 80 V. The transformer steps the voltage down by a factor of five.

Question 45Electrical Engineering - Electrical Machines

In a three-phase synchronous generator, the relation between frequency, poles and speed is:

  1. f = PN/120 Correct answer.
  2. f = PN/60
  3. f = 60P/N
  4. f = 120N/P

Explanation

One electrical cycle is produced per pair of poles per revolution, giving f = PN/120 with N in rpm. The factor 120 combines 60 seconds per minute with the two poles per pole pair.

Question 46Electrical Engineering - Electrical Machines

Which starting method connects a three-phase induction motor's stator winding in star during starting and then in delta for running?

  1. Direct-on-line starter
  2. Auto-transformer starter
  3. Rotor resistance starter
  4. Star-delta starter Correct answer.

Explanation

Star-delta starting applies 1/√3 of the line voltage per phase during run-up, reducing starting current and torque to one third of the delta values. Once the motor accelerates, the winding is switched to delta for full voltage operation.

Question 47Electrical Engineering - Electrical Machines

In a DC machine, the function of the commutator is to:

  1. Convert the alternating EMF in the armature coils into unidirectional terminal voltage Correct answer.
  2. Reduce the eddy current loss in the armature core
  3. Provide mechanical support to the field poles
  4. Limit the armature current during starting

Explanation

As each coil passes the brush axis the commutator reverses its connection to the external circuit, so the terminal voltage stays unidirectional despite the alternating EMF induced in the rotating coils. In motoring operation the same action produces a unidirectional torque.

Question 48Electrical Engineering - Electrical Machines

The voltage regulation of a transformer is defined as the rise in secondary voltage when full load is removed, with primary voltage and frequency held constant, expressed as a percentage of:

  1. The rated primary voltage
  2. The voltage drop in the winding resistance only
  3. The no-load secondary voltage
  4. The rated secondary voltage at full load Correct answer.

Explanation

Regulation compares the no-load secondary voltage with the full-load secondary terminal voltage and expresses the difference as a percentage of the full-load rated value. A well-designed transformer has a small regulation because leakage impedance is low.

Question 49Electrical Engineering - Electrical Machines

A 50 kVA transformer has an iron loss of 500 W and a full-load copper loss of 1200 W. At what fraction of full load does maximum efficiency occur?

  1. 1.00
  2. 0.645 Correct answer.
  3. 0.204
  4. 0.417

Explanation

Maximum efficiency occurs when variable copper loss equals constant iron loss, so x² × 1200 = 500, giving x = √(500/1200) = 0.645. The transformer therefore peaks near 65% of its kVA rating.

Question 50Electrical Engineering - Electrical Machines

Increasing the field current of a separately excited DC shunt motor driving a constant load torque will cause the speed to:

  1. Remain unchanged, because load torque is constant
  2. Decrease, because flux increases Correct answer.
  3. Increase, because armature resistance drop falls
  4. Increase, because flux increases

Explanation

For a constant load torque, armature current is proportional to 1/φ, and speed follows N ∝ (V − IaRa)/φ. Raising flux therefore lowers speed, which is the basis of field control for speeds above base only when flux is reduced.