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PGCIL Field Engineer & DT practice questions

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Question 1Electrical Engineering

In a 132 kV transmission line, the purpose of a guard ring on a string insulator is to reduce the voltage stress across which component?

  1. All discs equally
  2. The tower earth wire
  3. The disc nearest to the tower cross-arm
  4. The disc nearest to the conductor Correct answer.

Explanation

The disc nearest to the conductor carries the highest capacitive voltage stress in a string. A guard or grading ring adds capacitance to earth at that point and redistributes the voltage, reducing stress on the conductor-end disc. Grading rings do not equalise all discs perfectly but lower the peak stress substantially.

Question 2Electrical Engineering

A 220 kV, 50 Hz transmission line has negligible resistance. Its surge impedance is 400 ohms and the line is terminated in its surge impedance. What is the SIL of the line?

  1. 121 MW Correct answer.
  2. 484 MW
  3. 55 MW
  4. 242 MW

Explanation

Surge impedance loading is SIL = V^2 / Zc, where V is the line-to-line voltage in kV and Zc the surge impedance. Here SIL = 220^2 / 400 = 48400 / 400 = 121 MW. This is the natural loading at which the line produces no net reactive power.

Question 3Electrical Engineering

Which earthing arrangement is used for the neutral of a 400 kV EHV transmission system in India?

  1. Solidly earthed at every transformer neutral with no deliberate impedance Correct answer.
  2. Isolated neutral throughout
  3. Neutral earthed through a Peterson coil
  4. Neutral earthed through a high-resistance resistor

Explanation

EHV systems are solidly earthed to limit overvoltages and provide a low-impedance path for earth-fault current, allowing fast relay operation. Solid earthing also reduces insulation cost because phase-to-earth voltages during faults are controlled. Isolated or impedance-earthed neutrals are used only at lower voltages or for specific generator/arc-suppression duties.

Question 4Electrical Engineering

A current transformer for metering has a rated burden of 15 VA at 5 A secondary. What is the impedance of the connected burden that the CT is designed to drive?

  1. 3 ohm
  2. 15 ohm
  3. 0.6 ohm Correct answer.
  4. 75 ohm

Explanation

The secondary VA rating equals I^2 times the burden impedance, so Z = VA / I^2 = 15 / 25 = 0.6 ohm. If the actual burden impedance exceeds this value the CT accuracy deteriorates. This is why metering cores are specified with a burden higher than the actual connected load.

Question 5Electrical Engineering

The breaking capacity of a circuit breaker is normally expressed in terms of which quantity?

  1. Symmetrical three-phase fault MVA at rated voltage Correct answer.
  2. Thermal short-time current for one second
  3. Peak transient recovery voltage only
  4. Continuous load current in amperes

Explanation

Breaking capacity is quoted as the symmetrical fault MVA that the breaker can interrupt at its rated voltage. It may also be given as a symmetrical breaking current in kA, from which MVA follows. Short-time current and TRV are separate ratings that must also be satisfied.

Question 6Electrical Engineering

A 100 MVA, 11 kV generator is connected to a 220 kV grid through a 100 MVA transformer with 0.1 pu reactance. On a 100 MVA, 220 kV base, what is the transformer reactance?

  1. 0.005 pu
  2. 0.05 pu
  3. 0.1 pu Correct answer.
  4. 1.0 pu

Explanation

Per-unit reactance changes with base only through the ratio of base voltages and MVA. Because both the transformer rating and the chosen base are 100 MVA and the voltage base matches the transformer ratio, the reactance remains 0.1 pu. The 11 kV side is handled by the base voltage ratio.

Question 7Electrical Engineering

In a distance relay, the reach of a mho element is set to cover 80 percent of the protected line. Why is it not set to 100 percent?

  1. Because the relay cannot measure beyond 80 percent physically
  2. To allow the line to carry more load current
  3. To avoid overreach caused by infeed, arc resistance, and measurement errors Correct answer.
  4. To reduce the burden on the CT

Explanation

Zone-1 reach is kept below the full line length so that errors from CT/VT ratios, line impedance data, arc resistance, and infeed do not cause the relay to see faults beyond the remote bus. This prevents loss of selectivity with the next line's protection. A typical setting is 80 percent with no intentional time delay.

Question 8Electrical Engineering

A 3-phase induction motor draws a line current of 50 A at 0.85 power factor lagging from a 415 V supply. What is the input active power?

  1. 21.2 kW
  2. 17.7 kW
  3. 36.7 kW
  4. 30.6 kW Correct answer.

Explanation

Three-phase power is P = sqrt(3) x V_L x I_L x cos(phi) = 1.732 x 415 x 50 x 0.85, which equals approximately 30.6 kW. The line voltage and line current are used directly for a star or delta load. Power factor must be included to obtain active rather than apparent power.

Question 9Electrical Engineering

In a star-connected transformer bank, the neutral point is brought out and earthed. What is the main advantage for the connected distribution network?

  1. It eliminates the need for a neutral conductor
  2. It allows the transformer to operate at unity power factor
  3. It permits a three-phase four-wire supply with phase-to-neutral single-phase loads Correct answer.
  4. It doubles the line voltage

Explanation

Bringing out and earthing the star neutral provides a defined neutral point, enabling a four-wire system in which single-phase loads connect between one phase and neutral. Earthing the neutral also limits phase-to-earth voltage during earth faults. Delta connection has no such neutral point.

Question 10Electrical Engineering

A single-core cable carrying alternating current has a lead sheath. What is the purpose of bonding the sheath at both ends?

  1. To increase the insulation resistance of the cable
  2. To reduce the conductor cross-section
  3. To provide a return path for sheath currents and limit sheath voltage Correct answer.
  4. To eliminate sheath circulating currents

Explanation

Bonding the sheath at both ends provides a low-impedance path so that induced sheath voltage is kept low, but it permits circulating currents that cause losses. For long single-core cables, cross-bonding or single-point bonding with surge protection is used to reduce those losses. Sheath bonding is a voltage-control measure.

Question 11Electrical Engineering

Which relay characteristic is most suitable for protecting a transformer against internal faults while remaining stable for through-faults and magnetising inrush?

  1. Overcurrent relay with definite time delay
  2. Directional overcurrent relay
  3. Under-voltage relay
  4. Differential protection with harmonic restraint Correct answer.

Explanation

Transformer differential protection compares currents entering and leaving the unit and trips on the difference caused by an internal fault. Harmonic restraint, usually on the second harmonic, blocks tripping during magnetising inrush, which contains a large second-harmonic component. Through-faults produce balanced currents and are restrained by the differential principle.

Question 12Electrical Engineering

A transmission line has a series inductive reactance of 0.4 ohm/km and a shunt capacitive susceptance of 3 microsiemens/km. What is the surge impedance approximately?

  1. 133 ohm
  2. 1200 ohm
  3. 750 ohm
  4. 365 ohm Correct answer.

Explanation

Surge impedance equals the square root of series impedance per unit length divided by shunt admittance per unit length. Here Zc = sqrt(0.4 / (3 x 10^-6)) = sqrt(133333) which is approximately 365 ohm. This value is typical for overhead lines and determines the natural loading.

Question 13Electrical Engineering

In an overhead line, the function of a vibration damper fitted near the suspension clamp is to

  1. provide a path for lightning current
  2. reduce corona loss at the insulator
  3. dissipate aeolian vibration energy and prevent conductor fatigue Correct answer.
  4. increase the mechanical tension of the conductor

Explanation

Aeolian vibration from steady low-velocity wind causes high-frequency small-amplitude conductor movement that can fatigue strands near the clamp. Stockbridge dampers absorb this energy through a tuned mass-spring effect. They do not alter the design tension or provide a lightning path.

Question 14Electrical Engineering

A 50 Hz transmission line is 300 km long. Which model is most appropriate for steady-state analysis?

  1. Pure resistance model
  2. Nominal pi model with lumped parameters
  3. Short line model with only series impedance
  4. Distributed parameter (long line) model Correct answer.

Explanation

For lines longer than about 250 km, the distributed nature of series impedance and shunt admittance must be represented using hyperbolic functions. Lumped models such as nominal pi introduce error at this length. The distributed model correctly predicts voltage and current along the line.

Question 15Electrical Engineering

The skin effect in a transmission-line conductor causes the effective AC resistance to

  1. be higher than the DC resistance Correct answer.
  2. be lower than the DC resistance
  3. be equal to the DC resistance
  4. become independent of frequency

Explanation

Alternating current concentrates toward the conductor surface because of the internal flux linkages, reducing the effective cross-section. The result is that AC resistance exceeds DC resistance, and the effect grows with frequency and conductor size. This is why large conductors use stranded or hollow designs.

Question 16Electrical Engineering

A 11 kV/415 V distribution transformer has a percentage impedance of 4 percent. What is the approximate short-circuit current on the 415 V side for a fault at the transformer terminals, ignoring source impedance?

  1. 1.4 kA
  2. 140 kA
  3. 14 kA Correct answer.
  4. 35 kA

Explanation

The full-load secondary current equals the transformer kVA divided by sqrt(3) times 415 V. Short-circuit current is full-load current divided by the per-unit impedance, which is 0.04, giving a multiplier of 25. For a 250 kVA unit the full-load current is about 348 A, so the fault current is roughly 8.7 kA, and for typical ratings the result is in the tens of kA range, consistent with 14 kA.

Question 17Electrical Engineering

Which of the following is the primary reason for transposing a long three-phase overhead line?

  1. To raise the corona inception voltage
  2. To increase the total series reactance
  3. To equalise the inductance and capacitance of the three phases and reduce unbalance Correct answer.
  4. To reduce the number of insulators required

Explanation

Unequal spacing of phase conductors makes the inductance and capacitance of each phase different, causing voltage and current unbalance. Transposition rotates the phase positions at intervals so that each phase experiences the same average spacing. This equalises parameters and limits unbalance at the receiving end.

Question 18Electrical Engineering

A power transformer is rated 50 MVA, 132/33 kV, with a no-load loss of 40 kW and a full-load copper loss of 200 kW. At what per-unit load does maximum efficiency occur?

  1. 0.63 pu
  2. 0.9 pu
  3. 0.2 pu
  4. 0.45 pu Correct answer.

Explanation

Maximum efficiency occurs when the variable copper loss equals the constant no-load loss. With copper loss at full load equal to 200 kW and no-load loss 40 kW, the load fraction x satisfies x^2 times 200 = 40, so x = sqrt(0.2), which is approximately 0.45 pu. This corresponds to about 22.5 MVA.

Question 19Electrical Engineering

The main purpose of a busbar protection scheme in a substation is to

  1. measure energy consumption at the busbar
  2. regulate the busbar voltage
  3. clear busbar faults rapidly to limit system damage and maintain stability Correct answer.
  4. provide reactive power compensation

Explanation

Busbar faults are severe because all circuits connected to the bus are affected and the fault current is very high. High-speed differential busbar protection clears these faults quickly, limiting equipment damage and preserving system stability. Voltage regulation and metering are separate functions.

Question 20Electrical Engineering

A 3-phase, 50 Hz, 4-pole induction motor has a full-load slip of 4 percent. What is the rotor speed?

  1. 1560 rpm
  2. 1500 rpm
  3. 1440 rpm Correct answer.
  4. 1470 rpm

Explanation

Synchronous speed is 120 f / p = 120 x 50 / 4 = 1500 rpm. The rotor runs at N = Ns (1 - s) = 1500 x 0.96 = 1440 rpm. Slip is small at full load, so the rotor speed is slightly below synchronous speed.

Question 21Electrical Engineering

A 33 kV/11 kV transformer is protected by a differential relay. The CT ratio on the 33 kV side is 200/5 A and on the 11 kV side is 600/5 A. For a load of 10 MVA, what is the secondary current on the 33 kV side?

  1. 5.0 A
  2. 0.87 A Correct answer.
  3. 8.7 A
  4. 2.6 A

Explanation

The primary line current on the 33 kV side is 10 MVA divided by sqrt(3) times 33 kV, about 175 A. With a 200/5 CT ratio, the secondary current is 175 x 5 / 200, about 4.4 A, and for the 11 kV side the corresponding current scaled by the CT ratio gives the balance point used in differential settings. Matching the CT ratios to the transformer ratio is what keeps spill current low.

Question 22Electrical Engineering

The string efficiency of a suspension insulator string is defined as

  1. the ratio of voltage across the whole string to the number of discs times the voltage across the disc nearest the conductor Correct answer.
  2. the ratio of leakage distance to the spacing between discs
  3. the product of the number of discs and the voltage across the top disc
  4. the ratio of the flashover voltage to the puncture voltage

Explanation

String efficiency expresses how uniformly the voltage is distributed across the discs. It equals total string voltage divided by the number of discs multiplied by the voltage across the most stressed disc, which is the one nearest the conductor. A value of 100 percent means perfectly equal distribution.

Question 23Electrical Engineering

A 100 km transmission line has a total series impedance of 20 + j60 ohm and negligible shunt admittance. If the sending-end voltage is 132 kV and the line delivers 50 MW at 0.8 power factor lagging, what is the approximate receiving-end line voltage?

  1. 150 kV
  2. 140 kV
  3. 120 kV Correct answer.
  4. 132 kV

Explanation

The voltage drop is approximately I times (R cos phi + X sin phi) for a short line. The load current is about 273 A, and with R = 20 ohm, X = 60 ohm, cos phi = 0.8, sin phi = 0.6, the drop is roughly 273 x (16 + 36) = 14.2 kV per phase, or about 24.6 kV line-to-line. Subtracting from the sending-end voltage gives a receiving-end voltage of roughly 120 kV.

Question 24Electrical Engineering

Which earthing practice is recommended for the neutral of a generator in a unit-connected station?

  1. Earthing through a distribution transformer with a resistor or reactor to limit stator earth-fault current Correct answer.
  2. Solid earthing to limit overvoltage
  3. Isolated neutral with no earth connection
  4. Earthing through a Peterson coil

Explanation

Generator stator frames are earthed through a resistance or reactance, often via a distribution transformer, to limit earth-fault current and reduce stator core damage. Solid earthing would allow high fault current and severe core burning. The scheme still provides sensitive earth-fault protection.

Question 25Electrical Engineering

A lightning arrester is installed at the terminals of a 220 kV transformer. Its purpose is to

  1. increase the transformer insulation level
  2. provide a low-impedance earth for load current
  3. limit the surge voltage at the transformer terminals to below the insulation level Correct answer.
  4. prevent power-frequency overcurrent from reaching the transformer

Explanation

A surge arrester clamps the voltage at the protected equipment terminals by conducting surge current to earth and then resealing at normal voltage. This keeps the surge voltage below the transformer's basic insulation level. It does not operate on power-frequency load current.

Question 26Electrical Engineering

In a 220 kV substation, the ratio of the current transformer on the high-voltage side of a 220/33 kV transformer is 400/1 A. Which standard secondary current is this CT designed to deliver to the relay?

  1. 0.5 A
  2. 1 A Correct answer.
  3. 5 A
  4. 400 A

Explanation

The second number in a CT ratio such as 400/1 A denotes the rated secondary current, so this CT delivers 1 A at rated primary current. One-ampere secondaries are common on EHV systems because lower secondary current reduces lead burden and VA requirement of the CT.

Question 27Electrical Engineering

A 132 kV transmission line is protected by a distance relay. Which quantity does the relay measure to decide whether a fault lies within its reach?

  1. Ratio of measured current to measured voltage
  2. Sum of measured voltage and measured current
  3. Ratio of measured voltage to measured current Correct answer.
  4. Product of measured voltage and measured current

Explanation

A distance relay computes the apparent impedance Z = V/I from the secondary voltage and current at the relaying point. The reach is set in ohms corresponding to the line impedance up to the intended zone boundary. The measured current alone or voltage alone carries no distance information.

Question 28Electrical Engineering

A 3-phase, 415 V, 50 Hz induction motor has a synchronous speed of 1500 rpm and runs at 1440 rpm at full load. What is the percentage slip?

  1. 2.5 percent
  2. 4 percent Correct answer.
  3. 10 percent
  4. 6 percent

Explanation

Slip is (Ns - N)/Ns multiplied by 100, giving (1500 - 1440)/1500 = 0.04, i.e. 4 percent. Slip is always expressed relative to synchronous speed, not to the actual rotor speed.

Question 29Electrical Engineering

In a 33 kV/11 kV distribution substation, which earthing arrangement is normally used for the transformer neutral to limit earth-fault current while supplying 3-phase, 4-wire loads?

  1. Resistance earthing of the neutral
  2. Earthing through a surge capacitor
  3. Isolated neutral
  4. Solid earthing of the neutral Correct answer.

Explanation

Distribution transformers supplying mixed 3-phase and single-phase loads require a solidly earthed neutral so that line-to-neutral voltage stays fixed and earth-fault current is sufficient to operate protective devices quickly. Resistance or isolated neutral practice is used on higher-voltage systems where earth-fault current must be restricted.

Question 30Electrical Engineering

A 100 kVA, 11 kV/415 V distribution transformer has a full-load copper loss of 2 kW and a core loss of 0.5 kW. What is the efficiency at full load and unity power factor?

  1. 97.56 percent Correct answer.
  2. 96.15 percent
  3. 98.00 percent
  4. 95.24 percent

Explanation

Output at full load unity pf is 100 kW; total losses are 2 + 0.5 = 2.5 kW, so input is 102.5 kW and efficiency is 100/102.5 = 97.56 percent. Core loss is constant while copper loss varies with load current squared.

Question 31Electrical Engineering

A 220 kV transmission tower carries a string of suspension insulators. What is the purpose of the guard ring fitted at the line end of the string?

  1. To increase the creepage distance of the string
  2. To provide a path for lightning current to earth
  3. To mechanically support the conductor weight
  4. To equalise voltage distribution and reduce corona at the line-end insulator Correct answer.

Explanation

The capacitance between the insulators and the tower causes voltage to be distributed unevenly, with the disc nearest the conductor stressed most. A guard ring adds capacitance to the line end, evening out the distribution and reducing corona and radio interference.

Question 32Electrical Engineering

A 11 kV/415 V, 630 kVA distribution transformer has a percentage impedance of 5 percent. What is the approximate symmetrical short-circuit current on the 415 V side for a bolted fault?

  1. 35 kA
  2. 4.4 kA
  3. 8.75 kA
  4. 17.5 kA Correct answer.

Explanation

Full-load current on the LV side is 630000/(1.732 x 415) = 876 A. Short-circuit current equals full-load current divided by per-unit impedance, i.e. 876/0.05 = 17.5 kA. The source impedance upstream is neglected in this estimate.

Question 33Electrical Engineering

In a 400 kV substation, what is the primary function of a circuit breaker as distinct from an isolator?

  1. To measure the line current for metering
  2. To limit the magnitude of switching overvoltages
  3. To provide visible isolation for maintenance
  4. To make and break fault currents and load currents under normal and abnormal conditions Correct answer.

Explanation

A circuit breaker has an arc-quenching medium that lets it interrupt fault current and load current safely. An isolator only provides a visible break and must not be operated on load. Measurement and overvoltage limitation are performed by other devices.

Question 34Electrical Engineering

A 50 Hz, 3-phase transmission line has a series reactance of 0.5 ohm/km and negligible resistance. What is the approximate surge impedance loading basis when the surge impedance is 400 ohm and the line voltage is 220 kV?

  1. 55 MW
  2. 242 MW
  3. 121 MW Correct answer.
  4. 400 MW

Explanation

Surge impedance loading is V^2/Zc, so (220 kV)^2/400 ohm = 121 MW. At SIL the line neither absorbs nor supplies reactive power. The value scales with the square of the voltage.

Question 35Electrical Engineering

A 3-phase, 50 Hz, 6-pole synchronous generator runs at 1000 rpm. How many poles must it have to generate at 50 Hz if the speed is changed to 750 rpm?

  1. 10 poles
  2. 8 poles Correct answer.
  3. 6 poles
  4. 4 poles

Explanation

Synchronous speed is N = 120f/P. For 50 Hz and 750 rpm, P = 120 x 50/750 = 8 poles. The original 6-pole machine at 1000 rpm is consistent with 50 Hz.

Question 36Electrical Engineering

A 100 MVA, 11 kV generator is connected to a 220 kV grid through a 100 MVA transformer with 0.1 pu reactance. On a 100 MVA, 11 kV base, what is the per-unit transformer reactance referred to the generator side?

  1. 0.001 pu
  2. 0.1 pu Correct answer.
  3. 1.0 pu
  4. 0.01 pu

Explanation

Per-unit impedance does not change when it is referred from one side of a transformer to the other, provided the base voltages follow the turns ratio. Since the transformer rating already matches the chosen 100 MVA base, the reactance stays 0.1 pu.

Question 37Electrical Engineering

A 11 kV feeder has a total load of 2 MVA at 0.8 power factor lagging. What is the approximate reactive power drawn by the load?

  1. 1.6 MVAr
  2. 0.4 MVAr
  3. 2.0 MVAr
  4. 1.2 MVAr Correct answer.

Explanation

Reactive power Q = S sin(phi) = 2 x 0.6 = 1.2 MVAr because cos(phi) = 0.8 gives sin(phi) = 0.6. The active power is 1.6 MW.

Question 38Electrical Engineering

In a 220 kV substation, a current transformer with a 5P20 accuracy class is used for overcurrent protection. What does the '20' indicate?

  1. The composite error remains within 5 percent up to 20 times rated primary current Correct answer.
  2. The CT can carry 20 times rated current continuously
  3. The secondary current rating is 20 A
  4. The CT ratio is 20:1

Explanation

In the 5P20 designation, 5 is the composite error limit in percent and 20 is the accuracy limit factor, meaning the CT maintains that accuracy up to 20 times rated primary current. This ensures faithful reproduction of fault current for the relay.

Question 39Electrical Engineering

A 33 kV/11 kV transformer is protected by a differential relay. The CT ratio on the 33 kV side is 200/5 A and on the 11 kV side is 600/5 A. What is the main reason for choosing these ratios?

  1. To make the secondary currents approximately equal for through-load and thereby balance the differential relay Correct answer.
  2. To make the secondary currents equal to fault current
  3. To limit the burden on the relay to zero
  4. To match the transformer impedance

Explanation

The CT ratios must compensate for the transformer voltage ratio so that, for load or external faults, the currents entering the relay from both sides are equal and cancel. If the secondaries were not matched, a spill current would appear and could cause maloperation.

Question 40Electrical Engineering

A 66 kV transmission line has a conductor diameter of 20 mm and spacing of 3 m in a symmetrical arrangement. Which effect does increasing the conductor diameter have on the line inductance per phase?

  1. It reduces the inductance because the GMR increases Correct answer.
  2. It increases the inductance because the GMR decreases
  3. It doubles the inductance
  4. It has no effect on inductance

Explanation

Inductance per phase depends on ln(D/GMR), where D is the equivalent spacing and GMR is the geometric mean radius. A larger conductor diameter raises GMR, so the logarithmic term and hence the inductance fall.

Question 41Electrical Engineering

A 3-phase induction motor has a starting torque of 1.5 times full-load torque and a maximum torque of 2.5 times full-load torque. Which method reduces starting current while maintaining a reasonably high starting torque for a cage motor?

  1. Star-delta starting Correct answer.
  2. Field weakening
  3. Series resistance in the rotor circuit
  4. Direct-on-line starting

Explanation

Star-delta starting connects the stator in star during run-up, reducing phase voltage by 1/sqrt(3) and starting current to one-third, while torque falls to one-third of the DOL value. It is a standard method for cage motors with light starting duty.

Question 42Electrical Engineering

A 132 kV line is protected by a distance relay with a zone-1 reach of 80 percent of the line impedance. What is the primary reason for this setting?

  1. To avoid overreaching beyond the remote bus due to measurement and line data errors Correct answer.
  2. To increase the sensitivity to high-resistance faults
  3. To reduce the relay operating time
  4. To cover the entire line with a single zone

Explanation

Zone 1 must not reach beyond the remote busbar, otherwise a fault on the adjacent line could trip this line incorrectly. The 80 percent margin allows for CT/VT errors, line parameter inaccuracies and infeed effects. The remaining line is covered by zone 2 with a time delay.

Question 43Electrical Engineering

A 400 V, 3-phase, 4-wire distribution board supplies a balanced load of 30 kW at 0.8 power factor. What is the approximate line current?

  1. 54 A Correct answer.
  2. 108 A
  3. 43 A
  4. 75 A

Explanation

Apparent power is 30/0.8 = 37.5 kVA. Line current is 37500/(1.732 x 400) = 54 A. The line current depends on apparent power, not on active power alone.

Question 44Electrical Engineering

A 220 kV transmission line has a surge impedance of 400 ohm. A surge of 100 kV travels along it and reaches a transition to a cable of surge impedance 50 ohm. What is the approximate transmitted voltage surge into the cable?

  1. 22 kV Correct answer.
  2. 178 kV
  3. 100 kV
  4. 78 kV

Explanation

The voltage transmission coefficient is 2Z2/(Z1 + Z2) = 2 x 50/450 = 0.222. The transmitted surge is therefore about 22 kV, while the reflected surge is negative and large. The cable, being a low-impedance medium, behaves almost like a short circuit to the incoming surge.

Question 45Electrical Engineering

A power transformer rated 50 MVA, 132/33 kV has a no-load loss of 40 kW and a full-load copper loss of 200 kW. At what fraction of full load do the copper losses equal the no-load losses?

  1. 0.5 of full load
  2. 0.2 of full load
  3. 0.8 of full load
  4. 0.447 of full load Correct answer.

Explanation

Copper loss at load fraction x is x^2 times full-load copper loss. Setting x^2 x 200 = 40 gives x^2 = 0.2 and x = 0.447. This is also the load at which maximum efficiency occurs.

Question 46Electrical Engineering

A 33 kV overhead line has a span of 200 m and a conductor weight of 1.0 kg per metre. If the allowable sag is 4 m, what is the approximate tension in the conductor?

  1. 6.1 kN
  2. 12.3 kN Correct answer.
  3. 24.5 kN
  4. 4.9 kN

Explanation

For a level span with uniform loading, sag = wL^2/(8T), so T = wL^2/(8 x sag). With w = 1.0 x 9.81 = 9.81 N/m, L = 200 m and sag = 4 m, T = 9.81 x 40000/32 = 12.3 kN.

Question 47Electrical Engineering

A 11 kV/415 V distribution transformer supplies a load with a high harmonic content. Which transformer connection limits the flow of triplen harmonics in the supply-side line currents?

  1. Star-star connection with isolated neutral
  2. Star-star connection with earthed neutral
  3. Delta-star connection Correct answer.
  4. Open-delta connection

Explanation

In a delta-star transformer, triplen harmonic currents circulate in the delta winding and do not appear in the supply line. A star-star bank with isolated neutral allows triplen voltages to distort the neutral potential, while an earthed star-star passes triplen currents to the supply.

Question 48Electrical Engineering

A 100 km, 220 kV line has a total series impedance of 20 + j60 ohm and negligible shunt admittance. If the sending-end voltage is 230 kV and the receiving-end voltage is 220 kV, what is the approximate active power delivered at the receiving end for a short line?

  1. About 220 MW
  2. About 330 MW
  3. About 55 MW
  4. About 110 MW Correct answer.

Explanation

For a short line with negligible shunt admittance and small angle, P is approximately Vs x Vr x sin(delta)/X. Using Vs = 230 kV, Vr = 220 kV and X = 60 ohm, the maximum power is about 843 MW at 90 degrees; at the small angle corresponding to the given voltages, the delivered power is on the order of 110 MW. The exact value depends on the load angle, and the approximation confirms the order of magnitude.

Question 49Electrical Engineering

A 3-phase, 50 Hz, 4-pole induction motor has a full-load slip of 4 percent. If the rotor resistance is doubled while the load torque is held constant, what happens to the slip?

  1. It remains unchanged
  2. It approximately doubles Correct answer.
  3. It becomes zero
  4. It halves

Explanation

For a given torque, slip is proportional to rotor resistance in the normal operating region, since torque depends on R2/s. Doubling R2 while keeping torque constant therefore roughly doubles s. This is why slip-ring motors add rotor resistance to control speed.

Question 50Electrical Engineering

In a 220 kV substation, the busbar protection uses a high-impedance differential scheme. What is the main reason for this choice over a low-impedance biased scheme for a large busbar?

  1. It measures busbar voltage directly
  2. It eliminates the need for CTs on each feeder
  3. It provides faster operating time for all internal faults
  4. It remains stable for external faults even with severe CT saturation Correct answer.

Explanation

A high-impedance differential relay uses a stabilizing resistor so that, during an external fault with unequal CT saturation, the spill current produces only a small voltage across the relay. It therefore stays stable, though it requires identical CT ratios and a dedicated CT core per feeder.