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Egypt Engineering Consultant Electrical practice questions

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Question 1Power System Analysis and Design

In an Egyptian MV distribution network operating at 11 kV, a three-phase short circuit is stated as 250 MVA. What is the approximate initial symmetrical short-circuit current?

  1. About 13.1 kA Correct answer.
  2. About 4.4 kA
  3. About 1.3 kA
  4. About 22.7 kA

Explanation

The short-circuit current follows from I = S/(√3·V), so 250×10^6 divided by (1.732×11×10^3) gives roughly 13.1 kA. This is the standard first step in MV fault-level assessment before any impedance correction.

Question 2Power System Analysis and Design

Which quantity is used to express the fault level of a busbar in a power system study?

  1. Short-circuit MVA at the busbar Correct answer.
  2. Voltage regulation at the busbar
  3. Transmission angle at the busbar
  4. Load power factor at the busbar

Explanation

Fault level is conventionally quoted as short-circuit MVA, obtained from the prospective fault current and nominal voltage. It summarises the strength of the busbar for switchgear rating and protection coordination.

Question 3Power System Analysis and Design

A 500 kVA, 11/0.415 kV distribution transformer has 5% impedance. What is the approximate low-voltage three-phase short-circuit current for an infinite source?

  1. About 69.5 kA
  2. About 13.9 kA Correct answer.
  3. About 5.6 kA
  4. About 1.4 kA

Explanation

Full-load current on the LV side is 500000/(√3×415) ≈ 696 A, and dividing by the 0.05 per-unit impedance gives about 13.9 kA. This is the transformer-limited fault current before cable and source impedance are added.

Question 4Power System Analysis and Design

Why is the per-unit system preferred for short-circuit and load-flow studies of interconnected Egyptian networks?

  1. It guarantees unity power factor throughout the network
  2. It converts all impedances into ohmic values referred to one side
  3. It eliminates the need for base quantities in calculations
  4. It removes voltage-level differences so impedances across transformers can be combined directly Correct answer.

Explanation

Per-unit values referred to a common base allow impedances on either side of a transformer to be added without turns-ratio conversions. This simplifies multi-voltage network reduction and keeps numerical values in a convenient range.

Question 5Power System Analysis and Design

In a radial MV feeder, the protective device nearest the fault should clear it first. Which coordination principle does this describe?

  1. Stability
  2. Selectivity margin of the transformer only
  3. Selectivity Correct answer.
  4. Sensitivity

Explanation

Selectivity means only the device closest to a fault operates, leaving upstream devices closed so the rest of the network stays energised. It is achieved by grading time-current characteristics between successive devices.

Question 6Power System Analysis and Design

A relay with a very inverse characteristic is set at 200 A pickup. For a fault current of 1000 A, what is the multiple of pickup?

  1. 5 Correct answer.
  2. 2
  3. 0.2
  4. 10

Explanation

The multiple of pickup is fault current divided by pickup setting, so 1000/200 = 5. This multiple is then applied to the relay curve to obtain the operating time.

Question 7Power System Analysis and Design

Which earthing arrangement is typical for the MV side of an Egyptian distribution substation feeding an 11 kV network?

  1. Unearthed neutral with no earth-fault detection
  2. Resistance earthing of the transformer neutral to limit earth-fault current Correct answer.
  3. Direct earthing of all phase conductors
  4. Solid earthing of every MV neutral with no impedance

Explanation

MV distribution systems commonly use resistance or impedance earthing to limit earth-fault current and reduce damage and step potentials. Earth-fault protection is then set to detect the deliberately limited current.

Question 8Power System Analysis and Design

A load-flow study of a transmission network treats a generator bus as a PV bus. Which quantities are specified at that bus?

  1. Voltage magnitude and reactive power injection
  2. Voltage magnitude and active power injection Correct answer.
  3. Voltage angle and reactive power injection
  4. Active and reactive power injection

Explanation

A PV bus fixes the voltage magnitude through excitation control and the active power from the prime mover. The voltage angle and reactive power are then solved by the load-flow algorithm.

Question 9Power System Analysis and Design

In a load-flow solution, the slack bus is required mainly to:

  1. Balance the difference between scheduled generation and total demand plus losses Correct answer.
  2. Fix the reactive power of every generator in the network
  3. Determine the short-circuit level of each bus
  4. Set the tap position of all transformers

Explanation

The slack bus absorbs the mismatch between total generation and total load plus network losses, which are unknown before the solution. Its voltage magnitude and angle are specified to give the equations a reference.

Question 10Power System Analysis and Design

A 220 kV transmission line has series reactance of 0.4 Ω/km and is 100 km long. Using a 100 MVA, 220 kV base, what is the approximate per-unit reactance?

  1. About 8.3 pu
  2. About 0.83 pu
  3. About 0.0083 pu
  4. About 0.083 pu Correct answer.

Explanation

Base impedance is V²/S = (220×10^3)²/100×10^6 = 484 Ω. The line reactance is 40 Ω, so the per-unit value is 40/484 ≈ 0.083 pu.

Question 11Power System Analysis and Design

Which phenomenon limits the maximum power that can be transferred over a long transmission line and is assessed by a stability study?

  1. Ferroresonance in voltage transformers
  2. Corona discharge along the conductors
  3. Steady-state rotor-angle stability Correct answer.
  4. Skin effect in the phase conductors

Explanation

Power transfer across a line depends on the sine of the angle between sending and receiving end voltages, so the angle limit sets the steady-state stability margin. Stability studies determine the maximum transferable power and the margin to the limit.

Question 12Power System Analysis and Design

A synchronous generator is rated 60 MW at 0.8 power factor lagging. What is its apparent power rating?

  1. 60 MVA
  2. 96 MVA
  3. 75 MVA Correct answer.
  4. 48 MVA

Explanation

Apparent power equals active power divided by power factor, so 60/0.8 = 75 MVA. The MVA rating determines the machine current and is the basis for per-unit impedance data.

Question 13Power System Analysis and Design

In a short-circuit study, what does the subtransient reactance X''d of a synchronous generator represent?

  1. The leakage reactance of the stator winding only
  2. The reactance in the steady state after the fault current decays
  3. The reactance immediately after a fault, dominated by damper winding effects Correct answer.
  4. The reactance of the generator transformer referred to the generator side

Explanation

X''d applies in the first few cycles after a fault, when damper windings and rotor body oppose flux changes, giving the highest fault current. It is used for switchgear and high-speed protection duties.

Question 14Power System Analysis and Design

Which study determines whether bus voltages remain within statutory limits under maximum and minimum load conditions?

  1. Insulation coordination study
  2. Voltage-profile load-flow study Correct answer.
  3. Earthing grid study
  4. Short-circuit study

Explanation

A load-flow study computes bus voltage magnitudes and angles under specified generation and demand, allowing comparison against statutory limits. Running maximum and minimum load cases reveals weak buses and reactive compensation needs.

Question 15Power System Analysis and Design

A 50 Hz system has a total synchronous generation inertia constant H of 4 s on a 500 MVA base. If a 50 MW load is lost, what is the approximate initial rate of change of frequency?

  1. About 0.25 Hz/s increase
  2. About 0.0025 Hz/s increase
  3. About 0.025 Hz/s increase Correct answer.
  4. About 2.5 Hz/s increase

Explanation

Using df/dt = f·ΔP/(2H·S), the change is 50×50/(2×4×500) = 0.025 Hz/s. The sign is positive because load loss leaves surplus generation.

Question 16Power System Analysis and Design

Why is a high-voltage transmission network operated at 500 kV rather than 220 kV for bulk transfer in Egypt?

  1. Higher voltage increases the reactive power absorbed by series reactance
  2. Higher voltage eliminates the need for reactive compensation
  3. Higher voltage reduces current and I²R losses for the same power transfer Correct answer.
  4. Higher voltage reduces the insulation requirement of the line

Explanation

For a given power, raising voltage lowers current, and losses proportional to current squared fall accordingly. This also permits smaller conductors and larger transfer capability over long distances.

Question 17Power System Analysis and Design

A 220 kV line has a surge impedance of 400 Ω and a surge impedance loading basis. If the line is loaded to its surge impedance loading, the receiving-end voltage will be approximately:

  1. Twice the sending-end voltage
  2. Half the sending-end voltage
  3. Equal to the sending-end voltage in magnitude Correct answer.
  4. Zero at the receiving end

Explanation

At surge impedance loading the line behaves as a matched load, so reactive losses in the series reactance balance reactive generation in the shunt capacitance. The voltage profile stays flat and the receiving-end magnitude equals the sending-end magnitude.

Question 18Power System Analysis and Design

In protection grading, what is the purpose of a time-grading margin between two successive overcurrent relays?

  1. To synchronise the two relays to the same clock
  2. To allow the downstream relay and breaker to clear the fault before the upstream relay trips Correct answer.
  3. To increase the pickup current of the upstream relay
  4. To reduce the burden on current transformers

Explanation

The grading margin covers the downstream breaker interrupting time plus relay and current-transformer errors. Without it, both relays could operate for the same fault and defeat selectivity.

Question 19Power System Analysis and Design

A 11 kV feeder supplies a 2 MVA load at 0.9 power factor lagging. What is the approximate line current?

  1. About 51 A
  2. About 117 A Correct answer.
  3. About 105 A
  4. About 175 A

Explanation

Line current equals apparent power divided by √3 times voltage, so 2×10^6/(1.732×11×10^3) ≈ 105 A at unity power factor, and dividing by 0.9 gives about 117 A. The power factor raises current for the same active power.

Question 20Power System Analysis and Design

Which method is commonly used to improve the voltage profile at the end of a long radial MV feeder?

  1. Reducing the transformer impedance to zero
  2. Increasing the overcurrent relay pickup setting
  3. Lowering the source voltage at the substation
  4. Installing shunt capacitors along the feeder Correct answer.

Explanation

Shunt capacitors supply reactive power locally, reducing the reactive current and hence the voltage drop along the feeder. This raises the voltage at downstream buses without changing protection settings.

Question 21Power System Analysis and Design

A three-phase fault occurs at the terminals of a 30 MVA generator with X''d = 0.15 pu on its own base. Using a 100 MVA base, what is the per-unit reactance?

  1. 0.50 pu Correct answer.
  2. 0.045 pu
  3. 0.15 pu
  4. 0.05 pu

Explanation

Converting to a larger base multiplies by the new base divided by the old base, so 0.15×100/30 = 0.50 pu. The physical reactance is unchanged, but the per-unit value grows with the base.

Question 22Power System Analysis and Design

In an Egyptian distribution substation, why are current transformers on the MV side often specified with a high accuracy limit factor?

  1. To eliminate the need for a relay setting study
  2. To preserve accuracy during the high fault currents that flow before protection operates Correct answer.
  3. To allow the CT to operate as a voltage transformer
  4. To reduce the secondary burden to nearly zero

Explanation

The accuracy limit factor indicates how many times rated current the CT can transform while staying within its composite error. High fault levels on MV networks require this capability so relays see a faithful current during faults.

Question 23Power System Analysis and Design

What is the main purpose of an earth-fault relay connected to a core-balance current transformer on an MV feeder?

  1. To detect small residual currents caused by insulation failure to earth Correct answer.
  2. To control the tap changer of the distribution transformer
  3. To limit the through-fault current of the feeder transformer
  4. To measure the positive-sequence load current for billing

Explanation

A core-balance CT sums the three phase currents and outputs only the residual, which is small in healthy operation but rises on an earth fault. This gives sensitive earth-fault detection on impedance-earthed systems.

Question 24Power System Analysis and Design

A load-flow study shows a bus voltage of 0.92 pu at peak load. Which corrective measure most directly raises this voltage?

  1. Reducing the transformer tap to lower the secondary voltage
  2. Opening a parallel transmission line
  3. Adding a shunt capacitor bank at or near the bus Correct answer.
  4. Increasing the active power demand at the bus

Explanation

Shunt capacitors inject reactive power, reducing the reactive component of line current and the associated voltage drop. This directly raises the bus voltage magnitude in the load-flow solution.

Question 25Power System Analysis and Design

In a symmetrical-components analysis of an unbalanced fault, which sequence network is connected for a single line-to-ground fault?

  1. Positive and zero sequence networks in parallel only
  2. Positive, negative and zero sequence networks in series Correct answer.
  3. Negative and zero sequence networks in series only
  4. Positive and negative sequence networks in parallel only

Explanation

A single line-to-ground fault imposes equal sequence currents in all three networks and zero voltage on the faulted phase, so the three networks are connected in series. The fault current is then found from the sum of the sequence impedances.

Question 26Power System Analysis and Design

In per-unit short-circuit analysis of an Egyptian 220 kV network, a 150 MVA transformer has an impedance of 0.12 pu on its own base. What is its reactance in per unit on a 100 MVA system base?

  1. 0.120 pu
  2. 0.240 pu
  3. 0.080 pu Correct answer.
  4. 0.180 pu

Explanation

Per-unit impedance converts between bases in direct proportion to the base power, so X_new = X_old × (S_new/S_old) = 0.12 × (100/150) = 0.080 pu. The voltage base is unchanged, so no voltage ratio term is needed. This base conversion is essential before combining transformer and line impedances in a single network model.

Question 27Power System Analysis and Design

A 500 kV transmission corridor in Egypt is being planned and the planners want an approximate indication of the steady-state power transfer capability per circuit before detailed load-flow work. Which quantity provides this first estimate?

  1. The surge impedance loading of the line Correct answer.
  2. The charging current measured at no load
  3. The thermal rating of the conductor at 40 °C ambient
  4. The short-circuit level of the receiving busbar

Explanation

Surge impedance loading, SIL = V²/Zc, gives the power at which the line's reactive losses are balanced by its charging, so voltage stays nearly flat and this value is used as a first estimate of transfer capability. Thermal rating limits current but is not a voltage-stability indicator for long lines. Short-circuit level and charging current describe other aspects of network behaviour.

Question 28Power System Analysis and Design

A consultant is specifying the neutral earthing of a generator connected directly to an 11 kV busbar in an industrial plant. Which practice is normally adopted to limit stator earth-fault current to a few amperes?

  1. Earthing through a low resistance sized for full fault current
  2. Solid earthing of the generator neutral
  3. Leaving the generator neutral isolated with no earth connection at all
  4. Earthing through a distribution transformer with a resistor in its secondary Correct answer.

Explanation

A generator is normally earthed through a distribution transformer loaded by a resistor, which reflects a high resistance into the stator circuit and limits earth-fault current to a few amperes, protecting the core from burning. Solid earthing would permit very large stator earth-fault currents and severe core damage. An isolated neutral is avoided because it allows damaging overvoltages and makes earth-fault detection difficult.

Question 29Power System Analysis and Design

A 220 kV double-circuit line feeds a substation, and a study must quantify the risk of a fault that involves all three phases simultaneously at the sending end. Which modelling assumption is appropriate for the initial symmetrical fault current in a conventional short-circuit study?

  1. Use the generator subtransient reactance and neglect all load current Correct answer.
  2. Use the synchronous reactance and include the pre-fault voltage profile
  3. Use only the transformer impedance and ignore the generators
  4. Use the generator transient reactance and include full pre-fault load current

Explanation

For the initial symmetrical current, machines are represented by their subtransient reactance because the rotor flux cannot change instantly, and pre-fault load current is neglected while the internal EMF is set to the pre-fault voltage. This gives the highest and most conservative fault current for equipment rating. Transient or synchronous reactances apply to later instants, not the initial peak.

Question 30Power System Analysis and Design

An 11 kV radial feeder has a load of 4 MVA at 0.85 power factor lagging at its far end. What is the approximate reactive power demand of this load?

  1. 4.0 Mvar
  2. 1.2 Mvar
  3. 2.1 Mvar Correct answer.
  4. 3.4 Mvar

Explanation

Reactive power is Q = S × sin φ, and with power factor 0.85 the angle is about 31.8°, so sin φ ≈ 0.527. Thus Q ≈ 4 × 0.527 ≈ 2.1 Mvar. This reactive demand drives voltage drop along the feeder and is the basis for capacitor compensation studies.

Question 31Power System Analysis and Design

During commissioning of a new 66 kV substation, a three-phase short circuit is applied at the feeder terminals and the resulting current is compared with the calculated value. What is the main engineering purpose of such a test?

  1. To verify the protection settings and current-transformer ratios against the design study Correct answer.
  2. To check the phase rotation of the incoming supply
  3. To measure the insulation resistance of the busbar before energisation
  4. To confirm the transformer tap-changer range under load

Explanation

A primary injection or staged fault test confirms that current transformers, relay settings and trip circuits respond as predicted by the short-circuit study. It is a protection-verification activity, not an insulation or tap-changer test. Phase rotation is checked by other means and does not require a fault.

Question 32Power System Analysis and Design

A load-flow study of an Egyptian 220 kV network shows a bus voltage of 0.94 pu at peak load, below the statutory limit. Which measure most directly raises the voltage at that bus?

  1. Increasing the reactance of the feeding transformer
  2. Adding shunt capacitors at the bus or nearby Correct answer.
  3. Reducing the conductor cross-section on the feeding line
  4. Increasing the transformer tap position to reduce the secondary voltage

Explanation

Shunt capacitors inject reactive power locally, reducing reactive flow through the feeding impedance and raising the bus voltage. This is the standard corrective measure for a depressed voltage at peak load. Increasing reactance or reducing conductor size increases voltage drop, and the wrong tap direction lowers voltage further.

Question 33Power System Analysis and Design

In a 220 kV substation, the busbar rated short-time withstand current is 40 kA for 1 s. A study shows a fault current of 31.5 kA that is cleared in 0.5 s. Which statement about the busbar is correct?

  1. The busbar is inadequate because the clearing time exceeds 1 s
  2. The busbar is inadequate because the fault current exceeds 40 kA
  3. The busbar is adequate only if the fault current is below 20 kA
  4. The busbar is adequate because I²t = 496 kA²s is below the 1600 kA²s rating Correct answer.

Explanation

The thermal duty is compared using I²t, so the actual duty is 31.5² × 0.5 ≈ 496 kA²s against a rating of 40² × 1 = 1600 kA²s. Since 496 is well below 1600, the busbar is thermally adequate. The comparison is on the I²t product, not on current alone or time alone.

Question 34Power System Analysis and Design

A 50 Hz, 200 MVA synchronous generator has an inertia constant H of 5 s. If the generator suddenly loses 20 MW of load while its mechanical input remains unchanged, what is the approximate initial rate of rise of frequency?

  1. 0.10 Hz/s Correct answer.
  2. 0.20 Hz/s
  3. 0.02 Hz/s
  4. 1.00 Hz/s

Explanation

The swing equation gives df/dt = f0 × ΔP/(2H × S), so df/dt = 50 × 20/(2 × 5 × 200) = 0.10 Hz/s. The inertia constant and machine rating set the initial frequency ramp after a load rejection. This rate is the basis for setting under-frequency and over-frequency protection.

Question 35Power System Analysis and Design

In the design of an MV distribution network in Egypt, the neutral of the 11 kV source is earthed through a resistor. What is the main reason for this choice compared with solid earthing?

  1. To allow the network to operate with one phase permanently disconnected
  2. To eliminate the need for earth-fault protection on the feeders
  3. To increase the insulation level required on the phase conductors
  4. To reduce the earth-fault current and limit damage while still enabling detection of the fault Correct answer.

Explanation

Resistance earthing limits the earth-fault current to a controlled value, reducing damage at the fault point and touch-voltage hazards while still giving a measurable current for earth-fault relays. It does not remove the need for protection, and it does not permit operation with a phase open. Limiting fault current generally reduces stress rather than increasing insulation requirements.

Question 36Power System Analysis and Design

A 132 kV line is protected by distance relays at both ends. A fault occurs at 80 % of the line length from end A, and the relay at A has zones set to 80 %, 120 % and 200 % of the line impedance. Which zone should trip instantaneously for this fault?

  1. Zone 2
  2. Zone 1 Correct answer.
  3. Zone 3
  4. No zone, because the fault is beyond the first zone

Explanation

Zone 1 typically covers about 80 % of the protected line and operates instantaneously without intentional delay, so a fault at 80 % lies at the zone 1 reach boundary and is cleared by zone 1. Zone 2 covers the remainder of the line plus a margin into the next line and is time delayed. Zone 3 provides remote backup with a longer delay.

Question 37Electrical Installation and Distribution Systems

In a TN-S earthing arrangement serving an Egyptian industrial building, how are the neutral conductor and the protective conductor related from the origin of the installation onward?

  1. The neutral is earthed at every distribution board along the route
  2. The protective conductor is connected to a local earth electrode only and isolated from the neutral
  3. They are combined in a single PEN conductor throughout the installation
  4. They are run as separate conductors from the supply origin, joined only at the source neutral point Correct answer.

Explanation

TN-S means the neutral (N) and protective earth (PE) are separate conductors across the whole installation, being connected together only at the origin, typically the transformer or supply neutral point. Keeping them separate limits potential differences on exposed metalwork and avoids stray neutral current in the earth path.

Question 38Electrical Installation and Distribution Systems

A final circuit is protected by a 32 A miniature circuit breaker of type C. What is its instantaneous tripping range according to IEC 60898 characteristics?

  1. Above 20 times and up to 30 times rated current
  2. Above 3 times and up to 5 times rated current
  3. Above 5 times and up to 10 times rated current Correct answer.
  4. Above 10 times and up to 20 times rated current

Explanation

Type C MCBs are defined to trip magnetically between 5 and 10 times the rated current, making them suitable for loads with moderate inrush such as small motors and lighting. Type B covers 3 to 5 times and type D covers 10 to 20 times the rating.

Question 39Electrical Installation and Distribution Systems

Which conductor colour is used for the protective earth in a single-phase final circuit wired with harmonised colours in Egypt?

  1. Black
  2. Brown
  3. Green-and-yellow Correct answer.
  4. Blue

Explanation

Under the harmonised European colour code adopted widely in Egypt, the protective conductor must be green-and-yellow and this combination may not be used for any other purpose. Line is brown and neutral is blue in single-phase circuits.

Question 40Electrical Installation and Distribution Systems

What is the primary purpose of a residual current device (RCD) in a distribution board?

  1. To correct the power factor of inductive loads
  2. To limit short-circuit current to the equipment rating
  3. To detect imbalance between line and neutral currents and disconnect the circuit Correct answer.
  4. To step down the supply voltage to a safe level

Explanation

An RCD continuously compares the current in the line and neutral conductors; a difference above its sensitivity indicates leakage to earth, and it trips to protect people and property. It does not respond to overload or short-circuit current, which is the role of the overcurrent device.

Question 41Electrical Installation and Distribution Systems

In a three-phase four-wire distribution system, which conductor carries the greatest current when the load is a balanced star-connected resistive load?

  1. Each line conductor carries equal current and the neutral carries none Correct answer.
  2. The neutral carries the arithmetic sum of the three line currents
  3. The neutral carries one third of the line current
  4. One line conductor carries twice the current of the others

Explanation

With a balanced three-phase load the line currents are equal in magnitude and displaced by 120 degrees, so their phasor sum is zero and no neutral current flows. The neutral only carries current when the phases are unbalanced or harmonics are present.

Question 42Electrical Installation and Distribution Systems

Which document type defines the protective device ratings, cable sizes, and earth fault loop values for an electrical installation in Egypt?

  1. A single-line diagram with associated protective device schedule Correct answer.
  2. A cable drum delivery note
  3. A load survey questionnaire
  4. A bill of quantities

Explanation

The single-line diagram together with the protective device schedule records the ratings, cable sizes, and earth fault loop impedance values that demonstrate compliance with disconnection requirements. It is the reference document used by the consultant for verification and by the contractor for installation.

Question 43Electrical Installation and Distribution Systems

Which of the following is the standard rated residual operating current for an RCD intended to provide additional protection against electric shock in a final circuit?

  1. 300 mA
  2. 30 mA Correct answer.
  3. 10 mA
  4. 100 mA

Explanation

A 30 mA RCD is the recognised rating for additional protection against electric shock in final circuits, because it limits the let-through energy and duration of a shock to a level generally considered non-lethal for healthy adults. Higher ratings are used for fire protection or equipment isolation, not personal protection.

Question 44Electrical Installation and Distribution Systems

What does the term 'diversity factor' represent when sizing the main incoming supply for a building?

  1. The ratio of the maximum demand of the installation to the sum of the connected loads Correct answer.
  2. The ratio of the total connected load to the maximum demand
  3. The ratio of earth fault current to rated current
  4. The ratio of reactive power to active power

Explanation

Diversity factor is the ratio of the maximum demand actually drawn to the sum of all connected loads, and it is always less than or equal to one. Applying it prevents the incoming supply and main protective device from being grossly oversized because not all loads operate simultaneously.

Question 45Electrical Installation and Distribution Systems

A 400 V three-phase motor draws a line current of 50 A at a power factor of 0.8. What is the approximate active power drawn?

  1. 16 kW
  2. 34.6 kW
  3. 40 kW
  4. 27.7 kW Correct answer.

Explanation

Active power for a balanced three-phase load is P = √3 × V × I × cos φ. Substituting gives 1.732 × 400 × 50 × 0.8, which is approximately 27.7 kW. The √3 factor accounts for the line-to-line voltage relationship in a star-connected equivalent.

Question 46Electrical Installation and Distribution Systems

A 4 mm² copper cable carries 30 A and has a resistance of 4.61 mΩ/m. What is the voltage drop over a 40 m single run, neglecting reactance?

  1. 5.5 V Correct answer.
  2. 22.2 V
  3. 2.8 V
  4. 11.1 V

Explanation

Voltage drop equals current times total resistance: 30 A × (4.61 mΩ/m × 40 m) = 30 × 0.1844 Ω, giving approximately 5.5 V. For a single-phase circuit the drop would be doubled because the return neutral path must also be counted.

Question 47Electrical Installation and Distribution Systems

Which disconnection time does IEC 60364 require for a 230 V final circuit protected by a 32 A overcurrent device in a TN system under a nominal 0.4 s condition?

  1. 0.4 s Correct answer.
  2. 5 s
  3. 0.1 s
  4. 0.2 s

Explanation

For final circuits not exceeding 32 A in a TN system at 230 V, the maximum disconnection time is 0.4 s, and the earth fault loop impedance must be low enough for the device to operate within that time. Distribution circuits and higher voltages carry different limits specified in the same table.

Question 48Electrical Installation and Distribution Systems

A distribution board serves a 100 A three-phase supply. Which busbar arrangement is most appropriate to ensure balanced loading across phases?

  1. Distribute single-phase final circuits evenly among the three phases and record the schedule Correct answer.
  2. Connect all single-phase loads to one phase and reserve the others for future use
  3. Connect all single-phase loads to the phase with the largest spare capacity
  4. Connect single-phase loads between two phases only to avoid the neutral

Explanation

Spreading single-phase circuits evenly across the three phases keeps the neutral current low and limits voltage unbalance, and the phase allocation should be recorded on the board schedule for future maintenance. Concentrating loads on one phase causes excessive neutral current and voltage rise on that phase.

Question 49Electrical Installation and Distribution Systems

Which of the following is the correct method for verifying the continuity of a protective conductor on a final circuit after installation?

  1. Measure the earth fault loop impedance with the supply energised
  2. Measure resistance from the earth bar to the exposed conductive part with a low-resistance ohmmeter Correct answer.
  3. Perform a polarity check with a voltage tester
  4. Measure insulation resistance between line and neutral

Explanation

Continuity of the protective conductor is verified by measuring resistance end to end with a low-resistance ohmmeter while the circuit is isolated, confirming a low value with no open joints. Earth fault loop impedance testing is a separate live test that confirms the overall fault path, not the continuity of a single conductor.

Question 50Electrical Installation and Distribution Systems

In a TT system, why is an RCD usually required for earth fault protection?

  1. Because the neutral is combined with the protective conductor throughout
  2. Because the protective conductor carries the full load current
  3. Because the supply transformer neutral is unearthed
  4. Because the earth fault loop impedance is too high for an overcurrent device to disconnect within the required time Correct answer.

Explanation

In a TT system the fault path runs through the installation earth electrode and the supply earth electrode, giving a loop impedance typically too high for a fuse or MCB to clear the fault quickly. An RCD responds to the small residual current that flows and disconnects within the required time.