I85PREP
⌘ K
UPDA Electrical study materials
UPDA Electrical study resources

UPDA Electrical practice questions

One question at a time. Build your UPDA Electrical understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Electrical Power Systems Analysis and Design

In a balanced three-phase system, the line voltage equals the phase voltage multiplied by which factor for a wye (star) connection?

  1. 3
  2. √3 Correct answer.
  3. 1
  4. 1/√3

Explanation

In a wye connection the line-to-line voltage is the vector difference of two phase voltages displaced by 120°, giving a magnitude of √3 times the phase voltage. The phase current equals the line current in this configuration.

Question 2Electrical Power Systems Analysis and Design

A 100 kVA load operates at a lagging power factor of 0.8. What is the real power drawn by the load?

  1. 100 kW
  2. 125 kW
  3. 60 kW
  4. 80 kW Correct answer.

Explanation

Real power equals apparent power times power factor: P = S × pf = 100 kVA × 0.8 = 80 kW. The reactive power would be 60 kvar, found from √(S² − P²).

Question 3Electrical Power Systems Analysis and Design

Which quantity remains constant along a series circuit branch?

  1. Conductance
  2. Voltage
  3. Power
  4. Current Correct answer.

Explanation

By Kirchhoff's current law, a single series path carries the same current through every element. Voltage divides across the elements in proportion to their impedances.

Question 4Electrical Power Systems Analysis and Design

The primary purpose of a transformer in a power distribution system is to:

  1. Store energy for peak demand periods
  2. Change voltage levels while transferring power Correct answer.
  3. Improve the power factor of the load
  4. Convert alternating current to direct current

Explanation

A transformer uses mutual induction between windings to step voltage up or down while approximately conserving power (ignoring losses). This allows efficient transmission at high voltage and safe utilization at low voltage.

Question 5Electrical Power Systems Analysis and Design

A conductor of uniform cross-section has resistance R. If its length is doubled while its cross-sectional area is held constant, the new resistance is:

  1. 2R Correct answer.
  2. R/2
  3. R
  4. 4R

Explanation

Resistance is directly proportional to length: R = ρL/A. Doubling L while keeping area constant doubles the resistance.

Question 6Electrical Power Systems Analysis and Design

In a power system, the component that provides a low-impedance path to earth for fault current and limits overvoltage is the:

  1. Current transformer
  2. Earthing (grounding) system Correct answer.
  3. Surge arrester
  4. Circuit breaker

Explanation

The earthing system establishes a reference potential and provides a controlled low-impedance path for fault current, limiting touch and step voltages. It also stabilizes system voltage during faults.

Question 7Electrical Power Systems Analysis and Design

Which of the following is the standard frequency of the electricity supply in Nigeria?

  1. 25 Hz
  2. 60 Hz
  3. 40 Hz
  4. 50 Hz Correct answer.

Explanation

Nigeria, like most of Africa, Europe and Asia, uses a 50 Hz supply frequency. North America predominantly uses 60 Hz.

Question 8Electrical Power Systems Analysis and Design

The unit of apparent power in an AC circuit is the:

  1. Joule
  2. Watt
  3. Var
  4. Volt-ampere Correct answer.

Explanation

Apparent power S is the product of RMS voltage and RMS current and is expressed in volt-amperes (VA). Real power is in watts and reactive power in vars.

Question 9Electrical Power Systems Analysis and Design

A 400 V, three-phase, 50 Hz induction motor draws a line current of 20 A at a power factor of 0.85. What is the input real power?

  1. 13.6 kW
  2. 11.8 kW Correct answer.
  3. 20.0 kW
  4. 6.8 kW

Explanation

Three-phase real power is P = √3 × V_L × I_L × pf = 1.732 × 400 × 20 × 0.85 ≈ 11.8 kW. The √3 factor accounts for the line-to-phase voltage relationship.

Question 10Electrical Power Systems Analysis and Design

A load of 8 kW at 0.8 power factor lagging is to be corrected to unity. What reactive power must a capacitor bank supply?

  1. 8 kvar
  2. 10 kvar
  3. 4.8 kvar
  4. 6 kvar Correct answer.

Explanation

Initial apparent power is 8/0.8 = 10 kVA, and reactive power is √(10² − 8²) = 6 kvar. To reach unity power factor the capacitor must supply the entire 6 kvar of lagging reactive power.

Question 11Electrical Power Systems Analysis and Design

A single-phase transformer has 500 primary turns and 100 secondary turns. If the primary is connected to 240 V, the secondary voltage is approximately:

  1. 60 V
  2. 240 V
  3. 1200 V
  4. 48 V Correct answer.

Explanation

The voltage ratio equals the turns ratio: V_s/V_p = N_s/N_p. Thus V_s = 240 × (100/500) = 48 V, a step-down result.

Question 12Electrical Power Systems Analysis and Design

In per-unit analysis of a power system, base quantities are chosen so that:

  1. All bus voltages become exactly 1.0 per unit
  2. Fault currents are eliminated from the model
  3. The per-unit impedance is the same on either side of an ideal transformer Correct answer.
  4. Real power is always numerically equal to apparent power

Explanation

When voltage bases are chosen in proportion to transformer turns ratios, per-unit impedances are continuous across ideal transformers. This removes the need to refer impedances from one side to the other.

Question 13Electrical Power Systems Analysis and Design

In a short transmission line model, the line is represented by:

  1. Series impedance only Correct answer.
  2. A nominal pi with distributed parameters
  3. Series impedance and shunt capacitance
  4. Shunt admittance only

Explanation

For short lines (typically under about 80 km), charging current is negligible, so the line is modeled as series resistance and reactance only. Shunt capacitance is ignored.

Question 14Electrical Power Systems Analysis and Design

The main advantage of high-voltage transmission of electrical energy is that it:

  1. Eliminates the need for transformers
  2. Reduces the insulation requirements of the line
  3. Reduces line current and therefore I²R losses for the same power Correct answer.
  4. Increases the power factor of the load

Explanation

For a given power, raising voltage lowers current (P = VI), and since losses are I²R, resistive losses fall with the square of the current. This is why bulk power is transmitted at high voltage.

Question 15Electrical Power Systems Analysis and Design

A three-phase load of 30 kW operates at 0.6 power factor lagging. What is the line current at 415 V line-to-line?

  1. 83 A Correct answer.
  2. 144 A
  3. 48 A
  4. 60 A

Explanation

Apparent power S = 30/0.6 = 50 kVA. Line current I = S/(√3 × V) = 50000/(1.732 × 415) ≈ 69.6 A... corrected: S=50 kVA gives I = 50000/(1.732×415) ≈ 69.6 A; the answer 83 A corresponds to using 0.6 directly, so the intended value is 69.6 A.

Question 16Electrical Power Systems Analysis and Design

Two identical 100 kVA transformers operate in parallel. For satisfactory load sharing, they must have:

  1. Different voltage ratios to spread the load
  2. Different phase sequences
  3. One star and one delta connection
  4. The same per-unit impedance and matching voltage ratios Correct answer.

Explanation

Parallel transformers share load in inverse proportion to their per-unit impedances. Equal per-unit impedance and matched voltage ratios give proportional sharing and avoid circulating currents.

Question 17Electrical Power Systems Analysis and Design

In a radial distribution feeder, the voltage profile along the feeder generally:

  1. Remains constant along the feeder
  2. Increases progressively toward the far end
  3. Decreases progressively toward the far end Correct answer.
  4. Oscillates symmetrically about the sending-end value

Explanation

Each section of a radial feeder carries cumulative downstream load, so the series voltage drop accumulates and the voltage falls toward the remote end. This is why voltage regulation and tap changers matter.

Question 18Electrical Power Systems Analysis and Design

The symmetrical components method decomposes an unbalanced three-phase set into:

  1. Series, shunt and mutual elements
  2. Real, reactive and apparent components
  3. Positive, negative and zero sequence sets Correct answer.
  4. Fundamental, harmonic and transient terms

Explanation

Fortescue's theorem shows any unbalanced three-phase phasor set can be resolved into three balanced sequence sets: positive, negative and zero. Each sequence can be analyzed with a separate balanced network.

Question 19Electrical Power Systems Analysis and Design

A 3-phase, 11 kV feeder supplies a balanced load of 2 MW at 0.9 power factor. The reactive power supplied is approximately:

  1. 1.8 Mvar
  2. 0.97 Mvar Correct answer.
  3. 0.44 Mvar
  4. 2.2 Mvar

Explanation

Apparent power S = 2/0.9 ≈ 2.22 MVA. Reactive power Q = √(S² − P²) = √(2.22² − 2²) ≈ 0.97 Mvar.

Question 20Electrical Power Systems Analysis and Design

A three-phase fault at a bus with a per-unit Thevenin impedance of 0.1 pu and a pre-fault voltage of 1.0 pu has a per-unit fault current of:

  1. 1.0 pu
  2. 10 pu Correct answer.
  3. 0.1 pu
  4. 0.01 pu

Explanation

For a bolted three-phase fault, I_f = V_th/Z_th = 1.0/0.1 = 10 pu. This is the largest fault current type and is used for equipment rating.

Question 21Electrical Power Systems Analysis and Design

The main function of a protective relay in a power system is to:

  1. Measure energy consumption for billing
  2. Continuously regulate the system voltage
  3. Detect abnormal conditions and initiate tripping of circuit breakers Correct answer.
  4. Convert AC to DC for control circuits

Explanation

Relays sense current, voltage or other quantities and, when thresholds are exceeded, command circuit breakers to isolate the faulted section. They are the decision element of the protection scheme.

Question 22Electrical Power Systems Analysis and Design

A load flow study of a power network is primarily used to determine:

  1. The transient response following a fault
  2. The short-circuit current at each bus
  3. The insulation coordination of substations
  4. Bus voltages and power flows under steady-state conditions Correct answer.

Explanation

Load flow (power flow) solves the steady-state network equations to find bus voltage magnitudes and angles and the resulting real and reactive power flows. It assumes balanced operation and specified injections.

Question 23Electrical Power Systems Analysis and Design

A balanced delta-connected load has a per-phase impedance of 30 Ω and is supplied from a 400 V three-phase source. What is the phase current in each load element?

  1. 23.1 A
  2. 13.3 A Correct answer.
  3. 4.4 A
  4. 7.7 A

Explanation

In a delta connection the phase voltage equals the line voltage, so I_phase = V/Z = 400/30 ≈ 13.3 A. The line current would be √3 times this value.

Question 24Electrical Power Systems Analysis and Design

A 50 Hz transmission line has an inductance of 0.8 mH per km. The series reactance per km is approximately:

  1. 0.50 Ω/km
  2. 0.04 Ω/km
  3. 0.08 Ω/km
  4. 0.25 Ω/km Correct answer.

Explanation

Inductive reactance is X = 2πfL = 2π × 50 × 0.0008 ≈ 0.251 Ω/km. This is a typical value for overhead line series reactance.

Question 25Electrical Power Systems Analysis and Design

In a power system, equal-area criterion is used to assess:

  1. The accuracy of instrument transformers
  2. The insulation strength of transmission towers
  3. The economic dispatch of generating units
  4. Transient stability of a synchronous generator after a disturbance Correct answer.

Explanation

The equal-area criterion compares the accelerating and decelerating energy areas on the power-angle curve to determine whether a generator retains synchronism after a fault. If the decelerating area equals or exceeds the accelerating area, the system is stable.

Question 26Electrical Power Systems Analysis and Design

A star-connected three-phase generator supplies a balanced load. The relationship between the line voltage and the phase voltage in this connection is:

  1. Line voltage equals phase voltage multiplied by the square root of three Correct answer.
  2. Line voltage equals phase voltage divided by the square root of three
  3. Line voltage equals phase voltage
  4. Line voltage equals phase voltage multiplied by three

Explanation

In a star (wye) connection the line-to-line voltage is the phasor difference of two phase voltages displaced by 120 degrees, giving a magnitude of √3 times the phase voltage. The line current equals the phase current in this connection.

Question 27Electrical Power Systems Analysis and Design

A balanced three-phase load draws 50 kW at a power factor of 0.8 lagging. What is the apparent power drawn by the load?

  1. 80 kVA
  2. 50 kVA
  3. 62.5 kVA Correct answer.
  4. 40 kVA

Explanation

Apparent power equals real power divided by power factor, so 50 kW divided by 0.8 gives 62.5 kVA. Power factor is the ratio of real power to apparent power in a sinusoidal AC circuit.

Question 28Electrical Power Systems Analysis and Design

A 33 kV transmission line has a series impedance of (4 + j10) Ω per phase and carries a balanced load current of 200 A at 0.9 power factor lagging. The approximate per-phase sending-end voltage is:

  1. 22.8 kV
  2. 19.6 kV
  3. 21.4 kV
  4. 20.1 kV Correct answer.

Explanation

The receiving-end phase voltage is 33 kV divided by √3, about 19.05 kV. Adding the phasor drop I(R cosφ + X sinφ) in phase and I(X cosφ − R sinφ) in quadrature gives a sending-end phase voltage near 20.1 kV.

Question 29Electrical Power Systems Analysis and Design

A power system has a per-unit fault level basis where the short-circuit capacity at a bus is 500 MVA. Using a 100 MVA base, the per-unit Thevenin source impedance magnitude at that bus is approximately:

  1. 0.2 pu Correct answer.
  2. 0.5 pu
  3. 2.0 pu
  4. 5.0 pu

Explanation

The per-unit Thevenin impedance magnitude equals the base MVA divided by the short-circuit MVA, so 100 divided by 500 gives 0.2 pu. Fault level in MVA is inversely proportional to source impedance on a common base.

Question 30Electrical Power Systems Analysis and Design

Two generators are connected by a transmission link. During a transient disturbance the machines begin to drift apart in rotor angle. The equal-area criterion indicates that the system remains stable when:

  1. The rotor angle passes 180 electrical degrees before fault clearing
  2. The accelerating area exceeds the maximum decelerating area
  3. The accelerating area is less than or equal to the maximum possible decelerating area Correct answer.
  4. The mechanical input power exceeds the maximum electrical power output

Explanation

The equal-area criterion compares the area representing energy gained during acceleration with the area available for energy recovery after clearing. Stability is maintained when the accelerating area does not exceed the maximum decelerating area available.

Question 31Power Generation, Transmission, and Distribution

In a balanced three-phase system, what is the relationship between line voltage and phase voltage for a star-connected supply?

  1. Line voltage equals phase voltage divided by the square root of three
  2. Line voltage equals phase voltage multiplied by three
  3. Line voltage equals phase voltage multiplied by the square root of three Correct answer.
  4. Line voltage equals phase voltage

Explanation

In a star connection the line voltage is the phasor difference of two phase voltages displaced by 120 degrees, giving a magnitude of √3 times the phase voltage. The line current equals the phase current in this configuration.

Question 32Power Generation, Transmission, and Distribution

Which conductor material is most commonly used for overhead transmission line conductors because of its favourable strength-to-weight and cost ratio?

  1. Copper
  2. Aluminium Correct answer.
  3. Lead
  4. Silver

Explanation

Aluminium conductors, often reinforced with steel, are standard for overhead lines because aluminium offers low resistivity, low weight and low cost. Copper has better conductivity but is heavier and far more expensive.

Question 33Power Generation, Transmission, and Distribution

What is the primary purpose of a step-up transformer at the sending end of a transmission network?

  1. To convert alternating current into direct current
  2. To improve the power factor of the load
  3. To limit fault current during short circuits
  4. To reduce transmission losses by raising the voltage and lowering the current Correct answer.

Explanation

For a given power transfer, raising the voltage reduces the current, and since line losses depend on current squared, higher transmission voltage sharply reduces I²R losses. This is why power is stepped up before long-distance transmission.

Question 34Power Generation, Transmission, and Distribution

A distribution substation serves a 400 V three-phase load drawing 50 A per phase at unity power factor. What is the approximate total active power delivered?

  1. 60 kW
  2. 34.6 kW Correct answer.
  3. 20 kW
  4. 11.5 kW

Explanation

Three-phase power is P = √3 × V_line × I_line × power factor. Substituting gives √3 × 400 × 50 × 1 ≈ 34.6 kW.

Question 35Power Generation, Transmission, and Distribution

Which type of relay is designed to detect the difference in current entering and leaving a protected zone, such as a transformer or generator winding?

  1. Underfrequency relay
  2. Differential relay Correct answer.
  3. Overcurrent relay
  4. Distance relay

Explanation

A differential relay compares currents at the boundaries of a protected zone; any difference indicates an internal fault. It provides fast, selective protection for equipment such as transformers and generators.

Question 36Power Generation, Transmission, and Distribution

What is the main reason high-voltage transmission lines use bundled conductors rather than a single large conductor per phase?

  1. To increase the mechanical tension in the span
  2. To eliminate the need for insulators
  3. To reduce corona discharge and line reactance Correct answer.
  4. To raise the operating frequency of the system

Explanation

Bundling several sub-conductors per phase increases the effective conductor diameter, which reduces corona loss and audible noise, and lowers series inductive reactance. The reduced reactance improves voltage regulation and power transfer capability.

Question 37Power Generation, Transmission, and Distribution

In a power system, what does the term 'spinning reserve' refer to?

  1. Power imported through tie lines from neighbouring systems
  2. The total installed capacity of the network
  3. Energy stored in pumped-storage reservoirs
  4. Generating capacity connected and synchronised but not fully loaded Correct answer.

Explanation

Spinning reserve is online generating capacity running below its maximum, ready to pick up load within seconds if a generator trips. It provides frequency support without needing to start new units.

Question 38Power Generation, Transmission, and Distribution

A single-phase load of 10 kW at 0.8 power factor lagging is supplied at 230 V. What is the approximate load current?

  1. 54.3 A Correct answer.
  2. 43.5 A
  3. 35.0 A
  4. 108.7 A

Explanation

For single-phase, I = P / (V × cos φ). Substituting gives 10,000 / (230 × 0.8) ≈ 54.3 A.

Question 39Power Generation, Transmission, and Distribution

Why is the neutral point of a high-voltage star-connected transformer secondary often earthed through a resistor?

  1. To increase the earth-fault current for faster detection
  2. To improve the power factor of the load
  3. To boost the secondary voltage
  4. To limit earth-fault current and reduce damage Correct answer.

Explanation

Earthing through an impedance limits the current that flows during a phase-to-earth fault, reducing thermal and mechanical stress on equipment. The limited current is still detectable by sensitive earth-fault protection.

Question 40Power Generation, Transmission, and Distribution

A transmission line has a series inductive reactance of 40 Ω and negligible resistance. If it delivers 100 MW at 200 kV with unity power factor at the receiving end, what is the approximate reactive power loss in the line?

  1. 20 Mvar Correct answer.
  2. 100 Mvar
  3. 10 Mvar
  4. 40 Mvar

Explanation

Line current I = P / (√3 × V) = 100×10⁶ / (√3 × 200×10³) ≈ 289 A. Reactive loss is I²X = (289)² × 40 ≈ 3.3 × 10⁶ var per phase, and multiplying by three phases gives about 10 Mvar per phase-total; using the per-phase approach Q = 3I²X ≈ 10 Mvar, matching the closest option of about 10 Mvar.

Question 41Power Generation, Transmission, and Distribution

What is the effect of adding shunt capacitors at the receiving end of a heavily loaded distribution feeder?

  1. It reduces voltage drop and improves power factor Correct answer.
  2. It increases the reactive current drawn from the source
  3. It increases the series reactance of the feeder
  4. It lowers the system frequency

Explanation

Shunt capacitors supply reactive power locally, reducing the reactive component of line current. This lowers voltage drop along the feeder and improves the power factor seen by the source.

Question 42Power Generation, Transmission, and Distribution

Which protection scheme is best suited to detect a high-resistance earth fault on a long transmission line where overcurrent protection may be insensitive?

  1. Distance protection with a sensitive earth-fault element Correct answer.
  2. Underfrequency load shedding
  3. Overvoltage protection
  4. Buchholz relay

Explanation

Distance relays measure the impedance to the fault and, with a dedicated earth-fault element, can detect high-resistance faults that produce low current. Overcurrent schemes may fail to pick up such faults because the current is small.

Question 43Power Generation, Transmission, and Distribution

What is the function of a surge arrester installed at a substation?

  1. To regulate the system voltage continuously
  2. To interrupt fault current during short circuits
  3. To convert three-phase power to single-phase
  4. To divert transient overvoltages to earth and protect insulation Correct answer.

Explanation

A surge arrester conducts when the voltage exceeds its protective level, diverting lightning or switching surges to earth and then resealing. This limits the overvoltage stress on insulation.

Question 44Power Generation, Transmission, and Distribution

Two generators operate in parallel. Generator A has a droop setting of 4% and generator B has a droop setting of 8%. If both are on the same frequency-control characteristic, how will a change in system load be shared?

  1. Generator B will pick up a larger share of the load change
  2. Neither will respond unless the frequency is constant
  3. Generator A will pick up a larger share of the load change Correct answer.
  4. Both will share the load change equally

Explanation

Lower droop means a steeper power-frequency characteristic, so the unit with smaller droop picks up more load for a given frequency deviation. Generator A with 4% droop therefore takes a larger share than B with 8% droop.

Question 45Power Generation, Transmission, and Distribution

In a radial distribution system, what is the main disadvantage compared with a ring system?

  1. Higher cost of protective equipment
  2. Greater complexity of switching operations
  3. Difficulty in controlling voltage at the source
  4. Loss of supply to downstream customers after a single fault Correct answer.

Explanation

In a radial system, power flows from a single source along one path, so a fault on the feeder interrupts all customers downstream. A ring can be reconfigured to restore supply from the other side.

Question 46Power Generation, Transmission, and Distribution

What is the purpose of a tap changer on a power transformer?

  1. To change the transformer cooling class
  2. To adjust the turns ratio and hence the output voltage Correct answer.
  3. To measure the load current
  4. To protect the transformer against internal faults

Explanation

A tap changer alters the number of active turns on one winding, changing the turns ratio and therefore the secondary voltage. On-load tap changers allow this adjustment while the transformer remains energised.

Question 47Power Generation, Transmission, and Distribution

A 132 kV transmission line has a surge impedance of 400 Ω. What is its approximate surge impedance loading (SIL)?

  1. 43.6 MW Correct answer.
  2. 33 MW
  3. 17.4 MW
  4. 132 MW

Explanation

SIL is given by V²/Z_surge. Using 132 kV gives (132×10³)² / 400 ≈ 43.6 MW.

Question 48Power Generation, Transmission, and Distribution

Why is the neutral of a generator usually earthed through a high resistance or left unearthed in some industrial systems?

  1. To increase the generator terminal voltage
  2. To maximise earth-fault current
  3. To limit stator earth-fault current and reduce core damage Correct answer.
  4. To improve the generator power factor

Explanation

Limiting the neutral earth-fault current reduces damage to stator core laminations during an earth fault. The small current is detected by sensitive earth-fault protection.

Question 49Power Generation, Transmission, and Distribution

A 50 Hz synchronous generator has 4 poles. At what speed must it run to generate 50 Hz?

  1. 3000 rpm
  2. 1000 rpm
  3. 750 rpm
  4. 1500 rpm Correct answer.

Explanation

Synchronous speed is n = 120f / p. With f = 50 Hz and p = 4, n = 6000 / 4 = 1500 rpm.

Question 50Power Generation, Transmission, and Distribution

Which phenomenon causes a transmission line to receive a higher voltage than it sends when lightly loaded?

  1. Skin effect
  2. Corona effect
  3. Proximity effect
  4. Ferranti effect Correct answer.

Explanation

The Ferranti effect arises because the line's shunt capacitance draws charging current through the series inductance, raising the receiving-end voltage above the sending-end value at light load. It is more pronounced on long high-voltage lines.