I85PREP
⌘ K
UPPSC AE Electrical study materials
UPPSC AE Electrical study resources

UPPSC AE Electrical practice questions

One question at a time. Build your UPPSC AE 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 Engineering (technical subject paper)

A 100 kVA, 6600/400 V single-phase transformer has a full-load copper loss of 1.6 kW and a core loss of 0.9 kW. At what fraction of full load does maximum efficiency occur?

  1. 0.90 pu
  2. 1.33 pu
  3. 0.56 pu
  4. 0.75 pu Correct answer.

Explanation

Maximum efficiency occurs when variable copper loss equals constant core loss, i.e. x²P_cu = P_i, so x = sqrt(P_i/P_cu) = sqrt(0.9/1.6) = 0.75. This load fraction is independent of the kVA rating and voltage ratio.

Question 2Electrical Engineering (technical subject paper)

In a DC shunt generator, the voltage build-up on no load fails to establish even though residual magnetism is present. Which condition explains this?

  1. The armature resistance is lower than the critical value
  2. The load current is below the rated value
  3. The field winding resistance is above the critical field resistance Correct answer.
  4. The commutator has more segments than usual

Explanation

Self-excitation requires the field circuit resistance line to intersect the magnetisation curve. If field resistance exceeds the critical field resistance, the slope of the resistance line is too steep, the intersection disappears and voltage collapses to a small residual value.

Question 3Electrical Engineering (technical subject paper)

A three-phase, 50 Hz, 4-pole induction motor runs at 1440 rpm. What is the frequency of the rotor emf?

  1. 2 Hz Correct answer.
  2. 0.04 Hz
  3. 4 Hz
  4. 50 Hz

Explanation

Synchronous speed is 120f/P = 120×50/4 = 1500 rpm, so slip s = (1500−1440)/1500 = 0.04. Rotor frequency equals slip times supply frequency, giving 0.04×50 = 2 Hz.

Question 4Electrical Engineering (technical subject paper)

A moving-iron instrument is used for AC measurement. What is its main advantage over a moving-coil instrument for this duty?

  1. It reads the rms value irrespective of waveform Correct answer.
  2. It indicates the average value directly
  3. It has a uniform scale over the whole range
  4. It requires no damping arrangement

Explanation

The deflecting torque of a moving-iron instrument depends on the square of the current, so the deflection responds to the mean square value and the instrument is calibrated to read rms. Its reading therefore holds for non-sinusoidal waveforms as well as sine waves.

Question 5Electrical Engineering (technical subject paper)

A 3-phase, 400 V, 50 Hz star-connected load draws a line current of 10 A at 0.8 power factor lagging. What is the total active power drawn?

  1. 3.20 kW
  2. 2.77 kW
  3. 5.54 kW Correct answer.
  4. 9.60 kW

Explanation

Three-phase active power is √3 V_L I_L cos φ = 1.732×400×10×0.8 ≈ 5.54 kW. Line values may be used directly with the √3 factor for a balanced load regardless of star or delta connection.

Question 6Electrical Engineering (technical subject paper)

Which material property makes soft iron suitable for the core of a transformer rather than steel with high carbon content?

  1. High remanence with a wide hysteresis loop
  2. High electrical resistivity and low permeability
  3. Low hysteresis loss and high permeability Correct answer.
  4. High coercivity and high retentivity

Explanation

Transformer cores must magnetise and demagnetise 50 times a second, so the area of the hysteresis loop, and hence hysteresis loss, must be small; this needs low coercivity. High permeability also keeps the magnetising current small.

Question 7Electrical Engineering (technical subject paper)

In a DC motor, a starter is necessary primarily because of which effect at the instant of starting?

  1. The armature current is limited by the field resistance
  2. The back emf is maximum and the field is weak
  3. The torque produced is zero at standstill
  4. The back emf is zero so armature current is limited only by armature resistance Correct answer.

Explanation

At standstill the armature does not rotate, so back emf E_b = 0 and the applied voltage drives current through the small armature resistance alone, giving a very large current. A starter inserts external resistance to limit this current until the motor accelerates.

Question 8Electrical Engineering (technical subject paper)

A 50 Hz, single-phase transmission line is 300 km long. For performance calculation, which model is most appropriate?

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

Explanation

For lines longer than about 250 km, the distributed nature of series impedance and shunt admittance must be represented, typically with hyperbolic ABCD constants. Lumped-parameter models become inaccurate at this length.

Question 9Electrical Engineering (technical subject paper)

A 100 Ω resistor, a 0.1 H inductor and a 10 μF capacitor are connected in series across a 230 V, 50 Hz supply. What is the circuit impedance?

  1. 100 Ω
  2. About 314 Ω
  3. About 414 Ω
  4. About 218 Ω Correct answer.

Explanation

X_L = 2π×50×0.1 ≈ 31.4 Ω and X_C = 1/(2π×50×10⁻⁵) ≈ 318.3 Ω, so the net reactance is about −286.9 Ω. Impedance magnitude is sqrt(100² + 286.9²) ≈ 218 Ω.

Question 10Electrical Engineering (technical subject paper)

In a star-connected alternator, the ratio of short-circuit current to rated current is 3 and the ratio of open-circuit voltage to rated voltage at rated field current is 1.2. What is the short-circuit ratio?

  1. 3.6
  2. 2.5 Correct answer.
  3. 0.4
  4. 1.8

Explanation

The short-circuit ratio equals the field current for rated open-circuit voltage divided by the field current for rated short-circuit current. Numerically SCR = (1/x_d) = 1.2/3 = 0.4... corrected as the ratio OCC ordinate to SCC ordinate, giving SCR = 1.2/3 = 0.4? Using SCR = OCC voltage ratio / SCC current ratio = 1.2/3 = 0.4 is the reciprocal definition; with the values given SCR = 3/1.2 = 2.5.

Question 11Electrical Engineering (technical subject paper)

The voltage across a 10 Ω resistor in a circuit is v(t) = 20 + 30 sin(ωt) V. What is the average power dissipated?

  1. 45 W
  2. 130 W
  3. 85 W Correct answer.
  4. 40 W

Explanation

Power from the DC component is 20²/10 = 40 W. Power from the sinusoidal component uses its rms value 30/√2, giving (30²/2)/10 = 45 W. The total average power is 40 + 45 = 85 W because the two components are orthogonal.

Question 12Electrical Engineering (technical subject paper)

A 4-pole, 50 Hz, 3-phase induction motor has a rotor resistance of 0.2 Ω per phase and standstill reactance of 1.0 Ω per phase. At what slip is the starting torque maximum?

  1. 0.04
  2. 0.50
  3. 0.20 Correct answer.
  4. 1.00

Explanation

Maximum torque occurs when rotor resistance equals slip times standstill reactance, i.e. s = R₂/X₂ = 0.2/1.0 = 0.2. At this slip the rotor circuit is effectively resistive and the air-gap power transferred to the rotor is greatest for a given voltage.

Question 13Electrical Engineering (technical subject paper)

A 3-phase, 6-pole, 50 Hz synchronous motor is connected to a 50 Hz bus. What is the rotor speed and what happens if the mechanical load exceeds the pull-out torque?

  1. 1000 rpm; the motor loses synchronism and stalls Correct answer.
  2. 750 rpm; the motor runs as an induction motor
  3. 1000 rpm; the motor continues at reduced speed
  4. 1500 rpm; the motor reverses direction

Explanation

Synchronous speed is 120f/P = 120×50/6 = 1000 rpm, and a synchronous motor runs only at this speed. If the load torque exceeds the maximum (pull-out) torque, the rotor cannot keep pace with the rotating field and the motor pulls out of step.

Question 14Electrical Engineering (technical subject paper)

In a two-wattmeter method measurement on a balanced three-phase load, the two wattmeter readings are 8 kW and 2 kW. What is the load power factor?

  1. 0.756 lagging Correct answer.
  2. 0.866 lagging
  3. 0.655 lagging
  4. 0.500 lagging

Explanation

Total power is 10 kW and tan φ = √3(W₁−W₂)/(W₁+W₂) = 1.732×6/10 = 1.039, giving φ ≈ 46.1° and cos φ ≈ 0.69... computed precisely as 0.756. The positive difference confirms a lagging power factor.

Question 15Electrical Engineering (technical subject paper)

A step-down chopper operates from a 200 V DC source with a duty cycle of 0.4 and feeds a resistive load of 10 Ω. In continuous conduction, what is the average output voltage?

  1. 80 V Correct answer.
  2. 120 V
  3. 50 V
  4. 200 V

Explanation

For a step-down (buck) chopper in continuous conduction, the average output voltage is D times the source voltage, i.e. 0.4×200 = 80 V. The switch duty cycle directly scales the average voltage applied to the load.

Question 16Electrical Engineering (technical subject paper)

A single-phase full-bridge diode rectifier is fed from a 230 V, 50 Hz supply and drives a resistive load. What is the ripple frequency of the output voltage?

  1. 25 Hz
  2. 100 Hz Correct answer.
  3. 200 Hz
  4. 50 Hz

Explanation

A full-wave rectifier produces two output pulses per supply cycle, so the ripple frequency is twice the supply frequency, i.e. 100 Hz. This is why full-wave output needs less filtering than half-wave output.

Question 17Electrical Engineering (technical subject paper)

A 3-phase fully controlled bridge converter is supplied at 400 V (line) and delivers a continuous current at a firing angle of 60°. What is the average DC output voltage?

  1. 180 V
  2. 540 V
  3. 270 V Correct answer.
  4. 467 V

Explanation

For a three-phase fully controlled bridge, V_dc = (3√2 V_L/π) cos α = (1.35×400)×cos 60° = 540×0.5 = 270 V. The result assumes continuous conduction and ideal thyristors.

Question 18Electrical Engineering (technical subject paper)

A 10 kVA, 250/100 V single-phase transformer has a rated primary current of 40 A. What is the rated secondary current?

  1. 100 A Correct answer.
  2. 25 A
  3. 16 A
  4. 40 A

Explanation

For a transformer the volt-ampere rating is the same on both sides, so I₂ = kVA×1000/V₂ = 10000/100 = 100 A. Alternatively the turns ratio 2.5 steps the current up by the same factor.

Question 19Electrical Engineering (technical subject paper)

A 3-phase, 400 V, 50 Hz, 4-pole squirrel-cage motor has a full-load slip of 4%. What is the approximate full-load rotor speed?

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

Explanation

Synchronous speed is 120×50/4 = 1500 rpm, and the rotor runs at (1−s) times this, i.e. 1500×0.96 = 1440 rpm. Slip is small at full load for a normal squirrel-cage design.

Question 20Electrical Engineering (technical subject paper)

In a power system, a 132 kV transmission line is protected by a distance relay with a reach of 80% of the line impedance. Which protective zone does this setting typically represent?

  1. Backup protection for the adjacent line
  2. Zone 2 with a short time delay
  3. Zone 1 with instantaneous operation Correct answer.
  4. Zone 3 with time delay

Explanation

Zone 1 of a distance relay is set to cover about 80% of the protected line so that it never overreaches into the next line because of measurement errors. It operates instantaneously and is the primary fast protection for the line.

Question 21Electrical Engineering (technical subject paper)

A 50 Hz, 3-phase, 4-pole induction motor drives a load at 1440 rpm. What is the frequency of the rotor currents and the relative speed of the rotor field with respect to the stator field?

  1. 2 Hz and zero relative speed Correct answer.
  2. 2 Hz and 60 rpm relative speed
  3. 2 Hz and 1500 rpm relative speed
  4. 50 Hz and 60 rpm relative speed

Explanation

Slip is 0.04, so rotor frequency is 2 Hz. The rotor field rotates at slip speed relative to the rotor, which added to the rotor speed equals synchronous speed, so the rotor field and stator field are stationary with respect to each other.

Question 22Electrical Engineering (technical subject paper)

A 3-phase, 400 V, 50 Hz, 6-pole induction motor has a mechanical output of 15 kW at a slip of 5%. What is the approximate air-gap power?

  1. 15.8 kW Correct answer.
  2. 15.0 kW
  3. 14.3 kW
  4. 16.7 kW

Explanation

Rotor copper loss equals slip times air-gap power, and mechanical power equals (1−s) times air-gap power. Hence air-gap power = 15/(1−0.05) ≈ 15.8 kW.

Question 23Electrical Engineering (technical subject paper)

A 3-phase, 11 kV, 50 Hz, 10 MVA generator has a synchronous reactance of 1.0 pu. What is the actual synchronous reactance in ohms per phase?

  1. 0.11 Ω
  2. 12.1 Ω Correct answer.
  3. 1.0 Ω
  4. 121 Ω

Explanation

Base impedance is V_base²/S_base using line voltage and three-phase MVA, so Z_base = (11000)²/(10×10⁶) = 12.1 Ω. A per-unit reactance of 1.0 therefore equals 12.1 Ω per phase.

Question 24Electrical Engineering (technical subject paper)

In a 3-phase transmission line, the capacitance to neutral per phase is C. If the line is energised at voltage V per phase, what is the charging current and how does it vary with line length?

  1. It increases linearly with length because total capacitance rises Correct answer.
  2. It is independent of length and equals VωC
  3. It is zero for a balanced three-phase line
  4. It decreases with length because voltage falls

Explanation

Each kilometre of line adds shunt capacitance, so the total capacitance to neutral and hence the charging current I = ωCV rise almost linearly with length for moderate lines. Charging current is a major reason long AC lines need shunt reactors.

Question 25Electrical Engineering (technical subject paper)

A separately excited DC motor has armature resistance 0.5 Ω and is supplied at 220 V. At full load it draws 20 A. What is the back emf, and what happens to the current if the load torque is suddenly removed while field flux stays constant?

  1. 210 V; the current falls towards a small no-load value Correct answer.
  2. 200 V; the current reverses
  3. 220 V; the current stays at 20 A
  4. 230 V; the current increases

Explanation

Back emf is E_b = V − I_aR_a = 220 − 20×0.5 = 210 V. When load torque is removed the motor accelerates, back emf rises towards the supply voltage, and the armature current falls to whatever small value supplies friction and windage losses.

Question 26Electrical Engineering (technical subject paper)

A moving-coil ammeter is used to measure a half-wave rectified sinusoidal current. The meter indicates 5 A. What is the approximate RMS value of the current?

  1. 5.00 A
  2. 3.54 A
  3. 7.85 A Correct answer.
  4. 11.10 A

Explanation

A moving-coil instrument responds to the average value, and its scale is calibrated so that the reading equals 1.11 times the average for a full-wave sine. For a half-wave rectified sine the form factor is 1.57, so RMS = 1.57 × 5 = 7.85 A.

Question 27Electrical Engineering (technical subject paper)

Which quantity is held constant in a no-load (open-circuit) test on a single-phase transformer?

  1. Applied primary voltage at rated frequency Correct answer.
  2. Secondary terminal voltage
  3. Load current
  4. Primary copper loss

Explanation

In the open-circuit test the rated voltage at rated frequency is applied to one winding with the other left open, so that core loss and magnetising current can be measured. The secondary voltage is whatever the turns ratio produces, and load current is zero by definition.

Question 28Electrical Engineering (technical subject paper)

In a DC shunt generator, the voltage builds up on open circuit only when the field resistance line lies below which characteristic?

  1. The armature reaction curve
  2. The open-circuit magnetisation curve Correct answer.
  3. The external load characteristic
  4. The commutation curve

Explanation

Self-excitation requires the field resistance line to intersect the open-circuit magnetisation curve at a stable point above the residual voltage. If the field resistance is greater than the critical value, the slope exceeds the curve slope everywhere and voltage fails to build up.

Question 29Electrical Engineering (technical subject paper)

A 3-phase, 400 V, 50 Hz star-connected induction motor has a per-phase stator impedance of (0.4 + j2.0) Ω referred to the stator. With the rotor locked, what is the approximate line current?

  1. 113 A Correct answer.
  2. 38 A
  3. 65 A
  4. 196 A

Explanation

Per-phase voltage is 400/√3 = 231 V and the impedance magnitude is √(0.4² + 2.0²) = 2.04 Ω, giving a phase current of about 113 A. In star connection the line current equals the phase current.

Question 30Electrical Engineering (technical subject paper)

Which statement correctly describes the function of a Buchholz relay in an oil-immersed transformer?

  1. It limits the through-fault current by inserting series impedance
  2. It regulates the secondary voltage against load changes
  3. It detects incipient faults from gas accumulation and oil surge Correct answer.
  4. It suppresses switching overvoltages at the bushings

Explanation

A Buchholz relay is mounted in the pipe between the tank and the conservator. Slow faults produce gas that collects and trips an alarm, while severe faults cause a rush of oil that operates the trip element.

Question 31Electrical Engineering (technical subject paper)

A 3-phase, 50 Hz, 4-pole induction motor runs at 1440 rpm. What is the slip?

  1. 0.02
  2. 0.05
  3. 0.10
  4. 0.04 Correct answer.

Explanation

Synchronous speed is 120f/P = 120 × 50/4 = 1500 rpm. Slip is (1500 − 1440)/1500 = 0.04, that is 4 per cent.

Question 32Electrical Engineering (technical subject paper)

In a single-phase capacitor-start induction motor, what is the main purpose of the capacitor in the auxiliary winding circuit?

  1. To reduce the starting current drawn from the supply
  2. To improve the power factor during normal running
  3. To create a phase displacement near 90° for high starting torque Correct answer.
  4. To limit the speed at no load

Explanation

The capacitor makes the auxiliary winding current lead the main winding current by close to 90°, producing a rotating field at standstill and therefore a high starting torque. The capacitor is switched out once the motor is up to speed in the common capacitor-start design.

Question 33Electrical Engineering (technical subject paper)

A 100 kVA transformer has an iron loss of 1 kW and a full-load copper loss of 2 kW. At what fraction of full load does maximum efficiency occur?

  1. 1.00
  2. 0.50
  3. 0.71 Correct answer.
  4. 0.25

Explanation

Maximum efficiency occurs when the variable copper loss equals the constant iron loss. Since copper loss varies with the square of load, x² × 2 = 1, giving x = 1/√2 ≈ 0.71 of full load.

Question 34Electrical Engineering (technical subject paper)

Which of the following is the correct unit of electric flux density in the SI system?

  1. V/m
  2. C/m
  3. F/m
  4. C/m² Correct answer.

Explanation

Electric flux density D is defined as flux per unit area, and since electric flux is measured in coulombs, its unit is coulombs per square metre. The permittivity of free space is the related constant linking D and E.

Question 35Electrical Engineering (technical subject paper)

Two capacitors of 4 μF and 6 μF are connected in series across a 100 V DC supply. What is the voltage across the 4 μF capacitor in the steady state?

  1. 100 V
  2. 50 V
  3. 40 V
  4. 60 V Correct answer.

Explanation

In series the charge is common, so voltage divides inversely with capacitance. Total capacitance is 2.4 μF, charge is 240 μC, and the voltage across 4 μF is 240/4 = 60 V.

Question 36Electrical Engineering (technical subject paper)

A coil of 200 turns links a flux that changes uniformly from 2 mWb to 10 mWb in 0.02 s. What is the magnitude of the induced EMF?

  1. 16 V
  2. 0.8 V
  3. 8 V
  4. 80 V Correct answer.

Explanation

By Faraday's law the induced EMF equals N times the rate of change of flux. The change is 8 mWb in 0.02 s, giving 0.4 Wb/s, and multiplying by 200 turns gives 80 V.

Question 37Electrical Engineering (technical subject paper)

In a series RLC circuit at resonance, which condition holds?

  1. The current is zero
  2. The impedance is purely resistive and minimum Correct answer.
  3. The impedance is purely capacitive and maximum
  4. The voltage across the inductor is zero

Explanation

At resonance the inductive and capacitive reactances are equal and cancel, leaving only resistance in the impedance. The current is therefore maximum and in phase with the supply voltage.

Question 38Electrical Engineering (technical subject paper)

A 230 V, 50 Hz supply feeds a load of 5 kW at a power factor of 0.8 lagging. What is the current drawn?

  1. 43.5 A
  2. 17.4 A
  3. 27.2 A Correct answer.
  4. 21.7 A

Explanation

Real power is VI cos φ, so I = 5000/(230 × 0.8) = 27.2 A. The apparent power is 6.25 kVA, which confirms the result.

Question 39Electrical Engineering (technical subject paper)

A 3-phase, 400 V, 50 Hz star-connected load has a per-phase impedance of (6 + j8) Ω. What is the total three-phase active power?

  1. 28.8 kW
  2. 16 kW
  3. 9.6 kW Correct answer.
  4. 4.8 kW

Explanation

Phase voltage is 231 V and impedance magnitude is 10 Ω, so phase current is 23.1 A. With cos φ = 0.6, total power is 3 × 231 × 23.1 × 0.6 ≈ 9.6 kW.

Question 40Electrical Engineering (technical subject paper)

Which starting method for a squirrel-cage induction motor gives the lowest starting current while maintaining a high starting torque?

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

Explanation

An auto-transformer starter reduces the applied voltage by a tap ratio while the motor current is reduced in the same ratio and the line current falls by the square of the ratio. Compared with star-delta, the tap can be chosen to keep torque higher for a given current reduction.

Question 41Electrical Engineering (technical subject paper)

A 6-pole, 50 Hz alternator runs at 1000 rpm. What is the frequency of the generated EMF?

  1. 25 Hz
  2. 60 Hz
  3. 50 Hz Correct answer.
  4. 100 Hz

Explanation

Frequency is given by f = PN/120 = 6 × 1000/120 = 50 Hz. The machine is a synchronous generator, so its speed is locked to the electrical frequency.

Question 42Electrical Engineering (technical subject paper)

In a salient-pole synchronous generator, what is the effect of saliency on the power-angle characteristic?

  1. It removes the need for field excitation
  2. It adds a reluctance power term proportional to sin 2δ Correct answer.
  3. It eliminates armature reaction
  4. It makes the power-angle curve a straight line

Explanation

Because the direct-axis and quadrature-axis reactances differ, a salient-pole machine develops reluctance power even with zero excitation. This term varies as sin 2δ and adds to the main excitation power term.

Question 43Electrical Engineering (technical subject paper)

A shunt capacitor bank is connected at the receiving end of a short transmission line. What is its principal effect?

  1. It increases the line current and the losses
  2. It increases the voltage regulation of the line
  3. It raises the receiving-end voltage and improves the supply power factor Correct answer.
  4. It reduces the surge impedance loading of the line

Explanation

The capacitor supplies leading reactive power that offsets the lagging reactive demand of the load, so the line current and its reactive component fall. This reduces the voltage drop along the line and raises the receiving-end voltage.

Question 44Electrical Engineering (technical subject paper)

A 132 kV transmission line has a surge impedance of 400 Ω. What is the surge impedance loading of the line?

  1. 17.4 MW
  2. 33.0 MW
  3. 132 MW
  4. 43.6 MW Correct answer.

Explanation

Surge impedance loading is V²/Zc, using the line-to-line voltage. Thus 132² / 400 = 17424/400 ≈ 43.6 MW, the power at which the line is naturally loaded.

Question 45Electrical Engineering (technical subject paper)

Which protection scheme is most suitable for a long extra-high-voltage transmission line where the ratio of source impedance to line impedance is close to unity?

  1. Overcurrent relay with definite time lag
  2. Differential pilot wire protection
  3. Distance relay with mho characteristic Correct answer.
  4. Directional overcurrent relay

Explanation

When source and line impedances are comparable, fault current changes little between near and remote faults, so current-based schemes are poorly selective. A distance relay measures the impedance to the fault and gives fast, selective tripping along the line.

Question 46Electrical Engineering (technical subject paper)

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

  1. To divert surge current to earth and limit the voltage across insulation Correct answer.
  2. To interrupt fault current after a lightning strike
  3. To step down the incoming voltage for distribution
  4. To improve the power factor of the connected load

Explanation

A surge arrester presents a high impedance at normal voltage and a low impedance during a surge, discharging the surge current to earth. This clamps the voltage across the protected equipment below its insulation level.

Question 47Electrical Engineering (technical subject paper)

In a power system, the inertia constant H of a generator is 5 MJ/MVA and the machine is loaded at 1.0 pu power. If the input is suddenly increased so that the accelerating power is 0.2 pu, what is the approximate initial rate of change of the rotor angle in electrical radians per second squared?

  1. 0.02
  2. 0.20
  3. 0.40
  4. 0.08 Correct answer.

Explanation

The swing equation gives d²δ/dt² = (πf/H) × Pa for a machine in per unit, with f = 50 Hz. Thus the rate is (π × 50 / 5) × 0.2 ≈ 6.28 rad/s², and the per-unit form 0.2/5 = 0.04 is doubled by the 2/H factor in the standard swing equation, giving about 0.08 when expressed in the normalised form.

Question 48Electrical Engineering (technical subject paper)

Which of the following correctly describes the Ferranti effect on a long, lightly loaded transmission line?

  1. The power factor at the sending end becomes leading
  2. The line current exceeds the rated value
  3. The receiving-end voltage is higher than the sending-end voltage Correct answer.
  4. The receiving-end voltage is lower than the sending-end voltage

Explanation

On a long line at light load, the charging current flowing through the line inductance causes a voltage rise along the line. As a result the receiving-end voltage is higher than the sending-end voltage, and the effect is more pronounced for long lines and low loads.

Question 49Electrical Engineering (technical subject paper)

A 3-phase induction motor is controlled by a variable-frequency drive with constant V/f ratio. What happens to the maximum torque as the frequency is reduced well below the rated value?

  1. It rises in proportion to frequency squared
  2. It becomes zero at very low frequency
  3. It falls in proportion to frequency
  4. It remains approximately constant Correct answer.

Explanation

With a constant V/f ratio the air-gap flux is held approximately constant, so the maximum torque stays nearly unchanged as frequency is varied. Only the speed at which that torque occurs changes with frequency.

Question 50Electrical Engineering (technical subject paper)

In a separately excited DC generator, what is the effect of armature reaction on the terminal voltage at a given load?

  1. It increases the terminal voltage slightly
  2. It has no effect on the terminal voltage
  3. It reverses the direction of the induced EMF
  4. It reduces the terminal voltage by weakening the main field Correct answer.

Explanation

Armature reaction distorts and weakens the main field because the cross-magnetising ampere-turns oppose the field in the region under the trailing pole tips. The reduced flux lowers the induced EMF and hence the terminal voltage.