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KPTCL AE/JE practice questions

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Question 1Electrical Engineering (core technical subject)

A 3-phase, 50 Hz, 4-pole induction motor runs at a slip of 4%. What is its rotor speed?

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

Explanation

Synchronous speed is Ns = 120f/P = 120 × 50 / 4 = 1500 rpm. Rotor speed is N = Ns(1 − s) = 1500 × 0.96 = 1440 rpm. Slip always reduces speed below synchronous value in motoring mode.

Question 2Electrical Engineering (core technical subject)

In a transformer, the no-load current is primarily limited by which quantity?

  1. Leakage reactance
  2. Winding resistance
  3. Load impedance
  4. Magnetising reactance Correct answer.

Explanation

At no load the secondary is open, so the current drawn is the exciting current set mainly by the magnetising branch of the equivalent circuit. Magnetising reactance dominates because core flux must be established. Winding resistance and leakage reactance contribute only a small in-phase and drop component.

Question 3Electrical Engineering (core technical subject)

The RMS value of a purely sinusoidal voltage with peak value 325 V is approximately:

  1. 230 V Correct answer.
  2. 163 V
  3. 325 V
  4. 460 V

Explanation

For a sinusoid, Vrms = Vpeak/√2 = 325/1.414 ≈ 230 V. This is why 230 V domestic supply has a peak near 325 V. The average value over a half cycle would be 2Vpeak/π ≈ 207 V, which is a different quantity.

Question 4Electrical Engineering (core technical subject)

Kirchhoff's current law applied at a node is a direct consequence of which physical principle?

  1. Conservation of charge Correct answer.
  2. Faraday's law
  3. Lenz's law
  4. Conservation of energy

Explanation

KCL states that the algebraic sum of currents at a node is zero, meaning charge does not accumulate at the node. This is the principle of conservation of charge. Conservation of energy underlies Kirchhoff's voltage law instead.

Question 5Electrical Engineering (core technical subject)

A moving-coil ammeter is used to measure a sinusoidal alternating current. What does it indicate?

  1. Average value
  2. Peak value
  3. Form factor
  4. RMS value Correct answer.

Explanation

A moving-coil instrument responds to the average torque produced by current, and on AC the inertia gives a steady deflection proportional to the mean square current, so it reads the RMS value. Its scale is calibrated in RMS for sinusoidal quantities. Form factor is a ratio, not a reading.

Question 6Electrical Engineering (core technical subject)

The SI unit of magnetic flux density and the quantity it measures are:

  1. Henry, flux linkage per ampere
  2. Tesla, flux per unit area Correct answer.
  3. Weber, total flux through a surface
  4. Ampere-turn, magnetomotive force

Explanation

Magnetic flux density B is measured in tesla (T), equal to one weber per square metre. It describes flux concentration at a point. Flux itself is in webers and inductance in henries.

Question 7Electrical Engineering (core technical subject)

In a DC shunt generator, the field winding is connected:

  1. In parallel with the armature Correct answer.
  2. Across the armature through a series resistor only
  3. In series with the armature
  4. In series with the load and armature

Explanation

A shunt (parallel) generator has its field winding connected in parallel with the armature so it receives the full terminal voltage. This gives a high-resistance field carrying a small current. A series generator instead puts the field in series with the armature and load.

Question 8Electrical Engineering (core technical subject)

A 100 W lamp operates at 250 V. What current does it draw?

  1. 25 A
  2. 0.4 A Correct answer.
  3. 0.04 A
  4. 2.5 A

Explanation

For a resistive load, P = VI, so I = P/V = 100/250 = 0.4 A. The lamp resistance is then V/I = 625 Ω. Substituting 2.5 A would correspond to 625 W at 250 V.

Question 9Electrical Engineering (core technical subject)

A 3-phase star-connected load draws a line current of 10 A with a phase impedance angle of 30°. If the line voltage is 400 V, the total active power is approximately:

  1. 6.9 kW
  2. 10.4 kW
  3. 4.0 kW
  4. 6.0 kW Correct answer.

Explanation

Total 3-phase power is P = √3 × VL × IL × cos φ = 1.732 × 400 × 10 × 0.866 ≈ 6000 W. In star connection phase current equals line current, and cos 30° = 0.866. Reactive power would be about 3.46 kvar.

Question 10Electrical Engineering (core technical subject)

The voltage regulation of a transformer is defined as:

  1. The ratio of secondary to primary turns
  2. (Full-load secondary voltage − no-load secondary voltage) / no-load secondary voltage
  3. (No-load secondary voltage − full-load secondary voltage) / full-load secondary voltage Correct answer.
  4. The ratio of copper loss to iron loss at full load

Explanation

Regulation expresses the rise in secondary terminal voltage when full load is removed, referred to the full-load value. A small regulation means the terminal voltage stays nearly constant. Turns ratio is a design parameter, not regulation.

Question 11Electrical Engineering (core technical subject)

In a DC series motor, the field winding has very few turns of thick wire. The reason is:

  1. To allow the motor to run safely at no load
  2. To carry the full armature current with low resistance and produce strong flux Correct answer.
  3. To keep the field current small and reduce copper loss
  4. To reduce the induced back EMF at high speed

Explanation

In a series motor the field carries the entire armature current, so it must have low resistance, achieved with few turns of thick conductor. Few turns still give adequate flux because the current is large. This also makes the flux proportional to armature current, giving high starting torque.

Question 12Electrical Engineering (core technical subject)

A coil of 200 turns links a flux that changes uniformly from 2 mWb to 6 mWb in 0.05 s. The average induced EMF is:

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

Explanation

Faraday's law gives e = N × ΔΦ/Δt = 200 × (0.006 − 0.002)/0.05 = 200 × 0.08 = 16 V. The rate of change of flux linkage sets the EMF magnitude. Lenz's law would additionally fix the polarity opposing the increase.

Question 13Electrical Engineering (core technical subject)

In a balanced star-connected system with line voltage 415 V, the phase voltage across each element is approximately:

  1. 207 V
  2. 415 V
  3. 720 V
  4. 240 V Correct answer.

Explanation

In star connection the phase voltage is the line voltage divided by √3: 415/1.732 ≈ 240 V. Line current equals phase current in star. In delta connection the phase voltage would equal the line voltage.

Question 14Electrical Engineering (core technical subject)

The main purpose of a commutator in a DC machine is to:

  1. Convert the alternating EMF in the armature coils into unidirectional terminal voltage Correct answer.
  2. Reduce eddy currents in the armature core
  3. Provide mechanical balancing of the rotor
  4. Limit the armature reaction effect

Explanation

The commutator acts as a mechanical rectifier, reversing coil connections at the instant the coil EMF reverses so the brushes see a unidirectional voltage. In motoring it performs the reverse function, feeding alternating current into the coils. Eddy currents are reduced by laminations instead.

Question 15Electrical Engineering (core technical subject)

A 3-phase induction motor has a synchronous speed of 1000 rpm and runs at 960 rpm. The frequency of the rotor EMF is:

  1. 2 Hz Correct answer.
  2. 48 Hz
  3. 4 Hz
  4. 50 Hz

Explanation

Slip s = (1000 − 960)/1000 = 0.04. Rotor frequency fr = s × f = 0.04 × 50 = 2 Hz. At standstill the rotor frequency equals supply frequency, and it falls as the rotor speeds up.

Question 16Electrical Engineering (core technical subject)

A single-phase transformer has 500 primary turns and 100 secondary turns. With 230 V applied to the primary, the secondary no-load voltage is approximately:

  1. 23 V
  2. 46 V Correct answer.
  3. 1150 V
  4. 230 V

Explanation

The voltage ratio equals the turns ratio: V2 = V1 × N2/N1 = 230 × 100/500 = 46 V. This is a step-down transformer with a ratio of 5:1. No-load voltage is used because regulation reduces it slightly on load.

Question 17Electrical Engineering (core technical subject)

The superposition theorem is applicable to:

  1. Power calculations in any resistive network
  2. Non-linear networks with dependent sources only
  3. Networks containing only ideal current sources
  4. Linear bilateral networks only Correct answer.

Explanation

Superposition relies on linearity, so it applies only to linear bilateral networks with more than one independent source. Each source is considered alone with the others deactivated. It cannot be used for power, since power is proportional to current squared.

Question 18Electrical Engineering (core technical subject)

A series RLC circuit has R = 10 Ω, L = 0.1 H and C = 100 µF. Its resonant frequency is approximately:

  1. 1000 Hz
  2. 50 Hz Correct answer.
  3. 159 Hz
  4. 503 Hz

Explanation

At resonance, fr = 1/(2π√(LC)) = 1/(2π√(0.1 × 100 × 10⁻⁶)) = 1/(2π × 0.00316) ≈ 50 Hz. At this frequency the inductive and capacitive reactances cancel, leaving only R. The circuit current is then maximum and in phase with the supply.

Question 19Electrical Engineering (core technical subject)

In a PN junction diode under forward bias, the depletion region:

  1. Narrows and the barrier potential decreases Correct answer.
  2. Remains unchanged in width
  3. Disappears completely at any forward voltage
  4. Widens and the barrier potential increases

Explanation

Forward bias applies the P side positive, opposing the built-in field, so majority carriers are pushed toward the junction and the depletion width shrinks. The barrier potential is lowered from its equilibrium value, allowing a large diffusion current. The region does not vanish entirely.

Question 20Electrical Engineering (core technical subject)

A 3-phase, 400 V, 50 Hz induction motor has a full-load power factor of 0.85 and input power of 15 kW. The line current is approximately:

  1. 43.3 A
  2. 25.5 A Correct answer.
  3. 15.0 A
  4. 21.7 A

Explanation

For a 3-phase load, P = √3 × VL × IL × cos φ, so IL = 15000/(1.732 × 400 × 0.85) ≈ 25.5 A. The power factor must be included because the apparent power exceeds the active power. Omitting it would overstate the current.

Question 21Electrical Engineering (core technical subject)

A wattmeter in a single-phase circuit reads 800 W. The ammeter reads 5 A and the voltmeter reads 200 V. The power factor of the circuit is:

  1. 0.6
  2. 0.8 Correct answer.
  3. 1.0
  4. 0.4

Explanation

Apparent power is VI = 200 × 5 = 1000 VA, and the wattmeter gives true power 800 W. Power factor is P/(VI) = 800/1000 = 0.8. The phase angle is therefore about 36.9°.

Question 22Electrical Engineering (core technical subject)

The primary reason for using a higher transmission voltage in a power system is that:

  1. It increases the conductor resistance and reduces losses
  2. It increases the power factor of the load
  3. It eliminates the need for transformers
  4. For a given power, the current falls and I²R line losses are reduced Correct answer.

Explanation

For a fixed power transfer, raising voltage lowers current proportionally, and line loss I²R falls with the square of current. This also reduces voltage drop and conductor size. Load power factor is set by the load, not by transmission voltage.

Question 23Electrical Engineering (core technical subject)

A separately excited DC motor has armature resistance 0.5 Ω and is connected to 220 V. At full load it draws 20 A. The back EMF is:

  1. 220 V
  2. 200 V
  3. 210 V Correct answer.
  4. 230 V

Explanation

For a motor, the supply voltage balances back EMF plus the armature drop: Eb = V − IaRa = 220 − 20 × 0.5 = 210 V. Back EMF is always less than terminal voltage in motoring. The armature drop is 10 V here.

Question 24Electrical Engineering (core technical subject)

Two transformers of equal voltage ratio, rated 500 kVA with 4% impedance and 1000 kVA with 5% impedance, are operated in parallel. Their load sharing is determined by:

  1. The ratio of their kVA rating to percentage impedance Correct answer.
  2. Their impedance voltages only
  3. Their efficiency at full load
  4. Their kVA ratings only

Explanation

In parallel operation, load is shared in proportion to rating divided by per-unit impedance, so the 500 kVA unit carries 125 kA-equivalent share per unit while the 1000 kVA unit carries 200. The unit with lower percentage impedance tends to take a larger share of the total load. Ratings alone or impedances alone are insufficient.

Question 25Electrical Engineering (core technical subject)

A transmission line of surge impedance 400 Ω is terminated by a resistive load of 400 Ω. A surge of 100 kV travels along the line. The voltage of the transmitted wave at the load is:

  1. 50 kV
  2. 0 kV
  3. 100 kV Correct answer.
  4. 200 kV

Explanation

The reflection coefficient is (ZL − Z0)/(ZL + Z0) = (400 − 400)/(400 + 400) = 0, so no reflected wave is produced. The transmitted wave equals the incident wave, 100 kV. This is the matched-load condition used to avoid reflections.

Question 26Electrical Engineering (core technical subject)

The Hopkinson demand factor of a power system is defined as the ratio of the maximum demand to the sum of the individual maximum demands of the consumers. Its value is normally:

  1. Exactly equal to one
  2. Independent of load diversity
  3. Less than one Correct answer.
  4. Greater than one

Explanation

Because consumers reach their peaks at different times, the system maximum demand is always smaller than the arithmetic sum of consumer maximum demands. The ratio is therefore the reciprocal of the diversity factor and is always less than unity for a practical system.

Question 27Electrical Engineering (core technical subject)

A generating station has a maximum demand of 25 MW, a connected load of 40 MW and an annual energy output of 120 × 10^6 kWh. The annual load factor is approximately:

  1. 55% Correct answer.
  2. 62%
  3. 48%
  4. 75%

Explanation

Average load equals annual energy divided by 8760 hours, giving about 13.7 MW. Load factor is average load divided by maximum demand, 13.7/25, which is about 0.548 or 55%.

Question 28Electrical Engineering (core technical subject)

In a thermal power plant, the economiser is placed in the flue-gas path between the boiler and the air preheater. Its function is to:

  1. Remove ash particles from the flue gas
  2. Superheat the steam before it enters the turbine
  3. Heat the boiler feed water using the leaving flue gases Correct answer.
  4. Reduce the excess air supplied to the furnace

Explanation

An economiser is a heat exchanger that recovers sensible heat from the flue gases to raise the feed-water temperature before it enters the drum. This reduces fuel consumption for the same steam output and lowers the stack temperature.

Question 29Electrical Engineering (core technical subject)

A hydroelectric plant has a net head of 100 m and a discharge of 20 m^3/s. Taking water density as 1000 kg/m^3 and g as 9.81 m/s^2, the theoretical power available is approximately:

  1. 1.96 MW
  2. 196 MW
  3. 9.81 MW
  4. 19.6 MW Correct answer.

Explanation

Hydraulic power equals density × g × discharge × head, that is 1000 × 9.81 × 20 × 100 W. This gives about 19.6 × 10^6 W, or 19.6 MW before turbine and generator efficiencies are applied.

Question 30Electrical Engineering (core technical subject)

A nuclear reactor uses heavy water as moderator and natural uranium as fuel. The main function of the moderator is to:

  1. Cool the fuel rods and carry heat to the turbine
  2. Absorb neutrons so that the chain reaction stops
  3. Slow down fast neutrons to thermal energy without much absorption Correct answer.
  4. Reflect escaping neutrons back into the core

Explanation

Fission neutrons are born fast and are far more likely to cause further fission in U-235 after being slowed to thermal energies. The moderator reduces neutron energy mainly through elastic scattering, and heavy water is chosen because it scatters well while absorbing very few neutrons.

Question 31Electrical Engineering (core technical subject)

In a 3-core cable, the capacitance between any two conductors is measured as 0.6 µF. The capacitance of each conductor to neutral is:

  1. 1.2 µF Correct answer.
  2. 0.9 µF
  3. 0.3 µF
  4. 0.6 µF

Explanation

The measured capacitance between two conductors is the series combination of two conductor-to-neutral capacitances, so C_measured = C_n/2. Hence C_n = 2 × 0.6 = 1.2 µF per conductor to neutral.

Question 32Electrical Engineering (core technical subject)

The insulation resistance of a cable is 500 MΩ at 20 °C. When the temperature rises, the insulation resistance:

  1. Decreases because conduction in the dielectric increases Correct answer.
  2. Remains unchanged since resistance depends only on length
  3. First rises and then falls to zero at 50 °C
  4. Increases exponentially with temperature

Explanation

In an insulating material the number of thermally generated charge carriers grows with temperature, so leakage current rises and insulation resistance falls. The fall is roughly exponential, which is why insulation resistance is always referred to a stated temperature.

Question 33Electrical Engineering (core technical subject)

A 3-phase, 50 Hz, 4-pole induction motor runs at a slip of 4%. The rotor speed and the rotor frequency are respectively:

  1. 1440 rpm and 2 Hz Correct answer.
  2. 1500 rpm and 2 Hz
  3. 1500 rpm and 50 Hz
  4. 1440 rpm and 50 Hz

Explanation

Synchronous speed is 120f/P = 120 × 50/4 = 1500 rpm, and rotor speed is (1 − s) times this, giving 1440 rpm. Rotor frequency equals slip times supply frequency, 0.04 × 50 = 2 Hz.

Question 34Electrical Engineering (core technical subject)

A squirrel-cage induction motor is started by a star-delta starter. The starting current drawn from the line, compared with direct-on-line starting, becomes:

  1. One-third Correct answer.
  2. One over root three
  3. One-half
  4. The same

Explanation

In star connection each phase winding receives the line voltage divided by root three, so the phase current falls to 1/root3 of its delta value. Since the line current in star also equals the phase current, the overall reduction relative to delta starting is one-third.

Question 35Electrical Engineering (core technical subject)

A 220 V DC shunt motor has an armature resistance of 0.4 Ω and takes an armature current of 25 A. The back EMF developed is:

  1. 210 V Correct answer.
  2. 200 V
  3. 220 V
  4. 230 V

Explanation

For a motor, back EMF equals terminal voltage minus the armature resistance drop, E_b = 220 − (25 × 0.4) = 210 V. The back EMF is always less than the supply voltage in motoring operation.

Question 36Electrical Engineering (core technical subject)

In a DC machine, armature reaction causes the magnetic neutral axis to shift. In a generator this shift is:

  1. Zero at full load and maximum at no load
  2. Opposite to the direction of rotation
  3. In the direction of rotation Correct answer.
  4. Always exactly 90° electrical regardless of load

Explanation

The cross-magnetising armature flux distorts the main field so that the neutral axis moves along the direction of rotation in a generator. In a motor the shift is opposite to rotation. The shift grows with armature current, so brushes are often shifted to suit full load.

Question 37Electrical Engineering (core technical subject)

A 50 kVA transformer has a full-load copper loss of 800 W and an iron loss of 400 W. The load at which maximum efficiency occurs is:

  1. 50 kVA
  2. 70.7 kVA
  3. 35.4 kVA Correct answer.
  4. 25 kVA

Explanation

Maximum efficiency occurs when the variable copper loss equals the constant iron loss. Copper loss varies with the square of load, so the fraction of full load is the square root of 400/800 = 0.707, giving 0.707 × 50 = 35.4 kVA.

Question 38Electrical Engineering (core technical subject)

In a transformer, the no-load current is small compared with the full-load current mainly because:

  1. The core flux required is set by the applied voltage and the core has low reluctance Correct answer.
  2. The secondary is always short-circuited at no load
  3. The winding resistance is very high
  4. Copper loss at no load is very large

Explanation

At no load the applied voltage must balance the induced EMF, which depends only on flux and frequency; a small magnetising current is enough to establish that flux in a low-reluctance core. The no-load current supplies only magnetising and core-loss components, both small.

Question 39Electrical Engineering (core technical subject)

An ideal transformer with a 10:1 turns ratio supplies a 10 Ω resistive load from a 230 V source. The primary current is:

  1. 23 A
  2. 2.3 A Correct answer.
  3. 230 A
  4. 0.23 A

Explanation

Secondary voltage is 230/10 = 23 V, so secondary current is 23/10 = 2.3 A. The primary current is secondary current divided by the turns ratio, giving 0.23 A.

Question 40Electrical Engineering (core technical subject)

The core of a transformer is laminated rather than solid because lamination:

  1. Reduces eddy-current loss by interrupting circulating paths Correct answer.
  2. Increases the mechanical strength of the core
  3. Reduces hysteresis loss by softening the magnetic material
  4. Increases the permeability of the core material

Explanation

A changing flux induces EMFs within the core itself, and in a solid core these drive large eddy currents. Thin insulated laminations break the current paths and greatly reduce the eddy-current loss, which varies with the square of lamination thickness.

Question 41Electrical Engineering (core technical subject)

A transmission line has a series impedance of (10 + j40) Ω and negligible shunt admittance. If the receiving-end voltage is 100 kV and the load current is 100 A at unity power factor, the sending-end voltage is approximately:

  1. 100 kV
  2. 141.4 kV
  3. 104.4 kV Correct answer.
  4. 110 kV

Explanation

With unity power factor the current is in phase with the receiving voltage, so the voltage drop is I(R + jX) = 100(10 + j40) = (1000 + j4000) V. Adding this to 100 kV gives a magnitude of about 104.4 kV.

Question 42Electrical Engineering (core technical subject)

A 3-phase transmission line delivers 10 MW at 0.8 power factor lagging and 33 kV. The line current is approximately:

  1. 219 A Correct answer.
  2. 303 A
  3. 175 A
  4. 350 A

Explanation

Three-phase power is root3 × V_L × I_L × cos φ, so I_L = 10 × 10^6 / (1.732 × 33 × 10^3 × 0.8). This gives about 219 A.

Question 43Electrical Engineering (core technical subject)

Bundled conductors are used on EHV transmission lines primarily to:

  1. Eliminate the need for insulators
  2. Increase the surge impedance of the line
  3. Increase the skin-effect resistance at power frequency
  4. Reduce corona loss and line reactance Correct answer.

Explanation

Splitting each phase into several sub-conductors raises the effective conductor diameter, which lowers the surface voltage gradient and hence corona loss. The increased geometric mean radius also reduces series inductance and line reactance, improving power transfer.

Question 44Electrical Engineering (core technical subject)

A 3-phase overhead line has a capacitance to neutral of 0.01 µF per km per phase and operates at 132 kV, 50 Hz. The charging current per kilometre is approximately:

  1. 0.41 A
  2. 0.24 A Correct answer.
  3. 1.2 A
  4. 0.72 A

Explanation

Charging current is V_phase × ωC, with V_phase = 132/root3 = 76.2 kV. Thus I = 76.2 × 10^3 × 314 × 0.01 × 10^-6 ≈ 0.24 A per km per phase.

Question 45Electrical Engineering (core technical subject)

In a power system, the function of a lightning arrester installed at a substation is to:

  1. Increase the basic impulse insulation level of the equipment
  2. Prevent lightning from striking the transmission tower
  3. Provide a low-impedance path to earth for surge currents and then restore insulation Correct answer.
  4. Continuously earth the line during normal operation

Explanation

A surge arrester conducts heavily during an overvoltage, diverting surge energy to earth and limiting the voltage across the protected equipment. Once the surge passes it reseals and returns to its normal high-impedance state.

Question 46Electrical Engineering (core technical subject)

A relay has a plug setting of 125% and a current transformer ratio of 200/1 A. The relay pick-up current in primary terms is:

  1. 160 A
  2. 200 A
  3. 125 A
  4. 250 A Correct answer.

Explanation

The relay rated current on the secondary side is 1 A, so a 125% plug setting corresponds to 1.25 A in the relay coil. Referred to the primary through the 200/1 ratio, this is 1.25 × 200 = 250 A.

Question 47Electrical Engineering (core technical subject)

In a distance relay protecting a transmission line, the relay measures the ratio of voltage to current. The impedance seen by the relay for a fault at the reach point is:

  1. Independent of the fault resistance and source impedance Correct answer.
  2. Directly proportional to the fault current
  3. Equal to the surge impedance of the line
  4. Dependent only on the source impedance behind the relay

Explanation

The apparent impedance is the line impedance up to the fault, since voltage and current are measured at the relaying point. For a bolted fault at the reach point this depends on line length and line constants, not on fault resistance or source strength.

Question 48Electrical Engineering (core technical subject)

The main advantage of a static overcurrent relay compared with an electromagnetic relay is that it:

  1. Requires no auxiliary DC supply
  2. Can operate without a current transformer
  3. Is immune to all types of power system faults
  4. Has no moving contacts in its measuring element and offers faster, more repeatable operation Correct answer.

Explanation

A static relay performs measurement with semiconductor circuits, so there are no moving parts to wear or to introduce inertia in the measuring stage. This gives quicker response and consistent pick-up, with the tripping contact driven electronically.

Question 49Electrical Engineering (core technical subject)

A 3-phase, 400 V, 50 Hz, 4-pole squirrel-cage motor has a full-load efficiency of 88% and delivers 11 kW at the shaft. Its input line current at 0.85 power factor is approximately:

  1. 18.7 A
  2. 24.5 A
  3. 21.2 A Correct answer.
  4. 15.5 A

Explanation

Input power is 11 kW divided by 0.88, about 12.5 kW. Using P = root3 × V_L × I_L × cos φ, the line current is 12500/(1.732 × 400 × 0.85) ≈ 21.2 A.

Question 50Electrical Engineering (core technical subject)

A single-phase capacitor-start induction motor is reversed by:

  1. Interchanging the supply leads to the main winding only
  2. Reducing the capacitance of the starting capacitor
  3. Reversing the connections of the auxiliary (starting) winding relative to the main winding Correct answer.
  4. Increasing the applied voltage during starting

Explanation

Direction of rotation is set by the phase displacement between the main and auxiliary winding currents. Reversing only the auxiliary winding connections reverses that displacement and hence the rotating field.