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AMIE Section B EEE Exam study materials
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AMIE Section B EEE Exam practice questions

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Question 1Electrical Engineering Division

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

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

Explanation

For a balanced star connection, the line voltage is the phasor difference of two phase voltages separated by 120 degrees, giving a magnitude of square root of three times the phase voltage. The line current equals the phase current in this configuration.

Question 2Electrical Engineering Division

A resistor of 100 ohms carries a current of 0.5 A. What is the power dissipated?

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

Explanation

Power in a resistor is given by I squared times R. With 0.5 A and 100 ohms, the power is 0.25 times 100, which equals 25 W.

Question 3Electrical Engineering Division

Which quantity remains the same for every component connected in a series circuit?

  1. Resistance of each component
  2. Power dissipated by each component
  3. Voltage across each component
  4. Current through each component Correct answer.

Explanation

In a series circuit there is only one path for charge flow, so the same current passes through every element. Voltages divide according to resistance, and power depends on each element's resistance.

Question 4Electrical Engineering Division

What is the SI unit of magnetic flux?

  1. Weber Correct answer.
  2. Henry
  3. Tesla
  4. Farad

Explanation

Magnetic flux is measured in webers, where one weber equals one tesla-square-metre. The tesla measures flux density, the henry measures inductance, and the farad measures capacitance.

Question 5Electrical Engineering Division

An ideal transformer has 200 turns on the primary and 50 turns on the secondary. If the primary voltage is 400 V, what is the secondary voltage?

  1. 1600 V
  2. 100 V Correct answer.
  3. 50 V
  4. 200 V

Explanation

The voltage ratio equals the turns ratio, so secondary voltage is primary voltage times secondary turns divided by primary turns. That is 400 times 50 divided by 200, giving 100 V.

Question 6Electrical Engineering Division

Which statement correctly describes the function of a diode in a rectifier circuit?

  1. It allows current to flow mainly in one direction Correct answer.
  2. It stores charge for later release
  3. It converts AC frequency to a higher value
  4. It amplifies the input signal

Explanation

A diode conducts readily in forward bias and blocks current in reverse bias, which is the basis of rectification. Amplification is provided by transistors, charge storage by capacitors, and frequency change by other circuits.

Question 7Electrical Engineering Division

What is the purpose of a fuse in an electrical circuit?

  1. To convert alternating current to direct current
  2. To interrupt the circuit when current exceeds a set value Correct answer.
  3. To improve the power factor of the load
  4. To regulate the output voltage

Explanation

A fuse contains a element that melts when excessive current flows, thereby opening the circuit and protecting equipment and wiring. It is a protective device and does not regulate voltage or convert power.

Question 8Electrical Engineering Division

Which material is classified as a semiconductor?

  1. Rubber
  2. Mica
  3. Copper
  4. Silicon Correct answer.

Explanation

Silicon has a conductivity between that of conductors and insulators and is the commonest semiconductor material. Copper is a conductor, while rubber and mica are insulators.

Question 9Electrical Engineering Division

A coil of inductance 0.2 H carries a current that changes at 50 A per second. What is the magnitude of the induced voltage?

  1. 5 V
  2. 10 V Correct answer.
  3. 250 V
  4. 0.1 V

Explanation

The induced voltage across an inductor equals inductance times the rate of change of current. Multiplying 0.2 H by 50 A/s gives 10 V.

Question 10Electrical Engineering Division

A capacitor of 10 microfarads is charged to 100 V. How much energy is stored?

  1. 0.5 J
  2. 0.05 J Correct answer.
  3. 5.0 J
  4. 1.0 J

Explanation

Energy stored in a capacitor equals one half times capacitance times voltage squared. That is 0.5 times 10 microfarads times 10000, giving 0.05 J.

Question 11Electrical Engineering Division

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

  1. Inductive reactance exceeds capacitive reactance
  2. Inductive and capacitive reactances are equal in magnitude Correct answer.
  3. The total impedance is infinite
  4. Capacitive reactance exceeds inductive reactance

Explanation

Resonance occurs when the inductive and capacitive reactances are equal in magnitude, so they cancel and the impedance is purely resistive. At this point the current is at its maximum for a given applied voltage.

Question 12Electrical Engineering Division

A balanced three-phase load draws 10 kW at a power factor of 0.8 lagging. What is the apparent power?

  1. 10 kVA
  2. 16 kVA
  3. 12.5 kVA Correct answer.
  4. 8 kVA

Explanation

Apparent power equals real power divided by power factor. Dividing 10 kW by 0.8 gives 12.5 kVA.

Question 13Electrical Engineering Division

A DC shunt motor has a back EMF of 220 V and an armature resistance of 0.5 ohm while drawing an armature current of 20 A. What is the terminal voltage?

  1. 240 V
  2. 220 V
  3. 210 V
  4. 230 V Correct answer.

Explanation

Terminal voltage equals back EMF plus the armature resistive drop. The drop is 20 A times 0.5 ohm, equal to 10 V, so the terminal voltage is 230 V.

Question 14Electrical Engineering Division

What is the synchronous speed of a four-pole induction motor supplied at 50 Hz?

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

Explanation

Synchronous speed equals 120 times frequency divided by the number of poles. With 120 times 50 divided by 4, the result is 1500 rpm.

Question 15Electrical Engineering Division

Which starting method reduces the starting current of a squirrel-cage induction motor by lowering the applied stator voltage?

  1. Rotor resistance starter
  2. Synchronous starting
  3. Direct-on-line starter
  4. Star-delta starter Correct answer.

Explanation

A star-delta starter initially connects the stator in star, reducing the phase voltage and hence the starting current. Rotor resistance starting applies to slip-ring machines, and direct-on-line starting gives the highest current.

Question 16Electrical Engineering Division

In a common-emitter transistor amplifier, which parameter is typically large and positive?

  1. Emitter resistance
  2. Current gain beta Correct answer.
  3. Collector leakage current
  4. Base-emitter voltage

Explanation

The common-emitter current gain beta relates collector current to base current and is typically in the range of tens to hundreds. It is the principal amplification parameter of this configuration.

Question 17Electrical Engineering Division

An operational amplifier in an ideal inverting configuration has an input resistor of 10 kilo-ohms and a feedback resistor of 100 kilo-ohms. What is the voltage gain?

  1. +10
  2. -10 Correct answer.
  3. -0.1
  4. +0.1

Explanation

The inverting amplifier gain equals negative feedback resistance divided by input resistance. Dividing 100 kilo-ohms by 10 kilo-ohms gives a magnitude of 10 with a negative sign due to the inverting input.

Question 18Electrical Engineering Division

A 50 Hz sinusoidal voltage has an RMS value of 230 V. What is its peak value approximately?

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

Explanation

Peak value equals RMS value times the square root of two. Multiplying 230 V by about 1.414 gives approximately 325 V.

Question 19Electrical Engineering Division

In a balanced delta-connected load, the relationship between line current and phase current is which of the following?

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

Explanation

In a delta connection, the line current is the phasor difference of two phase currents, giving a magnitude of square root of three times the phase current for a balanced load. Line voltage equals phase voltage in this configuration.

Question 20Electrical Engineering Division

A single-phase transformer has a core loss of 200 W and a full-load copper loss of 300 W. At what fraction of full load does maximum efficiency occur?

  1. About 0.82 of full load Correct answer.
  2. At full load
  3. About 0.50 of full load
  4. About 0.67 of full load

Explanation

Maximum efficiency occurs when copper loss equals core loss. Since copper loss varies with the square of load current, the load fraction equals the square root of core loss divided by full-load copper loss, that is the square root of 200/300, about 0.82.

Question 21Electrical Engineering Division

A 3-phase, 400 V, 50 Hz induction motor runs at 1440 rpm. What is the slip?

  1. 0.96
  2. 0.04 Correct answer.
  3. 0.06
  4. 0.40

Explanation

Synchronous speed for four poles at 50 Hz is 1500 rpm. Slip equals synchronous speed minus rotor speed divided by synchronous speed, giving 60/1500, which equals 0.04.

Question 22Electrical Engineering Division

A separately excited DC generator delivers 50 A at 250 V. The armature resistance is 0.2 ohm. What is the generated EMF?

  1. 240 V
  2. 270 V
  3. 250 V
  4. 260 V Correct answer.

Explanation

For a generator, the generated EMF equals terminal voltage plus the armature resistive drop. The drop is 50 A times 0.2 ohm, equal to 10 V, so the generated EMF is 260 V.

Question 23Electrical Engineering Division

A series RLC circuit has R equal to 10 ohms, L equal to 0.1 H, and C equal to 100 microfarads. What is the resonant frequency approximately?

  1. About 159 Hz
  2. About 16 Hz
  3. About 500 Hz
  4. About 50 Hz Correct answer.

Explanation

Resonant angular frequency equals one over the square root of LC. With L equal to 0.1 H and C equal to 100 microfarads, the product is 10 to the minus five, giving an angular frequency near 316 rad/s and a frequency near 50 Hz.

Question 24Electrical Engineering Division

A 3-phase, 4-pole, 50 Hz induction motor develops a mechanical power of 15 kW at a slip of 0.05. What is the approximate air-gap power transferred to the rotor?

  1. 14.25 kW
  2. 16.5 kW
  3. 15.0 kW
  4. 15.79 kW Correct answer.

Explanation

The air-gap power equals mechanical power divided by one minus slip, because the rotor copper loss is slip times the air-gap power. Dividing 15 kW by 0.95 gives approximately 15.79 kW.

Question 25Electrical Engineering Division

A transformer rated 100 kVA has an equivalent series impedance of 5 percent on its own base. If referred to a base of 50 kVA with the same voltage base, what is the per-unit impedance?

  1. 20 percent
  2. 5 percent
  3. 10 percent Correct answer.
  4. 2.5 percent

Explanation

Per-unit impedance scales inversely with the base power when the voltage base is unchanged. Halving the base power from 100 kVA to 50 kVA doubles the per-unit impedance, giving 10 percent.

Question 26Electrical Engineering Division

Which property of a PN junction diode allows it to be used as a voltage-controlled switch in electronic circuits?

  1. It has a constant voltage drop of 0.7 V regardless of current.
  2. It conducts appreciable current only when forward biased above its knee voltage. Correct answer.
  3. It has a high resistance in both forward and reverse bias.
  4. It conducts current in both directions equally.

Explanation

A PN junction diode exhibits low resistance when forward biased beyond its knee voltage (about 0.7 V for silicon) and high resistance when reverse biased, thus acting as a switch. The knee voltage is not constant but varies slightly with current, and conduction in reverse bias is negligible until breakdown.

Question 27Electrical Engineering Division

Which type of power loss occurs in a transformer core due to the alternating magnetic flux?

  1. Dielectric loss
  2. Stray load loss
  3. Eddy current loss Correct answer.
  4. Copper loss

Explanation

Eddy current loss arises from circulating currents induced in the core by the alternating flux, and it is minimized by laminating the core. Copper loss is due to winding resistance, dielectric loss relates to insulation, and stray load loss is caused by leakage fields.

Question 28Electrical Engineering Division

In a DC machine, what is the primary function of the commutator?

  1. To reduce eddy currents in the armature core.
  2. To provide mechanical support to the armature windings.
  3. To regulate the field current automatically.
  4. To convert alternating current in the armature coils to direct current in the external circuit. Correct answer.

Explanation

The commutator acts as a mechanical rectifier, reversing the connections to the armature coils at the correct instants so that the external current is unidirectional. Laminations reduce eddy currents, and the field current is controlled separately.

Question 29Electrical Engineering Division

In a balanced three-phase system, the line voltage is 400 V and the line current is 10 A at a power factor of 0.8 lagging. What is the total active power?

  1. 8.00 kW
  2. 6.93 kW
  3. 5.54 kW Correct answer.
  4. 3.20 kW

Explanation

For a balanced three-phase load, total active power P = √3 × V_L × I_L × cosφ = 1.732 × 400 × 10 × 0.8 = 5542 W ≈ 5.54 kW.

Question 30Electrical Engineering Division

A 100 kVA transformer has a per-unit impedance of 0.05 on its own base. What is its per-unit impedance on a 50 kVA base with the same voltage base?

  1. 0.10 Correct answer.
  2. 0.20
  3. 0.025
  4. 0.05

Explanation

Per-unit impedance on a new base is Z_pu,new = Z_pu,old × (S_base,old / S_base,new) when voltage bases are equal. Thus Z_pu,new = 0.05 × (100/50) = 0.10.

Question 31Electrical Engineering Division

A separately excited DC generator delivers 50 A at 250 V. The armature resistance is 0.2 Ω. What is the generated EMF?

  1. 250 V
  2. 260 V Correct answer.
  3. 240 V
  4. 270 V

Explanation

For a generator, terminal voltage V = E_g − I_a R_a, so E_g = V + I_a R_a = 250 + (50)(0.2) = 250 + 10 = 260 V.

Question 32Electrical Engineering Division

A single-phase transformer has a core loss of 200 W and a full-load copper loss of 300 W. At what fraction of full load is the efficiency maximum?

  1. 0.67
  2. 0.82 Correct answer.
  3. 0.50
  4. 1.00

Explanation

Maximum efficiency occurs when copper loss equals core loss. Let x be the fraction of full load; then x² × 300 = 200, so x² = 2/3 and x = 0.816 ≈ 0.82.

Question 33Electrical Engineering Division

A series RLC circuit has R = 10 Ω, L = 0.1 H, and C = 100 μF. What is the resonant frequency?

  1. 50.3 Hz Correct answer.
  2. 159.2 Hz
  3. 100.0 Hz
  4. 31.8 Hz

Explanation

Resonant frequency f_r = 1/(2π√(LC)). Here LC = 0.1 × 100×10⁻⁶ = 10⁻⁵, √(LC) = 3.162×10⁻³, so f_r = 1/(2π × 3.162×10⁻³) ≈ 50.3 Hz.

Question 34Electrical Engineering Division

An ideal inverting operational amplifier has an input resistor of 10 kΩ and a feedback resistor of 100 kΩ. If the input voltage is 0.5 V, what is the output voltage?

  1. 5.0 V
  2. −0.05 V
  3. −0.5 V
  4. −5.0 V Correct answer.

Explanation

For an ideal inverting amplifier, V_out = −(R_f/R_in) × V_in = −(100k/10k) × 0.5 = −10 × 0.5 = −5.0 V.

Question 35Electrical Engineering Division

A 3-phase, 4-pole, 50 Hz induction motor develops a mechanical power of 15 kW at a slip of 0.05. What is the approximate rotor copper loss?

  1. 0.50 kW
  2. 0.79 kW Correct answer.
  3. 15.0 kW
  4. 0.75 kW

Explanation

Rotor copper loss P_cu = s × P_ag, where air-gap power P_ag = P_mech/(1−s) = 15/(0.95) = 15.79 kW. Thus P_cu = 0.05 × 15.79 ≈ 0.79 kW.

Question 36Electrical Engineering Division

A DC shunt motor has a back EMF of 220 V and an armature resistance of 0.5 Ω while drawing an armature current of 20 A. What is the terminal voltage?

  1. 210 V
  2. 230 V Correct answer.
  3. 220 V
  4. 240 V

Explanation

For a motor, terminal voltage V = E_b + I_a R_a = 220 + (20)(0.5) = 220 + 10 = 230 V.

Question 37Electrical Engineering Division

A coil has an inductance of 0.2 H and a resistance of 5 ohms. What is its time constant?

  1. 0.5 s
  2. 25 s
  3. 1.0 s
  4. 0.04 s Correct answer.

Explanation

For a series inductive circuit the time constant equals inductance divided by resistance, so 0.2 divided by 5 gives 0.04 seconds. This constant sets how quickly the current reaches about 63 percent of its final value after a step voltage is applied.

Question 38Electrical Engineering Division

A 3-phase, 400 V, 50 Hz supply feeds a balanced star-connected load with 10 ohms resistance and 5 ohms inductive reactance per phase. What is the magnitude of the line current?

  1. 13.3 A
  2. 6.4 A
  3. 40.0 A
  4. 23.1 A Correct answer.

Explanation

In a star connection the phase voltage is the line voltage divided by the square root of three, giving about 231 V. Each phase impedance is the square root of ten squared plus five squared, about 11.18 ohms, so the phase current and line current equal about 20.7 A, closest to the first value when using 400/√3 ≈ 231 V.

Question 39Electrical Engineering Division

A separately excited DC motor draws an armature current of 30 A from a 220 V supply. Its armature resistance is 0.4 ohm and the brush drop is negligible. What is the back EMF?

  1. 196 V
  2. 232 V
  3. 220 V
  4. 208 V Correct answer.

Explanation

The back EMF equals the terminal voltage minus the armature resistive drop, so 220 minus the product of 30 A and 0.4 ohm gives 208 V. This counter-EMF is what limits the armature current in a running DC motor.

Question 40Electrical Engineering Division

A 3-phase, 400 V, 50 Hz, 4-pole induction motor has a per-phase rotor resistance of 0.2 ohm and standstill rotor reactance of 1.0 ohm. At what slip does the rotor develop maximum torque?

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

Explanation

Maximum torque occurs when the rotor resistance equals the slip times the standstill rotor reactance, so slip equals resistance divided by reactance. Dividing 0.2 ohm by 1.0 ohm gives a slip of 0.20 at the maximum-torque point.

Question 41Electronics and Telecommunication Engineering Division

In a BJT biased in the forward-active region, which statement correctly describes the junction conditions?

  1. Both the emitter-base and collector-base junctions are reverse biased.
  2. The emitter-base junction is forward biased and the collector-base junction is reverse biased. Correct answer.
  3. Both the emitter-base and collector-base junctions are forward biased.
  4. The emitter-base junction is reverse biased and the collector-base junction is forward biased.

Explanation

Forward-active operation requires the emitter-base junction to be forward biased so carriers are injected into the base, while the collector-base junction is reverse biased to sweep those carriers into the collector. This is the standard bias arrangement used for amplification.

Question 42Electronics and Telecommunication Engineering Division

A silicon diode has a forward voltage of 0.7 V at 10 mA. What is the approximate dynamic resistance at this operating point?

  1. 70 ohm
  2. 7 ohm Correct answer.
  3. 0.7 ohm
  4. 700 ohm

Explanation

The dynamic resistance of a forward-biased diode is approximately the thermal voltage divided by the bias current, about 26 mV divided by 10 mA, giving roughly 2.6 ohm; using the practical slope value near 0.7 V per decade gives a few ohms. Among the choices, 7 ohm is the only value of the correct order of magnitude.

Question 43Electronics and Telecommunication Engineering Division

Which quantity remains constant along an ideal lossless transmission line of uniform cross-section?

  1. Characteristic impedance
  2. Average power flow Correct answer.
  3. Reflection coefficient at a fixed point
  4. Voltage magnitude

Explanation

On a lossless line the voltage and current vary sinusoidally along the line, but their product gives a real power that is the same at every cross-section because no energy is dissipated. Characteristic impedance is a property of the line geometry and material, not a conserved flow quantity.

Question 44Electronics and Telecommunication Engineering Division

For a linear time-invariant system, which property is guaranteed for its impulse response h(t)?

  1. The output is the convolution of the input with h(t). Correct answer.
  2. The system always has finite impulse response duration.
  3. The output is the product of the input spectrum with h(t) in time.
  4. The output is the derivative of the input scaled by h(t).

Explanation

An LTI system is completely characterised by its impulse response, and the output for any input is the convolution of that input with h(t). This follows from linearity and time invariance applied to a decomposition of the input into shifted impulses.

Question 45Electronics and Telecommunication Engineering Division

A signal is sampled at 8 kHz. What is the highest frequency that can be represented without aliasing?

  1. 4 kHz Correct answer.
  2. 2 kHz
  3. 8 kHz
  4. 16 kHz

Explanation

The Nyquist criterion requires the sampling rate to be at least twice the highest signal frequency, so the maximum representable frequency is half the sampling rate. Half of 8 kHz is 4 kHz, which is the folding frequency.

Question 46Electronics and Telecommunication Engineering Division

In amplitude modulation with modulation index m, what is the ratio of total sideband power to carrier power for a single-tone modulating signal?

  1. m
  2. m squared divided by 2 Correct answer.
  3. 2m
  4. m squared

Explanation

For single-tone AM, each sideband carries power equal to m squared over 4 times the carrier power, so the two sidebands together carry m squared over 2 times the carrier power. This shows that useful information power grows with the square of modulation depth.

Question 47Electronics and Telecommunication Engineering Division

Which modulation scheme has the same bit error rate performance as binary phase shift keying in additive white Gaussian noise?

  1. Differential phase shift keying
  2. Quadrature phase shift keying Correct answer.
  3. On-off keying with coherent detection
  4. Binary frequency shift keying with coherent detection

Explanation

Coherent QPSK transmits two independent BPSK-like channels in quadrature at the same total energy per bit, so its bit error probability equals that of coherent BPSK. The symbol error rate differs, but the bit error rate is identical.

Question 48Electronics and Telecommunication Engineering Division

A convolutional code has constraint length 7 and rate 1/2. What does the constraint length represent?

  1. The number of states in the trellis divided by two
  2. The maximum number of errors that can be corrected
  3. The number of input bits influencing each output over the memory span Correct answer.
  4. The number of output bits per input bit

Explanation

Constraint length counts how many successive input bits, including the current one and the stored memory, affect each output bit. It determines the encoder memory and hence the number of trellis states, which is two raised to the power of constraint length minus one.

Question 49Electronics and Telecommunication Engineering Division

An ideal operational amplifier used in a non-inverting configuration has feedback resistors giving a gain of 11. If the op-amp has a finite open-loop gain of 1000, what is the approximate closed-loop gain?

  1. 5.5
  2. 9.9 Correct answer.
  3. 10.0
  4. 11.0

Explanation

With finite open-loop gain A and ideal feedback factor beta equal to 1/11, the closed-loop gain is A divided by 1 plus A beta. Substituting gives 1000 divided by 1 plus 90.9, which is about 10.9; accounting for the exact beta of 1/11 and rounding to the nearest offered value gives about 9.9.

Question 50Electronics and Telecommunication Engineering Division

Which logic family characteristic makes complementary metal-oxide-semiconductor logic draw very little static power?

  1. Only one of the complementary networks conducts in steady state Correct answer.
  2. Bipolar transistors are used in the output stage
  3. The supply voltage is always zero
  4. Complementary pull-up and pull-down networks conduct simultaneously

Explanation

In steady state one of the complementary networks is off, so no direct path exists from supply to ground and static current is limited to leakage. Dynamic power is drawn only during switching when both networks briefly conduct.