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100+ Free IESL Part II Electrical Practice Questions

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2026 Statistics

Key Facts: IESL Part II Electrical Exam

100

Practice Questions

IESL Syllabus

3 hours

Exam Time Limit

IESL Rules

50%

Required Pass Mark

IESL Guidelines

LKR 15k

Approx. Exam Fee

IESL circulars

5 Topics

Core Subject Areas

Electrical Syllabus

AMIESL

Resulting Credential

IESL Membership

The IESL Qualifying Examination Part II - Electrical Engineering consists of 100 questions with a 3-hour time limit, requiring a passing score of 50%. The exam fee is approximately LKR 15,000 and is administered by the Institution of Engineers, Sri Lanka. The syllabus comprises five core areas: Electrical Machines, Power Systems, Electromagnetic Fields, Electronic Devices/Circuits, and Control Systems. Success in this exam is a prerequisite for Associate Membership (AMIESL).

Sample IESL Part II Electrical Practice Questions

Try these sample questions to test your IESL Part II Electrical exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A 50 kVA, 2400/240 V, single-phase transformer has an equivalent resistance of 1.5% and an equivalent reactance of 4.0%. What is the voltage regulation of this transformer when operating at full load with a power factor of 0.8 lagging?
A.2.4%
B.3.2%
C.3.6%
D.5.0%
Explanation: Voltage regulation (VR) in per-unit is given by VR = R_pu * cos(theta) + X_pu * sin(theta) for a lagging power factor. Given R_pu = 0.015, X_pu = 0.04, cos(theta) = 0.8, and sin(theta) = sqrt(1 - 0.8^2) = 0.6. Substituting these values: VR = 0.015 * 0.8 + 0.04 * 0.6 = 0.012 + 0.024 = 0.036 or 3.6%. Therefore, the correct voltage regulation is 3.6%.
2Which of the following describes the phase relationship between the primary and secondary line voltages in a three-phase transformer with a Dy11 vector group?
A.Secondary line voltage is in phase with the primary line voltage.
B.Secondary line voltage lags the primary line voltage by 30 degrees.
C.Secondary line voltage leads the primary line voltage by 30 degrees.
D.Secondary line voltage leads the primary line voltage by 180 degrees.
Explanation: In Dy11 vector notation, 'D' represents delta primary winding, 'y' represents star secondary winding, and '11' represents the clock face position. Since each hour on a clock represents 30 degrees, 11 o'clock means the secondary line voltage leads the primary line voltage by 11 * 30 degrees = 330 degrees, which is equivalent to leading by +30 degrees (or lagging by -330 degrees). Therefore, the secondary line voltage leads the primary line voltage by 30 degrees.
3A 4-pole, three-phase, 50 Hz induction motor runs at a speed of 1440 RPM at full load. What is the frequency of the rotor currents under this operating condition?
A.1.0 Hz
B.2.0 Hz
C.3.0 Hz
D.50.0 Hz
Explanation: First, calculate the synchronous speed (Ns): Ns = 120 * f / P = 120 * 50 / 4 = 1500 RPM. The slip (s) is calculated as s = (Ns - Nr) / Ns = (1500 - 1440) / 1500 = 60 / 1500 = 0.04 (or 4%). The rotor frequency (fr) is fr = s * f = 0.04 * 50 Hz = 2.0 Hz.
4For a three-phase induction motor, how does the maximum torque developed relate to the rotor resistance (R2) per phase?
A.The maximum torque is directly proportional to R2.
B.The maximum torque is inversely proportional to R2.
C.The maximum torque is independent of R2, but the slip at which it occurs is directly proportional to R2.
D.The maximum torque is independent of R2, and the slip at which it occurs is inversely proportional to R2.
Explanation: The maximum electromagnetic torque of an induction motor is independent of the rotor resistance R2. However, the slip at which the maximum torque occurs (s_max) is directly proportional to the rotor resistance R2 (s_max = R2 / X2). Adding external resistance in the rotor circuit of a slip-ring induction motor shifts the maximum torque to a lower speed (higher slip), allowing high starting torque, but the peak value remains the same.
5In a synchronous motor, the V-curves represent the relationship between which two operational parameters?
A.Armature current and torque at constant speed
B.Armature current and field current at constant load
C.Power factor and speed at constant excitation
D.Torque and speed at variable excitation
Explanation: V-curves of a synchronous motor plot the armature current (Ia) on the y-axis against the field excitation current (If) on the x-axis for different constant load levels. The curve has a characteristic V-shape. The lowest point on the V-curve represents the minimum armature current, which occurs at unity power factor. To the left of the minimum is the under-excited lagging region, and to the right is the over-excited leading region.
6An 8-pole DC machine has a duplex wave-wound armature. How many parallel paths (A) exist in this armature winding for current conduction?
A.2
B.4
C.8
D.16
Explanation: For a wave-wound armature, the number of parallel paths A is given by A = 2 * m, where m is the multiplicity of the winding. For a simplex wave winding, m = 1, so A = 2. For a duplex wave winding, m = 2, so A = 2 * 2 = 4 paths. In contrast, a lap-wound duplex machine would have A = 2 * m * P = 2 * 2 * 8 = 32 paths.
7A 220 V DC shunt motor runs at 1000 RPM while drawing an armature current of 50 A. The armature resistance is 0.2 ohms. If the field flux is reduced by 10% while keeping the armature current constant, what will be the new operating speed of the motor?
A.900 RPM
B.1000 RPM
C.1111 RPM
D.1222 RPM
Explanation: The back EMF is given by Eb = V - Ia * Ra. Initially, Eb1 = 220 - 50 * 0.2 = 220 - 10 = 210 V. The speed is proportional to Eb / flux (N = k * Eb / phi). Since the armature current is constant, Eb remains constant at Eb2 = 210 V. The new speed N2 is related by N2/N1 = (Eb2/Eb1) * (phi1/phi2) = 1 * (1 / 0.9) = 1.111. Therefore, N2 = 1000 * 1.111 = 1111 RPM.
8Two single-phase transformers, each rated at 100 kVA, are connected in an open-delta (V-V) configuration to supply a three-phase balanced load. What is the maximum total three-phase kVA that this connection can supply without overloading either transformer?
A.100.0 kVA
B.115.4 kVA
C.173.2 kVA
D.200.0 kVA
Explanation: In an open-delta connection, the total capacity is sqrt(3) times the rating of one transformer. So, Total Capacity = sqrt(3) * Rating = 1.732 * 100 kVA = 173.2 kVA. This represents 57.7% of the capacity of a full closed-delta bank of three identical 100 kVA transformers (300 kVA). Operating beyond 173.2 kVA will overload the windings of the open-delta transformers.
9A transformer has a core loss of 800 W and a full-load copper loss of 1600 W. At what percentage of full load will this transformer operate with maximum efficiency?
A.50.0%
B.70.7%
C.80.0%
D.100.0%
Explanation: Maximum efficiency in a transformer occurs when variable copper losses (P_cu) equal constant core losses (P_i). Let x be the fraction of full-load kVA. Then x^2 * P_cu_FL = P_i. So x = sqrt(P_i / P_cu_FL) = sqrt(800 / 1600) = sqrt(0.5) = 0.707 or 70.7% of full load.
10A synchronous generator is supplying a load at a lagging power factor. If the load is suddenly disconnected, what happens to the terminal voltage of the generator, assuming speed and excitation remain constant?
A.The terminal voltage remains completely constant.
B.The terminal voltage drops due to loss of load.
C.The terminal voltage rises significantly due to the removal of demagnetizing armature reaction.
D.The terminal voltage fluctuates erratically before settling at zero.
Explanation: Under a lagging power factor load, the armature reaction has a demagnetizing effect on the main field flux. Disconnecting the load removes this demagnetizing effect, causing the field flux to return to its full value, which results in a significant rise in terminal voltage (no-load voltage is higher than full-load terminal voltage). The voltage regulation is positive.

About the IESL Part II Electrical Exam

The IESL Qualifying Examination Part II in Electrical Engineering is a formal academic assessment conducted by the Institution of Engineers, Sri Lanka. It serves as a qualifying benchmark for engineering graduates seeking Associate Membership (AMIESL) and the Chartered Engineer (C.Eng) pathway. The exam tests candidates on advanced engineering topics including Electrical Machines, Power Systems, Electromagnetic Fields, Electronic Devices/Circuits, and Control Systems. Mastery of mathematical derivations, network analysis, and field theory calculations is essential for success.

Assessment

100 multiple-choice questions (including numeric calculations)

Time Limit

3 hours

Passing Score

50%

Exam Fee

~LKR 15,000 (Institution of Engineers, Sri Lanka (IESL))

IESL Part II Electrical Exam Content Outline

20%

Electrical Machines

Equivalent circuits, parameters, and regulation of single-phase and three-phase transformers; operation, torque-speed characteristics, and starters for induction motors; excitation, power factor control, and stability of synchronous generators; speed control and winding layout of DC machines.

20%

Power Systems

Modeling of transmission lines (pi and T models), efficiency, and voltage regulation; load flow studies (Newton-Raphson, Gauss-Seidel); fault analysis using symmetrical components for line-to-ground and line-to-line faults; power system stability and relay protection schemes.

20%

Electromagnetic Fields

Coulomb's Law, Gauss's Law, divergence theorem, and electrostatic boundary conditions; Biot-Savart Law, Ampere's Law, magnetic forces, and vector potential; Maxwell's equations in differential and integral forms; plane wave propagation in different media and transmission lines.

20%

Electronic Devices & Circuits

Characteristics of diodes, BJTs, JFETs, and MOSFETs; bias stability, small-signal models, and frequency response of single-stage amplifiers; operational amplifier applications (integrators, summing, instrumentation); logic gates, flip-flops, counters; rectifier and inverter topologies.

20%

Control Systems

Linear time-invariant systems, block diagrams, signal flow graphs, and Mason's gain formula; transient and steady-state error analysis; Routh-Hurwitz stability criterion, root locus technique, Nyquist stability criterion, and Bode plots; state-space representation and controllability/observability.

How to Pass the IESL Part II Electrical Exam

What You Need to Know

  • Passing score: 50%
  • Assessment: 100 multiple-choice questions (including numeric calculations)
  • Time limit: 3 hours
  • Exam fee: ~LKR 15,000

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

IESL Part II Electrical Study Tips from Top Performers

1Master three-phase power calculations and symmetrical component analysis — these are highly weighted in Power Systems.
2Practice drawing equivalent circuit diagrams for induction motors and synchronous generators to solve for voltage regulation and efficiency.
3Understand boundary conditions for electric and magnetic fields, as these are critical for solving Electromagnetic Fields problems.
4Become proficient in solving op-amp and BJT small-signal models under timed conditions.
5Memorize the relationship between transient response parameters (overshoot, settling time) and transfer function pole locations in Control Systems.
6Use Bode plots and Nyquist criteria to quickly determine system stability margins (gain margin and phase margin).
7Ensure you practice solving multi-step calculations using standard scientific calculators allowed by IESL.

Frequently Asked Questions

What is the IESL Part II Electrical Qualifying Examination?

The IESL Part II is an engineering qualifying examination administered by the Institution of Engineers, Sri Lanka. It validates academic competency in electrical engineering, serving as a route to Associate Membership (AMIESL) for individuals who did not graduate from a directly accredited four-year Sri Lankan engineering degree.

What is the passing score for the exam?

The minimum passing score required to pass the IESL Qualifying Examination Part II is 50% in the subject paper. Candidates must demonstrate proficiency in both descriptive/theoretical concepts and quantitative problem-solving.

How much does the IESL Part II exam cost?

The examination fee is approximately LKR 15,000 for the Part II papers. Candidates should verify the exact schedule and fees in the official circulars issued by the IESL Education and Evaluation Division.

What core subjects are covered on the Electrical Engineering paper?

The syllabus is evenly split across five main subject groups: (1) Electrical Machines (transformers and AC/DC motors/generators), (2) Power Systems (transmission, analysis, and protection), (3) Electromagnetic Fields (static fields and Maxwell's wave propagation), (4) Electronic Devices & Circuits (both analog op-amps and digital gates/sequential logic), and (5) Control Systems (frequency analysis and state-space).

How long is the actual exam, and what format is it?

The standard exam paper is a 3-hour written assessment. It incorporates both conceptual multiple-choice items and detailed numerical calculation problems requiring candidates to solve engineering formulas and draw circuit diagrams.

Can I sit the Part II exam if I haven't passed Part I?

Generally, you must either successfully pass the IESL Qualifying Examination Part I or have your foreign/non-recognized qualification formally assessed and exempted from Part I by the IESL Education Division before you are eligible to sit the Part II papers.