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Chapter 1

Basic Electrical Engineering

Lok Sewa Aayog (Public Service Commission) past exam questions

Past questions and answers

37 questions set from this chapter. Most repeated first.

  • 2081 · 1st · 1 mark

Which of the following is NOT a characteristic of a peak load power plant? A) Quick start-up time B) High fuel efficiency C) Ability to handle sudden load changes D) Operates only during high demand periods

Answer

(b) High fuel efficiency

A peak load power plant runs only for short periods during maximum demand, so it is designed for quick start-up and rapid response to load changes rather than for fuel economy. High fuel efficiency is a characteristic of base load plants, which run continuously at high load factor.

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  • 2081 · 1st · 1 mark

The speed of a DC generator is 1500rpm. The frequency of current in the armature winding with 4 pole on the stator would be ...... Hz. A) 0 B) 25 C) 50 D) 100

Answer

The frequency of the induced EMF in the armature winding of a DC generator depends on the number of poles and the speed of the armature. The formula is:

[ f = \frac{PN}{120} ]

where:

  • (P) = number of poles = 4
  • (N) = speed in rpm = 1500

Substituting:

[ f = \frac{4 \times 1500}{120} = \frac{6000}{120} = 50 \text{ Hz} ]

So the frequency is 50 Hz.

(c) 50

[ f = \frac{PN}{120} = \frac{4 \times 1500}{120} = 50\ \text{Hz} ]

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  • 2081 · 1st · 1 mark

The core of a transformer is laminated in order to ...... A) simplify its construction B) minimize eddy current loss C) reduce cost D) reduce hysteresis loss

Answer

(b) minimize eddy current loss

Laminating the core (thin sheets insulated from each other) breaks up the path for circulating eddy currents, which increases the resistance offered to them and thereby reduces the eddy current loss (which varies as the square of the thickness of the sheet).

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  • 2081 · 1st · 1 mark

Which of the following is NOT a pre-condition for synchronizing a synchronous generator to the system? A) Both the incoming generator and the system should be unloaded. B) The frequency of the incoming generator should be same as that of the system C) The voltages of both must be the same D) The phase rotation of the incoming generator and the system should be same

Answer

(a) Both the incoming generator and the system should be unloaded.

The conditions for synchronizing a synchronous generator are: equal voltage, equal frequency, same phase sequence (phase rotation), and zero phase difference at the instant of closing. The system is never unloaded — it carries its normal load — so requiring the system to be unloaded is not a valid pre-condition.

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  • 2081 · 1st · 1 mark

What is the synchronous speed of AC 3 phase induction motor having 6 poles at a frequency of 50Hz? A) 600rpm B) 800rpm C) 1000rpm D) 1200rpm

Answer

(c) 1000 rpm

Synchronous speed is given by (N_s = \dfrac{120f}{P} = \dfrac{120 \times 50}{6} = 1000) rpm.

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  • 2081 · 1st · 1 mark

A circuit breaker is a ...... A) current controlling device B) circuit interrupting device C) current limiting device D) none of the above

Answer

(b) circuit interrupting device

A circuit breaker is a protective device whose main function is to interrupt (break) the circuit — i.e., to open the contacts and stop the flow of current automatically under fault conditions (overload or short circuit), and it can also be closed manually to restore supply. It does not primarily control or limit current continuously like a rheostat or a fuse.

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  • 2081 · 1st · 1 mark

In comparison to Ni-Cd battery, lead acid battery has ...... A) better efficiency B) poorer efficiency C) equal efficiency D) maintenance free criterion

Answer

The model answer is incorrect. In general, Ni-Cd batteries have better energy efficiency (around 80–90%) than lead-acid batteries (around 70–85%, often lower in practice). Lead-acid batteries are cheaper and widely used, but they do not have better efficiency than Ni-Cd. Also, "maintenance free criterion" is not an efficiency comparison and is not generally true for all lead-acid batteries.

(b) poorer efficiency

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  • 2081 · 1st · 1 mark

Which battery is used in an electric power station? A) Aluminium-ion B) Lead-acid C) Zinc-carbon D) Nickel-cadmium

Answer

(b) Lead-acid

In an electric power station, the standby/backup DC supply (for switchgear control, protection relays, emergency lighting and SCADA) is provided by a lead-acid storage battery (usually a large vented lead-acid or VRLA bank). It is chosen because it is cheap, has a high current-delivering capacity, is rechargeable, and has a long service life.

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  • 2081 · 1st · 1 mark

All other things remaining same, the sag of the conductor of a 132kV overhead transmission line ...... A) increases in winter B) decreases in winter C) increases as well as decreases in winter D) does not change between summer and winter

Answer

Let me work through this.

Reasoning:

Sag of an overhead transmission line conductor is given by:

[ S = \frac{wL^2}{8T} ]

where (w) = weight per unit length, (L) = span, (T) = tension.

  • In winter, ambient temperature is low → conductor contracts → tension (T) increases → sag (S) decreases.
  • In summer, temperature is high → conductor expands → tension decreases → sag increases.

So the sag decreases in winter.

The model answer states (b) decreases in winter, with correct reasoning. This matches option B.

Checking the options:

  • A) increases in winter — wrong
  • B) decreases in winter — correct
  • C) increases as well as decreases in winter — wrong
  • D) does not change — wrong

The answer is correct and complete.

OK

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  • 2081 · 1st · 1 mark

Step-up substations are associated with ...... A) distributors B) concentrated load C) consumer location D) generating station

Answer

(d) generating station

Step-up substations are located at generating stations, where the generated voltage is stepped up to a high value for economical long-distance transmission.

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  • 2081 · 1st · 1 mark

Lightning arresters are used in power system to protect electrical equipment against ...... A) direct strokes of lightning B) over-voltages due to indirect lightning stroke C) power frequency over voltages D) over-currents due to lightning

Answer

(b) over-voltages due to indirect lightning stroke

A lightning arrester (surge diverter) is connected between the line and earth to discharge surge voltages caused by indirect lightning strokes (travelling waves) to ground and protect equipment insulation. Direct strokes are handled by the overhead earth wire/shield wire, not by the arrester.

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  • 2081 · 1st · 1 mark

A circuit is disconnected by isolator when ...... A) line is on full load B) circuit breaker is not open C) there is no current in the line D) line is energized

Answer

(c) there is no current in the line

An isolator is an off-load device — it has no arc-quenching arrangement, so it must only be operated when the circuit is dead. It is therefore opened only after the circuit breaker has already interrupted the current, i.e. when no current flows in the line.

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  • 2081 · 1st · 1 mark

Functionally a fuse is similar to ...... A) isolator B) earthing switch C) oil circuit breaker D) lightning arrester

Answer

(c) oil circuit breaker

A fuse is functionally similar to a circuit breaker (including an oil circuit breaker) because both are protective devices that automatically open/break the circuit under fault or overcurrent conditions. An isolator and an earthing switch are manually operated switching devices used for isolation/safety, not for automatic fault interruption, and a lightning arrester protects against overvoltage surges rather than interrupting fault current.

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  • 2081 · 1st · 1 mark

...... is a key characteristic of a good insulator for overhead lines. A) High conductivity B) Lighter weight C) High thermal conductivity D) High dielectric strength

Answer

(d) High dielectric strength

A good insulator for overhead lines must have high dielectric strength so that it can withstand the high voltage of the line without breaking down (i.e., without allowing current to leak through it). High conductivity and high thermal conductivity are properties of conductors, not insulators.

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  • 2081 · 1st · 1 mark

Isolators are used for disconnecting a circuit when ...... A) line is energized B) line is on full load C) line carries no current or at no load condition D) can be operated under any condition

Answer

(c) line carries no current or at no load condition

Isolators (off-load switches) have no arc-quenching arrangement, so they must only be opened or closed when the circuit is de-energized or carrying negligible current. Opening them under load would draw a dangerous arc and damage the contacts.

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  • 2081 · 1st · 1 mark

Which of the following motors is used in the ceiling fan? A) Universal motor B) Induction motor C) Series motor D) Synchronous motor

Answer

(a) Universal motor

A ceiling fan uses a single-phase universal motor (a small series-wound motor that can run on AC or DC), because it is compact, cheap, and provides high starting torque with speed control via a regulator.
An induction motor is not typical for ceiling fans; it is used in applications like water pumps and other fixed-speed AC drives. A series motor is used in traction, and a synchronous motor runs only at synchronous speed.

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  • 2081 · 1st · 1 mark

A large diversity factor of the load in a power system ...... the installation cost. A) reduces B) increases C) does not affect D) none of the above

Answer

(a) reduces

A large diversity factor means the maximum demand of the whole system is much smaller than the sum of the individual maximum demands. So the generating plant and distribution equipment need to be sized for a smaller peak load, which reduces the installation cost.

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  • 2081 · 1st · 1 mark

Which of the following instruments has same calibration for AC and DC? A) Moving iron B) Moving coil C) Hot wire D) Induction

Answer

The model answer is correct in its final choice, but the explanation contains a factual error: hot-wire instruments also have the same calibration for AC and DC (both respond to RMS/heating effect). The question asks which instrument has the same calibration for AC and DC — both moving iron and hot wire satisfy this. However, in standard objective exams, the intended answer is moving iron.

Let me verify:

  • Moving iron: Deflection depends on I2I^2 (or V2V^2), so it responds to RMS value → same calibration for AC and DC. ✓
  • Moving coil: Responds to average value; works only on DC (unless rectified). ✗
  • Hot wire: Deflection depends on heating effect (I2RI^2R), so it also responds to RMS → same calibration for AC and DC. ✓
  • Induction: Works only on AC. ✗

So both A and C are technically correct. But in Lok Sewa Aayog exams, the standard answer is (a) Moving iron, as hot-wire instruments are less common and the question likely expects the most typical answer.

The model answer's explanation incorrectly states that hot-wire instruments have "different thermal behaviour" leading to different calibration — this is wrong. Hot-wire instruments are inherently RMS-responding and have the same calibration for AC and DC.

However, since the final answer (a) is the intended correct option, and the question is 1 mark MCQ, the answer is acceptable.

OK

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  • 2081 · 1st · 1 mark

How is a voltmeter connected in a circuit? A) In series B) In parallel C) Either series or parallel D) In any way

Answer

(b) In parallel

A voltmeter is connected in parallel with the component whose voltage is to be measured, so that the same potential difference appears across both. It has a very high resistance so that it draws almost no current and does not disturb the circuit.

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  • 2081 · 1st · 1 mark

Which of the following lamps is suitable for highway lighting? A) Neon light B) Sodium vapor light C) Fluorescent light D) Incandescent light

Answer

(b) Sodium vapor light

Sodium vapor lamps have high luminous efficacy (about 100–150 lumens/watt) and a long service life, so they give the most economical illumination over long stretches of highway. Their yellowish light also penetrates fog and mist better than the whiter light of fluorescent or incandescent lamps.

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  • 2081 · 1st · 1 mark

Power factor of the industrial loads is generally ...... A) lagging B) unity C) leading D) zero

Answer

(a) lagging

Industrial loads are mostly induction motors and transformers, which draw a magnetising current and are inductive in nature. In an inductive circuit the current lags behind the voltage, so the power factor is lagging (typically 0.8–0.9 lagging).

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  • 2081 · 1st · 1 mark

A power electronic device that converts electrical energy from the DC source to AC form is called ...... A) Rectifier B) Silicon-controlled rectifier C) Filter circuit D) Inverter

Answer

(d) Inverter

An inverter is a power electronic device that converts DC (direct current) input into AC (alternating current) output. A rectifier does the opposite (AC to DC), an SCR is a switching device, and a filter circuit only smooths the output waveform.

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  • 2081 · 1st · 1 mark

A pn junction acts as a ...... switch. A) unidirectional B) bidirectional C) multidimensonal D) universal

Answer

(a) unidirectional

A pn junction diode conducts current easily in forward bias but blocks it in reverse bias, so it behaves as a one-way (unidirectional) switch.

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  • 2081 · 2nd · 5 marks

What are the differences between capacitance and inductance? Write.

Answer

Differences between Capacitance and Inductance

S.N.CapacitanceInductance
1It is the property of a circuit element (capacitor) to store energy in the electric field.It is the property of a circuit element (inductor) to store energy in the magnetic field.
2It opposes any change in voltage across the element.It opposes any change in current through the element.
3It is denoted by CC and measured in farad (F).It is denoted by LL and measured in henry (H).
4Current through a capacitor leads the voltage by 90∘90^\circ.Current through an inductor lags the voltage by 90∘90^\circ.
5Capacitive reactance is XC=12πfCX_C = \dfrac{1}{2\pi f C}; it decreases as frequency increases.Inductive reactance is XL=2πfLX_L = 2\pi f L; it increases as frequency increases.
6Energy stored is W=12CV2W = \dfrac{1}{2}CV^2.Energy stored is W=12LI2W = \dfrac{1}{2}LI^2.
7It blocks DC (acts as open circuit) and allows AC to pass.It allows DC to pass (acts as short circuit) and opposes AC.
8It is present in a capacitor, formed by two conducting plates separated by a dielectric.It is present in an inductor, formed by a coil or winding.
9It causes the current to lead, so it is used for power factor correction (in parallel).It causes the current to lag, so it is used in filters, chokes and transformers.
10Its effect is due to the dielectric between the plates.Its effect is due to self-induced EMF (Faraday's law) in the coil.

In short: Capacitance stores energy in an electric field and opposes change of voltage, while inductance stores energy in a magnetic field and opposes change of current.

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  • 2081 · 2nd · 5 marks

A resistor of 50Ω50\Omega is connected in series to a parallel combination of three resistances; 20Ω20\Omega, 45Ω45\Omega and 90Ω90\Omega. What is the value of power consumed by the 50Ω50\Omega resistor if the source voltage is 230V? Calculate.

Answer

Given:

  • Series resistor, Rs=50 ΩR_s = 50\,\Omega
  • Parallel resistors: R1=20 ΩR_1 = 20\,\Omega, R2=45 ΩR_2 = 45\,\Omega, R3=90 ΩR_3 = 90\,\Omega
  • Source voltage, V=230 VV = 230\,\text{V}

Equivalent resistance of parallel combination:

1Rp=120+145+190\frac{1}{R_p}=\frac{1}{20}+\frac{1}{45}+\frac{1}{90}

Taking LCM =180=180,

1Rp=9+4+2180=15180=112\frac{1}{R_p}=\frac{9+4+2}{180}=\frac{15}{180}=\frac{1}{12} Rp=12 ΩR_p = 12\,\Omega

Total circuit resistance:

RT=Rs+Rp=50+12=62 ΩR_T = R_s + R_p = 50 + 12 = 62\,\Omega

Circuit current:

I=VRT=23062=3.71 AI=\frac{V}{R_T}=\frac{230}{62}=3.71\,\text{A}

Power consumed by the 50 Ω50\,\Omega resistor:

P=I2RsP=I^2R_s P=(3.71)2×50P=(3.71)^2 \times 50 P=13.76×50≈688.1 WP=13.76 \times 50 \approx 688.1\,\text{W}

Answer:
The power consumed by the 50 Ω50\,\Omega resistor is approximately 688.1 W.

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  • 2081 · 2nd · 5 marks

Describe the working principles of a three phase induction motor with appropriate diagram in brief.

Answer

Three Phase Induction Motor — Working Principle

Construction (brief):

  • Stator: carries a 3-phase winding placed in slots, fed from a 3-phase supply.
  • Rotor: either squirrel cage (bars shorted by end rings) or slip-ring/wound type. No external supply is given to the rotor.

Working principle:

  1. When a 3-phase supply is given to the stator winding, a rotating magnetic field (RMF) of constant magnitude is produced. Its speed is the synchronous speed:

Ns=120fP rpmN_s = \frac{120f}{P} \text{ rpm}

where ff = supply frequency (Hz), PP = number of poles.

  1. The RMF cuts the stationary rotor conductors. By Faraday's law of electromagnetic induction, an emf is induced in the rotor conductors (transformer action — hence it is also called a rotating transformer).

  2. Since the rotor circuit is closed (cage bars shorted by end rings, or slip rings shorted), the induced emf drives a rotor current.

  3. This current-carrying rotor conductor lies in the stator's magnetic field, so a force acts on it. By Lenz's law, the rotor moves in the same direction as the RMF, so as to oppose the relative speed causing the induction. Hence the rotor runs in the direction of the rotating field, but always at a speed NN less than NsN_s.

  4. If the rotor ran at NsN_s, relative speed would be zero → no induced emf → no rotor current → no torque. So the rotor can never reach synchronous speed. The difference is the slip:

s=Ns−NNs,N=Ns(1−s)s = \frac{N_s - N}{N_s}, \qquad N = N_s(1-s)

Typical full-load slip = 3–5%.

  1. Torque is produced by the interaction of the RMF and rotor current; the motor is self-starting and runs as an induction (asynchronous) motor.

Diagram:

        3-phase supply
        R  Y  B
        |  |  |
   +----+--+--+----+
   |   STATOR      |  <- 3-phase winding in slots
   |  [ ROTOR ]    |  <- squirrel cage / wound rotor
   |   (shaft)     |
   +---------------+
        |     |
      Ns ->   N ->   (N < Ns, same direction)

(Label: stator core, stator winding, air gap, rotor bars, end ring, shaft, RMF direction, rotor rotation direction.)

Key points to write for full marks: RMF production → induced emf in rotor → rotor current → torque by Lenz's law → rotor runs below NsN_s → slip → self-starting.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 5 marks

Describe the basic operational principle of surge arresters with the explanation of its need in electrical safety system.

Answer

Surge Arrester — Basic Operational Principle and Need in Electrical Safety

1. Basic operational principle

A surge arrester is a protective device connected between the line (phase) and earth. Its operation depends on the property of a non-linear (voltage-dependent) resistance element — commonly a metal-oxide varistor (ZnO) disc, or in older types a spark gap in series with a silicon-carbide resistor.

  • Normal operating condition: At normal system voltage, the arrester presents a very high resistance (almost an open circuit). Only a small leakage current of the order of milliamperes flows to earth, so the arrester does not disturb normal power supply.
  • Surge (overvoltage) condition: When a lightning surge or switching overvoltage strikes the line, the voltage across the arrester rises sharply. The non-linear element instantly changes to a very low resistance state, and the surge current (of the order of kiloamperes) is diverted (bypassed) to earth instead of travelling to the equipment.
  • After the surge: As soon as the surge is discharged and the voltage falls back to normal, the element automatically returns to its high-resistance state, cutting off the follow (power-frequency) current. The arrester thus resets itself and is ready for the next surge.

In short: high resistance at normal voltage, low resistance at overvoltage, automatic self-restoring — this is the basic operational principle.

2. Need in the electrical safety system

  1. Protection of equipment insulation: Lightning surges and switching surges can raise line voltage to many times the rated value and puncture the insulation of transformers, generators, circuit breakers, insulators and cables. The arrester limits the surge voltage to a safe level (the protective level) below the insulation withstand strength of the equipment.
  2. Limiting overvoltage to a safe value: It clamps the peak surge voltage appearing across the terminal of the protected equipment, so that the basic insulation level (BIL) of the equipment is not exceeded.
  3. Prevention of fire, flashover and failure: By diverting surge energy to earth, it prevents flashover, arcing, insulation breakdown, fire and costly outage of the power system.
  4. Coordination with other protective devices: It forms the first line of defence in insulation co-ordination, working together with earthing systems and protective relays to ensure the safety of both personnel and apparatus.
  5. Economy and reliability: Since it is self-restoring and requires no maintenance after each operation (unlike a fuse or spark gap), it gives continuous, reliable and economical protection to the whole installation.

Thus the surge arrester is an essential safety device that protects costly electrical equipment from transient overvoltages by providing a low-resistance path to earth during surges and restoring itself immediately afterwards.

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  • 2081 · 2nd · 5 marks

What is power line carrier communication? How does it work? And, describe its importance.

Answer

Power Line Carrier Communication (PLCC)

Power Line Carrier Communication is a method of transmitting speech, telemetry and protection signals over the existing high-voltage transmission lines themselves, using the conductors as the communication channel. No separate communication cable is laid; the power line carries both the power frequency current (50 Hz) and the high-frequency communication signal at the same time.

Working principle

  1. Signal injection — At the sending end, the audio/telemetry signal is modulated onto a high-frequency carrier (typically 40 kHz to 500 kHz) by a transmitter. This HF signal is coupled to the power line through a coupling capacitor and a line trap (wave trap).
  2. Coupling capacitor — It blocks the 50 Hz power frequency from entering the communication equipment while passing the HF carrier signal to the line. It also serves as a potential divider for the CVT (capacitor voltage transformer).
  3. Line trap (wave trap) — A parallel LC tuned circuit connected in series with the line at each end. It offers very high impedance to the carrier frequency and very low impedance to 50 Hz, so the HF signal is confined to the protected line section and does not leak into the substation bus or adjacent lines.
  4. Transmission — The HF signal travels along the transmission line (in a phase-to-phase or phase-to-ground mode) to the receiving end, where an identical coupling capacitor and line trap tap it off.
  5. Reception — The received signal is filtered, amplified and demodulated to recover the original speech or data. The same line can carry several channels at different carrier frequencies (frequency-division multiplexing) without interfering with power transmission.

Importance

  1. Economical — Uses the existing transmission line, so no separate communication cable or repeater stations are needed; capital and maintenance cost is very low.
  2. Reliable and independent — The communication path is owned by the utility itself and does not depend on external agencies such as telecom operators or mobile networks.
  3. Protection signalling — Provides the carrier channel for distance/differential protection schemes (carrier tripping, transfer trip, inter-tripping) so that faults are cleared quickly at both ends of the line.
  4. Operational control — Enables speech communication between substations and load dispatch centres for switching, load management and coordination.
  5. Telemetry and SCADA — Carries data on voltage, current, power flow, breaker status and energy metering for remote monitoring and control.
  6. Coverage — Works over long distances in remote hilly areas where other communication media are unavailable or unreliable.
  7. Continuous availability — Remains available even during system disturbances and faults, when communication is needed most.

Limitations (brief) — Signal attenuation is high on long lines; noise and corona discharge interfere; coupling equipment is costly; and the channel is affected when the line itself is taken out of service.

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  • 2081 · 2nd · 10 marks

Draw schematic diagram of diesel power plant and show all the components in diagram with brief explanation of major components.

Answer

Diesel Power Plant — Schematic Diagram and Components

Schematic Diagram

                    ┌──────────────────────────────────────────────┐
                    │            DIESEL POWER PLANT                │
                    └──────────────────────────────────────────────┘

   ┌────────────┐      ┌──────────────────┐      ┌─────────────────┐
   │  Fuel Tank │─────▶│ Fuel Injection   │─────▶│    DIESEL       │
   │  (Storage) │      │ Pump & Injector  │      │    ENGINE       │
   └────────────┘      └──────────────────┘      │  (Cylinder +    │
                                                 │   Piston +      │
   ┌────────────┐      ┌──────────────────┐      │   Crankshaft)   │
   │ Air Filter │─────▶│  Air Intake      │─────▶│                 │
   │            │      │  Manifold        │      └────────┬────────┘
   └────────────┘      └──────────────────┘               │
                                                          │ Shaft
   ┌────────────┐      ┌──────────────────┐               ▼
   │  Cooling   │◀────▶│  Water Jacket /  │      ┌─────────────────┐
   │  Tower /   │      │  Radiator        │      │   COUPLING /    │
   │  Radiator  │      └──────────────────┘      │   FLYWHEEL      │
   └────────────┘                                 └────────┬────────┘
                                                           │
   ┌────────────┐      ┌──────────────────┐                ▼
   │  Lubricat- │─────▶│  Lubricating Oil │      ┌─────────────────┐
   │  ing Oil   │      │  Pump            │      │  ALTERNATOR /   │
   │  Tank      │      └──────────────────┘      │  GENERATOR      │
   └────────────┘                                 └────────┬────────┘
                                                           │
   ┌────────────┐      ┌──────────────────┐                ▼
   │  Starting  │─────▶│  Starting Motor  │      ┌─────────────────┐
   │  Battery   │      │  (Cranking)      │      │  SWITCHGEAR &   │
   └────────────┘      └──────────────────┘      │  TRANSFORMER    │
                                                 └────────┬────────┘
                                                          │
   ┌────────────┐      ┌──────────────────┐               ▼
   │  Exhaust   │◀─────│  Exhaust Valve / │      ┌─────────────────┐
   │  Silencer  │      │  Manifold        │      │  POWER SUPPLY   │
   │  & Chimney │      └──────────────────┘      │  TO CONSUMERS   │
   └────────────┘                                 └─────────────────┘

Major Components and Their Functions

1. Diesel Engine (Prime Mover) It is the heart of the plant. It is a compression-ignition (CI) engine in which air is compressed to a high pressure and temperature, and diesel fuel is injected into the hot compressed air, causing spontaneous ignition. The reciprocating motion of the piston is converted into rotary motion of the crankshaft. It may be two-stroke or four-stroke, and is usually multicylinder for higher capacity.

2. Fuel Supply System Consists of a fuel storage tank, fuel filter, fuel transfer pump, fuel injection pump and injectors. Its function is to store diesel, remove impurities, and supply the correct quantity of fuel at the correct time and pressure to each cylinder for combustion.

3. Air Intake System Comprises an air filter (cleaner), air intake manifold and, in larger plants, a supercharger/turbocharger. It supplies clean, dust-free air to the cylinders in the required quantity. Turbocharging increases the mass of air admitted, thereby increasing power output.

4. Exhaust System Includes exhaust valves, exhaust manifold, silencer (muffler) and chimney. It carries the burnt gases away from the engine, reduces noise, and discharges the gases to the atmosphere at a safe height.

5. Cooling System Consists of a water jacket around the cylinder, a radiator or cooling tower, a water pump and a fan. It removes the excess heat produced by combustion and friction so that the engine metal does not overheat and lose strength or seize. Cooling may be air-cooled or water-cooled.

6. Lubricating System Comprises a lubricating oil tank, oil pump, oil filter and piping. It supplies oil to bearings, cylinder walls and moving parts to reduce friction and wear, absorb heat, and carry away metal particles. It greatly increases the life of the engine.

7. Starting System Consists of a starting battery and a starting (cranking) motor, or a compressed-air starting arrangement. It rotates the crankshaft initially at the required speed so that the engine can start on its own power.

8. Governor It is a speed-regulating device fitted on the engine. It automatically controls the fuel supply according to the load so that the engine speed (and hence the generated frequency) remains constant even when the load varies.

9. Coupling and Flywheel The flywheel stores energy during the power stroke and releases it during other strokes, giving uniform torque. The coupling connects the engine shaft to the alternator shaft and transmits mechanical power.

10. Alternator (Generator) It is coupled to the engine shaft and converts the mechanical energy of the engine into electrical energy (usually 3-phase, 400 V or 11 kV).

11. Switchgear and Transformer Switchgear consists of circuit breakers, isolators, relays, meters and protective devices. It controls, protects and distributes the generated power. The transformer steps up the voltage for transmission or steps it down for local distribution.

12. Cooling Water and Auxiliary Pumps Auxiliary pumps circulate cooling water and lubricating oil, and supply fuel, ensuring continuous and reliable operation of the plant.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 10 marks

Explain the torque speed characteristics of 3-phase induction motor. Explain why squirrel cage rotor induction motor is called maintenance free motor.

Answer

Torque–Speed Characteristics of a 3-Phase Induction Motor

The torque developed by a 3-phase induction motor is given by:

T=K⋅s⋅R2⋅E22R22+(sX2)2T = \frac{K \cdot s \cdot R_2 \cdot E_2^2}{R_2^2 + (sX_2)^2}

where ss = slip, R2R_2 = rotor resistance, X2X_2 = rotor standstill reactance, E2E_2 = rotor induced emf at standstill, KK = constant.

The curve of torque against speed (or slip) has three distinct regions:

1. Starting region (s = 1, at standstill): At start, slip is maximum (s=1s = 1). The starting torque is:

Tst=K⋅R2⋅E22R22+X22T_{st} = \frac{K \cdot R_2 \cdot E_2^2}{R_2^2 + X_2^2}

Since X2≫R2X_2 \gg R_2 in a normal motor, the starting torque is low (about 1.5 to 2 times full-load torque for squirrel cage motors).

2. Maximum (breakdown) torque region: Torque rises as the motor accelerates (slip falls) and reaches a maximum value called the breakdown torque or pull-out torque when R2=sX2R_2 = sX_2, i.e. at slip:

sm=R2X2s_m = \frac{R_2}{X_2}

Tmax=K⋅E222X2T_{max} = \frac{K \cdot E_2^2}{2X_2}

This value is typically 2 to 3 times the full-load torque. Beyond this point, further increase in speed (decrease in slip) causes torque to fall.

3. Normal operating region (s = 0.02 to 0.05): Here the slip is very small, so sX2≪R2sX_2 \ll R_2 and the torque equation reduces to:

T∝sR2T \propto \frac{s}{R_2}

Thus torque is almost directly proportional to slip. The motor runs stably on the steep, nearly linear part of the curve between no-load and full-load, where a small change in load causes only a small change in speed. This is why an induction motor is called a constant-speed motor.

Key points of the characteristic:

  • At s=0s = 0 (synchronous speed), torque is zero — the motor cannot run at synchronous speed.
  • Torque increases from zero at synchronous speed to a maximum at sms_m, then decreases.
  • The stable operating range is 0<s<sm0 < s < s_m; the region sm<s<1s_m < s < 1 is unstable.
  • If load torque exceeds TmaxT_{max}, the motor stalls (breaks down).

Why a Squirrel Cage Rotor Induction Motor is Called a Maintenance-Free Motor

  1. No brushes and no slip rings: The rotor bars are permanently short-circuited by end rings. There is no sliding electrical contact, hence no brush wear, no sparking, and no carbon dust.

  2. No commutator: Unlike DC machines, there is no commutator to be cleaned or replaced.

  3. Rugged and simple construction: The rotor is a solid mass of laminated core with cast aluminium or copper bars — no windings to insulate, no soldered joints to loosen, and no moving parts other than the shaft.

  4. No external rotor connections: Since the rotor circuit is closed internally, there are no external resistances or connections that need periodic attention.

  5. Low losses and low heating: Absence of brush contact resistance means less heat and less wear.

  6. Cheap and easy replacement: Because of its simple, robust design, the motor needs only occasional bearing lubrication and cleaning — hence it is practically maintenance-free compared to slip-ring (wound rotor) motors, which need regular brush and slip-ring maintenance.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 5 marks

What is the importance of Periodic Maintenance? For a minimum oil circuit breaker, what periodic maintenance needs to be carried out? Clarify.

Answer

Importance of Periodic Maintenance

Periodic maintenance means the scheduled, routine inspection, testing, cleaning and servicing of equipment carried out at fixed intervals (daily, monthly, quarterly, yearly) throughout its service life. Its importance:

  1. Ensures reliability of supply — faults are detected before they cause breakdown, so unplanned outages are avoided.
  2. Increases equipment life — wear, corrosion, insulation deterioration and loose connections are corrected in time, delaying costly replacement.
  3. Reduces repair and replacement cost — minor defects are rectified cheaply at the preventive stage instead of expensive breakdown repairs.
  4. Improves safety — prevents electrical accidents (flashover, explosion, fire) to operating personnel and the public.
  5. Maintains efficiency and performance — the equipment continues to operate at its rated capacity and accuracy.
  6. Reduces downtime and revenue loss — planned shutdowns can be scheduled at light-load hours, avoiding forced outages.
  7. Provides record and trend data — test results help in condition monitoring and in deciding the replacement/overhaul time of the equipment.

Periodic Maintenance of a Minimum Oil Circuit Breaker (MOCB)

A minimum oil circuit breaker uses a small quantity of oil only for arc quenching; the oil does not insulate the live parts from earth. Hence maintenance is more frequent and more critical than for a bulk oil breaker. The following periodic maintenance is carried out:

A. External / routine checks (monthly or as per schedule)

  1. Clean the insulator bushings and check for cracks, chipping, dust or oil leakage.
  2. Check oil level in the oil chamber/indicator and top up with the correct grade of transformer oil if low.
  3. Check tightness of all terminal connections and busbar joints; tighten if loose.
  4. Inspect the tank, gaskets and seals for oil leakage and rusting; paint if necessary.
  5. Check the condition of the operating mechanism (spring/motor charging), linkages and bearings; lubricate them.
  6. Verify the tripping and closing coil operations, anti-pumping relay and local/remote control.
  7. Check the auxiliary contacts, indication lamps and interlocks.

B. Yearly / periodic detailed maintenance (after a fixed number of operations or annually)

  1. Oil testing — take an oil sample and test dielectric strength (BDV), acidity, moisture content and colour. Filter or replace the oil if the dielectric strength falls below the specified value (generally below 30 kV for 2.5 mm gap).
  2. Contact inspection — open the chamber and examine the arcing contacts and main contacts for burning, pitting or erosion; dress or replace them if worn beyond limits. Measure contact resistance/travel.
  3. Arc-quenching chamber (baffle/arc chute) check — clean or replace the fibre/insulating baffles and nozzles if carbonised or eroded.
  4. Insulation resistance (IR) test — measure IR of bushings, contacts to earth and across open contacts with a megger; compare with previous readings.
  5. Dielectric/over-voltage test of the oil chamber and bushings where required.
  6. Mechanism overhaul — check closing and opening times, spring charging, damper and dashpot action, and adjust the tripping linkage.
  7. Oil replacement — after the specified number of short-circuit interruptions the oil becomes carbonised; drain, flush and refill with fresh oil.
  8. Painting, earthing and general cleaning of the breaker, and updating the maintenance record card.

Conclusion: Since a MOCB carries only a small oil volume, its oil and contacts deteriorate quickly with each fault interruption. Regular periodic maintenance as above is therefore essential to keep the breaker ready to clear faults reliably and safely.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 5 marks

How the ranges of ammeters and voltmeters can be extended? Explain.

Answer

Extension of ranges of ammeters and voltmeters

1. Ammeter range extension An ammeter has low resistance and is designed to carry only a small current (its full-scale deflection current IgI_g). To measure larger currents, a low resistance called a shunt is connected in parallel with the ammeter.

Let II = current to be measured, IgI_g = full-scale current of ammeter, GG = ammeter resistance, SS = shunt resistance.

Since the shunt and ammeter are in parallel, the voltage across both is the same:

IgG=(I−Ig)SI_g G = (I - I_g) S

S=IgGI−IgS = \frac{I_g G}{I - I_g}

Also, multiplying factor m=IIgm = \dfrac{I}{I_g}, so S=Gm−1S = \dfrac{G}{m-1}.

Thus most of the current bypasses through the shunt and only IgI_g flows through the ammeter. By choosing a suitable small shunt resistance, the ammeter can read much larger currents.

2. Voltmeter range extension A voltmeter has high resistance and gives full-scale deflection at a small voltage VgV_g. To measure higher voltages, a high resistance called a multiplier is connected in series with the voltmeter.

Let VV = voltage to be measured, VgV_g = full-scale voltage of voltmeter, RmR_m = multiplier resistance, GG = voltmeter resistance.

V=Ig(G+Rm)V = I_g (G + R_m)

Rm=VIg−GR_m = \frac{V}{I_g} - G

Multiplying factor m=VVgm = \dfrac{V}{V_g}, so Rm=(m−1)GR_m = (m-1)G.

The multiplier limits the current through the voltmeter to IgI_g even at high voltage. By choosing a suitable high series resistance, the voltmeter can read much larger voltages.

Conclusion: An ammeter range is extended by connecting a low-resistance shunt in parallel; a voltmeter range is extended by connecting a high-resistance multiplier in series.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 5 marks

Discuss on single phase AC voltage controller and list out its applications.

Answer

Single Phase AC Voltage Controller

A single-phase AC voltage controller is a power electronic converter that converts a fixed AC input voltage into a variable AC output voltage of the same frequency, without changing the frequency. It uses phase control of thyristors (SCRs) or TRIACs to control the RMS value of the voltage applied to the load.

Circuit Configuration

1. Phase-controlled with two SCRs (anti-parallel)

Two SCRs, T1 and T2, are connected in anti-parallel (back-to-back) across the supply. T1 conducts during the positive half-cycle and T2 during the negative half-cycle. Each SCR is fired with a delay angle α measured from the zero crossing of its respective half-cycle.

2. TRIAC-based controller

A single TRIAC replaces the two anti-parallel SCRs. It conducts in both half-cycles and is fired by a gate pulse in each half-cycle. This is the most common configuration for low-power applications.

Working Principle

  • For 0 < α < 90°, the output voltage increases as α decreases; at α = 0 the controller behaves like a closed switch and the full supply voltage appears across the load.
  • For α = 90°, each SCR conducts for only half of its half-cycle and, for a resistive load, the RMS output voltage falls to Vs/2≈0.707VsV_s/\sqrt{2} \approx 0.707V_s.
  • For α > 90°, the conduction angle (π−α)(\pi-\alpha) is less than 90°, so the load receives only a small part of each half-cycle and the RMS output voltage falls further.
  • For a resistive load, the output voltage waveform is a chopped sinusoid and the current is in phase with the voltage.
  • For an inductive (R–L) load, the current lags the voltage; the SCR continues to conduct after the voltage zero crossing until the current falls to zero. The extinction angle β depends on the load power factor, and the range of α for successful control is limited (α > φ, the load impedance angle).

RMS Output Voltage (Resistive Load)

For a firing angle α, the RMS output voltage is:

Vrms=Vs1π[(π−α)+sin⁡2α2]V_{rms} = V_s \sqrt{\frac{1}{\pi}\left[(\pi - \alpha) + \frac{\sin 2\alpha}{2}\right]}

where VsV_s is the RMS supply voltage. At α = 0, Vrms=VsV_{rms} = V_s; at α = 90°, Vrms=Vs/2V_{rms} = V_s/\sqrt{2}; as α increases towards 180°, VrmsV_{rms} decreases towards zero.

Applications

  1. Lighting control (dimmers) — smooth variation of incandescent lamp brightness.
  2. Fan and motor speed control — speed regulation of single-phase induction motors and universal motors.
  3. Domestic heating control — electric heaters, ovens, geysers and room heaters.
  4. Transformer tap changing — on-load voltage regulation of transformers.
  5. AC voltage stabilizers — automatic correction of supply voltage fluctuations.
  6. Industrial heating and welding — resistance welding and induction heating control.
  7. Soft starting of motors — limiting the starting current of AC motors.
  8. Speed control of small appliances — mixers, blenders and drills.

Advantages and Limitations

Advantages: simple circuit, low cost, high efficiency (no resistive dropping), and smooth stepless control.

Limitations: low input power factor at large firing angles, generation of harmonics and radio-frequency interference (RFI), and poor performance with inductive loads due to limited control range.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2081 · 2nd · 5 marks

What causes a low power factor? Suggest a method for correcting it.

Answer

Causes of low power factor

  1. Inductive loads — Induction motors, transformers, arc lamps, welding sets and induction furnaces draw magnetising (reactive) current, which lags the voltage and lowers the power factor. This is the main cause in industry.
  2. Lightly loaded motors — An induction motor running at a small fraction of its rated load has almost the same magnetising current as at full load, so the power factor falls sharply (often below 0.5 at no load).
  3. Transformer magnetising current — Transformers on light load draw reactive current for core magnetisation, lowering the overall plant power factor.
  4. Arc and discharge lamps — Discharge lamps (fluorescent, mercury vapour) with magnetic chokes operate at a low lagging power factor.
  5. Improperly selected or oversized equipment — Motors and transformers rated much above the actual demand always run lightly loaded.
  6. Single-phase and intermittent loads — Small motors, air conditioners and fluctuating loads cause poor and varying power factor.
  7. Harmonics — Non-linear loads (rectifiers, electronic ballasts, VFDs) distort the waveform and reduce the true (displacement) power factor.

Method of correction

The most common and economical method is to connect static capacitors in parallel with the inductive load. The capacitor draws a leading reactive current that partly cancels the lagging reactive current of the load, so the net reactive power and the phase angle both decrease and the power factor improves.

Capacitor required:

Qc=P (tan⁡ϕ1−tan⁡ϕ2) kVARQ_c = P\,(\tan\phi_1 - \tan\phi_2)\ \text{kVAR}

where PP = active power in kW, ϕ1\phi_1 = angle before correction and ϕ2\phi_2 = angle after correction.

Other methods: synchronous condensers (over-excited synchronous motors) for large, varying loads; phase advancers for slip-ring induction motors; and proper motor/transformer sizing so that machines are not run lightly loaded.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 5 marks

Explain the consideration for choice of voltage in a transmission line.

Answer

Considerations for Choice of Voltage in a Transmission Line (5 marks)

The transmission voltage is selected after weighing the following technical and economic factors:

1. Amount of power to be transmitted The voltage level must match the power requirement. The power transmitted is (P = \sqrt{3},V_L I_L \cos\phi), so for a given power, a higher voltage reduces the line current. High power over long distances therefore demands high voltage.

2. Length of the transmission line Line length decides the voltage level in practice:

  • Short lines (up to about 50 km): 11–33 kV
  • Medium lines (50–150 km): 66–132 kV
  • Long lines (above 150 km): 220 kV and above

3. Economy (cost of conductor vs cost of insulation) Raising the voltage reduces current, so a smaller conductor cross-section is needed and copper/aluminium cost falls. But the cost of insulators, towers, transformers and switchgear rises sharply with voltage. The voltage chosen is the one at which the total annual cost (capital + transmission loss) is minimum — the most economical voltage.

4. Voltage drop and regulation A higher voltage keeps the per-unit voltage drop and (I^2R) loss within permissible limits (usually within ±6% at the far end), giving better regulation and efficiency.

5. Power loss and efficiency Loss (= I^2R). Since current falls as voltage rises, transmission efficiency improves at higher voltage.

6. Standardisation Only standard voltages (11, 33, 66, 132, 220, 400 kV) are used so that transformers, insulators and protective equipment are interchangeable and spare parts are easily available.

7. Corona and radio interference Very high voltage causes corona loss, radio interference and audible noise. The voltage must be kept below the level at which corona becomes excessive for the chosen conductor size and spacing.

8. Right of way and tower height Higher voltage needs larger conductor spacing and taller towers, requiring a wider right of way, which may be difficult and costly in hilly or populated terrain.

9. Future expansion Some provision for future load growth is kept so that the line need not be rebuilt soon.

Conclusion: The voltage is chosen so that the line gives satisfactory voltage regulation and efficiency at the lowest overall cost, using standard voltage levels suited to the power and distance involved.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 10 marks

What is an infinite bus-bar? Explain and sketch the double bus-bar scheme used in substation. Also, indicate its advantages that justify the additional cost compared to single bus-bar scheme.

Answer

Infinite Bus-Bar

An infinite bus-bar is an idealised bus-bar whose voltage and frequency remain constant regardless of the amount of active and reactive power drawn from or supplied to it. Its characteristics are:

  • Voltage magnitude and phase angle are fixed (usually taken as reference, V∠0°V\angle 0°).
  • Frequency is constant.
  • Its short-circuit capacity (fault level) is infinite, so its equivalent impedance is zero.
  • It can supply or absorb any amount of power without change in its terminal conditions.

In practice, a large interconnected power system (or a very large generating station) is treated as an infinite bus-bar because its capacity is so large compared to any single machine connected to it. It is used as the reference bus in load-flow and stability studies.

Double Bus-Bar Scheme

In this scheme there are two main bus-bars (Main Bus-I and Main Bus-II) to which every incoming and outgoing circuit is connected through two isolators (bus selectors) and one circuit breaker. A bus coupler (circuit breaker with isolators) is provided between the two bus-bars.

 Bus-I  ━━━━┯━━━━━━━━━━┯━━━━━━━━━━┯━━━━━━━━━━
            │          │          │
          [I1]       [I1']      [Ic]
            │          │          │
            │          │        [CB-C]   (bus coupler)
            │          │          │
            │          │        [Ic']
            │          │          │
 Bus-II ━━━━┿━━━┯━━━━━━┿━━━┯━━━━━━┷━━━━━━━━━━
            │   │      │   │
            │ [I2]     │ [I2']
            │   │      │   │
            └─┬─┘      └─┬─┘
            [CB-1]     [CB-2]
              │          │
           Feeder-1   Feeder-2

Working: Each circuit has one CB, with one bus-selector isolator (I1, I1') to Bus-I and another (I2, I2') to Bus-II; the bus coupler (CB-C with isolators on both sides) joins Bus-I to Bus-II. Each circuit is run on either Bus-I or Bus-II by closing the corresponding bus selector isolator. The bus coupler normally remains closed so that both bus-bars are energised and the load is shared. For maintenance or fault on one bus-bar, all circuits are transferred to the other bus-bar by the selector isolators, and the faulty bus is isolated — supply is maintained throughout.

Advantages (justifying the extra cost over single bus-bar)

  1. Continuity of supply: One bus-bar can be taken out for maintenance or repair while the whole substation keeps working on the other bus-bar.
  2. Flexibility in operation: Any feeder or transformer can be connected to either bus-bar, so circuits can be grouped or transferred as required.
  3. Easy maintenance without shutdown: Bus-bar maintenance, extension or repair is done without interrupting the supply — impossible in a single bus-bar scheme.
  4. Fault isolation: A fault on one bus-bar trips only that bus; the healthy bus continues to supply through the bus coupler.
  5. Higher reliability: Duplication of the main bus reduces the risk of a total station shutdown, which is very costly for a power utility.
  6. Future expansion: New bays and additional feeders can be added conveniently without disturbing existing connections.

Because of these operational and reliability benefits, the additional cost of the second bus-bar, extra isolators and bus coupler is justified for important and large substations.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2081 · 2nd · 10 marks

What are different kinds of supports and conductor used in secondary distribution system? Mention when they would be suitable.

Answer

Supports and Conductors Used in Secondary Distribution System

1. Supports (Poles) Used in Secondary Distribution

(a) Wooden Poles

  • Naturally available, cheap, and easy to erect.
  • Suitable for rural and hilly areas with light loads, low voltage (400/230 V), and short spans (up to about 30–40 m).
  • Not suitable where fire risk or decay is high; life is short unless treated.

(b) Steel Tubular (Mild Steel) Poles

  • Strong, long life, neat appearance, and low maintenance.
  • Suitable for urban and semi-urban areas carrying secondary lines along streets, where span and load are moderate.
  • Commonly used as service poles and for street-light fittings.

(c) RCC (Reinforced Cement Concrete) Poles

  • Strong, durable, fire-proof, and require almost no maintenance; life is long (25–30 years).
  • Suitable for urban, semi-urban and rural secondary distribution, especially where wooden poles are scarce or termite-prone.
  • Most widely used now for LT (400/230 V) lines.

(d) Rail Poles / RS Joists

  • Old railway rails used as poles; strong and cheap if available second-hand.
  • Suitable for temporary or rural LT lines and where salvage material is available.

(e) Lattice Steel Poles / H-Frame Structures

  • Used where a very long span or heavy loading (many service connections, transformer pole) is required.
  • Suitable at distribution transformer locations and road crossings.

2. Conductors Used in Secondary Distribution

(a) Bare Conductors

ConductorWhere Suitable
All Aluminium Conductor (AAC)Rural and semi-urban LT lines; cheap, light, good conductivity; short spans
All Aluminium Alloy Conductor (AAAC)Areas needing higher tensile strength than AAC; longer spans, hilly terrain
Aluminium Conductor Steel Reinforced (ACSR)Long spans, river/road crossings, areas with high wind/ice loading
Copper (bare)Rarely used now due to high cost; only where high conductivity and corrosion resistance needed

(b) Insulated / Covered Conductors

ConductorWhere Suitable
PVC-insulated copper or aluminium wireService connections from pole to consumer premises; congested areas
Aerial Bunched Cable (ABC)Densely populated urban areas, narrow lanes, theft-prone and tree-congested localities; reduces clearance problems
Weather-proof (WP) wireService drops in coastal or corrosive atmosphere

3. General Suitability Criteria

  • Voltage level: Secondary distribution is at 400 V (3-phase, 4-wire) / 230 V (1-phase).
  • Span: Short spans (30–50 m) → AAC/AAAC on RCC or wooden poles.
  • Load density: High density (urban) → ABC or insulated conductors on steel/RCC poles; low density (rural) → AAC on wooden/RCC poles.
  • Terrain and climate: Hilly/windy → AAAC/ACSR; corrosive → insulated or alloy conductors.
  • Economy and safety: RCC poles with AAC/ABC give the best compromise of cost, safety, and life for modern LT distribution.

Answer checked by Claude Haiku · Oct 21, 2026

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Questions from Lok Sewa Sub-Engineer Civil Group Exam (Sub-Engineer (Civil Group) - Non-Gazetted Third Class (Technical) · 2nd · 2082), Lok Sewa Nepal Engineering Service Sub (Sub-Engineer (Civil Group) - Non-Gazetted Third Class (Technical) · 2nd · 2081), Industrial District Management Limited (Sub-Engineer (Level 5) · 1st · 2081) and IDM Exam Question for Sub-Engineer (Sub-Engineer (Level 5) · 2nd · 2081). Answers are written for this site; check them against your class notes.