Chapter 1
Basic Electrical Concepts and Circuit Theory
Nepal Electricity Authority past exam questions
Past questions and answers
48 questions set from this chapter. Most repeated first.
- 2082 · 2nd · 5 marks
Describe Ohm's law. Explain the role of resistance in electric circuit in brief. 2+3=5
Answer
Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, provided temperature remains constant.
Ohm's law gives the relation between voltage, current and resistance in a circuit. It states that at constant temperature, the current (I) flowing through a conductor is directly proportional to the potential difference (V) applied across its ends.
where R is the resistance of the conductor. The unit of resistance is the ohm (Ω); one ohm is the resistance of a conductor in which a current of 1 ampere flows when a potential difference of 1 volt is applied across it. A conductor that obeys this relation at constant temperature is called an ohmic conductor (e.g. metals), while devices such as diodes and filament lamps do not obey it and are non-ohmic.
Role of resistance in an electric circuit:
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Controls the current: For a given supply voltage, resistance fixes the value of current, I = V/R. A higher resistance means a smaller current. This is how resistors protect components from excessive current.
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Converts electrical energy into heat: Resistance opposes the flow of electrons, and the work done against this opposition appears as heat (H = I²Rt). This is useful in heaters, irons, fuses and filament lamps, but it also causes energy loss in transmission lines.
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Produces required voltage drop: When current flows through a resistor, a voltage drop IR is developed across it. This is used in voltage dividers, biasing circuits and in series circuits to distribute voltage among components.
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Limits and divides current in circuits: Resistors connected in series increase total resistance (R = R₁ + R₂ + …) and reduce current, while those in parallel provide separate paths so that each branch gets current according to its resistance. This allows different components to work at their rated current from the same supply.
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Determines the time constant of circuits: In RC circuits, resistance together with capacitance decides the charging and discharging time (τ = RC), which is used in timers, oscillators and delay circuits.
Thus resistance is an essential circuit parameter: it limits current, converts electrical energy to heat, creates useful voltage drops and, along with capacitance and inductance, controls the behaviour of the whole circuit.
Answer checked by Claude Sonnet · Oct 21, 2026
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- 2082 · 2nd · 5 marks
Why transistor is called current controlled device? Explain. 5
Answer
A transistor (BJT) is called a current controlled device because its output current is controlled by its input current, not by an input voltage.
Explanation:
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Input is a current. In a bipolar junction transistor, the controlling quantity at the input is the base current . The output collector current is directly governed by through the relation where (current gain) is a constant for a given transistor in the active region. So the output current is set by the input current.
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Small input current controls large output current. A very small change in base current produces a large change in collector current (since is typically 20–500). For example, if and changes by , then changes by . Thus the base current "controls" the collector current.
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Input junction is forward biased. The base–emitter junction is forward biased, so the base current flows easily and its magnitude determines how many carriers are injected into the base and hence collected at the collector.
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Contrast with a voltage controlled device. In a FET (e.g., JFET, MOSFET), the output drain current is controlled by the gate–source voltage , so a FET is a voltage controlled device. A BJT, by contrast, needs a definite base current to set its operating point.
Conclusion: Since the collector (output) current of a transistor is determined and varied by the base (input) current, the transistor is called a current controlled device.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2082 · 2nd · 5 marks
What are the differences between DC generator and DC motor? Write. 5
Answer
Differences between DC Generator and DC Motor
| S.N. | DC Generator | DC Motor |
|---|---|---|
| 1. | It converts mechanical energy into electrical energy. | It converts electrical energy into mechanical energy. |
| 2. | It works on the principle of electromagnetic induction (Faraday's law) — when a conductor moves in a magnetic field, an emf is induced in it. | It works on the principle that a current-carrying conductor placed in a magnetic field experiences a force (F = BIL sin θ). |
| 3. | Its input is mechanical energy (from a prime mover such as a turbine or engine) and output is DC electrical energy. | Its input is DC electrical energy and output is mechanical energy (rotation of the shaft). |
| 4. | The direction of the induced emf is given by Fleming's Right-Hand Rule. | The direction of rotation (force) is given by Fleming's Left-Hand Rule. |
| 5. | A split-ring commutator with brushes acts as a rectifier, converting the alternating emf of the armature into unidirectional (DC) output. | The same commutator acts as an inverter, reversing the armature current at the correct instant so that the torque always acts in one direction. |
| 6. | The induced emf in the armature opposes the applied driving force; the machine converts mechanical work against this opposition into electrical energy. | A back emf is induced in the armature which opposes the applied voltage; the machine converts electrical energy into mechanical work. |
| 7. | Its emf equation is Eg = (ΦZN/60) × (P/A) volts. | Its back emf equation is Eb = (ΦZN/60) × (P/A) volts. |
| 8. | The generated emf is greater than the terminal voltage (Eg = V + IaRa + brush drop, i.e. V = Eg − IaRa). | The back emf is less than the applied voltage (V = Eb + IaRa). |
| 9. | It is used in power stations, battery charging, welding sets and as an exciter for alternators. | It is used in electric traction, cranes, hoists, fans, pumps and machine tools. |
Note: The construction of both machines is essentially the same; the same machine can work as a generator or as a motor depending on whether mechanical energy or electrical energy is supplied to it.
Answer checked by Claude Sonnet · Oct 21, 2026
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- 2082 · 2nd · 5 marks
Why rotor of three phase induction motor always moves in the direction of rotating field? Describe. 5
Answer
Why the rotor of a three-phase induction motor always moves in the direction of the rotating field
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Production of rotating magnetic field: When a three-phase supply is given to the three-phase stator winding (displaced 120° in space), it produces a magnetic field of constant magnitude that rotates at synchronous speed
in a definite direction. -
Relative speed and induced emf: At starting the rotor is stationary, so the rotor conductors cut the rotating field at synchronous speed. By Faraday's law of electromagnetic induction, an emf is induced in the rotor bars.
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Rotor current: The rotor circuit is closed (squirrel cage or slip-ring short-circuited), so the induced emf drives a current through the rotor conductors.
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Force by Lenz's law: The current-carrying rotor conductors lie in the stator's magnetic field, so a force acts on each conductor (F = BIL). By Lenz's law, the induced current always opposes the cause producing it — the cause is the relative speed between the rotating field and the rotor. The rotor therefore tries to reduce this relative speed by moving in the same direction as the field.
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Result: The rotor accelerates in the direction of the rotating field. It can never reach synchronous speed, because at the relative speed becomes zero, no emf is induced, no rotor current and no torque exist. Hence the rotor always runs at a speed slightly less than synchronous speed (slip ), i.e. it always rotates in the direction of the rotating field.
Note: If the rotor were made to run above synchronous speed by an external prime mover, the relative speed reverses, rotor current and torque reverse, and the machine works as an induction generator — again showing that the rotor direction is decided by the relative motion with respect to the rotating field.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2082 · 2nd · 5 marks
Mention the components of a typical solar power system and their functions in converting sunlight into electricity. 2+3=5
Answer
Components of a typical solar power system and their functions
A. Components (2 marks)
- Solar panels (PV modules)
- Solar charge controller
- Battery bank
- Inverter
- Mounting structure and wiring (balance of system)
B. Functions in converting sunlight into electricity (3 marks)
1. Solar panels (PV modules): They are the heart of the system. Each panel is made of many solar cells (usually silicon). When sunlight falls on the cells, the photovoltaic effect produces DC electricity. The panels are connected in series and parallel to give the required voltage and current.
2. Solar charge controller: The DC power from the panels is fed to the charge controller. It regulates the voltage and current going into the battery, prevents overcharging and deep discharge, and thus protects the battery life. In MPPT type controllers, it also extracts maximum power from the panels.
3. Battery bank: It stores the DC electricity produced during sunny hours so that power is available at night and during cloudy periods. It supplies DC power to the inverter and to DC loads.
4. Inverter: It converts the DC electricity from the battery (or directly from the panels) into AC electricity at 230 V, 50 Hz, which is required to run common household and office appliances such as lights, fans, TVs and computers.
5. Mounting structure and wiring (balance of system): The mounting structure holds the panels at the proper tilt and orientation (facing south in Nepal) to receive maximum sunlight. Cables, connectors, junction boxes, fuses and meters carry the power safely from one component to the next and to the load.
Summary of energy flow: Sunlight → Solar panels (DC) → Charge controller → Battery (storage) → Inverter (DC to AC) → Load
Answer checked by Claude Haiku · Oct 21, 2026
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- 2082 · 2nd · 5 marks
Describe the role of diesel power plant in load-shedding. Also, write down its demerits. 3+2=5
Answer
Role of diesel power plant in load-shedding (3 marks)
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Emergency / standby supply: Diesel plants are quick-starting (within minutes) and are kept as standby units. When the demand exceeds the supply from hydro and other sources, they are switched on immediately to bridge the gap and reduce load-shedding hours.
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Peak-load shaving: During peak hours (evening), demand rises sharply. Diesel generating sets are run to meet this peak demand, so that the load need not be shed on a large scale.
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Supply to essential and remote areas: In areas not connected to the main grid, or where the transmission line has failed, diesel plants supply power to hospitals, industries, communication stations and rural load centres, so that load-shedding does not affect essential services.
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Support during dry season: In winter, when river discharge falls and hydropower generation drops, diesel plants compensate for the deficit and shorten load-shedding periods.
Demerits of diesel power plant (2 marks)
- High operating cost: Diesel fuel is expensive; the cost of generation per unit (Rs/kWh) is much higher than that of hydropower.
- Fuel dependence and transport problem: Fuel must be transported to the site, which is difficult and costly in hilly and remote areas; supply may be interrupted.
- Pollution and noise: It emits smoke, CO, CO₂ and SO₂ and produces high noise, causing environmental and health problems.
- High maintenance and short life: Engine parts wear out quickly, requiring frequent maintenance, lubrication and spare parts; plant life is short (about 10–15 years).
- Limited capacity: It cannot supply large blocks of power for long durations, so it can only reduce, not fully remove, load-shedding.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2082 · 2nd · 10 marks
Draw the layout of a hydroelectric power plant and briefly explain the working of its main components. 5+5=10
Answer
Layout of a Hydroelectric Power Plant
Reservoir (Dam / Intake)
|
+-------------------+
| Trash Rack |
+-------------------+
|
+-------------------+
| Forebay / Surge |
| Tank |
+-------------------+
|
+-------------------+
| Penstock (Pipe) | <-- carries water under pressure
+-------------------+
|
+-------------------+
| Turbine (in Power| <-- water strikes runner
| House) |
+-------------------+
|
+-------------------+
| Generator | <-- coupled to turbine shaft
+-------------------+
|
+-------------------+
| Transformer | <-- steps up voltage
+-------------------+
|
+-------------------+
| Switchyard / |
| Transmission Line|
+-------------------+
|
Tailrace (water returned to river)
Working of Main Components
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Reservoir / Intake: Water from the river is stored behind a dam or diverted at the intake. The intake structure controls the amount of water entering the plant.
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Trash Rack: A screen of steel bars placed at the intake that stops leaves, stones and debris from entering and damaging the turbine.
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Forebay / Surge Tank: The forebay stores water temporarily and regulates its flow into the penstock. A surge tank absorbs sudden pressure changes (water hammer) when the turbine load changes.
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Penstock: A closed pipe (steel or concrete) that carries water from the forebay to the turbine under high pressure. The height difference (head) between the intake and the turbine gives the water its potential energy.
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Turbine: The high-pressure water strikes the turbine runner (Pelton, Francis or Kaplan type, chosen by head and discharge). The kinetic energy of water rotates the runner, converting hydraulic energy into mechanical energy.
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Generator: The turbine shaft is coupled to the generator. As the rotor spins inside the stator, electromagnetic induction produces electrical energy (AC).
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Transformer: The generated voltage (e.g. 11 kV) is stepped up to a high voltage (e.g. 132 kV or 220 kV) for efficient long-distance transmission.
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Switchyard and Transmission Line: The switchyard contains circuit breakers, isolators and protective equipment. Power is fed to the transmission line and distributed to consumers.
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Tailrace: After striking the turbine, the used water is discharged through the tailrace back into the river downstream.
Summary of energy conversion: Potential energy of water → Kinetic energy (penstock) → Mechanical energy (turbine) → Electrical energy (generator) → transmitted to load centres.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2082 · 2nd · 10 marks
Explain the working and starting method of three phase induction motor. 5+5=10
Answer
Three-Phase Induction Motor: Working and Starting Method
1. Working Principle (5 marks)
A three-phase induction motor works on the principle of electromagnetic induction and Lenz's law.
Construction: It has two main parts:
- Stator: A stationary hollow cylindrical core (laminated silicon steel) carrying a three-phase winding placed in slots 120° apart. When fed from a three-phase supply, it produces a rotating magnetic field (RMF) of constant magnitude, rotating at synchronous speed:
where = supply frequency (Hz), = number of poles.
- Rotor: Either squirrel-cage (copper/aluminium bars shorted by end rings) or slip-ring (wound rotor with slip rings and brushes) type. It carries no external supply.
Working:
- The three-phase stator supply creates an RMF rotating at .
- This RMF cuts the stationary rotor conductors, inducing an emf in them (Faraday's law). Since the rotor circuit is closed, currents flow in the rotor conductors.
- These rotor currents set up their own flux, which interacts with the stator RMF and produces a torque (Lenz's law: the rotor tries to reduce the relative speed).
- The rotor therefore rotates in the same direction as the RMF, but always at a speed less than . If , relative speed is zero, no emf, no current, no torque — hence the motor can never run at synchronous speed. This is why it is called an asynchronous motor.
Slip:
Rotor frequency: . At start ; at full load .
Torque: , maximum when .
2. Starting Method (5 marks)
At starting, , so rotor emf and rotor current are very large (5–8 times full-load current), while starting torque is low. Hence a starter is needed to limit the starting current. Common methods:
(a) Direct-on-Line (DOL) Starter — For small motors (up to about 5 HP). Full voltage is applied through a contactor with overload and no-volt protection. Simple and cheap, but starting current is very high.
(b) Star-Delta Starter — For squirrel-cage motors designed to run in delta. At start the winding is connected in star, so phase voltage becomes of line voltage and starting current is reduced to one-third. After the motor picks up speed, a changeover switch reconnects the winding in delta. Starting torque also falls to one-third.
(c) Auto-transformer Starter — A tapped auto-transformer applies reduced voltage () at start. Starting current and torque are reduced by . Used for large squirrel-cage motors; costly but gives several voltage taps.
(d) Rotor Resistance Starter (slip-ring motors) — External resistance is added in the rotor circuit through slip rings. This reduces starting current and simultaneously increases starting torque (since torque is maximum when ). Resistance is cut out in steps as speed rises. Best method for high starting torque requirements (hoists, mills, crushers).
(e) Stator Resistance / Reactance Starter — A resistance or reactor in series with the stator reduces applied voltage and hence current; used for small medium motors.
Summary: The three-phase induction motor is self-starting, robust and cheap; its speed is always slightly below synchronous speed, and the choice of starter depends on motor size and required starting torque.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2082 · 2nd · 5 marks
What is corona? What are the factors which affect corona? Write. 2+3=5
Answer
Corona is the phenomenon of partial ionization of the air surrounding a high-voltage conductor, accompanied by a faint violet glow, a hissing sound, and the smell of ozone. It occurs when the potential gradient at the conductor surface exceeds the dielectric strength of air (about 30 kV/cm peak, i.e. 21.1 kV/cm rms at NTP), causing the air in the immediate vicinity of the conductor to break down and conduct, while the rest of the gap remains insulating. Corona is therefore a localized breakdown, not a complete flashover, and it causes power loss and radio interference.
Factors affecting corona:
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Applied voltage / potential gradient — Corona starts only when the surface voltage gradient exceeds the disruptive value of air; the higher the voltage, the more intense the corona.
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Conductor diameter (size) — A larger conductor has a lower surface voltage gradient for the same voltage, so corona loss decreases. Corona is severe on small-diameter conductors.
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Conductor spacing (spacing between conductors) — Increasing the spacing reduces the electrostatic stress at the conductor surface and hence reduces corona, though the effect becomes small beyond a certain distance.
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Atmospheric conditions — Corona increases in humid, rainy, foggy, or stormy weather because moisture and dust lower the dielectric strength of air; it is minimum in dry, clear weather. Air density also matters — at high altitude (low density) corona increases.
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Surface condition of the conductor — Rough, dirty, or irregular surfaces (scratches, dust, water droplets) create local field concentrations and increase corona; smooth, clean, polished conductors reduce it.
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Line configuration — Bundle conductors, larger conductor diameter, and higher conductor height above ground reduce the surface gradient and hence corona loss.
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- 2082 · 2nd · 5 marks
How does earth fault relay work? Explain with necessary circuit diagram in brief. 5
Answer
Principle: An earth fault relay operates on the residual (zero-sequence) current that flows when a phase conductor touches earth. In a healthy balanced system the phasor sum of the three line currents is zero:
When a phase-to-earth fault occurs, part of the current returns through earth, so the sum is no longer zero. This residual current is what the relay senses.
Working (core-balance CT method):
- The three phase conductors (or the cable) pass together through the window of a ring-type core-balance current transformer (CBCT).
- Under normal load or a balanced fault, the fluxes of the three currents cancel, so the CBCT secondary carries no current and the relay stays idle.
- Under an earth fault, the residual current produces a net flux in the core and a secondary current flows through the operating coil of the relay.
- When this current exceeds the relay pick-up setting, the relay closes its contacts.
- The contacts energise the trip coil of the circuit breaker, which opens and isolates the faulty part.
Circuit diagram (schematic):
R ----+----------------+---- to load
Y ----+--[ CBCT ]------+----
B ----+----(o)---------+----
|
secondary winding
|
[ Earth fault relay coil ] --- relay contact --- [ Trip coil of CB ] --- DC supply
An alternative is the residual connection, where the secondaries of the three line CTs are connected in parallel and the relay is placed in the common return; it carries current only when the phase currents are unbalanced by an earth fault.
Earth fault relays are set with a low pick-up (often 10 to 20 percent of full-load current), so they detect small fault currents that overcurrent relays would miss.
Answer checked by Claude Sonnet · Oct 21, 2026
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- 2082 · 2nd · 5 marks
What are the uses of relays and isolators in electric system? Describe. 5
Answer
Uses of Relays and Isolators in an Electric System
1. Relays
A relay is an automatic protective device that senses abnormal conditions in the power system and gives a command to the circuit breaker to isolate the faulty section. Its main uses are:
- Fault detection and protection: It continuously monitors current, voltage, frequency and other quantities. When these exceed preset values (overcurrent, earth fault, overvoltage, undervoltage, reverse power, etc.), the relay operates and trips the circuit breaker, protecting generators, transformers, transmission lines and motors from damage.
- Selectivity and discrimination: By proper time and current grading, relays isolate only the faulty portion of the network while the healthy sections continue to supply power, thus maintaining continuity of supply.
- Equipment safety: It prevents overheating, insulation failure and mechanical damage in costly equipment such as transformers and generators.
- System stability: By clearing faults quickly (high-speed relays), it prevents loss of synchronism and cascade tripping of the grid.
- Indication, alarm and automation: Auxiliary relays give visual/audible indication of the type of fault, and relays are also used in control, interlocking and automatic switching schemes.
2. Isolators (Disconnecting Switches)
An isolator is an off-load switching device used to isolate a circuit or equipment from the live supply for maintenance and repair. Its main uses are:
- Isolation for maintenance: It disconnects the line, transformer or breaker from the live bus so that repair and inspection work can be carried out safely.
- Safety of personnel: By providing a visible air gap, it gives a clear indication that the circuit is dead, ensuring safety of working staff.
- Transfer of load/connections: It is used to transfer a line or transformer from one bus to another and to change over connections in a substation, but only when the circuit is de-energized.
- Switching of small charging currents: Some isolators can break small currents such as no-load transformer current and line charging current.
- Interlocking: It is interlocked with the circuit breaker so that it can be operated only after the breaker is opened, since an isolator has no arc-quenching arrangement and cannot break load or fault current.
Conclusion: Relays provide automatic protection by sensing faults and tripping breakers, while isolators provide safe, off-load isolation of equipment for maintenance. Both are essential for the safe, reliable and uninterrupted operation of the electric power system.
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- 2082 · 2nd · 5 marks
Highlight the significance of load factor and diversity factor. 5
Answer
Load factor is the ratio of the average load to the maximum (peak) demand over a given period:
Diversity factor is the ratio of the sum of individual maximum demands of consumers (or groups) to the maximum demand of the whole system:
Significance of load factor:
- Plant capacity and cost: A high load factor means the plant's installed capacity is utilised more fully, so the fixed cost (capital cost) is spread over more units of energy generated — the cost per unit (Rs/kWh) falls.
- Economy of operation: High load factor reduces the maximum demand on the plant, so smaller generating capacity and smaller transmission and distribution equipment are needed for the same energy supplied.
- Tariff design: Load factor is used to fix two-part and block tariffs; consumers with a better load factor get a lower average rate.
- Plant selection and reserve: It indicates how much reserve capacity is required and helps in deciding base-load versus peak-load plants.
- System planning: It guides the sizing of generators, transformers and feeders, and helps forecast future demand.
Significance of diversity factor:
- Reduced installed capacity: Since all consumers do not reach their peak at the same time, the system peak is less than the sum of individual peaks. A diversity factor greater than 1 allows smaller generating and distribution capacity, saving capital cost.
- Economical design of feeders, transformers and switchgear: Equipment is rated on the diversified (coincident) maximum demand, not on the arithmetic sum of consumer demands.
- Lower cost per unit: Reduced capital investment lowers fixed charges, so the cost of supply per kWh decreases.
- Load forecasting and expansion planning: It is essential for estimating the simultaneous maximum demand of a locality or feeder while planning new substations and lines.
- Better utilisation: It shows how effectively the connected load is used and guides load-management and demand-side measures.
In short: Load factor measures how steadily the load is used over time and mainly affects the cost of energy (running and fixed cost per unit), while diversity factor measures the non-coincidence of individual peaks and mainly affects the required installed capacity and capital cost of the system. Both are key indices for economical design and tariff fixing in a power utility like NEA.
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- 2082 · 2nd · 5 marks
How electrical safety is maintained during the power system operation and maintenance? Describe. 5
Answer
Electrical safety during power system operation and maintenance
Electrical safety in power system operation and maintenance is maintained through a combination of technical, procedural and administrative measures:
-
Isolation and switching (Lock-Out/Tag-Out) — Before any work, the concerned line/equipment is switched off from all possible sources of supply, and the isolators/breakers are locked and tagged so that no one can re-energise the circuit. The key or tag is kept personally by the person carrying out the work.
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Earthing (grounding) of the isolated section — After isolation, portable earths (grounding sets) are applied on all phases on both sides of the working zone. This discharges trapped/induced charge and protects the worker if the line is accidentally charged.
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Testing/proving dead — The line or equipment is tested with a voltage detector or neon tester to confirm that it is actually dead before earthing and before touching any conductor.
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Use of safety equipment and PPE — Insulated gloves, insulated tools, rubber mats, safety belts/harnesses (for pole and tower work), helmets, safety shoes and line-testers are mandatorily used. Rubber mats are laid in switchyards and control rooms.
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Clearance and safety distances — Minimum statutory clearances are maintained between live conductors and the ground, buildings and other lines. Approach to live parts is allowed only up to the prescribed safety distance; beyond that, the supply must be shut off.
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Permit-to-work (PTW) system — Work is started only after written permission from the authorised control/operating engineer, and the permit is surrendered only after the work is completed, tools removed and the section is ready for re-energisation. The "permit to work" and "permit to operate" registers are strictly maintained.
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Warning signs, barriers and guarding — Danger boards, "Men at work" boards, reflective tapes, barricades and fencing are placed around the work area and at the entry points of substations to prevent unauthorised entry.
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Trained and authorised personnel — Only competent, trained and authorised persons are allowed to operate switchgear or work on live/isolated lines. Regular safety training, tool-box talks and refresher courses are conducted.
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Protective devices and interlocks — Relays, fuses, MCBs, earth-leakage/earth-fault protection, surge arresters and mechanical/electrical interlocks in switchgear prevent unsafe operation and isolate faulty sections automatically.
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Maintenance practices — Preventive and predictive maintenance (thermography, oil testing, contact resistance measurement, cleaning of insulators) is carried out to keep equipment in safe condition. Work on live lines is done only by the "live-line" method with proper insulated tools and by trained personnel.
-
Emergency preparedness — First-aid training, artificial respiration (mouth-to-mouth/CPR), fire extinguishers (CO₂/dry powder for electrical fires), emergency contact numbers and rescue equipment are kept ready, and accidents are investigated to prevent recurrence.
Thus, safety in power system operation and maintenance is ensured by strict adherence to the isolation–earthing–testing–permit sequence, use of proper PPE and tools, authorised working, and continuous training and supervision.
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- 2082 · 2nd · 5 marks
What are the functions of lightening arrestor and overhead earth wire? Explain. 5
Answer
Functions of Lightning Arrester and Overhead Earth Wire
1. Lightning (Lightening) Arrester
A lightning arrester is a protective device installed at substations and at the terminals of equipment (transformers, generators, circuit breakers) to protect them from over-voltage surges caused by lightning or switching.
Functions:
- Diverts surge current to earth: When a high-voltage surge (e.g. from lightning) appears on the line, the arrester becomes conducting and provides a low-resistance path, discharging the surge current safely to the ground instead of allowing it to enter the equipment.
- Limits the voltage: It clamps the voltage at its terminals to a predetermined safe value (spark-over or discharge voltage), so the insulation of the protected equipment is not stressed beyond its withstand level.
- Reseals after discharge: After the surge passes, the arrester quickly restores its insulating property (resealing) and stops the flow of normal power-frequency current to earth, so the system continues to operate normally.
- Protects against both lightning and switching over-voltages: It safeguards transformer windings, bushings, cables and other apparatus from impulse voltages.
2. Overhead Earth Wire (Shield Wire)
It is a bare conductor run at the top of transmission line towers, above the phase conductors, and connected to the tower structure and to earth.
Functions:
- Shields phase conductors from direct lightning strokes: It intercepts the lightning stroke before it reaches the phase conductors, so the stroke current flows through the earth wire and tower to ground.
- Reduces voltage stress on line insulation: By diverting the stroke, it prevents excessive voltage build-up across the insulator strings, thereby reducing the chance of flashover and line outage.
- Provides a low-resistance earth path: Being bonded to the tower footing earth, it gives an easy path for discharge current, limiting tower top potential rise.
- Reduces induced over-voltages: It also reduces the electrostatically induced voltages on phase conductors due to nearby strokes.
- Improves reliability of supply: By minimising lightning-caused faults and outages, it increases the continuity of power supply.
Conclusion: The lightning arrester protects station equipment by discharging and limiting surges, while the overhead earth wire protects the transmission line itself by intercepting lightning strokes above the conductors. Together they form the main lightning protection of a power system.
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- 2082 · 2nd · 10 marks
What is power factor and how low power factor increase the cost of energy? How can it be corrected? Explain with example. 6+4=10
Answer
Power Factor and Its Effect on Energy Cost
1. Power Factor (6 marks)
Definition: Power factor (PF) is the ratio of active (real) power to apparent power in an AC circuit.
where is the phase angle between voltage and current.
Physical meaning: In AC circuits, current has two components:
- Active (in-phase) component — does useful work (kW)
- Reactive (quadrature) component — only builds magnetic fields in motors, transformers, etc. (kVAR)
Apparent power is the vector sum:
Types:
- Unity PF (1.0): Resistive loads (heaters, incandescent lamps)
- Lagging PF: Inductive loads (motors, transformers, induction furnaces) — most common in industry
- Leading PF: Capacitive loads (capacitor banks, synchronous condensers)
Typical values: Induction motors 0.7–0.9, transformers 0.8–0.9, resistive heaters ~1.0.
2. How Low Power Factor Increases the Cost of Energy (4 marks)
a) Higher current for the same useful power At low PF, current increases. For the same kW delivered, a load at 0.7 PF draws more current than at 0.95 PF.
b) Increased I²R losses Line losses . Since current rises inversely with PF, losses rise as . At PF 0.7, losses are about times those at PF 0.95 — wasted energy paid for by the consumer/utility.
c) Larger equipment and cables Cables, transformers, switchgear and generators must be rated in kVA. Low PF means more kVA capacity is installed for the same kW output — higher capital cost.
d) Voltage drop and poor regulation Higher current causes greater voltage drop, degrading supply quality and requiring additional regulation equipment.
e) Penalty / tariff charges NEA applies a power factor penalty (and sometimes a rebate for high PF). Consumers with PF below the prescribed limit (usually 0.9) pay extra demand charges, directly raising their electricity bill.
Example: A 100 kW load at 0.7 PF draws kVA. At 0.95 PF it draws only kVA — a saving of 38 kVA of capacity and the associated losses and penalty.
3. Correction of Low Power Factor (with example)
Methods:
- Static capacitors — most common; supply reactive power locally (kVAR), reducing line current. Cheap, low loss, but susceptible to harmonics and switching transients.
- Synchronous condensers — synchronous motor running over-excited; gives stepless, flexible leading kVAR; costly and needs maintenance.
- Phase advancers — connected to induction motor shaft; improves motor PF directly.
- Synchronous motors — run at leading PF to compensate plant reactive demand.
- Avoid oversizing motors / use proper-rated equipment — reduces inherent low-PF operation.
Example (capacitor sizing):
Given:
- Load kW
- Existing PF (lagging)
- Desired PF (lagging)
Step 1 — Find reactive powers:
Step 2 — Capacitor rating required:
Result: Installing a capacitor bank of about 69 kVAR raises the PF from 0.7 to 0.95, reducing the line current from 143 kVA to 105 kVA (a 26.6% reduction) and cutting I²R losses by about 46%.
Conclusion: Power factor correction reduces current, losses, kVA demand and penalty charges, thereby lowering the overall cost of energy and improving voltage regulation.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2082 · 2nd · 10 marks
Why SCADA system is implemented? Explain its various components. 5+5=10
Answer
SCADA stands for Supervisory Control and Data Acquisition. It is a computer-based control system used to monitor and control geographically spread-out electrical networks from a central location.
Why SCADA is Implemented (5 marks)
- Real-time monitoring — Gives the control centre a live picture of the power system (voltage, current, power flow, frequency, breaker status) without sending staff to each substation.
- Remote control — Operators can open/close circuit breakers, isolators and change transformer taps from the control centre, giving fast response during faults.
- Faster fault detection and restoration — Faults and abnormal conditions are detected immediately through alarms, reducing outage time and improving supply reliability.
- Efficient load management — Helps in load dispatch, load shedding and balancing generation with demand by providing accurate system data.
- Reduced manpower and cost — One control centre can supervise many substations, so unmanned or lightly manned substations become possible.
- Data logging and reporting — Historical data is stored for analysis, planning, energy accounting and post-fault investigation.
- Improved safety — Operators control equipment remotely, keeping personnel away from live high-voltage equipment.
Components of SCADA (5 marks)
1. Field Devices / Sensors and Actuators
- Sensors/Transducers — CTs, PTs, transducers measure voltage, current, power, frequency, temperature, tap position.
- Actuators — Circuit breakers, isolators, tap changers that carry out control commands.
2. Remote Terminal Unit (RTU) / PLC
- Located at the substation. Collects analog and digital data from field devices, converts them into digital form, and sends them to the master station. It also receives commands from the master and operates the actuators.
3. Communication System
- The link between RTUs and the master station. Media used are fibre optics, microwave, radio, leased telephone lines or power line carrier. Protocols such as IEC 60870-5-101/104, DNP3 and Modbus are used.
4. Master Station / SCADA Server (Control Centre)
- The central computer system that processes data from all RTUs, maintains the real-time database, and issues control commands. It includes the SCADA server, historical/archive server and application servers.
5. Human Machine Interface (HMI)
- Operator workstations with monitors and mimic diagrams (single-line diagrams) showing the live status of the network. Alarms, trends and control buttons are provided here.
6. Data Processing and Application Software
- Performs alarm processing, event logging, trending, load flow, state estimation and report generation.
6. Uninterruptible Power Supply (UPS)
- Provides backup power to the master station and RTUs so that monitoring continues even during supply failure.
Block diagram (concept):
In NEA, SCADA is used at the Load Dispatch Centre to supervise grid substations and hydro/thermal stations for reliable and economic operation of the power system.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
Discuss the role of components such as dams, forebay, penstock, turbines and generators in the operation of a hydroelectric power plant.
Answer
A hydroelectric power plant converts the potential energy of stored water into electrical energy. Each component performs a specific function in this chain:
1. Dam (or weir/barrage)
- Built across a river to raise the water level and create a storage reservoir.
- Creates the head (height difference) that gives water its potential energy.
- Regulates the flow of water supplied to the plant, storing excess water during the monsoon and releasing it in the dry season.
2. Forebay
- A small storage pond located at the end of the headrace canal, just before the penstock.
- Acts as a balancing reservoir: it absorbs sudden fluctuations in water demand when the turbine load changes, and settles sediment before water enters the penstock.
- Ensures a steady, regulated supply of water to the penstock.
3. Penstock
- A closed pipe (steel, concrete or HDPE) that carries water under pressure from the forebay down to the turbine.
- Its main role is to convert the potential energy of the water into kinetic energy as the water gains velocity while descending.
- Must be strong enough to withstand high internal water pressure; valves and gates control the flow.
4. Turbine
- The prime mover of the plant. Water strikes the runner blades and makes it rotate, converting the kinetic energy of water into mechanical energy.
- Types: Pelton (high head), Francis (medium head), Kaplan/propeller (low head).
- The turbine shaft is coupled to the generator, so its speed determines the frequency of the generated power.
5. Generator
- Coupled to the turbine shaft. When the rotor (field) rotates inside the stator, it converts mechanical energy into electrical energy by electromagnetic induction.
- Produces 3-phase AC at the required voltage (e.g., 6.6 kV or 11 kV), which is then stepped up by transformers for transmission.
Summary of the energy chain:
Thus, the dam and forebay store and regulate water, the penstock transports it under pressure, the turbine extracts mechanical power, and the generator finally delivers usable electricity to the grid.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
State the formula that relates electrical power, voltage and resistance. Explain how it is related to the heating effect of a circuit.
Answer
Formula relating power, voltage and resistance
From Ohm's law, (V = IR), and the power relation (P = VI), we get:
where (P) = power in watts (W), (V) = voltage in volts (V), (I) = current in amperes (A), and (R) = resistance in ohms (Ω).
Relation to the heating effect
When current (I) flows through a resistance (R), the electrical energy supplied is converted into heat. The heat produced in time (t) is given by Joule's law of heating:
This shows the heating effect depends on the square of the current and directly on the resistance. So:
- If the current is doubled, the heat produced becomes four times ((I^2) factor).
- For a fixed voltage, (P = V^2/R), so a lower resistance draws more current and produces more heat — this is why a heater coil of low resistance glows hot while a high-resistance filament in a small bulb gives less heat.
- For a fixed current, (P = I^2R), so a higher resistance in the path produces more heat — this is the principle of a fuse wire, which melts due to (I^2R) heating when excess current flows.
In short, the power equation (P = I^2R = V^2/R) directly gives the rate at which electrical energy is dissipated as heat in a resistive circuit.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
What is an ammeter? Briefly explain the role of shunt resistor in ammeter.
Answer
Ammeter: An ammeter is an electrical measuring instrument used to measure the magnitude of electric current flowing through a circuit. It is always connected in series with the circuit element whose current is to be measured. An ideal ammeter has zero (very low) resistance so that it does not appreciably change the circuit current.
Role of shunt resistor in an ammeter:
A basic moving-coil (galvanometer-type) movement can carry only a very small current (its full-scale deflection current ). To convert it into an ammeter of larger range, a low resistance called a shunt is connected in parallel with the galvanometer coil. Its roles are:
- Bypasses excess current: Most of the circuit current flows through the shunt, and only the safe fraction passes through the coil. Thus the instrument is protected from burning out.
- Extends the range: The range of the ammeter is increased from to .
- Reduces the effective resistance: The parallel combination gives a very low equivalent resistance, so the ammeter drops almost no voltage and disturbs the circuit only slightly.
Relation: If is the shunt resistance and the galvanometer resistance, then for full-scale deflection,
where is the desired maximum current to be measured.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
Describe how energy transfer occurs between the primary and secondary winding through mutual induction.
Answer
Mutual induction and energy transfer between windings (5 marks)
1. Setup. A transformer has two coils wound on a common laminated soft-iron core: the primary (connected to the AC supply) and the secondary (connected to the load). The core links the two windings magnetically.
2. Varying flux in the primary. When alternating voltage is applied to the primary, an alternating current flows in it. This current produces an alternating magnetic flux in the core. Since the supply is AC, the flux continuously changes in magnitude and direction.
3. Flux linkage with the secondary. The soft-iron core carries almost the whole flux and links it with every turn of the secondary winding. Thus the secondary is linked by the same alternating flux produced by the primary — this linking of one coil by the flux of another is mutual induction.
4. EMF induced in the secondary. By Faraday's law of electromagnetic induction, an alternating emf is induced in the secondary:
Similarly, a back emf is induced in the primary:
Dividing, the emf ratio equals the turns ratio:
5. Energy transfer. The electrical energy in the primary is first converted into magnetic field energy in the core, and this energy is transferred to the secondary through the mutual flux. The induced emf drives current through the secondary circuit, so electrical energy is delivered to the load. For an ideal transformer (no losses):
i.e. power is transferred without change in magnitude — only voltage and current are stepped up or stepped down according to the turns ratio. In practice a small part of the energy is lost as copper loss (), eddy current and hysteresis losses in the core, and leakage flux.
Conclusion. Energy passes from primary to secondary purely through the alternating magnetic flux in the common core — there is no electrical connection between the windings — and this transfer takes place by mutual induction.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
What is rectification in power electronics and how does it convert AC to DC? Explain.
Answer
Rectification is the process of converting alternating current (AC), which periodically reverses direction, into direct current (DC), which flows in only one direction. It is the first stage in most power electronic systems, since electronic circuits (ICs, microcontrollers, DC motors, battery chargers) require a steady DC supply while the mains supply available is AC (in Nepal, 230 V, 50 Hz).
How it converts AC to DC:
-
Rectifying element — A device that conducts current in one direction only is used. In modern practice this is a power diode (or thyristor/IGBT in controlled rectifiers). A diode conducts when forward biased (anode more positive than cathode) and blocks when reverse biased.
-
Half-wave rectification — A single diode in series with the load. During the positive half-cycle of the AC input the diode is forward biased and current flows through the load; during the negative half-cycle it is reverse biased and no current flows. The output is a pulsating DC that contains only one half of each input cycle. It is simple but has low efficiency, high ripple, and a DC output of only about 31.8% of the peak value.
-
Full-wave rectification — Both half-cycles are used:
- Centre-tap type: Two diodes with a centre-tapped transformer; each diode conducts on alternate half-cycles.
- Bridge type: Four diodes in a bridge arrangement; diodes D1–D2 conduct on the positive half-cycle and D3–D4 on the negative half-cycle, so current through the load is always in the same direction. The output frequency is twice the input frequency and the DC value is about 63.6% of the peak.
-
Smoothing (filtering) — The pulsating DC is passed through a capacitor (or inductor) filter, which charges during the peaks and discharges during the valleys, giving a nearly steady DC with small ripple. A voltage regulator may follow for a constant output.
Thus rectification uses the unidirectional conduction property of diodes to block the negative (or alternate) half of the AC wave and allow only one direction of current, producing DC. The basic relation for a full-wave bridge rectifier is:
where (V_m) is the peak value of the input AC voltage.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
Explain the principle of operation of an energy meter for measuring electrical energy consumption.
Answer
An energy meter (kWh meter) measures the total electrical energy consumed over a period, i.e. it integrates power with respect to time.
Basic principle: Energy = Power × Time. The meter continuously multiplies the instantaneous voltage and current (giving power) and integrates this product over time to register energy in kilowatt-hours (kWh).
Construction and working of an induction (electromechanical) type energy meter:
-
Driving system: Consists of a series (current) coil carrying the load current and a shunt (voltage/pressure) coil connected across the supply. These produce alternating fluxes that are out of phase.
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Rotating element: An aluminium disc is placed in the air gap between the coils. The two alternating fluxes induce eddy currents in the disc. The interaction between these eddy currents and the fluxes produces a driving torque proportional to the product of voltage, current and the power factor — i.e. proportional to the true power (P = VI cos φ).
-
Braking system: A permanent magnet provides a braking torque proportional to the speed of the disc. The disc attains a steady speed when driving torque equals braking torque, so the speed is directly proportional to power.
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Registering (counting) mechanism: Since speed ∝ power, the number of revolutions ∝ energy (power × time). A gear train drives the counter/register, which records energy directly in kWh.
Key relation:
Electronic (static) energy meter: Voltage and current are sensed by potential and current transformers, converted to digital form, and a microcontroller computes P = VI cos φ and integrates it over time to display energy in kWh. It is more accurate and has no moving parts.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 10 marks
Discuss methods for extending the measuring range of an ammeter and voltmeter. Explain the principles behind range extension and the techniques used to achieve wider measurement capabilities.
Answer
Methods for Extending the Measuring Range of an Ammeter and Voltmeter
1. Principle Behind Range Extension
A basic moving-coil (PMMC) instrument is a current-sensitive device. Its full-scale deflection (FSD) occurs at a small current ( I_g ) (typically 10 µA–50 mA) and it has a small internal resistance ( R_g ). Since the same coil must serve as both ammeter and voltmeter, the measuring range is limited. Range extension is therefore based on diverting or limiting the current through the meter using resistive networks:
- Ammeter → connect a low resistance in parallel (shunt) so that most of the current bypasses the meter.
- Voltmeter → connect a high resistance in series (multiplier) so that most of the voltage drops across it.
2. Range Extension of an Ammeter (Shunt Method)
Principle
A low resistance ( R_{sh} ) is connected in parallel with the meter. The line current ( I ) divides: ( I_g ) flows through the meter (causing FSD) and ( I_{sh} ) flows through the shunt.
Derivation
Let the required range be ( I ) (i.e., multiplying factor ( m = I/I_g )).
[ I_{sh} = I - I_g ]
Since ( R_{sh} ) and ( R_g ) are in parallel:
[ I_g R_g = I_{sh} R_{sh} = (I - I_g) R_{sh} ]
[ \boxed{R_{sh} = \frac{I_g R_g}{I - I_g} = \frac{R_g}{m - 1}} ]
Key Points
- Shunt resistance is very small compared to ( R_g ).
- Shunt is made of manganin (low temperature coefficient) to avoid error due to heating.
- Ayrton shunt (universal shunt) is used for multi-range ammeters — a chain of series resistors with tapping points, so the meter is never left without a shunt (prevents damage).
- Ammeter resistance becomes ( R_A = R_g \parallel R_{sh} ), which is very low → negligible voltage drop.
Example
A meter with ( I_g = 1\text{ mA} ), ( R_g = 100\ \Omega ) is to read up to ( 1\text{ A} ).
[ m = \frac{1}{0.001} = 1000,\quad R_{sh} = \frac{100}{1000-1} \approx 0.1\ \Omega ]
3. Range Extension of a Voltmeter (Multiplier Method)
Principle
A high resistance ( R_{se} ) (multiplier) is connected in series with the meter. The voltage ( V ) to be measured drops partly across ( R_{se} ) and partly across the meter (which must give FSD).
Derivation
Let the required range be ( V ) (multiplying factor ( m = V/V_g ), where ( V_g = I_g R_g )).
[ V = I_g (R_{se} + R_g) ]
[ \boxed{R_{se} = \frac{V}{I_g} - R_g = R_g(m - 1)} ]
Key Points
- Multiplier resistance is very high compared to ( R_g ).
- Made of manganin or constantan wire wound non-inductively to reduce frequency and temperature errors.
- Multi-range voltmeters use a series chain of tapped resistors; the multiplier value changes with the range.
- Voltmeter resistance becomes ( R_V = R_g + R_{se} ), which is very high → draws negligible current from the circuit.
Example
A meter with ( I_g = 1\text{ mA} ), ( R_g = 100\ \Omega ) is to read up to ( 100\text{ V} ).
[ R_{se} = \frac{100}{0.001} - 100 = 99{,}900\ \Omega \approx 100\ \text{k}\Omega ]
4. Comparison of the Two Methods
| Feature | Ammeter (Shunt) | Voltmeter (Multiplier) |
|---|---|---|
| Element connected | Low resistance in parallel | High resistance in series |
| Purpose | Bypass excess current | Drop excess voltage |
| Value | ( R_{sh} = R_g/(m-1) ) | ( R_{se} = R_g(m-1) ) |
| Resulting resistance | Very low | Very high |
| Material | Manganin | Manganin / constantan |
5. Wider Measurement Capabilities — Additional Techniques
- Multi-range instruments — A rotary switch selects different shunts (ammeter) or multipliers (voltmeter), giving several ranges in one meter.
- Current transformers (CT) — For AC ammeters measuring very large currents (e.g., 100 A, 1000 A); the meter reads a stepped-down current (typically 5 A or 1 A).
- Potential transformers (PT) — For AC voltmeters measuring very high voltages (e.g., 11 kV, 33 kV); the meter reads a stepped-down voltage (typically 110 V).
- Rectifier-type instruments — Combine a rectifier with a PMMC meter to extend AC measurement ranges.
- Electronic (digital) meters — Use amplifiers and shunts/multipliers internally to cover very wide ranges with high input impedance.
6. Conclusion
The measuring range of an ammeter is extended by connecting a low shunt resistance in parallel, and that of a voltmeter by connecting a high multiplier resistance in series. Both methods exploit the current-dividing and voltage-dividing properties of resistors respectively. For very large currents and voltages (especially AC), instrument transformers (CT and PT) are used, while multi-range switching and electronic meters provide flexible, wide-range measurement capability.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 10 marks
Describe base load power plants and peak load power plants, explaining their respective roles in meeting electricity demand. Discuss the characteristics and operational considerations of each type of power plant.
Answer
Base Load Power Plants and Peak Load Power Plants
1. Base Load Power Plants
Definition: A base load power plant is one that operates continuously, 24 hours a day, throughout the year, supplying the constant minimum demand (base load) of the power system.
Role in meeting electricity demand: It supplies the "base" portion of the load curve — the demand that never falls below a certain level. It runs at or near full capacity and forms the backbone of the grid.
Characteristics:
- High load factor (typically 80–90%) and high capacity factor.
- Constant, steady output — generation is not varied frequently.
- Low running (fuel) cost per kWh, but high capital cost.
- Long startup and shutdown times (hours to days).
- Large unit sizes (economies of scale).
- Long life and low maintenance cost per unit of energy.
Examples: Large hydroelectric plants (run-of-river with pondage and storage), nuclear plants, large coal-fired thermal plants, geothermal plants.
Operational considerations:
- Must be sited where fuel/water is reliably available.
- Scheduled maintenance is planned during low-demand seasons (e.g., monsoon for thermal plants in Nepal, dry season for hydro).
- Sudden tripping of a base load unit causes a large generation deficit, so spinning reserve must be maintained.
- In Nepal, run-of-river hydro plants act as base load during monsoon but their capacity falls sharply in the dry season, so load management and import from India become necessary.
2. Peak Load Power Plants
Definition: A peak load power plant is one that operates only during the periods of maximum demand (peak load) — usually a few hours a day — to meet the additional demand above the base load.
Role in meeting electricity demand: It supplies the "peak" portion of the load curve. It is started quickly when demand rises (typically morning and evening peaks) and shut down when demand falls.
Characteristics:
- Low load factor (10–30%) and low capacity factor.
- Output can be varied rapidly to follow the load curve.
- High running (fuel) cost per kWh, but low capital cost.
- Short startup time (minutes) — quick-start capability.
- Smaller unit sizes.
- Often remotely controlled and automated.
Examples: Gas turbine plants, diesel generator sets, pumped-storage hydro plants, small storage hydro plants, diesel/dual-fuel plants.
Operational considerations:
- Must be able to start and synchronise within minutes to meet the evening peak.
- High fuel cost means they are used only for limited hours; running them for long hours is uneconomical.
- Pumped-storage plants also serve as spinning reserve and provide frequency regulation.
- In Nepal, the evening peak (around 6–9 pm) is met partly by diesel/gas plants (e.g., Hetauda, Duhabi) and by importing power from India.
- Environmental and fuel-storage considerations apply for diesel/gas units.
3. Comparison
| Basis | Base Load Plant | Peak Load Plant |
|---|---|---|
| Operating period | Continuous (24 hours) | Only during peak hours |
| Load factor | High (80–90%) | Low (10–30%) |
| Capital cost | High | Low |
| Running cost per kWh | Low | High |
| Startup time | Long | Short |
| Typical examples | Large hydro, nuclear, coal | Gas turbine, diesel, pumped storage |
4. Conclusion
A reliable power system needs both types working together: base load plants supply the steady minimum demand cheaply and efficiently, while peak load plants (and pumped-storage units) cover the short, sharp rises in demand. In Nepal, the seasonal variation of river flow makes this combination especially important — storage hydro and pumped storage can act as both base and peak plants, while thermal (diesel/gas) plants and cross-border imports are used to bridge the dry-season and evening peak deficits. Proper planning of the generation mix, based on the load duration curve, keeps the cost of supply low and the system reliable.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
Why are relays used in electrical power system? Explain.
Answer
Relays in an electrical power system — purpose and explanation
A relay is an automatic sensing device that detects an abnormal condition in a power system and, through its contacts, trips the circuit breaker to isolate the faulty section. It is used because the following requirements cannot be met by the circuit breaker alone:
1. Fault detection. The circuit breaker is only a switching device; it cannot decide when to operate. The relay continuously measures current, voltage, frequency or impedance and identifies the abnormal condition (overcurrent, earth fault, over/under voltage, reverse power, etc.).
2. Discrimination (selectivity). The relay is set with a definite time or current grading so that only the relay nearest the fault operates. This isolates the smallest possible part of the network and keeps the healthy portion in supply.
3. Fast clearance. A relay senses the fault in a few milliseconds and gives the trip signal, so the breaker clears the fault before it can damage transformers, generators, cables or busbars, and before it can disturb the stability of the grid.
4. Protection of equipment and personnel. By disconnecting the faulty element quickly, relays limit thermal and mechanical stress on equipment, reduce fire and shock hazards, and protect costly plant such as transformers and generators.
5. Economy and reliability of supply. Relays allow the system to be operated close to its thermal limits without risk, reduce outage time and outage area, and make unattended substations possible through automatic tripping and alarm.
6. Backup protection. If the primary relay or its breaker fails, the backup relay at the upstream end operates after a time delay, so no fault is left uncleared.
Conclusion: A relay is the "brain" of the protection scheme — it senses the abnormality, decides whether it lies within its zone, and commands the circuit breaker to trip. Without relays, the power system could neither be operated safely nor supply reliable, uninterrupted power.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
Write the difference between circuit breaker and isolator. Also, mention their uses.
Answer
Difference between Circuit Breaker and Isolator
| S.N. | Circuit Breaker | Isolator |
|---|---|---|
| 1 | It is an on-load switching device; it can be opened or closed while current is flowing. | It is an off-load switching device; it must be operated only when no current flows. |
| 2 | It has an arc-quenching arrangement (arc chute, oil, air blast, SF₆, vacuum) to extinguish the arc formed on opening. | It has no arc-quenching arrangement; it is a simple knife/rotary contact. |
| 3 | It can interrupt fault currents (short-circuit currents) by itself. | It cannot interrupt fault currents; opening it on load causes a dangerous arc. |
| 4 | It is operated automatically by relays, and also manually. | It is operated manually only (by hand or motor), never automatically. |
| 5 | Its contacts are enclosed in a chamber (tank, pole, or vacuum bottle). | Its contacts are open to air and clearly visible. |
| 6 | It is costly and needs regular maintenance of the arc chamber. | It is cheap and needs very little maintenance. |
| 7 | It is used to make and break the circuit. | It is used only to isolate the circuit, i.e., to give a visible air gap. |
Uses
Circuit breaker:
- To switch a line or equipment on and off under normal load.
- To automatically clear faults (overload, short circuit, earth fault) and protect generators, transformers, feeders and motors.
- Used as the main protective switch in substations, switchyards and distribution boards.
Isolator:
- To isolate a circuit, equipment or section of a line from the live supply for maintenance and repair work, giving a visible break.
- To transfer load from one busbar to another (bus-bar selector) in a substation.
- To extend or sectionalise a line, and to break a small charging/no-load current only.
Note: In practice the isolator is always opened after the circuit breaker and closed before it, so that the isolator is never called upon to break or make load current.
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- 2081 · 2nd · 5 marks
Write the importance of earth wire and surge arrester in overhead transmission line.
Answer
Importance of earth wire and surge arrester in overhead transmission line
1. Earth wire (overhead ground wire)
- It is a bare conductor run at the top of the transmission towers, above the phase conductors, and is connected to earth at every tower through the tower body and footing earth.
- It shields the phase conductors from direct lightning strokes. A lightning stroke hits the earth wire instead of the phase conductors, so the line insulation is saved from flashover.
- The stroke current is conducted safely to the ground through the tower and its earthing, so no dangerous voltage appears on the phase conductors.
- It reduces the voltage surge that reaches the line, and hence lowers the risk of insulator flashover and line outage.
- It also provides a low-resistance path for induced charges and improves the shielding angle (usually about 30°) so that the phase conductors lie in the protected zone.
- It gives mechanical support and reduces the effect of wind and vibration to some extent, and provides a path for fault current during earth faults.
2. Surge arrester (lightning arrester)
- It is connected between each phase conductor and earth, usually near the substation end and at important towers.
- It provides a low-resistance path to earth for the travelling surge, so the surge is discharged to ground before it reaches the transformer or other equipment.
- It limits the surge voltage to a value below the basic impulse insulation level (BIL) of the equipment, so the insulation of transformers, circuit breakers and insulators is protected.
- After the surge is discharged, it restores its high resistance (non-linear characteristic of the valve-type/ZnO arrester), so normal power frequency current is not allowed to flow to earth.
- It protects the line and substation equipment from switching surges as well as lightning surges.
- It is cheaper and simpler than other protective devices and needs little maintenance.
Conclusion: The earth wire gives shielding against direct lightning strokes along the whole line, while the surge arrester discharges the remaining travelling surges to earth and clamps the voltage. Together they protect the insulators, transformers and other equipment, reduce outage time and increase the reliability of the power supply.
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- 2081 · 2nd · 5 marks
Describe power factor correction and the methods used for power factor correction.
Answer
Power factor correction is the process of improving the power factor of an electrical installation by adding devices that supply the reactive power (vars) required by inductive loads, so that the reactive power drawn from the supply is reduced. This brings the power factor closer to unity.
Why it is needed: Inductive loads such as induction motors, transformers, ballasts and arc furnaces draw lagging reactive power. This increases the line current for the same useful (active) power, causing:
- Higher (I^2R) losses in cables and transformers
- Larger voltage drop and poorer voltage regulation
- Reduced capacity of transformers, alternators and cables
- Penalty charges / poor tariff (kVA demand billing)
Methods of power factor correction:
-
Static capacitors (shunt capacitors) — Capacitors connected in parallel with the load supply leading reactive power, cancelling the lagging reactive power of the load. They are cheap, have low losses, need little maintenance, and can be switched in steps. Most widely used method in industries and substations.
Required capacitance per phase:
[ Q_C = P(\tan\phi_1 - \tan\phi_2), \qquad C = \frac{Q_C}{\omega V^2} ] -
Synchronous condensers — An over-excited synchronous motor running on no load behaves as a capacitor and supplies leading reactive power. It gives smooth, stepless control and can also be used for voltage regulation, but it is costly, has rotating parts and needs more maintenance. Used mainly in large transmission/substation applications.
-
Phase advancers — An AC exciter connected to the rotor circuit of a slip-ring induction motor injects emf into the rotor, improving the power factor of the motor itself. Suitable only for slip-ring motors; not common now.
-
Synchronous motors — Where a synchronous motor is already used as a drive, running it at leading power factor (over-excitation) improves the overall plant power factor at no extra cost.
-
Tap changing / automatic PF control panels — Automatic capacitor banks switched by a PF relay or microcontroller maintain the desired PF (usually 0.95–0.98 lagging) as the load varies.
Conclusion: Shunt capacitors are the most economical and commonly adopted method for power factor correction in NEA distribution systems and industries, while synchronous condensers are used where smooth control and voltage support are required.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
Explain the sectionalized bus bar arrangement in substations. Discuss how it enhances system reliability, fault isolation and maintenance flexibility.
Answer
Sectionalized Bus Bar Arrangement in Substations
1. What it is
A sectionalized (sectionalised) bus bar arrangement is a single bus bar that is divided into two or more independent sections by means of bus sectionalizing isolators or circuit breakers placed at intervals along the bus. Each section is fed by its own incoming source (transformer, generator or transmission line) and supplies its own group of outgoing feeders. The sections are normally operated coupled (sectionalizing breaker closed) so that the whole bus behaves as one node, but they can be split (breaker open) whenever required. A common form is the single bus bar with sectionalizing breaker, and the same principle is extended to double bus bar and one-and-a-half breaker schemes, where each main bus is also sectionalized.
Typical layout (single bus, two sections):
Source-1 Source-2
| |
[CB] [CB]
| |
======|=====[BS]=====|====== Bus Section 1 | Bus Section 2
| | | |
F1 F2 F3 F4 (outgoing feeders)
where BS = bus sectionalizing breaker/isolator.
2. Enhancement of system reliability
- Continuity of supply: If a fault occurs on one bus section, only that section is isolated; the healthy section continues to supply its feeders. Without sectionalizing, a single bus fault would trip the entire bus and black out all feeders.
- Source redundancy: Each section has an independent source. Loss of one source or one section does not deprive the other section of supply, so the station retains at least partial capability.
- Limitation of disturbance: The outage is confined to a part of the station instead of the whole substation, reducing the connected load lost and the severity of the system disturbance.
- N-1 contingency capability: The arrangement allows the system to survive the loss of any one transformer, line or bus section without complete loss of the station.
3. Fault isolation
- The bus is divided into zones, each protected by its own differential (or busbar) protection with the sectionalizing breaker as the zone boundary. A fault in Section 1 trips only the breakers of Section 1 and the sectionalizing breaker, leaving Section 2 undisturbed.
- The sectionalizing breaker acts as a barrier that prevents a fault from propagating to the healthy section, and it also limits the short-circuit current contributed by sources on the other side, reducing the fault level that breakers and equipment must withstand.
- Repair and restoration are faster because the faulty section can be de-energized, earthed and worked upon while the rest of the station remains energized.
4. Maintenance flexibility
- Any section, bus bar, isolator or connected breaker can be taken out for preventive maintenance, testing or repair while the other section continues in service.
- Load can be transferred from one section to the other before shutdown, so maintenance is carried out without interrupting consumer supply.
- In double bus bar sectionalized schemes, a feeder can be shifted to the reserve bus, giving still greater freedom for maintenance of main bus bars and sectionalizing equipment.
- It also permits phased extension of the substation: new bays and bus sections can be added without shutting down the existing sections.
5. Conclusion
Sectionalizing the bus bar divides the station into independent, switchable zones. This confines faults to a single zone, keeps the remaining sections in service, limits short-circuit levels, and allows maintenance without a station shutdown — thereby improving reliability, selectivity of protection and operational flexibility at modest additional cost.
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- 2081 · 2nd · 5 marks
Describe demand-side management and explain its significance in the energy sector.
Answer
Demand-Side Management (DSM) is the planning, implementation and monitoring of activities of a utility (here, the Nepal Electricity Authority) designed to influence customer use of electricity so as to produce desired changes in the utility's load shape — i.e. in the timing and magnitude of demand. It acts on the demand side (consumers) rather than on the supply side (generation, transmission), and is achieved through a mix of technologies, incentives and pricing.
Main techniques of DSM
- Load management — direct load control, time-of-use (ToU) tariffs, interruptible/curtailment contracts, and peak clipping, valley filling, load shifting, strategic conservation and strategic load growth.
- Energy efficiency (conservation) — efficient lamps (LED), efficient motors, pumps, fans, air conditioners, improved cooking and industrial processes, and loss-reduction at the consumer end.
- Demand response — voluntary reduction by consumers during system peak or emergencies, supported by smart meters and smart-grid signals.
- Awareness and incentives — consumer education, energy audits, subsidy/rebate on efficient appliances, and financing schemes.
Significance in the energy sector
- Reduces peak demand and hence the need for costly peaking plants and imports; NEA meets its evening peak largely by importing power from India, so peak reduction directly saves foreign exchange.
- Defers capital investment in new generation, transmission and distribution capacity — the cheapest "resource" is the negawatt saved.
- Improves system load factor and the utility's financial health by shifting energy use to off-peak hours (e.g. off-peak water heating, irrigation pumping, EV charging).
- Reduces energy imports and trade deficit, and improves energy security for a country with a seasonal supply–demand mismatch (surplus in monsoon, deficit in dry season).
- Environmental and social benefits — lower fossil-fuel generation, reduced CO₂ and local pollution, and lower bills for consumers.
- Supports national policy — the Nepal Electricity Regulatory Commission Act, 2074 and NEA's own DSM/energy-efficiency programmes, plus the White Paper on energy, water resources and irrigation (2075), which emphasise demand-side measures.
- Enables renewable integration — DSM and demand response help absorb variable solar/wind and manage the duck curve, complementing NEA's target of meeting demand through domestic hydro and renewables.
In short, DSM is significant because it meets growing demand at lower cost, with less investment, less import dependence and less environmental damage than building additional supply capacity.
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- 2081 · 2nd · 10 marks
A 3-phase 110km long transmission line is delivering 46MW of power at 0.8 power factor lagging at 132kV. The resistance and reactance per phase are 0.1 ohm/km and 0.3 ohm/km respectively. The shunt admittance is Siemens/km. Assuming the load to be balanced and neglecting the leakage current, calculate the sending end voltage, sending end power, sending end power factor, efficiency and regulation of the line.
Answer
Given: MW, lagging, kV (line), length 110 km. The shunt admittance is neglected, so the short-line model applies.
Receiving-end quantities (per phase)
Taking as reference, A.
Sending-end voltage
Line-to-line sending-end voltage 144.7 kV
Sending-end power
Sending-end power factor
leads by , and lags by , so the angle between and is .
Efficiency
Regulation
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- 2081 · 2nd · 10 marks
Write about safety precautions during maintenance of distribution line and importance of earthing of electrical system.
Answer
Safety precautions during maintenance of distribution lines
- Isolation and switching — The concerned feeder/section must be switched off from the controlling substation or switchgear, and the isolators opened so that the line is dead before any work begins.
- Permit-to-work (PTW) system — A written permit is issued by the authorised engineer to the maintenance crew, and the line is formally handed over only after the permit is signed. The permit is surrendered only after the crew reports clear.
- Earthing of the line — After isolation, portable earthing sets (earth rods with jumpers) are applied on all three phases on both sides of the working span, so that induced or accidental energisation is discharged to earth.
- Danger/notice boards and tags — "Men at work — do not switch on" boards and warning tags are fixed at the controlling switchgear and at both ends of the section.
- Testing before touching — The line is tested dead with a voltage detector or discharge rod before the crew touches the conductors.
- Use of PPE — Rubber gloves, insulated tools, safety belts, helmets, safety shoes and, where required, line hose and insulating mats must be used.
- Working at height — Ladders, poles and platforms must be sound; the lineman must use a safety belt with the strap secured to the pole, and never work alone.
- Clearance from live parts — Minimum statutory clearances from adjacent live lines must be maintained; if not possible, the adjacent line must also be isolated or guarded.
- Weather conditions — Work should be avoided during lightning, heavy rain or strong wind.
- Trained and authorised staff — Only trained, authorised and medically fit persons should carry out maintenance, under the supervision of a responsible person.
- First aid and rescue — First-aid box and artificial respiration (mouth-to-mouth) knowledge must be available; rescue should be done only after switching off the supply.
- Restoration procedure — Tools and earthing sets are removed, the crew is counted and reported clear, and only then is the supply restored by the authorised person.
Importance of earthing of an electrical system
- Protection of human life — Earthing provides a low-resistance path so that, on insulation failure, the equipment body is kept at earth potential. A person touching it does not receive a dangerous shock.
- Operation of protective devices — A low earth resistance allows a large fault current to flow, which quickly operates fuses, MCBs or relays and disconnects the faulty circuit.
- Protection of equipment — It limits the rise of potential and prevents damage to transformers, generators and machines from lightning surges and faults.
- Neutral point earthing — Earthing of the neutral fixes the system voltage with respect to earth, limits over-voltages during faults, and gives a return path for fault current in a 3-phase system.
- Lightning and surge protection — Earth electrodes and earth wires divert lightning strokes and surge currents safely to earth.
- Voltage stabilisation — It keeps the potential of the neutral and of non-current-carrying metal parts at or near zero, so that voltage does not float and damage appliances.
- Safety against static charge — It discharges static electricity from machines, tanks and structures, avoiding spark and fire hazards.
- Statutory requirement — Earthing is compulsory under the Electricity Rules and Nepal Electricity Authority standards; without proper earthing the system is unsafe and illegal to operate.
Conclusion — Maintenance of distribution lines must always follow the sequence isolate → earth → test dead → work → clear → restore, and a properly designed earthing system is the basic safeguard that makes the whole distribution network safe for both the public and the lineman.
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- 2081 · 2nd · 5 marks
State the properties of a series circuit and explain the difference between resistance connected in series and capacitance connected in series.
Answer
Properties of a series circuit
A series circuit is one in which components are connected end to end so that the same current flows through each of them. Its main properties are:
- Same current: The same current flows through every component, i.e.
- Voltage divides: The total supply voltage is the sum of the voltage drops across individual components, i.e.
- Single path: There is only one path for current, so if any component breaks, the whole circuit is broken and no current flows.
- Total resistance adds up: The equivalent resistance is the sum of individual resistances, , so it is greater than the largest resistance in the circuit.
- Common switches and fuses: One switch or fuse connected anywhere in the series path controls the entire circuit.
- Power adds up: Total power consumed equals the sum of powers of individual components,
Difference between resistance in series and capacitance in series
| Resistance in series | Capacitance in series |
|---|---|
| Equivalent resistance is the sum of individual resistances: | Equivalent capacitance is the reciprocal sum: |
| The equivalent resistance is greater than the largest resistance in the circuit. | The equivalent capacitance is smaller than the smallest capacitance in the circuit. |
| The same current flows through each resistor and voltage divides in proportion to resistance. | The same charge is stored on each capacitor and voltage divides in inverse proportion to capacitance. |
| Energy is dissipated as heat in each resistor. | Energy is stored in the electric field of each capacitor. |
| For two resistors: | For two capacitors: |
Thus, resistances in series behave like a single larger resistance, while capacitors in series behave like a single smaller capacitor.
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- 2081 · 2nd · 5 marks
Describe different types of electric power associated with alternative current system and explain the significance of term 'power factor'.
Answer
Types of Electric Power in an AC System
In an AC circuit the voltage and current are sinusoidal and generally out of phase by an angle φ. Three types of power are defined:
1. Active (Real) Power (P) The power actually consumed or converted into useful work (heat, light, mechanical energy) by the resistive part of the circuit. It is measured by a wattmeter and is the power for which the consumer is billed.
2. Reactive Power (Q) The power that oscillates between the source and the reactive elements (inductors and capacitors) without being consumed. It is required to build up magnetic fields in motors, transformers and to charge capacitors. It is measured by a VAr meter.
3. Apparent Power (S) The product of the rms voltage and rms current, i.e. the total power that appears to flow in the circuit. It is measured by a VA meter and determines the rating (size) of generators, transformers, cables and switchgear.
The three are related by the power triangle:
Significance of Power Factor
Power factor is defined as the cosine of the phase angle between voltage and current:
Its significance:
-
Utilisation of equipment capacity: For a given VA rating of a generator or transformer, the useful (active) power delivered is . A low power factor means the same machine supplies less useful power, so its capacity is poorly utilised.
-
Line losses and voltage drop: For the same active power, current . A low p.f. increases the current, hence increases losses in transmission and distribution lines and causes greater voltage drop and poorer voltage regulation.
-
Cost of supply: Larger currents demand thicker conductors, bigger transformers and larger switchgear, raising capital and operating cost. Hence utilities penalise consumers having low power factor (below about 0.85–0.9 lagging) through power-factor surcharge or tariff incentives.
-
Nature of load: Unity p.f. (resistive load) is ideal; lagging p.f. indicates inductive load (motors, transformers, induction furnaces) and leading p.f. indicates capacitive load. Power factor is improved by connecting capacitors or synchronous condensers in parallel with the load, which supply the reactive power locally and reduce the line current.
Thus power factor is a measure of the effectiveness with which the installed electrical capacity is used, and its improvement leads to reduced losses, better voltage regulation and lower cost of power supply.
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- 2081 · 2nd · 5 marks
Describe about the classification of hydroelectric plants based on availability of head, nature of load and quantity of water available.
Answer
Hydroelectric plants can be classified in several ways. The three common bases are given below.
1. Based on availability of head
| Type | Head | Turbine | Example |
|---|---|---|---|
| Low head | below about 30 m | Kaplan / propeller | run-of-river plants on large flows |
| Medium head | 30 to 100 m | Francis | Marsyangdi (about 100 m class) |
| High head | above 100 m | Pelton | Kulekhani I (about 550 m) |
Low head plants need large discharge, while high head plants work with small discharge.
2. Based on nature of load
- Base load plants: run continuously at near-constant output to meet the minimum system demand, e.g. run-of-river plants with steady flow.
- Peak load plants: run only during peak demand hours, e.g. storage and pumped storage plants.
- Combined plants: serve as base load normally and take peak load when required.
3. Based on quantity of water available
- Run-of-river without pondage: no storage; output varies with river flow (e.g. Chilime, Trishuli).
- Run-of-river with pondage: a small daily or weekly pond lets the plant meet hourly load changes.
- Storage (reservoir) plants: a large reservoir stores monsoon water to give firm power through the dry season (e.g. Kulekhani I).
- Pumped storage plants: water is pumped back to the upper reservoir during off-peak hours and used for generation at peak hours.
These classes overlap; for example a high head plant can be a storage plant that supplies peak load.
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- 2081 · 2nd · 5 marks
Explain different types of hydro turbines and ways of selecting the appropriate one.
Answer
Hydro turbines convert the energy of water into mechanical energy of a rotating shaft. They are classified by the way water acts on the runner.
1. Impulse turbines
Water is converted into a high-velocity jet at atmospheric pressure, and only kinetic energy is used.
- Pelton wheel: high head (above about 100 m) and low discharge; runner has split buckets and is not submerged. Suitable for hilly Nepal sites such as Kulekhani.
- Turgo and cross-flow (Banki): used for medium head and small flows, common in micro and small hydro.
2. Reaction turbines
Water enters under pressure and both pressure and kinetic energy are used. The runner is enclosed in a casing.
- Francis (mixed flow): medium head (about 10 to 100 m) and medium discharge.
- Propeller / Kaplan (axial flow): low head (below about 10 m) and large discharge. Kaplan blades are adjustable, giving good part-load efficiency.
- Bulb / tubular: very low head variant of Kaplan.
Selecting the appropriate turbine
- Find the net head and design discharge of the site, then power .
- Calculate specific speed
and choose the runner type from its range (Pelton lowest, Francis medium, Kaplan highest). 3. Check part-load efficiency, since flow in Nepali rivers varies sharply between monsoon and dry season. 4. Check cavitation (suction head / NPSH) for reaction turbines. 5. Consider silt content (Pelton tolerates sediment better than Francis), cost, number of units and ease of maintenance.
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- 2081 · 2nd · 5 marks
Describe transformer. Why same transformers are not suitable for the distribution and transmission system? Explain.
Answer
Transformer — A transformer is a static (stationary) electrical device that transfers electrical energy from one circuit to another by electromagnetic induction, without changing the frequency, at increased or decreased voltage. It works on the principle of mutual induction between two windings wound on a common laminated soft-iron core. It has no moving parts, so its efficiency is very high (above 95%). It consists of:
- Primary winding — connected to the supply (input).
- Secondary winding — connected to the load (output).
- Laminated core — provides a low-reluctance path for the mutual flux and reduces eddy-current loss.
If and are the primary and secondary turns and , the corresponding voltages, then
where is the transformation ratio. If it is a step-up transformer; if it is a step-down transformer.
Why the same transformer is not suitable for both distribution and transmission:
-
Voltage level (transformation ratio): Transmission uses very high voltage (132 kV, 220 kV, 400 kV) to reduce loss over long distances, while distribution works at 11 kV / 400 V / 230 V near consumers. A transformer is designed for one fixed turns ratio, so one transformer cannot serve both voltage levels.
-
Insulation requirement: Transmission transformers need heavy insulation (oil-impregnated paper, large clearances) to withstand impulse and switching overvoltages at high voltage. Distribution transformers need only light insulation. Using a distribution transformer on a transmission line would cause insulation breakdown.
-
Core and winding design: Transmission transformers are designed for high flux density and are operated near full load continuously, so they are made with better core material (CRGO) and larger cross-section. Distribution transformers are designed for maximum efficiency at about 50–75% load, since load varies widely and is not continuous.
-
Cooling system: Transmission transformers use forced oil and forced air cooling (ONAF/OFAF) because of large losses and size. Distribution transformers are usually self-cooled (ONAN) and pole-mounted or plinth-mounted.
-
Size, weight and cost: Transmission transformers are large, heavy and costly; distribution transformers are small, light and cheap so they can be installed on poles in towns and villages.
-
Connection and protection: Transmission transformers are connected in star–star or star–delta with tap changers and sophisticated protection (differential, Buchholz). Distribution transformers are usually delta–star with simple protection (fuse, lightning arrester).
Hence, because of different voltage levels, insulation, cooling, core design, loading pattern and cost, the same transformer cannot be used in both the transmission and distribution systems.
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- 2081 · 2nd · 5 marks
Why is DC system necessary in power stations and substations? Explain.
Answer
Why DC system is necessary in power stations and substations
A reliable DC (direct current) supply is an essential auxiliary system in every power station and substation. It is necessary for the following reasons:
-
Operation of protection and control equipment — Protective relays, trip coils of circuit breakers, and control circuits are DC-operated. DC gives a steady, independent supply so that faults are cleared instantly and selectively.
-
Availability during AC failure — When a major fault, blackout or station shutdown occurs, the AC auxiliary supply itself fails. The DC system (battery bank) continues to supply control, protection and indication circuits, so the station can still be operated and restored.
-
Circuit breaker tripping and closing — Spring-charge motors, closing coils and trip coils need a strong, instantaneous current. A battery bank can deliver this high short-time current, which a small AC supply cannot do reliably.
-
Emergency lighting and auxiliaries — DC emergency lamps, annunciators, alarms and communication equipment remain working during AC failure, allowing safe evacuation and operation.
-
Supply to SCADA, telemetry and signalling — Remote control, indication and communication systems in substations require an uninterrupted DC supply for correct data transmission.
-
Independence from the power system — Since the battery is charged from the station auxiliaries but does not depend on the main supply for its stored energy, the DC system gives a "fail-safe" source. Even if the whole station is dead, the battery can trip breakers and save equipment.
-
Starting of emergency equipment — DC is used for starting diesel generators (emergency supply), fire-fighting pumps and other critical drives.
Conclusion: The DC system acts as the "brain and nerve" supply of a power station or substation. It ensures that protection, control, indication and emergency services operate correctly at all times, especially when the AC supply fails, thereby maintaining the safety and reliability of the power system.
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- 2081 · 2nd · 10 marks
Explain the working principle of synchronous generator with a suitable figure. Mention the conditions of their parallel operation.
Answer
Working Principle of a Synchronous Generator
A synchronous generator (alternator) converts mechanical energy into electrical energy at a constant frequency that is locked to (synchronised with) its speed of rotation. Its name comes from the fact that the rotor runs at exactly synchronous speed, given by:
where f = supply frequency (Hz), P = number of poles, N_s = speed in rpm.
Construction (essential parts):
- Stator (armature): stationary part carrying a 3-phase winding in slots, usually on the inner periphery of the core.
- Rotor (field): rotating part carrying field poles excited by DC (field winding or permanent magnets). Two types — salient pole (projecting poles, low speed, hydro turbines) and cylindrical/non-salient (smooth cylindrical rotor, high speed, steam/gas turbines).
- Prime mover: turbine (water, steam or gas) that supplies mechanical input.
- Exciter: DC source feeding the rotor field winding.
Figure (schematic):
Prime mover (turbine)
|
v
+---------------------------+
| ROTOR (field poles) | <-- DC excitation
| N S N S ... |
+---------------------------+
| rotating magnetic field
v
+---------------------------+
| STATOR (3-phase armature |
| winding in slots) |
+---------------------------+
|
v
3-phase output to busbar
Working principle:
- The prime mover rotates the rotor at synchronous speed.
- The DC-excited rotor produces a rotating magnetic field in the air gap.
- This rotating field cuts the stationary 3-phase stator conductors, so by Faraday's law of electromagnetic induction an EMF is induced in the stator winding.
- The frequency of the induced EMF is f = PN_s/120, i.e. it is directly proportional to speed and number of poles.
- The magnitude of the induced EMF depends on the field excitation (flux) and speed; it is controlled by the field rheostat.
- When the stator is connected to a load, current flows and the machine delivers electrical power. The rotor must keep turning at synchronous speed; if it slows down or speeds up, the rotor "falls out of step" (loss of synchronism).
Conditions for parallel operation (synchronising) of synchronous generators:
Two or more alternators are connected in parallel to share load. Before closing the synchronising switch, the incoming machine must satisfy:
- Equal voltage: The terminal voltage (magnitude) of the incoming generator must equal the busbar voltage. Adjusted by field excitation.
- Equal frequency: The frequency of the incoming generator must equal the busbar frequency. Adjusted by prime-mover speed.
- Same phase sequence: The phase sequence of the incoming machine must match that of the busbar (R-Y-B order).
- Zero phase difference (in phase): The incoming voltage must be in exact phase with the busbar voltage at the instant of closing the switch — checked by synchronising lamps (dark or bright lamp method), a synchroscope or a voltmeter.
If these conditions are not met, a large circulating current flows at the instant of paralleling, causing mechanical shock, heating and possible damage to the windings. After paralleling, load sharing is controlled by the prime-mover governors (for active power) and field excitation (for reactive power).
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- 2081 · 2nd · 10 marks
What are the effects of low power factor in the power system? Why describe any two methods to improve the power factor.
Answer
Power Factor
Power factor is the cosine of the angle between voltage and current in an a.c. circuit:
Most industrial and domestic loads (induction motors, transformers, arc lamps, furnaces) are inductive, so they draw a lagging current and give a low lagging power factor. A low power factor means that for the same useful kW, a larger current must flow through the system.
Effects of Low Power Factor
1. Increased current for the same power If falls, the current rises in inverse proportion. For example, at 0.5 p.f. the current is double that at unity p.f. for the same kW load.
2. Increased copper (I²R) losses Line, transformer and generator winding losses are proportional to . Since the current rises sharply, the losses rise as the square of the current, so efficiency of transmission and distribution falls and more energy is wasted as heat.
3. Poor voltage regulation The larger current causes a larger voltage drop () in lines and transformers. The voltage at the consumer's terminals falls, causing dim lamps, sluggish motors and overheating of equipment.
4. Reduced effective capacity (kW capacity) of plant Generators, transformers, cables and switchgear are rated in kVA. Their useful output is kW = kVA × cos φ. At 0.6 p.f. a 100 kVA transformer can deliver only 60 kW, so the same plant serves fewer consumers and the capacity is wasted.
5. Larger and costlier equipment To carry the extra current, larger conductor sizes, bigger transformers, generators, cables and switchgear are required — increasing capital cost and the cost of installation.
6. Higher tariff and power factor penalty Nepal Electricity Authority and most utilities bill industrial consumers on maximum demand in kVA and levy a power factor penalty (or refuse a p.f. rebate) when the p.f. falls below about 0.85–0.9 lagging. The consumer therefore pays more for the same energy consumed.
7. Increased losses and reduced system stability Extra reactive current causes additional losses in the whole chain (generation, transmission, distribution), reduces the active power that can be transmitted over a given line, lowers the stability margin of the system and may cause overloading and tripping of lines and transformers.
Methods of Improving Power Factor
The basic principle is to supply the lagging reactive power (kVAr) required by the load from a local source, so that the line current — and hence the line losses and voltage drop — is reduced. The capacitor (or compensator) rating needed is:
where = existing p.f. angle and = desired (improved) p.f. angle.
Method 1: Shunt (Static) Capacitors
A bank of static capacitors is connected in parallel (shunt) with the inductive load, either at the load terminals or at the substation busbar. The capacitor draws a leading current which is exactly opposite to the lagging reactive component of the load current, so the two cancel and the resultant line current falls and comes closer to the supply voltage.
Advantages
- Very low losses (about 0.5–1 kW per 100 kVAr) and high efficiency.
- No moving parts, so no wear, no lubrication and almost no maintenance.
- Simple, light and can be installed in small units and extended step by step as load grows.
- Low capital cost and quick payback through reduced p.f. penalty and lower losses.
- No foundation or special cooling required; can be switched in steps by contactors as the load varies.
Limitations
- The reactive power supplied varies as the square of the voltage (), so compensation falls when the voltage dips.
- Cannot be used where harmonics are present without detuning reactors, as it may cause resonance.
- Ageing and failure of capacitors; faulty units are difficult to locate.
- Not suitable for loads that vary widely unless automatic switching is provided.
Method 2: Synchronous Condenser (Over-excited Synchronous Motor)
A synchronous motor running at no load (or lightly loaded) and over-excited behaves like a capacitor: it draws a leading current and supplies reactive power to the system. The field excitation can be adjusted with the field rheostat, so the power factor can be varied smoothly from lagging through unity to leading.
Advantages
- Continuously and smoothly variable reactive power — the p.f. can be set to any desired value.
- Also improves voltage regulation, since it can supply or absorb reactive power.
- Overload capacity is high and it can be used for power factor correction at large substations and at the sending end of long lines.
- No problem of resonance with harmonics as in capacitors.
Limitations
- High capital cost, and it needs its own excitation system, starting arrangement and cooling.
- Higher losses than a capacitor bank, so running cost is more.
- Requires skilled operation and regular maintenance; not economical for small or medium loads.
Other methods (brief)
- Synchronous motors used as normal drives run at over-excitation to supply reactive power at no extra cost.
- Phase advancers connected to the rotor circuit of induction motors inject an emf that cancels part of the rotor mmf, improving the p.f. of large slip-ring motors.
- Automatic p.f. correction relays with switched capacitor banks are used where the load fluctuates, so that only the required kVAr is connected at any time.
Conclusion: A low power factor increases current, losses, voltage drop, equipment size and electricity cost, and reduces the useful capacity and stability of the system. Shunt capacitors are the cheapest and most widely used remedy for steady loads, while a synchronous condenser is preferred where smooth, adjustable compensation and voltage support are needed.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
What is the function of a galvanometer? How would it work as a voltmeter? Explain.
Answer
Function of a galvanometer: A galvanometer is a sensitive instrument used to detect the presence and direction of a small electric current in a circuit. It works on the principle that a current-carrying coil placed in a magnetic field experiences a torque; the deflection of the pointer (or coil) is proportional to the current passing through it. It is also used to measure small currents and to indicate the null point (zero deflection) in experiments such as a potentiometer or Wheatstone bridge.
Working as a voltmeter: A galvanometer by itself cannot be used as a voltmeter because:
- it has low resistance, so it would draw a large current and disturb the circuit, and
- it gives full-scale deflection for a very small current.
To convert a galvanometer into a voltmeter, a high resistance (called the multiplier) is connected in series with it. Then the same current flows through both the galvanometer and the multiplier, and the whole combination is connected in parallel (across) the two points whose potential difference is to be measured.
If = resistance of galvanometer, = current for full-scale deflection, and = maximum voltage to be measured, then:
Thus the series resistance limits the current to when the voltage is , and the scale of the galvanometer is calibrated to read the potential difference in volts directly. The voltmeter so formed has very high resistance, so it draws negligible current from the circuit and gives an accurate reading of the potential difference.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
What is synchronization in power system? Why is it required? Describe.
Answer
Synchronization is the process of connecting an incoming alternator (generator) to an already energised busbar or to another running alternator so that the two systems operate in parallel as a single unit. The incoming machine is brought to the same voltage, frequency, phase sequence and phase angle as the running system, and only then is its circuit breaker closed.
Conditions to be satisfied before closing the breaker:
- Equal voltage — the terminal voltage of the incoming generator must equal the busbar voltage (usually adjusted by the field excitation).
- Equal frequency — the frequency of the incoming machine must match the busbar frequency (adjusted by the prime mover speed).
- Same phase sequence — the three-phase sequence of the incoming machine must be identical to that of the busbar.
- Zero phase difference — the incoming voltage phasor must be in phase with the busbar voltage phasor at the instant of closing.
Why synchronization is required:
- To avoid heavy circulating current: If the breaker is closed when the voltages are out of phase, the resultant voltage difference drives a large circulating current through the generator and busbar. This current can reach several times the rated current and may damage the stator winding, the prime mover shaft and the breaker itself.
- To prevent mechanical shock and damage: A large synchronising current produces a sudden electromagnetic torque (shock) on the rotor and shaft, which can cause mechanical damage to the coupling and bearings.
- To ensure stable parallel operation: Generators must run in synchronism so that they share the load in proportion to their ratings and the system frequency and voltage remain within limits. Loss of synchronism causes hunting and possible system collapse.
- To maintain power quality: Proper synchronization keeps frequency and voltage constant, which is essential for the quality of supply to consumers.
- To allow load sharing and continuity of supply: It permits additional generating units to be brought on load as demand rises, and allows one unit to be taken out for maintenance without interrupting supply.
Methods of synchronization:
- Dark lamp method — lamps connected across the synchronising switch remain dark when the voltages are in phase.
- Bright lamp method — lamps are brightest at the instant of synchronism.
- Two bright, one dark method — one lamp dark and the other two equally bright indicates correct phase sequence and synchronism.
- Synchroscope — the pointer rotates; at the 12 o'clock position the frequencies and phase match, and the breaker is closed.
Thus synchronization is essential for safe, stable and efficient parallel operation of alternators in a power system.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
What is secondary distribution system? Explain the secondary distribution system which is in use in Nepal distribution practices.
Answer
Secondary Distribution System
Definition: The secondary distribution system is the final stage of the electric power supply network that delivers electrical energy from the distribution transformer (secondary side of the transformer) to the actual consumers. It operates at low voltage (LV) — in Nepal, at 400 V (3-phase, 4-wire) and 230 V (single-phase, 2-wire) — and consists of the LV feeder, the distribution mains, the service mains, and the energy meter at the consumer's premises.
Main components
- Distribution transformer — steps down the primary voltage (11 kV or 33 kV) to 400/230 V.
- LV feeder / distributor — carries power from the transformer to the locality.
- Service mains — the tapping from the distributor to each consumer's meter board.
- Energy meter and main switch — at the consumer's point of supply.
Secondary distribution system in use in Nepal (NEA practice)
NEA follows a radial secondary distribution system with the following features:
- Voltage levels: 400 V, 3-phase, 4-wire for mixed (domestic + commercial + small industrial) loads; 230 V, 1-phase, 2-wire for small domestic consumers. The neutral is earthed at the transformer and at intervals along the line.
- Configuration: A radial system — each distribution transformer feeds its own LV network, and there is no interconnection between the LV networks of different transformers. This is simple, cheap and easy to protect, though reliability is lower than a ring system.
- Conductor: Mostly bare ACSR / AAAC conductors on 11 kV/0.4 kV pole-mounted distribution transformers (generally 25, 50, 100, 200, 315 kVA). In dense urban areas (Kathmandu, Pokhara, Biratnagar), ABC (Aerial Bundled Conductor) and underground cables are increasingly used to reduce theft and faults.
- Loading: Transformers are usually loaded to about 60–80% of rating, with LV feeders designed for a voltage drop within permissible limits (about 5–6%).
- Protection: Drop-out fuse (DO fuse) on the 11 kV side, LT fuse / MCCB on the secondary side, and MCB at each consumer's service mains.
- Metering: Single-phase and three-phase energy meters (now mostly smart/prepaid meters) installed at the consumer's premises.
- Earthing: Pipe or plate earthing at the transformer neutral and at consumer installations as per NEA standards.
Conclusion
Thus, the secondary distribution system is the LV part of the network that directly serves consumers. In Nepal, NEA uses a 400/230 V radial secondary distribution system with pole-mounted transformers, bare or bundled conductors, DO fuses and LT protection, and prepaid/smart metering — chosen for its simplicity, low cost and ease of maintenance.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
Explain the common insulators used in distribution system.
Answer
Common insulators used in distribution systems (NEA — Supervisor, Level 5)
Insulators support and insulate the live conductors from earth and from each other. In the distribution system (11 kV and 400/230 V), the following insulators are commonly used:
1. Pin insulators (11 kV / 33 kV lines)
- Used on straight (tangent) poles of 11 kV and 33 kV lines.
- Made of wet-process porcelain or toughened glass; the conductor sits in a groove on top and is tied with a binding wire.
- Available as single-piece (up to 11 kV) or two/three-piece (for 33 kV and above).
- Cheap, robust, and gives high mechanical strength; but cannot be used on sharp turns or at dead-ends.
2. Shackle (spool) insulators
- Used on straight poles of low-voltage (400/230 V) distribution lines.
- Porcelain spool shape; conductor is tied in the groove with a soft binding wire.
- Cheap and easily mounted on cross-arms; not suitable for high voltages or high tension.
3. Pin-type / shackle-type stay insulators
- Egg-shaped porcelain insulator inserted in the stay wire of a pole.
- Insulates the stay wire from earth so that leakage current or a broken stay does not become a shock hazard.
4. Disc (suspension) insulators
- Used on 11 kV / 33 kV line dead-ends, angle poles, and river crossings where tension is high.
- Porcelain or toughened-glass discs (usually 254 mm × 146 mm) connected in a string by a ball-and-socket or clevis joint.
- Each disc is rated for about 11 kV; the number of discs is increased with voltage. Any damaged disc can be replaced individually.
5. Reel / strain insulators
- Used at dead-ends and on sharp turns of LV lines to take the full conductor tension.
- Two or more grooves; conductor is wound around the grooves.
6. LT / HT pin insulators and bushing insulators
- LT pin insulators for 400/230 V lines; HT pin insulators for 11 kV.
- Porcelain bushings are used where conductors pass through pole tops, transformers, and switchgear.
Desirable properties of a good insulator
- High dielectric strength and high insulation resistance.
- High mechanical (tensile, compressive, bending) strength.
- Heat and moisture resistant; free from cracks, pores, and impurities.
- Not porous, so that leakage current and surface flashover are avoided.
Conclusion In NEA's distribution practice, shackle and LT pin insulators are used on LV lines, pin insulators on 11 kV/33 kV straight poles, and disc (suspension) insulators at dead-ends, angles, and crossings — each chosen according to voltage level and mechanical duty.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 5 marks
What is a substation? Explain different types of electrical substations based on the operating voltage.
Answer
Substation
A substation is a part of the power system where electrical energy is received, and its voltage is changed (stepped up or down), controlled, protected, switched or metered before being sent on to the next stage. It contains transformers, circuit breakers, isolators, busbars, lightning arresters, CTs/PTs, protection relays and control equipment. In Nepal, the Nepal Electricity Authority (NEA) operates substations such as Hetauda, Bhaktapur, Balaju and Duhabi.
Types of substation based on operating voltage
| Type | Typical voltage | Function |
|---|---|---|
| Step-up (generating) substation | Generator voltage (6.6, 11 kV) stepped up to 132, 220 or 400 kV | Located at the power station. It raises the voltage so that bulk power can be transmitted with low current and low loss. |
| Primary grid (transmission) substation | 220 kV, 132 kV, 66 kV | Receives power over the transmission lines and steps it down to the sub-transmission level (66 kV, 33 kV). It also interconnects the grid. |
| Secondary (sub-transmission) substation | 66 kV, 33 kV | Steps the voltage down to 33/11 kV for the load centres of towns and industries. |
| Distribution substation | 11 kV, 6.6 kV down to 400/230 V | Steps down to the utility voltage (400 V three-phase, 230 V single-phase) for consumers. It is mounted on a plinth, pole or in a pad-mounted kiosk. |
| Bulk supply / industrial substation | 33 kV or 11 kV | Supplies large industries (cement, steel) directly at the consumer's own voltage. |
In short, a substation is classified as step-up, step-down (primary, secondary, distribution) according to the voltage level at which it operates. A step-up substation raises the voltage, while the others lower it progressively, ending at the 400/230 V level supplied to consumers.
Answer checked by Claude Sonnet · Oct 21, 2026
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- 2081 · 2nd · 5 marks
Write any five differences between overhead lines and underground cablings in transmission system.
Answer
Differences between overhead lines and underground cabling in a transmission system:
| S.N. | Overhead Lines | Underground Cabling |
|---|---|---|
| 1 | Conductors are supported in open air on poles or towers. | Conductors (cables) are laid in trenches below the ground surface. |
| 2 | Insulation is provided by air; bare conductors are used. | Insulation is provided by the cable's insulating sheath (paper, XLPE, PVC); insulated cables are used. |
| 3 | Initial (capital) cost is low. | Initial cost is much higher due to excavation, ducts and cable cost. |
| 4 | Fault location and repair are easy and quick. | Fault location and repair are difficult and time-consuming. |
| 5 | Highly affected by weather, storms, lightning, trees and external objects; more chances of faults. | Not affected by weather, lightning or external objects; fewer chances of faults. |
| 6 | Requires wide right-of-way and tall towers; spoils the scenery. | Requires no right-of-way on the surface; neat appearance. |
| 7 | Line capacitance is small, so charging current and voltage drop are small; suitable for long distance. | Large capacitance to earth, so high charging current; not suitable for long distance. |
| 8 | Maintenance cost is higher and inspection is easy. | Maintenance cost is low but inspection is difficult. |
(Any five points are sufficient for full marks.)
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- 2081 · 2nd · 10 marks
What is the purpose of using Megger in electrical measurement system? Explain its working principle in detail.
Answer
Purpose of using a Megger
A Megger (insulation resistance tester) is used to measure the insulation resistance of electrical equipment such as cables, transformers, motors, generators and switchgear. Its purposes are:
- To check insulation health — it detects whether the insulation between conductors, or between a conductor and earth, has deteriorated due to moisture, dust, heat, ageing or mechanical damage.
- To detect faults — it reveals earth faults, short circuits between turns or phases, and leakage paths before the equipment is energised.
- To ensure safety — by confirming that insulation resistance is above the safe minimum, it protects personnel from electric shock and prevents failure/ fire hazards.
- For maintenance and testing — it is used for commissioning tests, periodic preventive maintenance and after repairs, to decide whether equipment is fit for service.
- To measure very high resistance — ordinary ohmmeters cannot measure resistances in the megohm range; a Megger is designed for this.
Working principle
A Megger works on the principle of comparison/deflection by a high DC voltage, and essentially consists of a hand-driven or motor-driven DC generator (or a rectified high-voltage DC source) producing typically 500 V, 1000 V, 2500 V or 5000 V, together with a moving-coil ohmmeter having two coils — a current (deflecting) coil and a pressure (control) coil — mounted on the same spindle in the field of a permanent magnet.
- The current coil is connected in series with the unknown insulation resistance under test, so the current through it is I = V / R<sub>x</sub>, i.e. it is inversely proportional to the insulation resistance.
- The pressure (voltage) coil is connected directly across the generator output (in series with a fixed resistance), so the current through it depends only on the supply voltage and is practically constant.
- The two coils produce opposing torques. The pointer comes to rest at a position where the two torques balance. Since the torque of the current coil depends on R<sub>x</sub> while that of the pressure coil depends on V, the deflection is a function of R<sub>x</sub> only and is independent of the generator voltage. The scale is therefore calibrated directly in ohms/megohms, being non-linear — crowded near zero (low resistance) and open at infinity.
Working (test procedure):
- The equipment under test is isolated and discharged to earth.
- The line terminal (L) of the Megger is connected to the conductor whose insulation is to be tested, and the earth terminal (E) to the earth/body of the equipment. The guard terminal (G), if present, is connected to the surface leakage path to eliminate surface leakage current.
- The handle (or motor) is rotated at rated speed (about 120–160 rpm) so that the generator builds up the full test voltage.
- The DC voltage drives a small leakage current through the insulation; the current coil carries this current while the pressure coil carries the reference current. The resulting deflection directly indicates the insulation resistance on the scale.
- The reading is compared with the minimum permissible value (e.g. for LV equipment, ≥ 1 MΩ; for HV equipment, much higher, often in hundreds of MΩ). A low reading indicates faulty insulation requiring drying, repair or replacement.
Thus the Megger gives a quick, direct and reliable indication of insulation condition without dismantling the equipment.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · 2nd · 10 marks
What is maintenance in an electrical system? Explain the preventive maintenance for distribution transformers.
Answer
Maintenance in an Electrical System
Maintenance is the set of planned and unplanned activities carried out on electrical equipment to keep it in proper working condition, to prevent failures, and to restore it to a safe and reliable operating state when a fault occurs. Its main objectives are:
- To ensure safety of personnel and equipment
- To maintain reliability and continuity of power supply
- To extend the service life of equipment
- To reduce failure rate, downtime and repair cost
- To maintain efficiency (losses, voltage, insulation strength) within limits
Maintenance is broadly of three types:
- Preventive (routine) maintenance — carried out periodically before a fault occurs.
- Breakdown (corrective) maintenance — carried out after a failure has occurred.
- Predictive maintenance — condition-based, using measurements and tests to decide when maintenance is due.
Preventive Maintenance of Distribution Transformers
Distribution transformers are the most numerous and costly-to-replace items in the distribution network, so a fixed periodic schedule is followed. The typical activities are:
1. Visual / External Inspection (monthly or quarterly)
- Check oil level in the conservator / oil gauge; top up if low.
- Check for oil leakage from the tank, gasket, bushings and drain valve.
- Inspect bushings for cracks, dirt, chipping and flashover marks; clean them.
- Check the silica gel breather — the gel should be blue; replace or reactivate when it turns pink.
- Verify that the explosion vent / pressure relief device and the breather pipe are clear.
- Check the earthing (grounding) connections of the tank, neutral and lightning arrester.
- Inspect the transformer body for rust, and repaint if necessary.
- Check the condition of the HT/LT cable boxes, glands and connections.
2. Oil Testing and Treatment (annually, or earlier if moisture is suspected)
- Dielectric strength (breakdown voltage) test of transformer oil — should generally be ≥ 30 kV for a 11/0.4 kV distribution transformer.
- Acidity test — should be below 0.2 mg KOH/g.
- Moisture (water content) test — should be below 20–25 ppm.
- If oil fails the tests, filter the oil (oil filtration) or replace it. For high moisture, carry out vacuum dehydration or oil replacement.
3. Electrical Tests (annually)
- Insulation resistance (IR) test with a megger — between HV–earth, LV–earth and HV–LV. Compare with previous readings; a falling trend indicates insulation deterioration.
- Winding resistance test — to detect loose or broken connections and winding deformation.
- Turns ratio test — to detect shorted turns.
- Earth resistance test of the earthing system — should be within the permissible value (generally ≤ 5 Ω for distribution substations).
- Polarity/phase sequence check on the LT side.
4. Cleaning and Tightening (annually)
- Clean bushings, insulators and the tank top with a dry cloth or dry-air blow.
- Tighten all HT and LT terminal connections to the specified torque to avoid hot spots.
- Clean the breather and replace the oil in the oil cup of the breather.
5. Load and Temperature Monitoring (continuous/periodic)
- Record the load current on the LT side and compare with the rated capacity; redistribute load if the transformer is overloaded.
- Check the oil temperature (dial thermometer) and winding temperature; ensure it does not exceed the permissible limit (oil ≈ 90 °C, winding ≈ 95–100 °C for ONAN distribution transformers).
- Verify the operation of the Buchholz relay (where provided) and the alarm/trip contacts.
6. Protection and Accessories
- Test the lightning arresters and replace damaged ones.
- Check the HT fuse (D.O. fuse / dropout fuse) rating and element condition; replace with the correct rating only.
- Check the LT main switch/MCCB, its contacts and tripping.
- Verify the neutral earthing and the earth pit condition.
7. Record Keeping
Maintain a transformer history card / log book recording oil test results, IR values, load readings, oil top-up quantity, and every maintenance action with date — this helps in condition-based decisions and in planning replacement.
Conclusion: Preventive maintenance of distribution transformers is a scheduled, periodic programme of inspection, testing, cleaning, oil treatment and record keeping. Followed properly, it prevents sudden failures, reduces outage time and repair cost, and gives the transformer its full designed life.
Answer checked by Claude Haiku · Oct 21, 2026
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Questions from NEA Exam Question for Supervisor (Supervisor / Electrical 5th Level · 2nd · 2082), NEA Exam Question for Meter Reader (Meter Reading Supervisor / Accountant 5th Level · 1st · 2082), NEA Nepal Bidhut Pradikaran Lok Sewa (Meter Reading Supervisor/ Accountant (5th Level) · 1st · 2081), NEA Supervisor Lok Sewa Exam Question (Supervisor 5th Level, Electrical/ Technical · 2nd · 2081) and NEA Lok Sewa Supervisor Internal Exam (Supervisor 5th Level, Electrical/Technical (Internal) · 2nd · 2081). Answers are written for this site; check them against your class notes.