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Chapter 1

Basic Electrical Engineering

Lok Sewa Aayog (Public Service Commission) 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

Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions remain constant.

If VV is the potential difference across a conductor and II is the current flowing through it, then

V∝I⇒V=IRV \propto I \quad \Rightarrow \quad V = IR

where RR is the constant of proportionality, called the resistance of the conductor.

From this law:

  • I=VRI = \dfrac{V}{R} — current is inversely proportional to resistance
  • R=VIR = \dfrac{V}{I} — resistance equals voltage divided by current

The SI 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 its ends.

Role of Resistance in an Electric Circuit

  1. Controls the current: Resistance opposes the flow of electric current. For a fixed supply voltage, a larger resistance allows a smaller current to flow (I∝1/RI \propto 1/R). Thus resistance is used to limit the current to a safe value and protect components from damage.

  2. Produces required potential drop: When current flows through a resistor, a voltage drop equal to IRIR appears across it. By choosing a suitable resistance, any desired portion of the supply voltage can be obtained across a part of the circuit. This is the basis of a potential divider.

  3. Converts electrical energy into heat: The electrical energy lost in overcoming resistance appears as heat (H=I2RtH = I^2Rt). This property is made use of in heaters, irons, fuses and filament lamps. A fuse wire of low resistance and low melting point melts and breaks the circuit when excess current flows, thus protecting appliances.

  4. Determines the current distribution in parallel branches: In a parallel circuit, current divides in the inverse ratio of the branch resistances. Hence resistance decides how much current each branch receives.

  5. Sets the time constant and damping in circuits: In circuits containing capacitors and inductors, resistance decides the charging/discharging rate (τ=RC\tau = RC) and the damping of oscillations.

In short, without resistance the current in a circuit would become excessively large; resistance makes the current in a circuit controllable, measurable and useful.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 5 marks

Why transistor is called current controlled device? Explain.

Answer

Why a transistor is called a current controlled device

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:

  1. In a bipolar junction transistor, the base current ( I_B ) is the input quantity and the collector current ( I_C ) is the output quantity.

  2. The relationship between them is: [ I_C = \beta I_B ] where ( \beta ) (current gain) is the ratio ( I_C / I_B ).

  3. Since the magnitude of the output (collector) current depends directly on the magnitude of the input (base) current, the device is controlled by current. A small change in ( I_B ) produces a large proportional change in ( I_C ).

  4. The base–emitter voltage ( V_{BE} ) only serves to forward bias the junction so that this base current can flow; the actual control parameter is the current ( I_B ).

Contrast: In a FET (field effect transistor), the output current (drain current) is controlled by the input voltage (gate–source voltage), so a FET is called a voltage controlled device. Since the BJT is controlled by base current, it is called a current controlled device.

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  • 2082 · 2nd · 5 marks

What are the differences between DC generator and DC motor? Write.

Answer

Differences between DC Generator and DC Motor

SNDC GeneratorDC Motor
1Converts mechanical energy into electrical energy.Converts electrical energy into mechanical energy.
2Works on the principle of electromagnetic induction (Faraday's law).Works on the principle of force on a current-carrying conductor in a magnetic field (Lorentz force).
3Input is mechanical (prime mover); output is electrical.Input is electrical; output is mechanical (rotation).
4The armature is rotated by an external prime mover.The armature rotates on its own due to electromagnetic torque.
5EMF is induced in the armature conductors.Back EMF is induced in the armature conductors.
6Direction of induced EMF is given by Fleming's Right-Hand Rule.Direction of rotation is given by Fleming's Left-Hand Rule.
7The commutator acts as a rectifier, converting AC in the armature to DC at the terminals.The commutator acts as an inverter, converting DC from the supply to AC in the armature.
8Generated EMF equation: Eg=ϕZNP60AE_g = \dfrac{\phi ZNP}{60A} (generated EMF is greater than terminal voltage).Back EMF equation: Eb=ϕZNP60AE_b = \dfrac{\phi ZNP}{60A} (back EMF is less than applied voltage).
9Used in power generation — e.g., DC power plants, battery charging, excitation of alternators.Used for driving loads — e.g., fans, pumps, cranes, electric traction, machine tools.

Key point: The construction of both machines is essentially the same; the difference lies in the direction of energy conversion. The same machine can act as either a generator or a motor (principle of reversibility of electrical machines).

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  • 2082 · 2nd · 5 marks

Why rotor of three phase induction motor always moves in the direction of rotating field? Describe.

Answer

Why the rotor of a three-phase induction motor always moves in the direction of the rotating field

  1. Production of rotating field: When a three-phase supply is given to the three-phase stator winding (displaced 120° electrically), a magnetic field of constant magnitude is produced which rotates at synchronous speed
    Ns=120fPN_s = \frac{120f}{P}

  2. Relative speed and induced emf: The rotor conductors are initially stationary, so they experience a relative speed with respect to the rotating field. By Faraday's law of electromagnetic induction, an emf is induced in the rotor bars.

  3. Rotor current: The rotor circuit is closed (squirrel cage or slip-ring short-circuited), so induced emf drives a current through the rotor conductors.

  4. Force (torque) production: This current-carrying rotor conductor lies in the stator's rotating magnetic field. By Lorentz force law (F=BILsin⁡θF = BIL\sin\theta), a force acts on each rotor bar, producing torque.

  5. Direction by Lenz's law: Lenz's law states that the induced current always opposes the cause producing it. The cause here is the relative motion between the rotating field and the rotor. To reduce this relative speed, the rotor must move in the same direction as the rotating field, so that relative speed decreases.

  6. Result: The rotor accelerates in the same direction as the rotating field. It can never reach synchronous speed, because at N=NsN = N_s the relative speed becomes zero, induced emf and rotor current become zero, torque falls to zero, and the rotor slips back. Hence it always runs at a speed slightly less than NsN_s (N<NsN < N_s), i.e. it always moves in the direction of the rotating field.

Conclusion: The rotor direction is decided by Lenz's law — the rotor follows the rotating field to minimise relative speed, so a three-phase induction motor is self-starting and always rotates in the direction of the rotating magnetic field.

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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

A typical solar power system (Solar PV system) consists of the following main components:

Components and their functions (2 marks for listing):

  1. Solar Panel (PV Module/Array)
  2. Charge Controller
  3. Battery (Battery Bank)
  4. Inverter
  5. Mounting Structure and Wiring (Balance of System)

Functions in converting sunlight into electricity (3 marks):

  1. Solar Panel (PV Module): It is made of many solar cells (usually silicon). When sunlight falls on the cells, the photovoltaic effect generates Direct Current (DC) electricity. It is the main component that actually converts solar energy into electrical energy.

  2. Charge Controller: The DC electricity from the panel is fed to the charge controller. It regulates the voltage and current going to the battery, prevents overcharging and deep discharge, and thus protects the battery and extends its life.

  3. Battery Bank: It stores the DC electricity produced during sunny hours so that power is available at night or during cloudy weather. It supplies DC power to the inverter or directly to DC loads.

  4. Inverter: It converts the DC electricity stored in the battery (or coming directly from the panel) into Alternating Current (AC) at 220 V, 50 Hz, which is required to run household appliances like bulbs, fans, TV and motors.

  5. Mounting Structure and Wiring: The mounting structure holds the panels at the proper tilt and orientation to receive maximum sunlight, while the wiring (cables, junction boxes, switches, protection devices) safely carries the generated electricity from one component to another up to the load.

Working summary: Sunlight → Solar Panel (DC) → Charge Controller → Battery (storage) → Inverter (DC to AC) → Load. In this way, solar energy is finally converted into usable electricity.

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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)

  1. Emergency/standby supply: During load-shedding hours, diesel generating sets are started immediately to supply power to essential loads (hospitals, industries, banks, telecom towers, water pumping stations) so that work is not interrupted.
  2. Peak load shaving: When the demand exceeds the supply from hydro/thermal plants, diesel plants are run to meet the shortfall and reduce the duration of load-shedding.
  3. Quick start and easy installation: A diesel plant needs only a few minutes to start and can be installed near the load centre, so it is the most practical source of power during load-shedding.
  4. Grid support / isolated operation: In places not connected to the national grid or where the grid supply fails, diesel plants supply power independently, acting as a reliable backup source.

Demerits (2 marks)

  1. High operating cost: Diesel fuel is expensive, so the cost of generation per unit (Rs/kWh) is much higher than that of hydro or solar plants.
  2. Fuel dependence and pollution: It depends on a continuous supply of diesel and emits smoke, noise and harmful gases (CO, CO₂, NOₓ), causing environmental pollution.
  3. High maintenance and short life: The engine has many moving parts, so frequent maintenance, lubrication and spare parts are needed, and its useful life is shorter than that of hydro plants.
  4. Limited capacity: Diesel sets are available only in small capacities (usually a few kW to a few MW), so they cannot replace large power plants for bulk supply.

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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 and Working of Its Main Components

1. Layout of a Hydroelectric Power Plant

        Reservoir (Dam / Intake)
                 │
        ┌────────┴────────┐
        │  Trash Rack &   │
        │  Intake Gate    │
        └────────┬────────┘
                 │
        ┌────────┴────────┐
        │  Headrace Canal │  (open channel / tunnel)
        │  / Penstock     │
        └────────┬────────┘
                 │
        ┌────────┴────────┐
        │  Surge Tank     │
        └────────┬────────┘
                 │
        ┌────────┴────────┐
        │  Penstock       │  (pressure pipe)
        │  (with valve)   │
        └────────┬────────┘
                 │
        ┌────────┴────────┐
        │  Turbine        │◄─── Turbine Shaft
        │  (in Power House)│
        └────────┬────────┘
                 │
        ┌────────┴────────┐
        │  Generator      │
        └────────┬────────┘
                 │
        ┌────────┴────────┐
        │  Transformer    │
        └────────┬────────┘
                 │
        ┌────────┴────────┐
        │  Switchyard /   │
        │  Transmission   │
        │  Lines          │
        └────────┬────────┘
                 │
              Tailrace  →  River (downstream)

Schematic block diagram:

Water Source → Intake → Headrace → Surge Tank → Penstock → Turbine → Generator → Transformer → Transmission
                                                                  ↓
                                                              Tailrace → River

2. Working of Main Components

1. Reservoir / Intake Water from a river is stored behind a dam to create a head (height difference). The intake structure with trash rack (to stop debris) and intake gate (to control flow) lets water enter the system.

2. Headrace Canal / Tunnel Carries water from the intake to the forebay or surge tank. It may be an open channel or a closed tunnel, depending on the terrain.

3. Surge Tank A vertical shaft provided between the headrace and penstock. It absorbs sudden pressure changes (water hammer) when the turbine load changes, protecting the penstock from damage.

4. Penstock A closed pressure pipe that carries water from the surge tank to the turbine. It converts the potential energy of water into kinetic energy as water flows down the slope.

5. Turbine Water strikes the turbine blades (Pelton, Francis or Kaplan type) with high velocity. The kinetic energy of water rotates the turbine, which is coupled to the generator shaft. After doing work, water leaves through the tailrace back to the river.

6. Generator The turbine shaft rotates the generator rotor inside a magnetic field. By electromagnetic induction, mechanical energy is converted into electrical energy (AC).

7. Transformer The generated voltage (e.g., 11 kV) is stepped up to a high voltage (e.g., 132 kV or 220 kV) to reduce transmission losses.

8. Switchyard and Transmission Lines The switchyard contains circuit breakers, isolators and protective relays to control and protect the supply. Power is then sent through transmission lines to load centres.

9. Tailrace The channel that carries water away from the turbine back to the river downstream.


Summary: Water stored at a height (potential energy) flows through penstock (kinetic energy) → rotates turbine (mechanical energy) → drives generator (electrical energy) → stepped up by transformer → transmitted to consumers.

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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)

The 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° electrical apart. It is connected to a three-phase AC supply.
  • Rotor: Either squirrel cage type (copper or aluminium bars shorted by end rings) or slip-ring (wound) type. It is mounted on a shaft inside the stator with a small air gap, and carries no external supply.

Working:

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

Ns=120fPN_s = \frac{120f}{P}

where ff = supply frequency (Hz), PP = number of poles. For example, at f=50f = 50 Hz and P=4P = 4, Ns=1500N_s = 1500 rpm.

  1. The RMF cuts the stationary rotor conductors. By Faraday's law, an EMF is induced in them. As the rotor circuit is closed (short-circuited), a current flows in the rotor conductors.

  2. By Lenz's law, the rotor current opposes the cause producing it — i.e. the relative speed between the RMF and the rotor. Hence the rotor experiences a force and starts rotating in the same direction as the RMF.

  3. The rotor can never reach synchronous speed. If it did, there would be no relative speed, no induced EMF, no rotor current, no torque. So the rotor always runs at a speed NN slightly less than NsN_s.

Slip:

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

Slip at full load is about 2–5%. Rotor frequency fr=sff_r = sf.

Torque: The torque developed is proportional to the product of stator flux and rotor current in phase, giving the torque–slip characteristic — torque rises from zero at s=1s=1, reaches a maximum (breakdown torque) at s=R2/X2s = R_2/X_2, and falls thereafter.

2. Starting Method (5 marks)

At starting, the rotor is stationary, so s=1s = 1. The rotor frequency equals supply frequency, rotor reactance is maximum, and the rotor current is very high (5–8 times full-load current) while the starting torque is poor. Hence a starter is necessary to limit the starting current. The main methods are:

(a) Direct-On-Line (DOL) Starter — For small motors (up to about 5 HP). The motor is connected directly to the supply 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 the voltage per phase becomes 1/31/\sqrt{3} of the line voltage and the starting current becomes one-third of the DOL value. 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 reduces the voltage applied to the stator. If the tapping gives xx times the line voltage, the starting current and torque are reduced by xx and x2x^2 respectively. Used for large squirrel cage motors.

(d) Rotor Resistance Starter — For slip-ring motors only. External resistance is inserted in the rotor circuit through slip rings at start. This limits rotor current and simultaneously increases the starting torque (since torque depends on rotor power factor). The resistance is gradually cut out as the motor accelerates.

(e) Stator Resistance / Reactance Starter — A resistance or reactor in series with the stator winding drops the voltage and limits the current; cheap but causes power loss and low starting torque.

Conclusion: DOL is used for small motors, star-delta and auto-transformer for medium and large squirrel cage motors, and rotor resistance starting for slip-ring motors where high starting torque is required.

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  • 2082 · 2nd · 5 marks

What is corona? What are the factors which affect corona? Write. 2+3=5

Answer

Corona (also called corona effect or corona discharge) is the phenomenon of partial electrical breakdown of the air surrounding a conductor when the potential difference between the conductors exceeds a certain critical value (the disruptive critical voltage). The air near the conductor surface becomes ionised, producing a faint violet glow, a hissing sound, a characteristic odour of ozone, and sometimes visible luminous streams called corona. It occurs mainly on high-voltage transmission lines, especially where the conductor surface has sharp points or irregularities.

Factors affecting corona:

  1. Applied voltage (potential difference): Corona forms only when the voltage exceeds the disruptive critical voltage. The higher the voltage above this value, the more intense the corona.

  2. Conductor size and shape: Corona is greater on thin conductors and on conductors with sharp edges, points or rough surfaces, because the electric field (potential gradient) is concentrated there. A larger conductor diameter reduces the potential gradient at the surface and hence reduces corona.

  3. Spacing between conductors: If the distance between conductors is increased, the electrostatic stress (potential gradient) at the conductor surface decreases, so corona is reduced. Corona is more likely when conductors are close together.

  4. Atmospheric conditions: Corona increases in humid, rainy, foggy or stormy weather and in dusty air, because moisture and dust particles lower the breakdown strength of air. It is least in dry, clear weather.

  5. Air density (temperature and pressure): Corona decreases with increasing air density. At high altitude, where air density is low, the breakdown voltage of air falls and corona increases. High temperature lowers air density and thus increases corona.

  6. Frequency of supply: At higher frequencies the corona loss increases, since the air around the conductor gets less time to de-ionise between successive cycles.

  7. Conductor surface condition: Rough, dirty or weathered conductor surfaces cause higher local field gradients and hence more corona than smooth, clean surfaces.

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  • 2082 · 2nd · 5 marks

How does earth fault relay work? Explain with necessary circuit diagram in brief.

Answer

Earth Fault Relay — Working Principle

An earth fault relay is a protective device that detects an abnormal flow of current to earth (ground) caused by insulation failure, a broken conductor touching the ground, or a live part coming into contact with the earth. It operates on the principle of residual (zero-sequence) current — under healthy conditions the vector sum of the three phase currents is zero, but during an earth fault this sum is no longer zero, and the relay senses this imbalance and trips the circuit breaker.

Working

  1. Healthy condition: The three line currents are equal and 120° apart, so IR+IY+IB=0I_R + I_Y + I_B = 0 The residual current is zero and the relay does not operate.

  2. Earth fault condition: Part of the current returns through the earth path, so the vector sum is no longer zero: IR+IY+IB=Iresidual≠0I_R + I_Y + I_B = I_{\text{residual}} \neq 0 This residual current flows through the relay operating coil.

  3. Core-balance (CBCT) type: All three phase conductors pass through a single toroidal core. In the healthy state the fluxes cancel; during an earth fault the net flux induces an EMF in the secondary winding, which energizes the relay.

  4. Operation: When the residual current exceeds the relay pickup setting (typically 10–20% of rated current), the relay operates its contacts, which energize the trip coil of the circuit breaker and isolate the faulty section. A time delay is often added to avoid maloperation during transient conditions.

Circuit Diagram (Core-Balance / Residual Connection Type)

        R  ───────────────┐
                          │
        Y  ───────────────┤  CBCT (toroidal core)
                          │
        B  ───────────────┘
                          │
                          ├────► Relay coil (R)
                          │
        Earth  ───────────┘

   Relay contact ──► Trip coil ──► Circuit Breaker (trips on fault)

In the residual connection method, the three CT secondaries are connected in star, and the relay is connected across the star point and the neutral:

   CT-R ──┐
   CT-Y ──┼──► Star point ──► Relay ──► Trip coil ──► CB
   CT-B ──┘

Key points

  • Sensitive to earth faults only; does not respond to phase-to-phase faults.
  • Pickup setting is low (10–20% of rated current) for high sensitivity.
  • Used in LV/MV distribution feeders, motors, and transformers.
  • Core-balance type is more sensitive and is preferred where the earth fault current is small.

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  • 2082 · 2nd · 5 marks

What are the uses of relays and isolators in electric system? Describe.

Answer

Uses of Relays and Isolators in an Electric System

Relay

A relay is an automatic protective device that senses abnormal conditions in the power system and gives a tripping command to the circuit breaker. Its main uses are:

  1. Fault detection and protection — It continuously monitors current, voltage, frequency and impedance, and detects faults such as short circuits, overloads and earth faults.
  2. Tripping the circuit breaker — On sensing a fault, it energises the trip coil of the circuit breaker and disconnects the faulty section quickly, limiting damage to equipment.
  3. Selective isolation — It isolates only the faulty part of the system while the healthy sections continue to supply power.
  4. Preventing equipment damage — It protects generators, transformers, motors, feeders and transmission lines from thermal and mechanical damage due to abnormal currents and voltages.
  5. System stability and safety — By clearing faults in minimum time, it maintains stability of the grid and protects operating personnel.
  6. Signalling and indication — It operates alarms, flags and indicators so that the operator knows the type and location of the fault.

Isolator (Disconnecting Switch)

An isolator is an off-load switching device used to isolate a circuit from the live supply for maintenance and repair work. Its main uses are:

  1. Isolation for maintenance — It disconnects the line, transformer or breaker from the supply so that repair and inspection work can be done safely.
  2. Off-load operation only — It is operated only when the circuit is de-energised (no current flowing); it is never used to break load or fault current.
  3. Sectionalising the system — It divides the network into sections so that a part can be taken out of service without disturbing the rest.
  4. Transfer of load — In double bus-bar arrangements, isolators are used to transfer a feeder or transformer from one bus to another.
  5. By-passing and earthing — It is used to by-pass a circuit breaker during maintenance and, with earthing blades, to connect the line to earth for safety.
  6. Safety interlocking — It provides a visible air gap in the circuit, giving a clear indication that the circuit is isolated.

Conclusion

The relay is the "brain" of the protection scheme, which detects the fault and commands the breaker to trip, while the isolator provides a visible, safe air gap 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.

Answer

Significance of Load Factor and Diversity Factor

Load Factor

Load factor is the ratio of the average load to the maximum (peak) demand over a given period:

Load Factor=Average LoadMaximum Demand\text{Load Factor} = \frac{\text{Average Load}}{\text{Maximum Demand}}

Its significance:

  1. Economy of generation: A high load factor means the plant's capacity is used more uniformly, so the fixed cost (capital cost) is spread over more units of energy generated, lowering the cost per unit (Rs/kWh).
  2. Plant selection and sizing: It helps decide the rated capacity of generating units and whether base-load or peak-load plants are needed.
  3. Tariff design: Consumers with a high load factor (steady demand) are charged a lower rate, since they utilise the system better.
  4. Operating efficiency: A high load factor indicates fewer idle hours and better utilisation of installed machinery, reducing wastage of capacity.

Diversity Factor

Diversity factor is the ratio of the sum of individual maximum demands to the simultaneous maximum demand of the whole system:

Diversity Factor=Sum of Individual Maximum DemandsMaximum Demand of the System\text{Diversity Factor} = \frac{\text{Sum of Individual Maximum Demands}}{\text{Maximum Demand of the System}}

Its significance:

  1. Reduced installed capacity: Because consumers' peaks do not occur at the same time, the system peak is less than the sum of individual peaks. A higher diversity factor means smaller plant capacity is required, saving capital investment.
  2. Economical distribution network: It allows the use of smaller conductors, transformers and switchgear, since the design load (coincident demand) is lower.
  3. Lower cost of supply: Reduced investment in plant and network lowers the fixed charges, which in turn lowers the tariff per unit.
  4. Load forecasting and planning: It guides the planning and expansion of generation, transmission and distribution systems.

Conclusion: Both factors measure how effectively the installed capacity is utilised. A high load factor reduces the cost per unit of energy, while a high diversity factor reduces the required installed capacity; together they lead to economical and efficient power system design and operation.

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  • 2082 · 2nd · 5 marks

How electrical safety is maintained during the power system operation and maintenance? Describe.

Answer

Electrical safety during power system operation and maintenance

Electrical safety in power system operation and maintenance is maintained through a combination of administrative procedures, technical measures and personal protective practices. The main aspects are:

1. Safety rules and permit system

  • Work is carried out only under a work permit (clearance) issued by the authorised control officer after the line/equipment is isolated and earthed.
  • The "permit to work" system ensures that no one re-energises the circuit until the permit is formally cancelled.
  • Only authorised and competent persons are allowed to operate switchgear; every panel is provided with a danger board/notice and locked.

2. Isolation, earthing and discharge

  • The concerned circuit breaker is opened, the isolator is opened and locked out/tagged out (LOTO).
  • The line is discharged to earth to remove trapped/induced charge, and portable earthing rods (PEGs) are applied on all three phases on both sides of the working zone.
  • "Test before touch" — the circuit is proved dead with a voltage detector before earthing.

3. Protective equipment and tools

  • Use of insulated gloves, rubber mats, insulated shoes, safety helmets, full-body harness and insulated ladders.
  • Use of insulated hand tools (screw drivers, pliers, discharge rods) tested periodically.
  • Arc flash suits while operating or switching high-voltage equipment.

4. Technical safety devices

  • Earthing (grounding) of equipment bodies, neutral earthing, and earth continuity conductors.
  • Fuses, MCBs, ELCBs/RCCBs, surge arresters and lightning arresters to clear faults and limit over-voltage.
  • Interlocks on isolators and breakers, safety fences/barriers, and danger boards at substations.
  • Double insulation and proper cable sizing to avoid overheating and short circuits.

4. Safe operating practices

  • Switching operations carried out strictly as per the switching programme/sequence (open breaker first, then isolator).
  • Live-line maintenance only with proper insulated tools and trained linemen.
  • Maintaining safe clearance from live conductors as per voltage level, and observing minimum approach distance.
  • First aid and artificial respiration (mouth-to-mouth/CPR) training to all staff; fire extinguishers (CO₂/dry powder) kept ready.

5. Organisational measures

  • Regular safety training, tool-box talks and refresher courses.
  • Periodic inspection, testing and maintenance of lines, transformers, breakers and protective relays.
  • Accident reporting and investigation to prevent recurrence.
  • Display of safety slogans, warning boards and emergency contact numbers.

Conclusion: Thus, electrical safety during operation and maintenance is maintained by combining isolation and earthing, permit-to-work and LOTO systems, protective devices and PPE, safe switching sequences, and continuous training and inspection — thereby protecting both personnel and equipment from electric shock, arc flash and fire hazards.

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  • 2082 · 2nd · 5 marks

What are the functions of lightening arrestor and overhead earth wire? Explain.

Answer

Functions of lightning (lightning) arrestor and overhead earth wire

1. Lightning arrestor (surge diverter)

  • It is installed on the top of the equipment/substation to protect transformers, insulators and other apparatus from high-voltage surges caused by lightning or switching.
  • It provides a low-resistance path to earth for the surge current, so the travelling wave is discharged to the ground instead of entering the equipment.
  • It limits the voltage appearing across the equipment to a safe value (its spark-over/breakdown value) and then restores normal insulation immediately after the surge, so the supply continues without interruption.
  • In short: it diverts the surge to earth and clamps the over-voltage to a safe level.

2. Overhead earth wire (shield wire / sky wire)

  • It is a bare conductor run above the phase conductors along the transmission line, earthed at every tower.
  • It shields the phase conductors from direct lightning strokes: the stroke falls on the earth wire instead of on the live conductors.
  • It reduces the voltage stress on the line insulation and hence reduces the chance of flashover and line outage.
  • It also provides a path for the lightning current to flow to earth through the tower footing, protecting the line from damage.

Difference in brief: the overhead earth wire prevents the lightning stroke from reaching the conductors (shielding), while the lightning arrestor discharges the surge that reaches the equipment (diversion). Both ultimately send the surge current safely to earth.

Answer checked by Claude Haiku · Oct 21, 2026

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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

Power factor is the ratio of active (real) power to apparent power in an AC circuit.

Power Factor=Active Power (kW)Apparent Power (kVA)=cos⁡ϕ\text{Power Factor} = \frac{\text{Active Power (kW)}}{\text{Apparent Power (kVA)}} = \cos\phi

where φ is the phase angle between voltage and current. It is a dimensionless quantity whose value lies between 0 and 1 (ideally 1, i.e., unity).

  • Unity power factor (1): purely resistive load (heaters, incandescent lamps)
  • Lagging power factor: inductive load (motors, transformers, ballasts)
  • Leading power factor: capacitive load (capacitor banks, synchronous condensers)

How Low Power Factor Increases the Cost of Energy

1. Higher apparent power for the same useful power Since kVA = kW / cosφ, a low power factor means more kVA must be supplied to deliver the same kW. For example, to deliver 100 kW at 0.5 pf requires 200 kVA, but at 0.9 pf only 111 kVA.

2. Larger equipment size and capital cost Generators, transformers, alternators, cables and switchgear must be rated in kVA. Low pf forces larger ratings, increasing capital investment.

3. Higher transmission and distribution losses Current I = P / (V cosφ). At low pf, current increases, so I²R losses in lines and cables rise sharply. For the same load, halving the pf doubles the current and quadruples the losses.

4. Poor voltage regulation Excessive current causes larger voltage drop, degrading supply quality and requiring additional voltage-correction equipment.

5. Penalty / tariff charges Utilities bill industrial consumers on maximum demand (kVA) and impose a power factor penalty (or pf surcharge) when pf falls below a prescribed limit (commonly 0.85–0.9). Consumers thus pay more for the same energy consumed.


Correction of Low Power Factor

Methods:

  1. Static capacitors — most common; connected in parallel with inductive loads to supply reactive power locally.
  2. Synchronous condensers — over-excited synchronous motors supplying leading reactive power; used for large loads.
  3. Phase advancers — connected to induction motor shafts to improve pf at the source.
  4. Synchronous motors — operated at leading pf to compensate lagging pf of other loads.
  5. Proper load management — avoiding lightly loaded motors and oversized equipment.

Capacitor bank sizing:

Qc=P(tan⁡ϕ1−tan⁡ϕ2)Q_c = P(\tan\phi_1 - \tan\phi_2)

where P = active power (kW), φ₁ = initial angle, φ₂ = desired angle.


Example

A factory has a load of 100 kW at 0.6 pf lagging. It is desired to raise the pf to 0.9 lagging. Find the capacitor kVA required and the reduction in apparent power.

Given: P = 100 kW, cosφ₁ = 0.6, cosφ₂ = 0.9

Solution:

ϕ1=cos⁡−1(0.6)=53.13°,tan⁡ϕ1=1.333\phi_1 = \cos^{-1}(0.6) = 53.13°,\quad \tan\phi_1 = 1.333 ϕ2=cos⁡−1(0.9)=25.84°,tan⁡ϕ2=0.484\phi_2 = \cos^{-1}(0.9) = 25.84°,\quad \tan\phi_2 = 0.484

Reactive power to be supplied by capacitor:

Qc=P(tan⁡ϕ1−tan⁡ϕ2)=100(1.333−0.484)=100×0.849Q_c = P(\tan\phi_1 - \tan\phi_2) = 100(1.333 - 0.484) = 100 \times 0.849

Qc=84.9 kVArQ_c = 84.9 \text{ kVAr}

Apparent power before correction:

S1=Pcos⁡ϕ1=1000.6=166.67 kVAS_1 = \frac{P}{\cos\phi_1} = \frac{100}{0.6} = 166.67 \text{ kVA}

Apparent power after correction:

S2=Pcos⁡ϕ2=1000.9=111.11 kVAS_2 = \frac{P}{\cos\phi_2} = \frac{100}{0.9} = 111.11 \text{ kVA}

Reduction in apparent power:

ΔS=166.67−111.11=55.56 kVA\Delta S = 166.67 - 111.11 = 55.56 \text{ kVA}

Result: A capacitor bank of about 85 kVAr raises the pf from 0.6 to 0.9, reducing the apparent power demand by 55.56 kVA (about 33%). This lowers the maximum demand charge, reduces line losses and avoids the pf penalty — thereby reducing the overall cost of energy.

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 (Supervisory Control and Data Acquisition) is implemented in power systems, water supply, industries and transmission networks to monitor and control geographically spread-out field equipment from a central point in real time. Its main reasons for implementation are:

Reasons for implementing SCADA:

  1. Real-time monitoring — gives the operator a live picture of the whole network (voltage, current, power flow, breaker status, water level, pressure, etc.) on one screen.
  2. Remote control — circuit breakers, isolators, pumps, valves and tap changers can be opened/closed from the control room without sending a crew to the site.
  3. Fast fault detection and restoration — faults are detected within milliseconds and the faulty section is isolated quickly, reducing outage time and revenue loss.
  4. Data acquisition and storage — measured values are logged automatically, so historical trends, energy accounting and reports are available for planning.
  5. Alarm and event handling — abnormal conditions raise audible/visual alarms so the operator acts before the situation becomes a major failure.
  6. Reduced manpower and cost — one control centre replaces many local operators; unmanned substations become possible.
  7. Improved safety — operators stay away from high-voltage or hazardous areas.
  8. Better system efficiency and reliability — load dispatch, optimal switching and coordinated operation improve power quality and system stability.
  9. Interoperability — standard protocols (IEC 60870-5-101/104, DNP3, Modbus) allow equipment from different manufacturers to work together.

Components of SCADA system:

  1. Field devices / Sensors and actuators (RTU level)

    • Sensors/transducers: CTs, PTs, RTDs, pressure and flow sensors — convert physical quantities into electrical signals.
    • Actuators: circuit breaker coils, motor-operated valves, relays — carry out the control command.
    • RTU (Remote Terminal Unit) / IED: placed at the substation or plant; collects analog and digital data from sensors, converts them into digital form, and sends them to the master station. It also executes control commands received from the centre. It works even when communication to the master fails (store-and-forward).
  2. Communication system

    • The link between RTUs and the master station — fibre optic, microwave, radio, leased telephone line, GSM/GPRS or PLC.
    • Uses standard protocols (IEC 60870-5-101/104, DNP3, Modbus, IEC 61850) so that data is exchanged reliably and securely.
  3. Master Station / Control Centre (MTU)

    • Front-end computers / communication servers: poll the RTUs and receive data.
    • SCADA server (data server): processes the data, maintains the real-time database and historical database.
    • HMI (Human Machine Interface): operator consoles with mimic diagrams, alarms, trends and control buttons.
    • Application servers: load flow, state estimation, load shedding, energy management (EMS/DMS).
    • Redundant servers: hot-standby arrangement so that operation continues if one server fails.
  4. Field instrumentation and control equipment

    • Panel meters, protection relays, PLCs and local control panels that interface the process with the RTU.
  5. Software

    • Operating system, SCADA application software, database (real-time + historical), communication drivers, report generation and cyber-security software (firewall, authentication, encryption).
  6. Peripheral devices

    • Printers for event/alarm logs and reports, UPS for backup power, GPS clock for time synchronisation of RTUs and servers.

Conclusion: SCADA is implemented to achieve centralised, real-time supervision and control of a wide network with minimum manpower, quick fault clearance and safe, economic and reliable operation. Its four essential building blocks are the field devices/RTU, the communication system, the master station and the software, supported by peripherals and cyber-security measures.

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

In a hydroelectric power plant, water is led from a high level to a low level, and the energy of falling water is converted step by step into electrical energy. Each component performs a specific function in this chain:

1. Dam (or weir/barrage)

  • Raises the water level of the river and creates a storage reservoir, thus developing the head (height of fall) needed for power generation.
  • Stores water so that generation can continue even in the dry season, and regulates the flow released downstream.
  • The stored water is the source of potential energy of the plant.

2. Forebay

  • A small storage basin located at the end of the headrace canal/tunnel and just before the penstock.
  • Its functions are to:
    • provide a steady, sediment-free supply of water to the penstock,
    • absorb sudden fluctuations in flow (when the turbine load changes),
    • allow settling of silt and provide a place for trash-rack cleaning and overflow arrangements.
  • It acts as a buffer between the low-capacity canal and the high-velocity penstock.

3. Penstock

  • A closed pipe (steel, RCC or HDPE) that carries water under pressure from the forebay to the turbine.
  • Its main role is to convert the potential energy of the water at the forebay level into kinetic energy at the turbine inlet, since the full head acts along its length.
  • It is designed to withstand high internal water pressure and is provided with anchor blocks, expansion joints and a valve at the turbine end.

4. Turbine

  • The prime mover of the plant. Water strikes the runner blades and rotates the shaft.
  • It converts the hydraulic energy (pressure and velocity of water) into mechanical energy in the form of shaft rotation.
  • Selection depends on head and discharge: Pelton for high head, Francis for medium head, Kaplan/propeller for low head and large flow.
  • Its speed is regulated by the governor (through the needle/servo mechanism) to keep frequency constant.

5. Generator

  • Coupled directly to the turbine shaft (or through a gearbox).
  • Converts the mechanical energy of the rotating shaft into electrical energy by electromagnetic induction.
  • Usually a synchronous AC generator (alternator); its excitation is controlled by the automatic voltage regulator to keep voltage constant.
  • The generated power is stepped up by transformers and transmitted through the switchyard.

Overall role: The dam creates and stores the head; the forebay regulates and cleans the flow; the penstock conveys water under pressure and builds up its energy; the turbine changes hydraulic energy into mechanical energy; and the generator changes mechanical energy into electrical energy. Thus each component is an essential link in the chain: potential energy of stored water → kinetic energy → mechanical energy → electrical energy.

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

P=V2RP = \frac{V^2}{R}

where P = power in watts (W), V = voltage in volts (V), R = resistance in ohms (Ω).

(Equivalently, P = VI and P = I²R, since V = IR.)

Relation to the heating effect of a circuit

When current I flows through a resistor R for time t, the electrical energy converted into heat is

H=I2Rt(joules)H = I^2Rt \quad \text{(joules)}

This is Joule's law of heating. Substituting V = IR into P = I²R gives P = V²/R, so the same power expression describes the rate at which heat is produced:

  • For a fixed voltage supply (as in domestic wiring), P = V²/R shows that a low-resistance element draws more power and therefore produces more heat per second — this is why heater coils are made of low-resistance nichrome wire.
  • For a fixed current (as in a series circuit), P = I²R shows heat production rises with the square of the current and in direct proportion to resistance — this is why a loose or high-resistance joint in a line becomes dangerously hot.

Thus the power formula P = V²/R is the direct measure of the heating rate of a circuit: the greater the power dissipated in the resistance, the greater the heat produced per unit time.

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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 electric current flowing through a circuit. It is always connected in series with the circuit so that the whole current to be measured passes through it. An ideal ammeter has very low (ideally zero) resistance so that it does not appreciably change the current it is measuring. It is calibrated in amperes (A), milliamperes (mA), or microamperes (µA).

Role of shunt resistor in an ammeter:

A basic moving-coil (galvanometer-type) meter can carry only a very small current (its full-scale deflection current, I_g). To convert it into an ammeter of larger range, a low resistance called a shunt is connected in parallel with the meter. Its roles are:

  1. Bypasses excess current: The shunt provides an alternative low-resistance path, so most of the circuit current (I − I_g) flows through the shunt and only the safe fraction I_g flows through the meter coil. This protects the coil from burning.
  2. Extends the range: By choosing the shunt value, the same meter can be made to read much larger currents. The required shunt is S=Ig GI−IgS = \frac{I_g \, G}{I - I_g} where G is the meter resistance, I_g the full-scale current of the meter, and I the desired maximum current to be measured.
  3. Reduces the effective resistance: Since the shunt is in parallel, the total resistance of the ammeter becomes very small (R = GS/(G+S)), which is the desired property of an ammeter so that it causes minimum disturbance to the circuit.

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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 between primary and secondary windings

When an alternating voltage is applied to the primary winding, an alternating current flows through it. This alternating current produces a changing magnetic flux in the soft-iron core. The same core links the primary and the secondary windings, so the changing flux passes through the secondary winding also.

According to Faraday's law of electromagnetic induction, a changing magnetic flux linked with a coil induces an emf in it. Since the flux through the secondary winding is continuously changing (because the primary current is alternating), an emf is induced in the secondary winding. This phenomenon, in which an emf is induced in one coil due to the changing current in a neighbouring coil, is called mutual induction.

If the secondary circuit is closed, this induced emf drives a current in the secondary winding. In this way, electrical energy is transferred from the primary to the secondary winding without any electrical connection between them — the transfer takes place through the magnetic field of the common core.

The magnitude of the induced emf depends on the rate of change of flux and the number of turns in the secondary winding:

Es=−NsdϕdtE_s = -N_s \frac{d\phi}{dt}

where NsN_s is the number of turns in the secondary and dϕdt\dfrac{d\phi}{dt} is the rate of change of magnetic flux. The ratio of secondary to primary voltage equals the ratio of their turns:

EsEp=NsNp\frac{E_s}{E_p} = \frac{N_s}{N_p}

Thus, energy is transferred from the primary to the secondary winding purely by mutual induction, with the iron core serving to link the magnetic flux between the two windings.

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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 in Power Electronics

Rectification is the process of converting alternating current (AC) into direct current (DC) using a device called a rectifier. The key property that makes this possible is unidirectional conduction — a diode (or thyristor) allows current to flow in only one direction, blocking the reverse half of the AC wave.

How AC is Converted to DC

  1. AC input: The supply voltage alternates in polarity — positive half-cycle and negative half-cycle.
  2. One-way conduction: A diode conducts only when it is forward-biased (anode more positive than cathode) and blocks current when reverse-biased.
  3. Half-wave rectification: A single diode conducts only during the positive half-cycle and blocks the negative half-cycle. The output is a pulsating DC — current flows in one direction but with gaps.
  4. Full-wave rectification: A centre-tap transformer with two diodes, or a bridge rectifier with four diodes, makes current flow through the load in the same direction during both half-cycles. The output is smoother pulsating DC with double the frequency of the input.
  5. Filtering: A capacitor (or inductor) filter is connected across the output to store charge during peaks and release it during valleys, reducing the ripple and giving steady DC.

Key Points

  • Rectifier = AC → DC; inverter = DC → AC.
  • Output of a rectifier is pulsating DC, not pure DC; a filter is needed for smooth DC.
  • Ripple factor: half-wave ≈ 1.21, full-wave ≈ 0.48 (before filtering).
  • Efficiency: half-wave ≈ 40.6%, full-wave ≈ 81.2%.
  • Applications: DC power supplies, battery chargers, welding, electroplating, HVDC transmission.

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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. power integrated over time. Its operation is based on the induction principle — a rotating aluminium disc driven by eddy currents induced by two alternating magnetic fluxes.

Construction:

  • Series (current) coil — carries the load current; connected in series with the load.
  • Shunt (voltage/pressure) coil — connected across the supply; carries a current proportional to supply voltage.
  • Aluminium disc — mounted on a spindle, free to rotate between the electromagnets.
  • Braking magnet (permanent magnet) — provides a damping torque proportional to speed.
  • Registering (counting) mechanism — gear train and dials to record revolutions as kWh.

Principle of operation:

  1. The shunt coil produces a flux Φ_v proportional to supply voltage V, and the series coil produces a flux Φ_s proportional to load current I.
  2. These two alternating fluxes are displaced in phase (by design, nearly 90°) and pass through the aluminium disc, inducing eddy currents in it.
  3. The interaction between the eddy currents and the alternating fluxes produces a driving torque proportional to the product of the two fluxes:

Td∝ΦsΦv∝V⋅I∝PowerT_d \propto \Phi_s \Phi_v \propto V \cdot I \propto \text{Power}

  1. The disc therefore rotates at a speed proportional to the power consumed.
  2. The permanent braking magnet induces eddy currents in the disc as it moves, producing a braking torque proportional to the speed N:

Tb∝NT_b \propto N

  1. At steady speed, driving torque = braking torque, so:

N∝VI∝PN \propto V I \propto P

  1. Since speed is proportional to power, the total number of revolutions is proportional to energy (P × t). The counter records these revolutions, calibrated so that 1 kWh corresponds to a fixed number of revolutions (the meter constant, e.g. 600 rev/kWh).

Result: The meter directly indicates energy consumption in kilowatt-hours (kWh), which is the basis of electricity billing.

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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. Introduction

A basic moving-coil (PMMC) instrument can measure only a small current (the full-scale deflection current, IgI_g) and a small voltage (the full-scale voltage, Vg=IgRgV_g = I_g R_g). To measure larger currents and voltages in practice, the range of the instrument must be extended. This is done by connecting an external resistance of the correct value and type in the correct manner — a low shunt resistance in parallel for an ammeter, and a high series resistance (multiplier) in series for a voltmeter.

2. Principle of Range Extension

The basic principle is to divert or limit the current through the meter coil so that only the rated full-scale current IgI_g flows through the coil when the actual quantity being measured is much larger.

  • Ammeter: A low resistance (shunt) is connected in parallel with the meter. Most of the line current bypasses the coil through the shunt, and only IgI_g passes through the coil. The deflection is thus proportional to the total line current.
  • Voltmeter: A high resistance (multiplier) is connected in series with the meter. Most of the measured voltage drops across the multiplier, and only VgV_g appears across the coil. The deflection is thus proportional to the total voltage.

The scale is then calibrated so that the pointer reads the actual (extended) value.

3. Extending the Range of an Ammeter (Shunt Method)

Let IgI_g = full-scale current of the meter, RgR_g = meter resistance, and II = desired (extended) range.

A shunt RshR_{sh} is connected in parallel, so the voltage across both is the same:

IgRg=(I−Ig)RshI_g R_g = (I - I_g) R_{sh}

Rsh=IgRgI−Ig\boxed{R_{sh} = \frac{I_g R_g}{I - I_g}}

The multiplying factor m=I/Igm = I/I_g, so:

Rsh=Rgm−1R_{sh} = \frac{R_g}{m - 1}

Key points:

  • The shunt must have a very low resistance and a low temperature coefficient (usually manganin or constantan) so that its resistance does not change with temperature and the reading stays accurate.
  • The shunt carries the major part of the current, so it must be thick enough to dissipate heat.
  • For AC measurements, a current transformer (CT) is used instead of a shunt.

4. Extending the Range of a Voltmeter (Multiplier Method)

Let VgV_g = full-scale voltage of the meter and VV = desired (extended) range. A high resistance RmR_m (multiplier) is connected in series:

V=Ig(Rg+Rm)V = I_g (R_g + R_m)

Rm=VIg−Rg\boxed{R_m = \frac{V}{I_g} - R_g}

The multiplying factor m=V/Vgm = V/V_g, so:

Rm=Rg(m−1)R_m = R_g (m - 1)

Key points:

  • The multiplier must have a high resistance and a low temperature coefficient (manganin) so that its value is stable.
  • The multiplier limits the current through the coil to IgI_g at full-scale voltage, protecting the meter from burning.
  • For AC measurements, a potential transformer (PT) is used instead of a multiplier.

5. Comparison of the Two Methods

FeatureAmmeter (Shunt)Voltmeter (Multiplier)
ConnectedIn parallel with meterIn series with meter
Resistance valueVery lowVery high
PurposeDiverts excess currentDrops excess voltage
MaterialManganin / constantanManganin
AC alternativeCurrent transformerPotential transformer

6. Numerical Illustration

Ammeter: Ig=10I_g = 10 mA, Rg=50 ΩR_g = 50\ \Omega, range to be extended to 11 A.

Rsh=10×10−3×501−0.01=0.50.99≈0.505 ΩR_{sh} = \frac{10 \times 10^{-3} \times 50}{1 - 0.01} = \frac{0.5}{0.99} \approx 0.505\ \Omega

Voltmeter: Ig=1I_g = 1 mA, Rg=100 ΩR_g = 100\ \Omega, range to be extended to 1010 V.

Rm=101×10−3−100=10000−100=9900 ΩR_m = \frac{10}{1 \times 10^{-3}} - 100 = 10000 - 100 = 9900\ \Omega

7. Conclusion

The range of an ammeter is extended by connecting a low-resistance shunt in parallel, and the range of a voltmeter by connecting a high-resistance multiplier in series. Both methods work on the same principle — keeping the current through the meter coil at its full-scale value IgI_g while the shunt or multiplier absorbs the excess current or voltage. The shunt and multiplier must be made of manganin or constantan to keep the reading accurate despite temperature changes. For AC circuits, current and potential transformers replace the shunt and multiplier respectively.

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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 key differences between base load power plants and peak load power plants in terms of their operational characteristics and their operational consideration.

Answer

Base Load Power Plants and Peak Load Power Plants

1. Introduction

The demand for electricity varies continuously over a day. It is lowest during the early morning hours and rises during daytime, reaching a maximum in the evening. To meet this varying demand economically and reliably, power plants are classified according to the portion of the load curve they serve — base load and peak load.

2. Base Load Power Plants

Definition: A base load power plant is one that operates continuously at a high, nearly constant output to supply the minimum (base) demand that exists 24 hours a day throughout the year.

Role in meeting demand:

  • It supplies the "base" of the load curve — the demand that never falls below a certain level.
  • It runs continuously (except for maintenance) and forms the backbone of the power system.
  • It maintains system frequency and voltage stability by providing a steady supply.

Examples: Large hydroelectric plants (run-of-river with pondage), nuclear plants, large coal-fired thermal plants, and geothermal plants.

Operational characteristics:

  • Very high load factor (80–90%) and high capacity factor.
  • Operates at or near full rated capacity for long periods.
  • Low running (fuel) cost per kWh, but high capital cost.
  • Long starting time (hours), so it cannot be started or stopped quickly.
  • Slow response to load changes; not suitable for frequent load variation.

3. Peak Load Power Plants

Definition: A peak load power plant is one that operates only during the period of maximum demand (peak hours), usually a few hours a day, to supply the additional demand above the base load.

Role in meeting demand:

  • It supplies the top portion of the load curve during peak hours (typically evening).
  • It fills the gap between base load supply and maximum demand.
  • It may also serve as a spinning reserve or standby unit for emergencies.

Examples: Gas turbine plants, diesel engine plants, pumped storage hydro plants, and small hydropower plants with storage.

Operational characteristics:

  • Low load factor (10–30%) and low capacity factor.
  • Operates only for a short duration each day.
  • High running (fuel) cost per kWh, but low capital cost per kW.
  • Short starting time (minutes), so it can be started and stopped quickly.
  • Quick response to load changes; suitable for frequent start–stop operation.

4. Key Differences

BasisBase Load PlantPeak Load Plant
Load servedMinimum continuous demandMaximum demand during peak hours
Load factorHigh (80–90%)Low (10–30%)
Operating durationContinuous, round the clockShort duration, few hours daily
Capital cost per kWHighLow
Running (fuel) cost per kWhLowHigh
Starting timeLong (hours)Short (minutes)
Response to load changeSlowFast
Efficiency at part loadPoorGood
Typical examplesNuclear, large coal, large hydroGas turbine, diesel, pumped storage

5. Operational Considerations

For base load plants:

  • Must be sited where fuel/water is available and transmission cost is low.
  • Designed for high efficiency at full load; maintenance scheduled during low-demand seasons.
  • Must be highly reliable, since failure causes large supply loss.
  • Unit commitment planned well in advance.

For peak load plants:

  • Must be sited near load centres to reduce transmission losses and voltage drop.
  • Must have fast start–stop capability and quick load-following response.
  • Fuel storage and quick availability of fuel are essential.
  • Economically justified only if the cost of peak energy is lower than the cost of buying/importing it or of load shedding.

6. Conclusion

Base load plants supply the steady minimum demand at low cost and high reliability, while peak load plants supply the short-duration maximum demand at higher cost but with quick response. A proper mix of both — base load for economy and peak load for flexibility — is essential for an economical, stable and reliable power supply system.

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

Why relays are used in electrical power system:

A relay is an automatic protective device that senses abnormal conditions in a power system and gives a command to the circuit breaker to isolate the faulty section. Relays are used for the following reasons:

  1. Fault detection: Relays continuously monitor current, voltage, frequency and other quantities. When these exceed preset values (overcurrent, overvoltage, earth fault, etc.), the relay operates and trips the circuit breaker.

  2. Protection of equipment: Generators, transformers, transmission lines and motors are very costly. Relays disconnect the faulty part quickly, so that equipment is saved from damage due to overheating, short circuit or insulation failure.

  3. Isolation of faulty section only: Relays with proper discrimination (time grading or directional feature) trip only the nearest breaker, so that the healthy portion of the system keeps supplying power. This maintains continuity of supply.

  4. Prevention of fault spreading: A fault left uncleared may spread to the whole system and cause a major blackout. Fast-acting relays clear the fault within milliseconds and prevent cascade tripping.

  5. Safety of personnel: By quickly removing the faulty line or equipment, relays reduce the danger of electric shock, fire and explosion to operating staff and the public.

  6. Economy and reliability: Relays allow the system to be operated near its thermal limit without fear of damage, and they replace costly continuous manual supervision. Thus they make the power supply reliable, safe and economical.

In short, relays act as the "senses and brain" of the protection system — they detect the abnormality and command the circuit breaker (the "muscle") to act, thereby protecting equipment, ensuring continuity of supply and safeguarding personnel.

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

Circuit Breaker vs Isolator

BasisCircuit BreakerIsolator
FunctionMakes and breaks the circuit on load as well as on no-loadMakes and breaks the circuit only on no-load (off-load)
Arc quenchingHas an arc-quenching arrangement (air, oil, SF₆, vacuum)No arc-quenching arrangement; cannot interrupt arc
OperationCan be operated manually or automatically (relay/trip coil)Operated manually only
ProtectionProtects the system against fault/over-current by tripping automaticallyProvides no protection; only isolates for safety
LocationInstalled on the load side of the isolator, in the main circuitInstalled on the supply side of the circuit breaker
Cost and constructionCostly and complexCheap and simple
Interlocking—Interlocked with the breaker so that it opens only after the breaker opens

Uses

Circuit breaker:

  1. To switch the circuit on and off under normal load conditions.
  2. To automatically disconnect the circuit during faults (short circuit, overload, earth fault), thereby protecting generators, transformers, motors and lines.
  3. Used in substations, distribution boards, transmission lines and industrial installations.

Isolator:

  1. To isolate the circuit breaker, transformer or line from the live supply for maintenance and repair work, so that the maintenance staff works safely.
  2. To give a visible air gap (confirmation of isolation) in the circuit.
  3. Used on the incoming side of a circuit breaker in substations and on the h.t. side of distribution transformers.

Answer checked by Claude Haiku · Oct 21, 2026

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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 Lines

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 falls on the earth wire instead of the line conductors, so the line is protected from direct hits.
  • It provides a low-resistance path to earth, so the lightning surge current is quickly discharged into the ground.
  • It reduces the voltage surge appearing across the line insulation (insulator strings), thereby preventing insulator flashover and line outages.
  • It also provides some protection against induced surges and reduces the number of trippings, improving the reliability and continuity of power supply.

2. Surge arrester (lightning arrester)

  • It is connected between the phase conductor and earth at substations and at vulnerable points of the line.
  • It provides a low-impedance path to earth for travelling surges (lightning or switching surges) coming along the line, so the surge is diverted to earth before reaching the equipment.
  • It limits (clamps) the surge voltage to a value below the basic impulse insulation level (BIL) of the equipment, thus protecting transformers, insulators, circuit breakers and other apparatus from insulation failure.
  • After the surge is discharged, it restores its normal high resistance automatically, so no follow-up power-frequency current flows to earth and the supply continues without interruption.
  • It thus prevents damage to costly equipment and avoids long outages, reducing repair cost and improving system stability.

Conclusion: The earth wire gives overall shielding of the line against direct lightning strokes, while the surge arrester protects terminal equipment by discharging the residual travelling surges to earth. Together they ensure safe, reliable and uninterrupted operation of the overhead transmission system.

Answer checked by Claude Haiku · Oct 21, 2026

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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 suitable devices so that the phase angle between voltage and current is reduced, bringing the power factor close to unity.

Why it is needed: A low power factor means a larger current is drawn for the same useful (active) power. This causes:

  • Higher copper losses (I²R) in cables and transformers
  • Larger conductor and switchgear sizes
  • Poor voltage regulation
  • Penalty or higher tariff from the utility
  • Reduced effective capacity of transformers and generators

Methods of power factor correction:

  1. Static capacitors (shunt capacitors) — Capacitors connected in parallel with the load supply leading reactive power (kVAr), which cancels the lagging reactive power of inductive loads. They are cheap, have low losses, require little maintenance, and can be switched in steps. Widely used in industries and distribution systems. Drawback: they are sensitive to harmonics and provide only fixed, stepwise correction.

  2. Synchronous condensers (synchronous motors running over-excited) — A synchronous motor run at no load or light load with over-excitation behaves like a capacitor and supplies reactive power. Its excitation can be varied continuously, so stepless correction is possible. It is suitable for large, high-voltage installations. Drawbacks: high cost, higher losses, more maintenance, and starting problems.

  3. Phase advancers — These are connected to the rotor circuit of induction motors and supply exciting ampere-turns to the rotor, thereby improving the power factor of the motor itself. They are useful where the load is a large induction motor, but are rarely used now.

  4. Static VAR compensators (SVC) / automatic capacitor banks — Thyristor-switched capacitor banks or SVCs automatically adjust the reactive power according to the changing load, giving dynamic and accurate correction. Used in modern industrial and transmission systems.

In practice: For a load of active power P (kW) at power factor cos φ₁ to be improved to cos φ₂, the required capacitor rating is:

Qc=P(tan⁡ϕ1−tan⁡ϕ2) kVArQ_c = P(\tan\phi_1 - \tan\phi_2)\ \text{kVAr}

Capacitors are normally connected in steps (e.g., 5, 10, 25 kVAr) through contactors controlled by a power factor relay, so that over-correction (leading power factor) is avoided.

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

Definition: A sectionalized bus bar arrangement is a scheme in which the main bus bar of a substation is divided into two or more separate sections by means of isolators (sectionalizing switches) or circuit breakers, while each section remains connected to the incoming and outgoing feeders. A common form is the single bus bar with sectionalization, where the bus is split into two halves, each half fed by its own transformer or incoming line.

Basic arrangement:

  • The main bus is split into Section A and Section B by a sectionalizing isolator (or breaker).
  • Each section has its own incoming supply (transformer/line) and a set of outgoing feeders.
  • Loads are distributed between the two sections so that neither section is overloaded.
  • The sectionalizing switch is normally closed (bus operated as one unit) and is opened only during a fault or maintenance.

Enhancement of system reliability:

  • If a fault occurs on one section of the bus, only that section is de-energized; the other section continues to supply its feeders. Thus a complete station shutdown is avoided.
  • Each section can be fed independently from a separate transformer or line, so failure of one source does not black out the whole substation.
  • Continuity of supply to important loads is maintained, raising overall availability of the system.

Fault isolation:

  • On occurrence of a fault on a bus section, the protective relays trip the breakers of that section and the sectionalizing breaker, isolating the faulty section from the healthy one.
  • The healthy section remains in service and continues to feed its loads.
  • Fault current is limited because only the sources connected to the faulty section contribute to it, reducing damage to equipment.
  • The faulty section can be identified and repaired while the rest of the station operates normally.

Maintenance flexibility:

  • Any section of the bus, or any feeder, transformer or breaker connected to it, can be taken out for preventive maintenance while the other section remains in operation.
  • The sectionalizing switch permits one half of the bus to be isolated completely, allowing safe working on bus bar, insulators, isolators and instrument transformers.
  • Loads on the section under maintenance can be transferred to the healthy section (where capacity permits), so supply is not interrupted.
  • Maintenance can be planned without shutting down the entire substation, reducing outage time and revenue loss.

Comparison (advantage over simple single bus):

FeatureSimple single busSectionalized single bus
Fault on busWhole station tripsOnly faulty section trips
Maintenance of busFull shutdownSection-wise shutdown
ReliabilityLowHigh
CostLowModerate

Conclusion: Sectionalization of the bus bar is a simple, economical and effective method of improving reliability, limiting fault effects and providing maintenance flexibility. It is widely used in 11 kV, 33 kV and 66 kV substations where a compromise between cost and continuity of supply is required. For still higher reliability, duplicate bus bar or one-and-a-half breaker schemes are adopted.

Answer checked by Claude Haiku · Oct 21, 2026

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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 an electric utility (or energy supplier) designed to influence the amount and timing of electricity (or energy) consumed by its customers. In simple terms, it means managing how much energy is used and when it is used, on the consumer's side, rather than building new generation plants on the supply side. DSM measures are broadly of two types:

  1. Energy efficiency (permanent reduction): using efficient devices — LED lighting, IE3/IE4 motors, star-rated appliances, improved cooking stoves, efficient boilers — so that the same service is obtained with less energy.
  2. Load management (shifting/levelling): time-of-use tariffs, peak clipping, valley filling, load shifting, strategic conservation and strategic load growth, so that demand is spread more evenly over the day.

Significance in the energy sector:

  1. Reduces the peak demand and the need for new capacity: every kW of peak demand avoided saves the utility the cost of building a peaking plant and its associated transmission and distribution (T&D) reinforcement. In Nepal, where load-shedding and dry-season deficits have been chronic, DSM is the cheapest "source" of power.
  2. Defers capital investment: DSM costs far less per kW saved than new hydro, thermal or import capacity, so scarce capital can be used elsewhere.
  3. Improves the load factor and system efficiency: flattening the load curve raises the utilisation of existing plants and reduces the need for costly peaking generation and imports from India.
  4. Reduces energy bills for consumers: efficient appliances and off-peak use lower the consumer's monthly bill and protect low-income households from tariff shocks.
  5. Energy security and reduced imports: lower demand means less dependence on imported electricity and fossil fuels, improving the balance of payments.
  6. Environmental benefits: avoided generation means avoided emissions of CO₂, SO₂ and particulate matter; DSM also supports clean cooking and reduces deforestation and indoor air pollution.
  7. Supports renewable integration: by shifting flexible loads (water pumping, EV charging, cold storage) to hours of surplus solar/wind or run-of-river hydro, DSM helps balance variable renewables.
  8. Demand-side response and reliability: during contingencies, curtailable loads act as a virtual reserve, improving grid stability and reducing forced outages.
  9. Equity and job creation: DSM programmes (LED distribution, efficient stove and pump programmes, energy audits) create local employment and reach remote consumers.
  10. Complements supply-side planning: in Nepal's context — hydropower-dominated, dry-season deficit, rising peak from cooling and irrigation — DSM is the least-cost option under the Integrated Resource Planning (IRP) approach, and is promoted through the Electricity Regulatory Commission's directives, Time-of-Day tariffs and NEA's efficient lighting and appliance programmes.

In short, DSM treats negawatts (saved watts) as a resource. It lowers cost, raises reliability, cuts emissions and defers investment, making it a cornerstone of a sustainable and affordable energy strategy.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 10 marks

A 3-phase 110km long transmission line is delivering 48MW 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 (3×10^-6) 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: 3-phase, 110 km line (medium line, nominal-π\pi method), VR=132V_R = 132 kV (line), PR=48P_R = 48 MW, cos⁡ϕR=0.8\cos\phi_R = 0.8 lagging.

Line constants (per phase)

R=0.1×110=11 Ω,X=0.3×110=33 Ω,Z=11+j33 ΩR = 0.1\times110 = 11\ \Omega,\quad X = 0.3\times110 = 33\ \Omega,\quad Z = 11 + j33\ \Omega

Y=3×10−6×110=3.3×10−4 S,Y2=1.65×10−4 SY = 3\times10^{-6}\times110 = 3.3\times10^{-4}\ \text{S},\quad \frac{Y}{2} = 1.65\times10^{-4}\ \text{S}

Leakage (conductance) is neglected, so the shunt admittance is purely capacitive.

Receiving end

VR=1320003=76210 V (phase),IR=48×1063×132000×0.8=262.4 AV_R = \frac{132000}{\sqrt3} = 76210\ \text{V (phase)},\qquad I_R = \frac{48\times10^6}{\sqrt3\times132000\times0.8} = 262.4\ \text{A}

IˉR=262.4(0.8−j0.6)=209.9−j157.5 A\bar I_R = 262.4(0.8 - j0.6) = 209.9 - j157.5\ \text{A}

Charging current at the receiving end

IˉC1=jY2VR=j12.58 A\bar I_{C1} = j\frac{Y}{2}V_R = j12.58\ \text{A}

Current in the series impedance

IˉL=IˉR+IˉC1=209.9−j144.9 A\bar I_L = \bar I_R + \bar I_{C1} = 209.9 - j144.9\ \text{A}

(i) Sending-end voltage

VˉS=VR+IˉLZ=76210+(209.9−j144.9)(11+j33)=83301+j5334 V\bar V_S = V_R + \bar I_L Z = 76210 + (209.9 - j144.9)(11 + j33) = 83301 + j5334\ \text{V}

∣VS∣=83471 V (phase)  ⇒  VS(line)=3×83471≈144.6 kV|V_S| = 83471\ \text{V (phase)} \;\Rightarrow\; V_{S(\text{line})} = \sqrt3\times83471 \approx \mathbf{144.6\ kV}

The angle of VSV_S with VRV_R is δ=tan⁡−1(5334/83301)=3.66∘\delta = \tan^{-1}(5334/83301) = 3.66^\circ.

Sending-end current

IˉC2=jY2VˉS=−0.88+j13.75 A\bar I_{C2} = j\frac{Y}{2}\bar V_S = -0.88 + j13.75\ \text{A}

IˉS=IˉL+IˉC2=209.1−j131.1 A,∣IS∣=246.8 A,∠IS=−32.1∘\bar I_S = \bar I_L + \bar I_{C2} = 209.1 - j131.1\ \text{A},\quad |I_S| = 246.8\ \text{A},\quad \angle I_S = -32.1^\circ

(ii) Sending-end power factor

ϕS=3.66∘−(−32.1∘)=35.8∘  ⇒  cos⁡ϕS≈0.81 lagging\phi_S = 3.66^\circ - (-32.1^\circ) = 35.8^\circ \;\Rightarrow\; \cos\phi_S \approx \mathbf{0.81\ lagging}

(iii) Sending-end power

PS=3 ∣VS∣∣IS∣cos⁡ϕS=3×83471×246.8×0.8117≈50.15 MWP_S = 3\,|V_S||I_S|\cos\phi_S = 3\times83471\times246.8\times0.8117 \approx \mathbf{50.15\ MW}

Check: line loss =3IL2R=3×(255.1)2×11=2.15= 3I_L^2R = 3\times(255.1)^2\times11 = 2.15 MW, and 48+2.15=50.1548 + 2.15 = 50.15 MW.

(iv) Efficiency

η=PRPS×100=4850.15×100≈95.7%\eta = \frac{P_R}{P_S}\times100 = \frac{48}{50.15}\times100 \approx \mathbf{95.7\%}

(v) Regulation

With A=1+YZ2=0.9946+j0.0018A = 1 + \dfrac{YZ}{2} = 0.9946 + j0.0018, the no-load receiving-end voltage is

VR0=∣VS∣∣A∣=834710.9946=83927 VV_{R0} = \frac{|V_S|}{|A|} = \frac{83471}{0.9946} = 83927\ \text{V}

Regulation=VR0−VRVR×100=83927−7621076210×100≈10.1%\text{Regulation} = \frac{V_{R0} - V_R}{V_R}\times100 = \frac{83927 - 76210}{76210}\times100 \approx \mathbf{10.1\%}

QuantityResult
Sending-end voltage144.6 kV (line)
Sending-end power50.15 MW
Sending-end power factor0.81 lagging
Efficiency95.7%
Regulation10.1%

Answer checked by Claude Sonnet · Oct 21, 2026

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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 Line

Distribution lines carry high voltage and are often located in public places, so maintenance work must be carried out with strict safety precautions. The main precautions are:

  1. Switching off and isolation (Lock-out/Tag-out): The concerned feeder or section must be switched off from the substation, and a "Men at Work" tag or danger board must be placed on the switch. The switch must be locked so that no one can re-energize the line accidentally.

  2. Testing for dead (Proving dead): Before touching any conductor, the line must be tested with a voltage detector or neon tester to confirm that it is actually dead. Testing must be done on all phases and between phases and earth.

  3. Earthing of the line: Temporary safety earths (short-circuiting jumpers) must be connected between all phases and to earth on both sides of the working section. This protects workers if the line is accidentally charged, as the fault current will flow to earth and trip the breaker.

  4. Use of personal protective equipment (PPE): Linemen must wear rubber gloves, rubber-soled boots, safety helmets, safety belts and full-body harness while climbing poles. Insulated tools (lineman's pliers, screwdrivers, discharge rods) must be used.

  5. Maintaining safe clearance: The worker must keep the minimum statutory clearance from live conductors and from other live lines crossing the section. Working on live lines should be avoided; if unavoidable, it must be done only with proper insulated platforms and under supervision.

  6. Barricading and warning signs: The work site must be barricaded and warning boards/red flags placed on both sides to warn the public and traffic. A flagman may be posted on busy roads.

  7. Weather precautions: Maintenance should not be done during rain, lightning, strong wind or when the ground is wet, as the risk of electric shock and slipping is high.

  8. Trained and authorized personnel: Only trained, authorized and medically fit linemen should do the work. At least two persons should work together — never alone — so that help is available in an emergency.

  9. First aid and rescue readiness: A first-aid box must be kept at the site. If a worker gets an electric shock, the supply must be switched off first (or the victim removed with an insulated material) before giving artificial respiration or CPR.

  10. Proper tools and equipment: Ladders, ropes, buckets and tools must be checked before use; damaged or wet equipment must not be used. After the work is over, all temporary earths and tools must be removed and only then should the line be re-energized.

Importance of Earthing of Electrical System

Earthing (grounding) means connecting the non-current-carrying metallic parts of an installation and the neutral point of the supply system to the general mass of earth through a low-resistance path. Its importance is as follows:

  1. Protection of human life: If a live conductor touches the metal body of a machine, the body becomes charged. With proper earthing, the fault current flows to earth and the fuse or MCB operates quickly, disconnecting the supply and preventing electric shock to the user.

  2. Protection of equipment: Earthing limits the voltage rise on equipment during faults, lightning surges, or switching surges, thereby preventing damage to machines, transformers, and appliances.

  3. Stability of the supply system: Earthing the neutral point of a transformer or generator provides a return path for fault current and keeps the phase voltages balanced and limited to a safe value with respect to earth.

  4. Fast operation of protective devices: A low earth resistance ensures a large fault current, which makes fuses, MCBs, and relays operate within the shortest possible time, reducing the duration of the fault.

  5. Safety against lightning and static charge: Lightning arresters and surge diverters discharge the surge energy into the earth. Earthing also discharges static electricity accumulated on machines and tanks, preventing sparks and fire.

  6. Voltage stabilization and reference: Earth provides a common reference (zero potential) for the system, so that insulation levels and protective settings can be designed properly.

  7. Reduction of fire and explosion risk: In places like petrol pumps and chemical plants, proper earthing prevents sparking, thus avoiding fire and explosion hazards.

Conclusion: Maintenance of distribution lines must always follow the "isolate–test–earth–work" sequence with proper PPE and supervision. Earthing is equally essential because it safeguards human life, protects costly equipment, and ensures reliable and safe operation of the electrical system.

Answer checked by Claude Haiku · Oct 21, 2026

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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

  1. Same current: The same current flows through every component, since there is only one path for the current.
  2. Voltage divides: The total supply voltage is the sum of the voltage drops across each component: V=V1+V2+V3+…V = V_1 + V_2 + V_3 + \dots
  3. Total resistance adds: The equivalent resistance is the sum of individual resistances: Req=R1+R2+R3+…R_{eq} = R_1 + R_2 + R_3 + \dots
  4. Single path: If any one component breaks or is removed, the whole circuit is broken and no current flows.
  5. Power adds: Total power is the sum of the power consumed by each component: P=P1+P2+P3+…P = P_1 + P_2 + P_3 + \dots
  6. Current is unaffected by the order of components, but the voltage drop across each depends on its resistance (larger resistance → larger voltage drop).

Difference between Resistance in Series and Capacitance in Series

Resistance in SeriesCapacitance in Series
Equivalent resistance is the sum of individual resistances: Req=R1+R2+R3R_{eq} = R_1 + R_2 + R_3Equivalent capacitance is the reciprocal sum: 1Ceq=1C1+1C2+1C3\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}
The equivalent resistance is greater than the largest resistance.The equivalent capacitance is smaller than the smallest capacitance.
Same current flows through each resistor; voltage divides in proportion to resistance.Same charge is stored on each capacitor; voltage divides in inverse proportion to capacitance (V=Q/CV = Q/C).
Energy is dissipated as heat (P=I2RP = I^2R).Energy is stored in the electric field (E=12CV2E = \frac{1}{2}CV^2); no heat loss in an ideal capacitor.
Adding more resistors in series increases the total resistance.Adding more capacitors in series decreases the total capacitance.

Key point: In series, resistors add directly while capacitors add reciprocally — this is because resistance opposes current (so resistances in the same path add up), whereas capacitance stores charge (so the charge is common and the voltages add as V=Q/CV = Q/C).

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 2+2+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, because voltage and current are sinusoidal and may differ in phase, 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 in watts (W) or kilowatts (kW).

P=VIcos⁡ϕ(single-phase)P = V I \cos\phi \quad \text{(single-phase)}

P=3 VLILcos⁡ϕ(three-phase)P = \sqrt{3}\, V_L I_L \cos\phi \quad \text{(three-phase)}

2. Reactive Power (Q) The power that oscillates back and forth between the source and the reactive elements (inductors and capacitors) without being consumed. It is needed to build magnetic fields in motors and transformers. It is measured in volt-ampere reactive (VAR) or kilovar (kVAR).

Q=VIsin⁡ϕQ = V I \sin\phi

3. Apparent Power (S) The product of the rms voltage and rms current, representing the total power that appears to flow in the circuit. It is measured in volt-ampere (VA) or kilovolt-ampere (kVA).

S=VI=P2+Q2S = V I = \sqrt{P^2 + Q^2}

The three are related by the power triangle:

S2=P2+Q2,cos⁡ϕ=PSS^2 = P^2 + Q^2, \qquad \cos\phi = \frac{P}{S}


Significance of Power Factor

Power factor is defined as the cosine of the phase angle between voltage and current:

Power factor=cos⁡ϕ=PS=Active powerApparent power\text{Power factor} = \cos\phi = \frac{P}{S} = \frac{\text{Active power}}{\text{Apparent power}}

Its significance:

  1. Measure of utilisation efficiency: It shows what fraction of the apparent power supplied is actually converted into useful work. A low power factor means a large part of the current merely shuttles back and forth as reactive power.

  2. Determines current for a given load: For the same active power, a lower power factor draws a larger current (I = P / V cos φ). This increases I²R losses in transmission lines and transformers, reducing efficiency.

  3. Affects voltage regulation: Large reactive currents cause greater voltage drop along feeders, leading to poor voltage at the consumer's terminals.

  4. Determines equipment and cable rating: Generators, transformers, alternators and cables must be rated in kVA. A low power factor forces larger, costlier equipment for the same useful output.

  5. Economic penalty: Utilities charge a penalty or demand a higher tariff when the power factor falls below a prescribed limit (usually about 0.9 lagging), because the supplier must generate and transmit extra reactive power.

  6. Indicates load nature: Unity power factor means purely resistive load; lagging power factor means inductive load (motors, transformers); leading power factor means capacitive load. Power factor correction (using capacitors) is therefore applied to bring it near unity.

Answer checked by Claude Haiku · Oct 21, 2026

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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

Classification of Hydroelectric Plants

Hydroelectric plants are classified in several ways. Three important bases are given below.

1. Based on availability of head

TypeHeadExample
Low head plantBelow about 30 mRun-of-river plants on plain rivers
Medium head plantAbout 30 m to 100 mPlants on foothill rivers
High head plantAbove about 100 mKulekhani (Nepal), mountain streams

2. Based on nature of load

  • Base load plant: Runs continuously at nearly full capacity to supply the steady, round-the-clock demand. It has a high load factor and is usually a large plant with a big reservoir.
  • Peak load plant: Operates only during the hours of maximum demand (morning and evening peaks). It has a low load factor and needs quick starting and stopping, so it is often a pumped-storage or storage plant.
  • Standby (emergency) plant: Remains idle normally and is started only when the main supply fails or demand suddenly rises.

3. Based on quantity of water available

  • Run-of-river plant: Uses the natural flow of the river as it comes, without storing water. Output varies with the seasonal discharge; surplus water is wasted during the monsoon.
  • Storage (reservoir) plant: A dam stores water in a reservoir during the high-flow season and releases it in the dry season or during peak hours. It can supply firm power throughout the year.
  • Pumped-storage plant: During off-peak hours, surplus thermal or hydro energy is used to pump water up to a high reservoir; this water is released back through the turbines during peak hours. It acts as a large storage battery.

Conclusion: The choice of type depends on the available head, the pattern of load demand and the river discharge, so that power is supplied reliably and economically.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 5 marks

Explain different types of hydro turbines and ways of selecting the appropriate one.

Answer

Different types of hydro turbines and their selection

1. Classification by action of water

(a) Impulse turbines — water pressure is converted into kinetic energy by a nozzle before striking the runner; the runner runs in air at atmospheric pressure.

  • Pelton wheel — one or more jets strike bucket-shaped vanes on a single runner. Used for high head (above about 250 m) and small to medium discharge. It is the only impulse turbine in common use.

(b) Reaction turbines — water enters the runner under pressure and leaves with both pressure and velocity reduced; the runner is fully enclosed and submerged in water.

  • Francis turbine — water enters radially inward and leaves axially; used for medium head (about 30–250 m) and medium discharge. It may be open-flume or closed (spiral casing) type.
  • Kaplan / propeller turbine — propeller-type runner with adjustable (Kaplan) or fixed (propeller) blades; used for low head (below about 30 m) and large discharge. Kaplan is the adjustable-blade version for varying load.

2. Classification by head

  • High head: Pelton (H > 250 m)
  • Medium head: Francis (30 m < H < 250 m)
  • Low head: Kaplan/propeller (H < 30 m)

3. Classification by specific speed (Ns)

  • Low specific speed (10–35): Pelton
  • Medium specific speed (60–300): Francis
  • High specific speed (300–1000): Kaplan/propeller

Ways of selecting the appropriate turbine

  1. Available head — the primary criterion. Measure the net head; choose Pelton for high head, Francis for medium, Kaplan for low head.
  2. Discharge (flow) available — from the river/stream flow-duration curve, take the design (firm) flow. Large discharge with low head points to Kaplan; small discharge with high head to Pelton.
  3. Specific speed — compute the required Ns from head, discharge and speed; match it with the turbine type that gives best efficiency at that Ns.
  4. Power output — estimate P = ρgQHη; the size and number of units follow from the plant capacity.
  5. Head variation — if head varies widely, prefer Kaplan (adjustable blades) or Pelton (multiple jets) for good part-load efficiency.
  6. Runner speed and generator matching — select speed so that the turbine can be directly coupled to the generator at synchronous speed.
  7. Cavitation and setting — check the Thoma's cavitation factor σ and set the turbine above the tailrace level accordingly; Francis and Kaplan are more cavitation-prone than Pelton.
  8. Site and economic factors — cost, ease of fabrication, transport and maintenance, sediment in water (Pelton is more abrasion-resistant), and availability of spare parts.

Conclusion — Head and discharge are the decisive parameters; specific speed links them to the turbine type. Final selection is made by comparing efficiency, cost and operating range of the candidate turbines at the design head and flow.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 1+4+5 marks

Describe transformer. Why same transformers are not suitable for the distribution and transmission system? Explain.

Answer

Transformer

Transformer is a static (stationary) electrical device that transfers electrical energy from one circuit to another by electromagnetic induction, without changing the frequency, at the same time stepping up or stepping down the voltage and current. It works on the principle of mutual induction and consists of two windings — primary and secondary — wound on a common laminated soft-iron core. When alternating current flows in the primary winding, an alternating magnetic flux is produced in the core, which links the secondary winding and induces an EMF in it. The voltage ratio is given by:

V2V1=N2N1=I1I2=K\frac{V_2}{V_1} = \frac{N_2}{N_1} = \frac{I_1}{I_2} = K

where N1,N2N_1, N_2 = turns of primary and secondary, and KK = transformation ratio. If N2>N1N_2 > N_1 it is a step-up transformer; if N2<N1N_2 < N_1 it is a step-down transformer. Since there are no rotating parts, transformer efficiency is very high (95–99%).

Why the same transformer is not suitable for both distribution and transmission

The requirements of the two systems are opposite, so one transformer cannot serve both:

  1. Voltage level: Transmission needs very high voltage (132 kV, 220 kV, 400 kV) to reduce I2RI^2R loss over long distances. Distribution needs low, safe voltage (400 V / 230 V) at the consumer end. A transformer designed for one voltage cannot give the other.

  2. Insulation: A transmission transformer must have heavy insulation (oil-impregnated paper, large clearance) to withstand high voltage and surges. A distribution transformer needs only light insulation. Using a transmission transformer in distribution is uneconomical; using a distribution transformer in transmission causes insulation breakdown.

  3. Current and conductor size: For the same kVA, a transmission transformer carries small current (large voltage), so thin conductors and small core are enough. A distribution transformer carries large current at low voltage, needing thick conductors and a bigger core.

  4. Cooling: Transmission transformers are large and use forced cooling (oil-forced, air-forced). Distribution transformers are small and use natural oil/air cooling. The cooling arrangement cannot be interchanged.

  5. Load and duty: Distribution transformers face fluctuating, unbalanced, single-phase-dominated loads and frequent overloading; transmission transformers carry balanced, steady three-phase loads. Their design (taps, winding arrangement) differs accordingly.

  6. Size and cost: For the same rating, a transmission transformer is bigger and costlier due to insulation and cooling; a distribution transformer is compact and cheap. Interchanging them wastes money and space.

  7. Protection: Transmission transformers need elaborate protection (differential, Buchholz, distance relays); distribution transformers use simple fuses and lightning arresters. The protection scheme is matched to the system, not transferable.

Conclusion: Since voltage, insulation, current, cooling, load nature and protection requirements of transmission and distribution systems are different, the same transformer cannot serve both purposes.

Answer checked by Claude Haiku · Oct 21, 2026

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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

In power stations and substations, the DC (direct current) system is an auxiliary supply that is kept independent of the main AC supply. It is necessary for the following reasons:

  1. Reliable control supply: Protection relays, control circuits, indicating lamps and annunciators need a supply that does not fail when the main AC supply fails. A station battery gives this continuity.

  2. Tripping and closing of circuit breakers: Circuit breakers must trip (open) and close even during a total AC failure or a system fault. The trip coil and closing coil are DC operated, so the DC system ensures the breaker can always be operated.

  3. Emergency lighting and essential auxiliaries: During blackout or station auxiliaries failure, DC supply runs emergency lighting, DC motors of critical drives and communication equipment.

  4. Independence from AC failure: Since the DC source is a storage battery (with charger), it is not affected by AC supply interruption, voltage dips or frequency variation. Thus protection and control remain healthy during faults.

  5. No interruption and simple storage: DC can be stored in batteries, so energy is available instantly without any starting delay (unlike a diesel generator), and there is no problem of synchronising or phase sequence.

  6. Operation of isolators, tap changers and annunciation: Motor-operated isolators, OLTC (on-load tap changer) controls and alarm/annunciation circuits in substations are commonly DC operated for safe and reliable operation.

Conclusion: The DC system acts as the "heart" of the control and protection scheme of a power station/substation, ensuring safe, reliable and uninterrupted operation of breakers, relays and emergency services even when the main AC supply fails.

Answer checked by Claude Haiku · Oct 21, 2026

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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 constant speed, producing an alternating EMF whose frequency is locked to the rotor speed. Its frequency is given by:

f=P⋅N120f = \frac{P \cdot N}{120}

where PP = number of poles, NN = speed in rpm.

Construction (essential parts)

  • Stator (armature): stationary part carrying a 3-phase winding in slots, from which output is taken.
  • Rotor (field): rotating part carrying field poles wound with DC-excited field winding.
  • Slip rings / exciter: supply DC to the rotor field.

Working Principle (Faraday's law of electromagnetic induction)

  1. The rotor field winding is excited by DC (from a DC exciter or slip rings), creating a rotating magnetic field of alternate N and S poles.
  2. The prime mover (turbine, diesel engine) rotates the rotor at synchronous speed Ns=120fPN_s = \dfrac{120f}{P}.
  3. The rotating magnetic field sweeps past the stationary stator conductors, so the flux linking each stator coil changes continuously.
  4. By Faraday's law, an EMF is induced in the stator winding: e=−Ndϕdte = -N\dfrac{d\phi}{dt}.
  5. Since the field poles alternate N–S, the induced EMF reverses every half revolution → alternating EMF of frequency f=PN/120f = PN/120.
  6. The magnitude of induced EMF per phase is:

Eph=4.44 f Φ Tph KwE_{ph} = 4.44 \, f \, \Phi \, T_{ph} \, K_w

where Φ\Phi = flux per pole, TphT_{ph} = turns per phase, KwK_w = winding factor.

  1. When the stator is connected to a load, current flows and the machine delivers electrical power. The rotor must keep turning at exactly synchronous speed — hence the name synchronous generator.

Figure

        Stator (armature) winding
        ┌───────────────────────────┐
        │   N                       │
        │      ┌─────────────┐      │
        │      │   ROTOR     │      │   ← DC field (excited)
        │  S   │  (field)    │  N   │
        │      └─────────────┘      │
        │   (rotates at Ns rpm)     │
        │      S                    │
        └───────────────────────────┘
             3-phase output → Load

The stator carries the 3-phase armature winding; the rotor carries the DC-excited field poles and is driven by the prime mover at synchronous speed.


Conditions for Parallel Operation of Synchronous Generators

Two or more alternators are connected in parallel to share load. Before closing the paralleling switch, the incoming machine must satisfy:

1. Equal voltage The terminal voltage (magnitude) of the incoming generator must equal the busbar voltage. This is adjusted by the field excitation.

2. Equal frequency The frequency of the incoming machine must equal the busbar frequency. This is adjusted by the prime-mover speed.

3. Same phase sequence The phase sequence of the incoming generator must match that of the busbar (R–Y–B order), otherwise the machines will not synchronise.

4. Zero phase difference (in phase) The incoming voltage must be exactly in phase with the busbar voltage at the instant of closing the switch. If not, a large circulating current flows and damages the machine.

4a. Waveform condition The voltage waveforms of both machines should be approximately sinusoidal (same shape), so that no circulating current flows due to harmonics.

Methods of checking these conditions

  • Three dark lamps method — lamps go dark when voltages are in phase.
  • Two bright, one dark lamps method — preferred, as it gives a clear indication.
  • Synchroscope — pointer at 12 o'clock position indicates exact synchronism.

After paralleling

  • Load sharing is controlled by the prime-mover governor (speed/frequency control).
  • Reactive power (power factor) sharing is controlled by the field excitation.

If the above conditions are not satisfied, heavy circulating currents flow between the machines, causing overheating, mechanical shock and possible damage to the windings.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2081 · 2nd · 10 marks

What are the effects of low power factor in the power system? Also describe any two methods to improve the power factor.

Answer

Effects of Low Power Factor in the Power System

Power factor, cos⁡ϕ=P/(VI)\cos\phi = P/(VI), is the ratio of active power to apparent power. When it is low, the same active power must be delivered at a much larger current, and this produces the following harmful effects:

  1. Increased line current. For a fixed load P=VIcos⁡ϕP = VI\cos\phi, a low cos⁡ϕ\cos\phi means a larger current II. All conductors, cables and windings must therefore carry more current than necessary.

  2. Higher I2RI^2R losses and heating. Since copper loss ∝I2\propto I^2, a low power factor sharply increases losses in transmission lines, transformers and generators, raising their temperature and reducing life.

  3. Poor voltage regulation and larger voltage drop. The increased current increases the voltage drop IRcos⁡ϕ+IXsin⁡ϕIR\cos\phi + IX\sin\phi along the line, so the consumer's terminal voltage falls and fluctuates badly.

  4. Reduced capacity (overloading) of plant. A generator, transformer or alternator is rated in kVA. At low power factor it delivers less useful kW for the same kVA rating, so its effective capacity is wasted and it appears overloaded.

  5. Increased size and cost of equipment. Cables, switchgear, circuit breakers and transformers must be designed for the larger current, so capital cost rises.

  6. Higher tariff / power factor penalty. Supply authorities charge on maximum kVA demand and impose a penalty (or refuse supply) below a specified power factor, usually 0.85–0.95 lagging.

  7. Poor efficiency and higher energy cost. Because of extra losses and larger kVA demand, the overall efficiency of the system falls and the cost of energy per unit rises.

  8. Reduced system stability and poor utilisation of installed capacity in long transmission lines.


Methods to Improve Power Factor

The basic principle is to supply the lagging reactive power (kVAr) required by the load locally, so that the line carries only active power. Any two of the following may be described.

1. Shunt (static) capacitor bank

Capacitors are connected in parallel with the load, either at the substation busbar or at the individual load terminals. 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.

If the load is PP kW at power factor cos⁡ϕ1\cos\phi_1 and it is to be raised to cos⁡ϕ2\cos\phi_2, the capacitor rating required is

Qc=P(tan⁡ϕ1−tan⁡ϕ2) kVArQ_c = P(\tan\phi_1 - \tan\phi_2)\ \text{kVAr}

Advantages: no moving parts, very low losses (about 0.5 %), low cost, simple installation, needs almost no maintenance, can be switched in steps as load varies.

Limitations: the reactive power supplied is fixed and falls as V2V^2 when voltage drops; switching causes transients; capacitors may resonate with supply harmonics.

2. Synchronous condenser (synchronous phase modifier)

An over-excited synchronous motor running on no load behaves as a synchronous condenser. By adjusting its field excitation it can be made to draw a leading current and thus supply reactive power to the system; under-excitation makes it absorb reactive power. The reactive output can be varied smoothly and continuously by merely changing the field current.

Advantages: continuously variable kVAr, gives fine voltage control, can be used for very large ratings (MVAr), improves system stability, and the same machine can be used as a motor when required.

Limitations: high capital cost, needs auxiliary starting equipment, has rotating parts so losses and maintenance are higher than for capacitors.

Other methods (for reference): over-excited synchronous motors used as drives, phase advancers for slip-ring induction motors, tap-changing transformers and boosters, and proper selection of motor rating so that motors are not lightly loaded.

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

(a) Function of a galvanometer — 2 marks

A galvanometer is a sensitive electrical instrument used to detect the presence and direction of a small electric current in a circuit. It also measures the magnitude of small currents. Its pointer deflects from the zero position when current flows through its coil, the direction of deflection indicating the direction of current.

(b) Working as a voltmeter — 3 marks

A galvanometer cannot be used directly as a voltmeter because:

  1. It has low resistance, so it would draw a large current and get damaged.
  2. Its deflection is proportional to current, not to potential difference.

So a galvanometer is converted into a voltmeter by connecting a high resistance R in series with it.

Working: Let G be the resistance of the galvanometer and I_g the current required for full-scale deflection. To measure a potential difference V, a series resistance R is added such that

V=Ig(G+R)V = I_g (G + R)

∴R=VIg−G\therefore R = \frac{V}{I_g} - G

The same current I_g flows through both the galvanometer and R. Since R is very large, the current drawn from the circuit is very small, so the voltmeter does not disturb the circuit. The deflection of the galvanometer is then directly proportional to the potential difference across the combination, so the scale can be calibrated in volts.

The voltmeter is connected in parallel with the component whose potential difference is to be measured.

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  • 2081 · 2nd · 5 marks

What is synchronization in power system? Why is it required? Describe.

Answer

Synchronization in Power System

Synchronization is the process of connecting an alternator (generator) to an already running power system (busbar) or connecting two separate power systems together, such that the incoming machine's voltage, frequency, phase sequence and phase angle match those of the running system before the circuit breaker is closed.

Conditions required for synchronization

  1. Equal voltage — the terminal voltage of the incoming generator must equal the busbar voltage (usually adjusted by the field excitation).
  2. Equal frequency — the frequency of the incoming generator must equal the system frequency (adjusted by the prime mover speed).
  3. Same phase sequence — the phase sequence of the incoming generator must match that of the busbar.
  4. Zero phase difference — the phase angle between the incoming voltage and busbar voltage must be zero at the instant of closing the breaker.

Why synchronization is required (necessity)

  1. To avoid heavy short-circuit current: If the generator is connected out of phase, a large circulating current flows between the machine and the busbar, which can damage the stator winding and the prime mover shaft.
  2. To avoid mechanical shock and damage: Wrong synchronization produces severe torque pulsations that stress the shaft, coupling and foundation of the alternator.
  3. To maintain power quality: Proper synchronization keeps voltage and frequency within permissible limits, protecting consumer equipment.
  4. For reliable and continuous supply: It allows additional generators to be paralleled so that load demand can be met and units can be taken out for maintenance without interruption.
  5. For economic and efficient operation: Generators can be loaded according to their most efficient (economic) loading, reducing generation cost.
  6. For system stability: Parallel operation through proper synchronization improves stability and reliability of the grid.

Methods of synchronization

  • Dark lamp method — lamps connected across the breaker glow dark at the instant of synchronizing.
  • Bright lamp method — lamps glow brightest at the instant of synchronizing.
  • Two bright, one dark lamp method — one lamp dark and two bright indicates correct phase sequence and synchronizing instant.
  • Synchroscope — indicates the phase difference directly; the breaker is closed when the pointer is at 12 o'clock position.

Conclusion: Synchronization is essential to safely parallel generators or grids so that voltage, frequency and phase sequence match, thereby preventing damage and ensuring stable, economic and reliable power supply.

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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

Secondary distribution system is that part of the electric power distribution system which supplies electric energy from the distribution transformer (secondary side of the transformer) to the ultimate consumers at utilisation voltage (400 V three-phase / 230 V single-phase in Nepal). It consists of the LT (low tension) lines, service mains, poles, insulators, and associated protective and metering equipment.

Secondary Distribution System in Use in Nepal (NEA Practice)

Nepal Electricity Authority (NEA) uses the following arrangement for secondary distribution:

  1. Voltage levels: 11 kV or 33 kV primary distribution is stepped down by 11/0.4 kV or 33/0.4 kV distribution transformers to a 3-phase, 4-wire, 400 V (line-to-line) / 230 V (line-to-neutral) secondary system. This is the standard secondary voltage in Nepal.

  2. System configuration: The LT network is mostly radial in nature. From each distribution transformer, LT feeders run along the streets on 9 m or 11 m PSC (pre-stressed concrete) or steel poles, supplying a group of consumers in that locality. This is essentially a radial secondary distribution system, which is simple and cheap but has poor reliability (a fault on the feeder interrupts all downstream consumers).

  3. Three-phase, four-wire distribution: The LT line has three phase conductors and one neutral conductor. Single-phase consumers (domestic loads) are connected between one phase and neutral (230 V), while three-phase consumers (small industries, commercial establishments) are connected across all three phases (400 V). Loads are distributed among the three phases to keep the system balanced.

  4. Service connection: From the nearest LT pole, a service main (service drop) is taken to the consumer's premises, where energy is measured by a single-phase or three-phase energy meter and protected by a main switch/MCB.

  5. Conductor and protection: LT lines generally use bare ACSR/AAAC or insulated conductors; service drops use insulated copper/aluminium conductors. Protection is provided by LT fuses, MCBs/MCCBs, and earthing at poles and at the consumer's premises. Pole-mounted distribution boxes with fuses are used for branching and isolation.

  6. Load management: To reduce voltage drop and losses, NEA uses multiple distribution transformers (small capacity, e.g. 50, 100, 200, 315 kVA) placed close to load centres, and reconfigures/extends LT feeders so that the secondary feeder length is kept short (generally within about 500 m).

Thus, the secondary distribution system in Nepal is a 400 V, 3-phase, 4-wire radial LT network fed from pole-mounted distribution transformers, with single-phase and three-phase service connections to consumers.

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  • 2081 · 2nd · 5 marks

Explain the common insulators used in distribution system.

Answer

Common Insulators Used in Distribution System

An insulator supports the conductor on the pole or cross-arm and isolates it from earth. It must have high dielectric and mechanical strength and resist moisture, dust and temperature change. Porcelain is the usual material; toughened glass and polymer (composite) insulators are also used.

1. Pin insulator Mounted on a steel pin fixed to the cross-arm; the conductor is tied in a groove on top (or on the side). Used on straight runs of 11 kV and 33 kV lines. It becomes bulky and costly above about 33 kV.

2. Suspension (disc) insulator Several porcelain or glass discs are joined in series by metal links to form a string, with the conductor hanging from the lowest disc. The number of discs increases with voltage (about one per 11 kV). It is used for 33 kV and above, and a damaged disc can be replaced individually. The disc nearest the conductor carries the highest voltage.

3. Strain insulator A suspension-type string, or a single disc, used horizontally at dead ends, sharp corners and long river or road crossings, where it takes the full conductor tension.

4. Shackle (spool) insulator A spool-shaped insulator fixed to the pole by a shackle (bracket), with the conductor tied in its groove. It is used on low-voltage lines (400/230 V) and on service connections, and works with either horizontal or vertical conductors.

5. Reel insulator A small cylindrical, drum-shaped insulator with a hole through the centre, used on low-voltage lines and on service entrances and dead ends. It is not used on EHV lines.

6. Stay (guy) insulator An egg-shaped insulator placed in the stay wire to insulate the lower part of the stay from the pole, so a live conductor touching the stay does not endanger people near the ground.

7. Post insulator A rigid pedestal-type insulator for 11-33 kV lines and substations, where it can be mounted vertically or horizontally and has higher strength than a pin insulator.

InsulatorTypical use
Pin11/33 kV straight line
Suspension33 kV and above
StrainDead end, corner, crossing
Shackle / ReelLV distribution, service connection
StayGuy wire
PostSubstation, 11-33 kV

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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 an assembly of electrical equipment (transformers, circuit breakers, isolators, busbars, protective relays, instrument transformers, etc.) that is used to:

  • step the voltage up or down,
  • switch, control and protect transmission and distribution circuits,
  • interconnect generating stations, transmission lines and distribution networks.

It is the link between generation, transmission and distribution.

Types Based on Operating Voltage

1. High Voltage (HV) substations: 11 kV to 66 kV Used in sub-transmission and distribution to supply towns, industries and large consumers. Examples: 33/11 kV and 66/11 kV substations, such as the 33/11 kV substations of the Nepal Electricity Authority (NEA).

2. Extra High Voltage (EHV) substations: 132 kV to 400 kV Form the backbone of the transmission system and connect power stations and load centres. They are mostly outdoor because of the large clearances needed. Examples: 132/33 kV, 220/132 kV and 400/220 kV substations. In Nepal, 132 kV and 220 kV substations are the main transmission nodes, and 400 kV (such as Dhalkebar) is being developed.

3. Ultra High Voltage (UHV) substations: above 400 kV Used for bulk power transfer over very long distances with low losses, for example 765 kV and 1200 kV systems. They need special insulation, large clearances and advanced protection.

Other classifications (for reference)

  • By function: step-up, step-down, switching, distribution, power-factor-correction and frequency-changer substations.
  • By construction: indoor, outdoor, underground and pole-mounted substations.

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 LinesUnderground Cabling
1Conductors are supported in air on poles or towers.Conductors (cables) are laid directly in the ground below the surface.
2Insulation is provided by air, so bare conductors can be used.Requires costly insulating material such as paper, PVC or XLPE around the conductor.
3Initial (capital) cost is low.Initial cost is very high — several times that of an overhead line.
4Fault location and repair are easy and quick.Fault location is difficult and repair needs excavation, so it is time-consuming.
5Highly exposed to lightning, storms, trees and public contact; chance of faults and accidents is high.Not affected by lightning, storms or external contact; safer and more reliable in operation.
6Requires wide right-of-way and is unsightly; voltage drop and corona loss are present.Needs no right-of-way, is neat in appearance, and has negligible voltage drop and no corona loss.

(Any five of the above may be written.)

Answer checked by Claude Haiku · Oct 21, 2026

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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

(a) Purpose of a Megger (2 marks)

A Megger is an insulation resistance tester. It is used to measure the very high resistance (in megohms) of insulation in cables, motors, generators, transformers and switchgear. Its purposes are:

  1. To check whether insulation is healthy or has deteriorated through moisture, dirt, ageing or damage.
  2. To test new or repaired equipment before it is energised.
  3. To prevent earth leakage, short circuits, electric shock and fire.
  4. To confirm that the insulation resistance is above the minimum value required by standards.

(b) Working principle (8 marks)

1. Construction

A Megger has two main parts:

  • A hand-driven (or motor-driven) DC generator that gives a high test voltage (500 V, 1000 V, 2500 V or 5000 V). A centrifugal clutch keeps the speed (about 160 rpm) and so the voltage steady.
  • A permanent magnet cross-coil (ratio) moving system with two coils fixed on the same spindle at an angle to each other in the field of the magnet. There is no controlling spring. The pointer moves over a scale marked in megohms, with infinity at one end and zero at the other.

The two coils are:

  • Current (deflecting) coil, connected in series with the insulation under test (and a protective resistor), so it carries the leakage current through the insulation.
  • Pressure (voltage or control) coil, connected across the generator in series with a high fixed resistance RpR_p, so it carries a current proportional to the test voltage.

2. Working

The test voltage VV from the generator is applied across the insulation resistance RR between the Line (L) and Earth (E) terminals. Then

Icurrent coil=VR+R1≈VR,Ipressure coil=VRpI_{\text{current coil}} = \frac{V}{R + R_1}\approx\frac{V}{R},\qquad I_{\text{pressure coil}} = \frac{V}{R_p}

The two coils produce torques in opposite directions. The current coil tends to turn the pointer toward the zero end, and the pressure coil tends to turn it toward the infinity end. The pointer comes to rest where the two torques balance. This position depends only on the ratio of the currents:

θ∝Icurrent coilIpressure coil=V/RV/Rp=RpR\theta \propto \frac{I_{\text{current coil}}}{I_{\text{pressure coil}}} = \frac{V/R}{V/R_p} = \frac{R_p}{R}

The test voltage VV cancels, so the reading depends only on RR and does not change with the speed at which the generator is cranked. The scale is therefore calibrated directly in MΩ, and it is non-linear.

  • If the insulation is good, RR is very large and the leakage current is tiny, so the pointer rests at infinity.
  • If the insulation is faulty or the terminals are shorted, the current coil current is large, so the pointer goes to zero.

3. Guard terminal

Surface leakage over dirty or damp insulation would give a falsely low reading. The Guard (G) terminal is connected to a conducting wrapping on the surface, which carries the surface leakage current straight back to the generator without passing through the current coil. The Megger then measures only the true volume resistance of the insulation.

4. Precautions

  1. Disconnect and de-energise the equipment before testing.
  2. Rotate the crank at the rated speed and take the reading after about 60 seconds, when the pointer is steady.
  3. Discharge the equipment to earth after the test to remove stored charge.
  4. Keep the test leads clean and well insulated.

Significance: A regular insulation test is a key part of preventive maintenance. A falling insulation resistance warns of a fault before it causes breakdown.

Answer checked by Claude Sonnet · 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, machines, and installations to keep them in proper working condition, to prevent breakdowns, to restore them when they fail, and to ensure safe, reliable, and efficient operation throughout their useful life. Its main objectives are:

  • To ensure continuous and reliable supply/service
  • To prevent premature failure and extend equipment life
  • To reduce downtime and repair cost
  • To ensure safety of personnel and equipment
  • To maintain efficiency (reduce losses) and compliance with standards

Maintenance is broadly of three types:

  1. Preventive maintenance – carried out at fixed intervals before failure occurs.
  2. Breakdown (corrective) maintenance – carried out only after a fault has occurred.
  3. Predictive (condition-based) maintenance – carried out based on measured condition (temperature, oil test, vibration, etc.).

Preventive Maintenance of Distribution Transformers

A distribution transformer is the costliest and most critical equipment of a distribution system. Its preventive maintenance is done periodically (usually every 6 months to 1 year, and more frequently for heavily loaded or outdoor units) as follows:

A. External inspection (visual checks)

  1. Check the transformer body, tank, and radiator for oil leakage, rust, or damage.
  2. Check oil level in the conservator/gauges; top up with tested transformer oil if low.
  3. Check silica gel in the breather — replace or reactivate when it turns pink.
  4. Check bushings (HT and LT) for cracks, dust, and carbon deposits; clean them.
  5. Check tightness of all terminal connections, nuts, and bolts; tighten if loose.
  6. Check earthing (body earth and neutral earth) — earth resistance should be within limits (generally ≤ 5 Ω for body earth).
  7. Check the breather pipe, vent pipe, and explosion vent for blockage.
  8. Check the tap changer position and lock it; operate it only off-load.
  9. Check the nameplate, danger boards, and fencing of the transformer.

B. Electrical tests and measurements

  1. Measure insulation resistance of HT–earth, LT–earth and HT–LT with a megger (should be high, generally in MΩ range).
  2. Measure winding resistance and earth resistance.
  3. Measure no-load and load current/voltage to check loading and unbalance.
  4. Check oil dielectric strength (BDV) — oil should be filtered/replaced if BDV falls below the specified value (about 30 kV for 11 kV class).
  5. Check oil acidity, moisture, and sludge; test oil at intervals.
  6. Check the transformer ratio and polarity if required.

C. Protective devices and accessories

  1. Test the Buchholz relay (for oil-immersed transformers above 500 kVA) and clean its contacts.
  2. Check explosion vent diaphragm, pressure relief device, and oil temperature gauge.
  3. Check HT fuse, LT fuse/MCCB, lightning arresters and replace blown/damaged ones.
  4. Verify overload and earth-fault protection settings and relay operation.
  5. Check the cooling arrangement (radiators, fans) and clean the fins for free air circulation.

D. Load and operating condition

  1. Keep loading within rated capacity; avoid sustained overload and severe unbalance.
  2. Record maximum demand, oil temperature, and winding temperature regularly.
  3. Ensure adequate ventilation and clearance around the transformer.
  4. Avoid loose connections that cause sparking and local heating.

E. Record keeping

Maintain a maintenance register/history card for each transformer showing date of inspection, oil test results, insulation resistance, faults found, and action taken. This helps in condition-based replacement and in planning maintenance.

Benefits

Preventive maintenance of distribution transformers reduces failure rate, avoids long and costly outages, extends transformer life (20–25 years), lowers oil and energy losses, and ensures safety of the public and operating staff.

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 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.