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Chapter 1

Electrical Engineering Fundamentals

Nepal Electricity Authority past exam questions

Past questions and answers

13 questions set from this chapter. Most repeated first.

  • 2082 · 2nd · 5 marks

State Thevenin's theorem and Norton's theorem and their applications.

Answer

Thevenin's Theorem

Any linear, bilateral, active two-terminal network containing one or more voltage/current sources and resistances can be replaced by a single equivalent voltage source VthV_{th} in series with a single equivalent resistance RthR_{th}, where:

  • VthV_{th} = open-circuit voltage across the terminals
  • RthR_{th} = equivalent resistance seen from the terminals with all independent sources killed (voltage sources short-circuited, current sources open-circuited)

Norton's Theorem

Any linear, bilateral, active two-terminal network can be replaced by a single equivalent current source INI_N in parallel with a single equivalent resistance RNR_N, where:

  • INI_N = short-circuit current through the terminals
  • RNR_N = equivalent resistance seen from the terminals with all independent sources killed

Relation between the two: Rth=RNR_{th} = R_N and Vth=INRNV_{th} = I_N R_N (source transformation).

Applications

  1. Simplifies analysis of complex networks by reducing them to a simple two-element equivalent.
  2. Used to find the current/voltage in a single variable branch (load) without solving the whole network.
  3. Maximum power transfer analysis: maximum power is delivered to the load when RL=Rth=RNR_L = R_{th} = R_N.
  4. Used in circuit design, matching networks, and analysis of power system networks (e.g., finding fault current and equivalent source impedance at a bus).
  5. Basis for modelling active devices and networks as equivalent sources in electronic circuit analysis.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 5 marks

Briefly explain operating principle and application of maximum demand meter.

Answer

Maximum Demand Meter — Operating Principle and Application

Operating Principle

A maximum demand (MD) meter measures and records the highest average power demand (usually in kW or kVA) drawn by a consumer over a fixed integration period, typically 15 or 30 minutes. Its operation rests on the fact that electricity tariffs are based not on instantaneous load but on the peak average demand, so the meter must "remember" the largest such average.

The classical thermal (thermal-type) MD meter works as follows:

  1. A current transformer (and potential transformer) feeds a current proportional to the load into a small heating coil wound around a bimetallic strip or a thermal element.
  2. The heat produced is proportional to I2RI^2R, i.e. to the power demand. The bimetallic strip deflects in proportion to this demand.
  3. A pusher (friction) mechanism drags a pointer (the "maximum" indicator) along a scale. The pointer moves forward as demand rises but cannot return when demand falls, because the pusher slips back over the pointer.
  4. Hence the pointer stays at the highest position reached — the maximum demand — until it is manually reset by the meter reader after recording.

In modern electronic (static) MD meters, the same logic is implemented digitally:

  • The meter samples voltage and current, computes instantaneous power, and averages it over each integration period (sliding or fixed block of 15/30 min).
  • A register stores the largest average value encountered in the billing cycle; the value is displayed and reset each month.

The demand is often expressed in kVA to account for power factor, using MD (kVA)=MD (kW)cos⁡ϕ\text{MD (kVA)} = \frac{\text{MD (kW)}}{\cos\phi}

Application

  1. Tariff billing: Used by utilities (including NEA) to levy the demand charge (Rs/kVA or Rs/kW per month) on industrial, commercial and bulk consumers, in addition to energy charge. This recovers the cost of generation, transmission and distribution capacity kept ready to serve the consumer's peak.
  2. Load management and power factor improvement: A high recorded MD warns the consumer to stagger loads or install capacitors, thereby reducing the demand charge and improving the system power factor.
  3. System planning: MD data from various feeders and substations help the utility forecast peak load, size transformers, conductors and generating units, and decide augmentation schedules.
  4. Consumer monitoring: Helps detect unauthorized load growth or inefficient use of connected load.

Thus the MD meter is an essential instrument for both revenue collection and efficient planning of the power system.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

Draw the schematic diagram of a hydroelectric power plant and explain its various components in detail.

Answer

(b) Schematic diagram of a hydroelectric power plant

   Reservoir / Intake
   ┌───────────────────┐
   │   (Forebay)       │
   └─────────┬─────────┘
             │  Penstock (with surge tank)
             ▼
        ┌─────────┐
        │ Turbine │◄─────── Generator ───────► Transformer ──► Transmission line
        └────┬────┘
             │
        ┌────▼────┐
        │ Draft   │
        │  Tube   │
        └────┬────┘
             ▼
        Tailrace / River downstream

Components and their functions

  1. Catchment area / reservoir — The watershed that collects rain and snow-melt; the reservoir stores water to ensure a regulated, continuous supply and to create the head.

  2. Intake structure — The opening with trash racks and gates that admits water from the reservoir into the conduit while screening out debris and allowing flow to be shut off for maintenance.

  3. Headrace / conveyance system — Canal, tunnel or flume that carries water from the intake to the forebay with minimum loss of head.

  4. Forebay and surge tank — The forebay provides a small storage to match sudden load changes; the surge tank absorbs the pressure surge (water hammer) set up when the turbine gates open or close rapidly.

  5. Penstock — A closed pipe (steel or concrete) that conveys water under pressure from the forebay to the turbine, converting the available head into kinetic energy at the turbine inlet. It is designed for the full static plus surge pressure.

  6. Turbine — The prime mover. Water strikes the runner and rotates it, converting hydraulic energy into mechanical shaft energy. Choice depends on head and discharge: Pelton (high head), Francis (medium head), Kaplan/propeller (low head, large flow).

  7. Generator — Coupled to the turbine shaft; converts mechanical energy into electrical energy (usually a synchronous 3-phase alternator at 11 kV or so).

  8. Transformer and switchgear — Step-up transformer raises the voltage for economical long-distance transmission; switchgear (circuit breakers, isolators, protection relays) controls and protects the plant.

  9. Draft tube — A gradually expanding conduit that carries water from the turbine outlet to the tailrace, recovering part of the kinetic energy and allowing the turbine to be set above tailwater level.

  10. Tailrace — The channel that returns the used water to the river downstream.

  11. Auxiliaries — Governing system (controls speed/frequency by regulating water flow), excitation system, cooling, lubrication, and control/monitoring equipment.

Working in brief: Water stored at a head H flows through the penstock, strikes the turbine runner, and leaves via the draft tube to the tailrace. The turbine drives the generator, producing electrical energy that is stepped up and fed to the grid. The available power is approximately

P=ρgQHηP = \rho g Q H \eta

where Q is the discharge, H the net head and η the overall efficiency.

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  • 2082 · 2nd · 10 marks

Explain the functions of circuit breaker, isolator and earthing switch. Why interlocking is necessary between them? Describe differential protection and state its various applications.

Answer

Functions of Circuit Breaker, Isolator and Earthing Switch

1. Circuit Breaker (CB)

  • It is a switching as well as protective device that makes, carries and breaks the circuit under normal and fault conditions.
  • On receipt of a tripping signal from a relay, it automatically interrupts the fault current within a few cycles, thereby isolating the faulty section and protecting equipment.
  • Its arc-quenching medium may be air blast, oil, SF₆ or vacuum. It can break full-load current and short-circuit current.

2. Isolator (Disconnecting Switch)

  • It is a no-load (off-load) switching device used only to isolate a circuit or equipment from the live supply for maintenance, by providing a visible air gap.
  • It cannot break or make load current or fault current; it is operated only after the circuit breaker has interrupted the circuit, and closed before the CB is closed.
  • It gives the statutory safety clearance (visible break) to the maintenance crew.

3. Earthing Switch

  • It is a switch that connects a circuit or equipment solidly to earth after isolation, to discharge trapped/induced charges and to keep the section at earth potential during maintenance.
  • It protects the working personnel against accidental re-energisation and against induced or lightning surges on the isolated line.
  • Like the isolator, it is a no-load device and is closed only after the isolator is opened and the line is de-energised.

Necessity of Interlocking

Interlocking is a mechanical/electrical safety arrangement that enforces the correct sequence of operation:

CB open→Isolator open→Earthing switch closed\text{CB open} \rightarrow \text{Isolator open} \rightarrow \text{Earthing switch closed}

and the reverse sequence while restoring supply. It is necessary because:

  1. Isolators and earthing switches have no arc-quenching capability. If opened or closed on load, the arc cannot be extinguished, causing flashover, equipment damage and danger to the operator.
  2. It prevents earthing a live circuit (which would cause a direct short circuit) and prevents closing the isolator while the earthing switch is closed.
  3. It prevents the CB from being closed while the isolator or earthing switch is in the wrong position.
  4. It eliminates human error in switching sequences, ensuring safety of personnel and plant as required by safety rules.

Differential Protection

Principle: It is a unit protection scheme based on Kirchhoff's current law — the current entering a protected zone must equal the current leaving it. CTs of identical ratio are placed on both sides of the protected equipment and their secondaries are connected in a circulating-current arrangement to a relay.

Iˉrelay=Iˉ1−Iˉ2\bar{I}_{\text{relay}} = \bar{I}_1 - \bar{I}_2

  • Normal/through-fault condition (external fault): Iˉ1=Iˉ2\bar{I}_1 = \bar{I}_2, so the relay current is (ideally) zero and the relay does not operate.
  • Internal fault: the sum of currents is no longer zero, a differential current flows through the relay, which operates and trips the CB.

Merits: Very fast (operates in a few ms), selective (responds only to faults inside its zone), and sensitive (can detect low-magnitude internal faults).

Limitations: Requires matched CTs, is affected by CT saturation, ratio error and magnetising inrush (in transformers), so harmonic restraint (2nd harmonic) and bias/percentage differential characteristics are used.

Applications:

  1. Power transformers — biased differential with harmonic restraint.
  2. Generators and large motors — stator winding phase-fault protection.
  3. Busbars — high-impedance or biased differential busbar protection.
  4. Transmission lines and feeders — pilot-wire / carrier-current differential protection.
  5. Cables — differential (or core-balance) protection against phase faults.
  6. Generator-transformer unit (GT) protection — overall unit differential protection.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

What are the methods of testing of transformer? Explain the principle of operation of parallel operation of transformers.

Answer

1. Methods of Testing of Transformers

Transformers are tested to verify losses, efficiency, voltage regulation, temperature rise and insulation strength.

(a) Performance tests

  • Open circuit (no-load) test: Rated voltage is applied to the LV side with the HV side open. Wattmeter, voltmeter and ammeter readings give the iron (core) loss and the magnetising branch parameters R0R_0 and X0X_0. Copper loss is negligible because the no-load current is small.
  • Short circuit test: The LV side is short-circuited and a reduced voltage (about 5-10% of rated) is applied to the HV side so that rated current flows. The wattmeter reading gives the full-load copper loss, and the readings give ReqR_{eq}, XeqX_{eq} and ZeqZ_{eq}. Core loss is negligible at this low voltage.
  • Sumpner's (back-to-back) test: Two identical transformers are connected back-to-back and loaded by circulating current. Full-load efficiency, regulation and temperature rise are found while drawing only the losses from the supply.
  • Direct load test: The transformer is loaded directly. It is used only for small units.

(b) Other routine and type tests

  • Ratio and polarity/vector-group test: checks the turns ratio, winding polarity and phase displacement.
  • Insulation resistance (megger) test: measures insulation between windings and to earth.
  • Dielectric tests: the separate-source (applied) voltage test, the induced over-voltage test and the lightning impulse test (1.2/50 μ\mus wave).
  • Transformer oil test: measures breakdown voltage (BDV) and dissolved gases.
  • Temperature rise test: the unit is run at rated load until the temperature is steady.

2. Principle of Parallel Operation

Two or more transformers are in parallel when their primaries are connected to the same supply bus and their secondaries to the same load bus. Because both secondaries sit across the same bus voltage, the load divides according to their impedances.

Conditions for satisfactory operation

  1. Same voltage ratio, otherwise a circulating current flows even at no load.
  2. Same polarity, otherwise there is a dead short circuit.
  3. Same phase sequence and phase angle (vector group) for three-phase units.
  4. Equal per-unit (percentage) impedance on their own ratings, with similar X/RX/R ratios, so that the load is shared in proportion to kVA rating and the currents are in phase.

Load sharing

Let the transformers have equivalent impedances ZAZ_A and ZBZ_B in ohms, referred to the same side. They carry load currents IAI_A and IBI_B at the same terminal voltage, so

IAZA=IBZB⇒IAIB=ZBZAI_A Z_A = I_B Z_B \quad\Rightarrow\quad \frac{I_A}{I_B}=\frac{Z_B}{Z_A}

The currents are inversely proportional to the ohmic impedances. In terms of kVA ratings SA,rS_{A,r}, SB,rS_{B,r} and per-unit impedances zAz_A, zBz_B (each on its own rating), the load shared is

SA=Stotal SA,r/zASA,r/zA+SB,r/zB,SB=Stotal SB,r/zBSA,r/zA+SB,r/zBS_A = S_{total}\,\frac{S_{A,r}/z_A}{S_{A,r}/z_A+S_{B,r}/z_B},\qquad S_B = S_{total}\,\frac{S_{B,r}/z_B}{S_{A,r}/z_A+S_{B,r}/z_B}

If zA=zBz_A=z_B, then SA:SB=SA,r:SB,rS_A:S_B=S_{A,r}:S_{B,r}, so the load is shared in proportion to the ratings. If the per-unit impedances differ, the one with the lower zz takes more than its share and overloads first, which limits the total load that can be taken.

Advantages: continuity of supply, since one unit can be maintained while the other runs. Units can be run near full load for better efficiency, and capacity can be added later.

Disadvantages: circulating currents and extra heating if the conditions are not met. The fault level at the bus is higher, and protection and synchronising are more complex.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

The efficiency of a 100kVA, 110/220V, 50Hz single phase transformer is 98.5% at half full load at 0.8 p.f. leading and 98.8% at full load unity p.f. Determine iron loss, full load copper loss and maximum efficiency at unity p.f.

Answer

Let
(W_i) = iron loss,
(W_{cu,FL}) = full-load copper loss.

At half full load, (x=0.5), p.f. (=0.8) leading:

[ P_{o1}=0.5\times 100\times 0.8=40\text{ kW} ]

[ \eta_1=\frac{P_{o1}}{P_{o1}+W_i+x^2W_{cu,FL}} ]

[ 0.985=\frac{40}{40+W_i+0.25W_{cu,FL}} ]

Total losses at half load:

[ 40\left(\frac{1}{0.985}-1\right)=0.60914\text{ kW}=609.14\text{ W} ]

So,

[ W_i+0.25W_{cu,FL}=609.14 \quad (1) ]

At full load, unity p.f.:

[ P_{o2}=100\text{ kW} ]

[ 0.988=\frac{100}{100+W_i+W_{cu,FL}} ]

Total losses at full load:

[ 100\left(\frac{1}{0.988}-1\right)=1.21457\text{ kW}=1214.57\text{ W} ]

So,

[ W_i+W_{cu,FL}=1214.57 \quad (2) ]

Subtracting (1) from (2):

[ 0.75W_{cu,FL}=1214.57-609.14=605.43 ]

[ W_{cu,FL}=807.24\text{ W} ]

Then from (2):

[ W_i=1214.57-807.24=407.33\text{ W} ]

Hence,

[ \boxed{W_i \approx 407.3\text{ W}} ]

[ \boxed{W_{cu,FL} \approx 807.2\text{ W}} ]

For maximum efficiency,

[ W_i=x^2W_{cu,FL} ]

[ x=\sqrt{\frac{W_i}{W_{cu,FL}}} =\sqrt{\frac{407.33}{807.24}} =0.7103 ]

At unity p.f., output at maximum efficiency:

[ P_{o,\max}=x\times 100\times 1=71.03\text{ kW} ]

Total loss at maximum efficiency:

[ 2W_i=2\times 407.33=814.66\text{ W} ]

[ \eta_{\max}=\frac{71030}{71030+814.66} =0.98866 ]

[ \boxed{\eta_{\max}\approx 98.87%} ]

So, the required values are:

[ \boxed{W_i=407.3\text{ W},\quad W_{cu,FL}=807.2\text{ W},\quad \eta_{\max}=98.87%} ]

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  • 2082 · 2nd · 5 marks

What is calibration of testing device? Explain its significance in maintaining the accuracy and precision of electrical measurement instruments.

Answer

Calibration of a testing device is the process of comparing the readings of an instrument (the device under test) against a reference standard of known, higher accuracy, and then adjusting or documenting the deviation between them. The reference standard is traceable to a national/international standard (in Nepal, through the Nepal Bureau of Standards and Metrology, NBSM). Formally, calibration establishes the relationship between the indicated value and the true value of the quantity being measured, usually by preparing a calibration curve or a table of corrections.

Significance in maintaining accuracy and precision:

  1. Ensures measurement accuracy (closeness to true value): It detects and quantifies systematic errors such as zero error, gain (scale) error, and nonlinearity in voltmeters, ammeters, wattmeters, energy meters, CTs, PTs, and relays. Corrections or adjustments are then applied so that readings agree with the standard.

  2. Ensures precision and consistency (repeatability): Periodic calibration confirms that repeated measurements of the same quantity give closely grouped results, so different instruments and different laboratories give comparable readings.

  3. Traceability and legal validity: Calibrated instruments are traceable to national/international standards. This is essential for NEA's commercial metering, where energy bills must be legally defensible and disputes settled reliably.

  4. Consumer and revenue protection: In energy meters and instrument transformers, even a small percentage error causes large financial loss or overcharging over time. Calibration keeps errors within the permissible limits prescribed by NEA/NBSM (e.g., ±0.5% for static meters, ±1% for CTs/PTs of class 1).

  5. Safety and reliability of the power system: Protective relays, circuit breakers, and protection test sets must operate at correct pick-up and time settings. Uncalibrated devices can cause maloperation or failure to trip, endangering personnel and equipment.

  6. Quality control and standards compliance: It satisfies quality management requirements (e.g., ISO/IEC 17025 for testing laboratories) and NEA's internal maintenance schedules, reducing downtime and rework.

  7. Detection of drift and degradation: Regular calibration detects ageing, temperature drift, and wear of components, allowing timely repair or replacement before errors exceed limits.

  8. Cost-effective operation: Accurate instruments reduce losses, avoid penalty for inaccurate billing, and prevent costly system disturbances.

In short, calibration is the backbone of reliable electrical measurement: without it, accuracy and precision cannot be guaranteed, and NEA's metering, protection, and testing functions would lose both technical and commercial credibility.

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  • 2082 · 2nd · 5 marks

Briefly explain tariff schemes in Nepal.

Answer

Tariff schemes in Nepal

The Nepal Electricity Authority (NEA) sells electricity under tariff rates approved by the Electricity Regulatory Commission (ERC) after a public hearing. The main features are:

1. Consumer categories. Separate rates apply to domestic, commercial, industrial, irrigation, water supply, temple, street-light, transport and other consumers. The rates reflect the cost of supply and social policy.

2. Block (slab) tariff. Domestic consumers pay an energy charge that rises with consumption. For example, a low rate applies to the first few units (such as 0-20 units), a higher rate to the next block, and progressively higher rates beyond that. This protects small consumers and discourages wasteful use.

3. Two-part tariff. Larger industrial and commercial consumers pay two components:

  • Demand charge: based on the maximum demand (kVA) or contract demand, and payable whether or not energy is used.
  • Energy charge: based on the units (kWh) consumed.

This recovers the fixed cost of capacity and encourages a better load factor.

4. Time-of-Day (TOD) tariff. Consumers with TOD meters pay different rates for peak hours (evening), off-peak hours (night) and normal hours. Cheaper off-peak energy shifts load away from the peak and improves the system load factor.

5. Minimum and service charge. A monthly minimum or service charge is levied even if little energy is used, to cover metering, billing and connection costs.

6. Other charges and subsidy. These include meter rent, a power factor penalty or rebate, and a penalty for exceeding the sanctioned load. Lifeline domestic consumers and irrigation are cross-subsidised by commercial and industrial consumers.

Conclusion. Nepal's tariff is a mix of block rates for households, two-part and TOD rates for bulk consumers, and service and penalty charges. It is designed to recover cost, promote equity and improve the load factor.

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  • 2082 · 2nd · 10 marks

What is the significance of earthing of equipment in substation? Explain the role of cooling system in substation with its significance.

Answer

Significance of Earthing of Equipment in a Substation

Earthing (grounding) means connecting the non-current-carrying metallic parts of equipment — transformer tanks, switchgear enclosures, structures, cable sheaths — to the general mass of earth through a low-resistance path. Its significance:

  1. Safety of personnel. If insulation fails, the equipment frame rises to line potential. A low-resistance earth path keeps touch and step voltages within safe limits, preventing electric shock to operators.
  2. Protection of equipment. Earthing provides a definite, low-impedance return path for fault current so that relays and circuit breakers operate quickly and isolate the faulty section, limiting damage to transformers, CTs, PTs and busbars.
  3. System neutral earthing. Earthing the transformer neutral stabilises the phase-to-earth voltage, limits overvoltages during arcing faults, and gives a path for earth-fault current so earth-fault relays can sense it.
  4. Lightning and surge protection. Surge arresters, lightning masts and shield wires must be earthed to divert lightning and switching surges safely to earth, protecting insulation.
  5. Voltage reference and stability. Earth provides a zero-potential reference for the whole system, keeps voltages balanced, and prevents the accumulation of static charge.
  6. Fire and fault containment. By clearing faults rapidly, earthing reduces the risk of fire, explosion of oil-filled equipment and spread of faults to healthy phases.

In short, earthing is the backbone of substation safety — it protects people, plant and the continuity of supply.

Role of the Cooling System in a Substation

Substation equipment — mainly power transformers, but also reactors, rectifier sets and control rooms — generates heat from copper loss (I2RI^2R) and core loss (eddy current and hysteresis). This heat must be removed continuously. The cooling system performs the following roles:

  1. Removal of generated heat. It transfers heat from windings and core to the surrounding air or water so that the temperature rise stays within the limits fixed by the insulation class.
  2. Preservation of insulation life. Transformer insulation (paper, oil) ages roughly by the Arrhenius law — life is halved for every 6–8 °C rise above rated hot-spot temperature. Effective cooling directly extends insulation life.
  3. Maintaining rated capacity. Without adequate cooling the transformer must be derated. Cooling allows the same unit to carry full load, and with forced cooling (ONAF/OFAF) it can be overloaded during peak demand.
  4. Preventing oil degradation and failure. Excess heat causes oil oxidation, sludge formation and gas evolution, which reduce dielectric strength and may lead to breakdown or fire.
  5. Ensuring reliability and economy. By avoiding premature failure and forced outages, cooling reduces maintenance cost and improves supply reliability.

Significance

Cooling is not an accessory but an essential auxiliary of a substation. It fixes the continuous rating of the transformer, governs its life expectancy, prevents catastrophic failure, and allows economic loading. A failure of the cooling system is treated as a serious contingency — protection is arranged to trip or alarm the transformer before the winding temperature reaches a dangerous value. Thus the cooling system is vital for the safe, reliable and economical operation of the substation.

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  • 2082 · 2nd · 10 marks

Briefly mention the factor governing the choice of economic span length of a transmission line. Why high voltage transmission system is required in modern power system? Explain.

Answer

Factors Governing the Choice of Economic Span Length of a Transmission Line

The economic span is that distance between two consecutive supports at which the total annual cost of the line (capital cost of supports, insulators and foundations plus the cost of conductors and the cost of energy lost) is minimum. The main factors governing this choice are:

  1. Cost of supports (towers/poles): Taller or stronger supports cost more. For a given ground clearance and sag, a longer span requires taller supports, so support cost per km falls as span increases but the cost per support rises.

  2. Cost of conductors: For a fixed sag, the conductor tension varies inversely with span. Longer spans need greater tension and hence stronger (more expensive) conductors and larger cross-sections to keep stress within limits.

  3. Cost of insulators and fittings: Number of insulator strings per km decreases as span increases (fewer supports per km), but each string must withstand higher mechanical loading.

  4. Cost of foundations: Fewer foundations per km with longer spans, but each foundation must carry greater vertical and transverse loads.

  5. Sag and ground clearance: Sag is proportional to the square of the span (for equal tension). A longer span therefore needs greater clearance or taller supports, which adds to cost.

  6. Wind and ice loading: In exposed terrain, longer spans attract greater wind/ice loads, requiring heavier towers and conductors.

  7. Right-of-way and terrain: Hilly or forested terrain favours longer spans to reduce the number of towers and clearing cost; flat terrain may favour shorter spans.

  8. Cost of energy losses: Longer spans with higher tension may use larger conductors, reducing I²R loss; this energy-saving must be balanced against higher capital cost.

  9. Maintenance and accessibility: More supports mean more inspection and maintenance; longer spans reduce this but make faults harder to locate and repair.

The optimum is found by equating the annual fixed charges (support + conductor + insulator + foundation) to the annual cost of energy lost, i.e. minimising total annual cost.


Why High Voltage Transmission is Required in Modern Power Systems

Modern power systems transmit large blocks of power over long distances from generating stations (often remote hydro/thermal sites) to load centres. High voltage (HV) and extra-high voltage (EHV) transmission is essential for the following reasons:

1. Reduction in transmission losses Power loss in a line is Ploss=I2R=P2RV2cos⁡2ϕP_{\text{loss}} = I^2 R = \frac{P^2 R}{V^2 \cos^2\phi} For the same power PP and conductor resistance RR, loss varies inversely as the square of the voltage. Raising voltage from 132 kV to 400 kV reduces losses to about (132/400)2≈11%(132/400)^2 \approx 11\% of the original value. This is the single most important reason for HV transmission.

2. Reduction in conductor material (copper/aluminium) required The volume of conductor is proportional to 1/V21/V^2 for a given power and loss. Higher voltage therefore means much lighter and cheaper conductors, reducing tower loading and cost.

3. Increase in power transmission capacity The power transmitted is P=3 VLILcos⁡ϕP = \sqrt{3}\, V_L I_L \cos\phi so for the same current-carrying capacity, doubling the voltage doubles the power that can be sent. This is vital as load demand grows.

4. Better voltage regulation Line voltage drop is ΔV=IRcos⁡ϕ±IXsin⁡ϕ\Delta V = I R \cos\phi \pm I X \sin\phi Since current falls as voltage rises, the drop and hence regulation improve, keeping consumer voltage within permissible limits.

5. Economy over long distances Although insulation, tower height and terminal equipment cost more at HV, the savings in conductor material and losses outweigh these beyond a certain distance (the "break-even distance"). Hence HV is economical for long-haul bulk transfer.

6. Interconnection of regional grids HV lines allow interconnection of separate power systems, enabling sharing of reserves, economic load dispatch, and improved reliability — a key feature of the modern grid.

7. Reduced number of circuits Fewer HV circuits can carry the same power as many LV circuits, reducing right-of-way requirements and total line cost.

8. Future expansion and stability EHV lines (220 kV, 400 kV, and in Nepal 400 kV and proposed 765 kV) provide the backbone needed for growing demand, large hydro projects, and cross-border power trade.

Thus high voltage transmission is adopted because it simultaneously reduces losses, saves conductor material, increases capacity, improves regulation, and makes long-distance bulk power transfer technically feasible and economically viable.

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  • 2082 · 2nd · 10 marks

Explain the different types of lighting sources used in illumination. Describe the construction and working principle of thyristor.

Answer

Lighting Sources Used in Illumination

1. Incandescent Lamp (Filament Lamp)

Construction: A tungsten filament enclosed in a glass bulb, either vacuum or filled with inert gas (argon/nitrogen). The filament is supported on lead wires connected to a screw base.

Working: When current passes through the high-resistance tungsten filament, it heats up to incandescence (about 2500–3000 K) and emits light. Only 5–10% of energy is converted to light; the rest is heat.

Features: Cheap, simple, instant start, warm colour, but low efficacy (10–20 lm/W) and short life (~1000 hours).

2. Halogen Lamp

A tungsten filament lamp with a small amount of halogen (iodine or bromine) added inside a quartz envelope. The halogen-tungsten cycle redeposits evaporated tungsten back onto the filament, giving longer life (~2000–4000 hours), higher efficacy (~20–25 lm/W) and whiter light. Used in floodlighting and vehicle headlamps.

3. Fluorescent Lamp (Tube Light)

Construction: A long glass tube coated internally with phosphor, containing argon gas and a small amount of mercury. Electrodes (coated with electron-emitting material) are fitted at both ends; a choke/ballast and starter are connected externally.

Working: When supply is switched on, the starter momentarily completes the circuit, heating the electrodes. When the starter opens, the choke induces a high voltage surge, striking an arc. Mercury vapour emits ultraviolet (UV) radiation, which strikes the phosphor coating and is converted into visible light.

Features: High efficacy (40–80 lm/W), long life (~5000–7500 hours), low heat, but requires ballast and gives a slightly harsh light. Widely used in offices and industries.

4. Compact Fluorescent Lamp (CFL)

A folded/miniature fluorescent tube with an integrated electronic ballast and screw base. It replaces incandescent lamps directly, giving 4–5 times higher efficacy (~50–70 lm/W) and life of about 8000–10000 hours. Contains mercury, so disposal needs care.

5. Mercury Vapour Lamp

A high-intensity discharge (HID) lamp with mercury and argon in an inner quartz tube, enclosed in an outer phosphor-coated glass bulb, with a ballast. The arc in mercury vapour gives bluish-green light; phosphor corrects colour. Efficacy ~40–60 lm/W. Used in street lighting and industrial areas. Takes a few minutes to reach full brightness.

6. Sodium Vapour Lamp

Low Pressure Sodium (LPS): U-shaped tube with sodium and neon/argon, inside a vacuum jacket. Emits nearly monochromatic yellow light (589 nm). Highest efficacy (~100–200 lm/W) but very poor colour rendering. Used on highways.

High Pressure Sodium (HPS): Sodium, mercury and xenon in a ceramic arc tube. Golden-white light, efficacy ~80–140 lm/W, good life. Widely used for street lighting.

7. Metal Halide Lamp

An HID lamp with metal halides added to mercury in an arc tube. Gives white light with good colour rendering (Ra > 65) and high efficacy (~80–100 lm/W). Used in stadiums, showrooms and floodlighting.

8. LED (Light Emitting Diode) Lamp

Construction: A semiconductor p-n junction chip (GaAs, GaP, InGaN) mounted on a heat sink, with driver circuitry and diffuser.

Working: When forward biased, electrons recombine with holes at the junction and release energy as photons (electroluminescence). Colour depends on band gap.

Features: Highest efficacy (~100–150 lm/W), very long life (~50000 hours), instant start, no mercury, dimmable. Now the preferred source for domestic, street and industrial lighting.

9. Neon Sign Lamp

A glass tube filled with neon (red-orange) or other noble gases with different phosphors, operating at high voltage. Used for advertisement signs and decoration.

10. Arc Lamp

Two carbon or metal electrodes struck to form a bright arc in air. Very high intensity; used in cinema projectors and searchlights (now largely obsolete).

Comparison Summary

SourceEfficacy (lm/W)Life (hours)ColourTypical Use
Incandescent10–201000Warm whiteDomestic (old)
Fluorescent40–805000–7500Cool whiteOffices, industry
CFL50–708000–10000Warm/coolDomestic
Mercury vapour40–608000BluishStreet, industry
LPS100–20010000–15000YellowHighway
HPS80–14015000–24000GoldenStreet
Metal halide80–1006000–10000WhiteStadium
LED100–15050000AnyAll applications

Thyristor (SCR) — Construction and Working Principle

Construction

A thyristor, commonly the Silicon Controlled Rectifier (SCR), is a four-layer, three-junction semiconductor device with layers arranged as p-n-p-n. It has three terminals:

  • Anode (A) — from the outer p-layer
  • Cathode (K) — from the outer n-layer
  • Gate (G) — from the inner p-layer near the cathode

The three junctions are J₁ (between anode p and inner n), J₂ (between inner n and inner p), and J₃ (between inner p and cathode n). The device is made of silicon and is designed to carry large currents; the gate controls the instant of conduction. A thyristor is represented by the symbol of a diode with a gate lead.

Working Principle (Two-Transistor Analogy)

The p-n-p-n structure can be treated as two transistors: a p-n-p transistor T₁ and an n-p-n transistor T₂, with the collector of each feeding the base of the other.

  1. Forward blocking mode: When anode is positive with respect to cathode (forward biased) but gate is open, junctions J₁ and J₃ are forward biased while J₂ is reverse biased. Only a small leakage current flows; the device is OFF.

  2. Turn ON (latching): If a small positive gate current is applied, it injects electrons into the inner p-layer, increasing the collector current of T₂. This current feeds the base of T₁, whose collector current in turn feeds the base of T₂. This regenerative (positive feedback) action rapidly drives both transistors into saturation, and the thyristor switches ON. Once ON, the gate loses control — the device latches and remains conducting even if the gate signal is removed. The anode current is then limited only by the external load resistance. The minimum anode current needed to sustain conduction is called the holding current (I_H).

  3. Turn OFF (commutation): The thyristor can be turned OFF only by reducing the anode current below the holding current, or by applying a reverse voltage across anode and cathode (natural or forced commutation). This restores the reverse-biased J₂ and the device returns to the blocking state.

V-I Characteristics

  • Forward characteristics: In forward blocking region, a small forward leakage current flows until the forward breakover voltage (V_BO) is reached; beyond this, the device switches ON and the voltage drops to about 1–2 V. A lower breakover voltage can be achieved by applying gate current.
  • Reverse characteristics: With anode negative, junctions J₁ and J₃ are reverse biased and only a small reverse leakage current flows until reverse breakdown voltage is reached, which may destroy the device.

Applications

Controlled rectifiers, AC voltage regulators, inverters, choppers, DC motor speed control, dimmer circuits, HVDC transmission, and overvoltage protection circuits.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 10 marks

Explain why load curve and load duration curves are important in a power system. Also, describe preventive maintenance of a transformer.

Answer

Load Curve and Load Duration Curve — Importance in a Power System

Load curve: A graph of load (demand) against time, plotted chronologically over a day (daily load curve), month (monthly) or year (annual load curve). It shows how demand varies with time.

Load duration curve: The same load data rearranged in descending order of magnitude, so that the curve shows for how many hours (or what percentage of time) a given load or higher is demanded. It is obtained by re-plotting the load curve from highest to lowest load.

Importance of the load curve

  1. Deciding plant capacity and installed reserve — the peak of the load curve fixes the maximum demand the system must meet, hence the total generating capacity required.
  2. Economic loading / merit-order scheduling — knowing when demand rises and falls allows the cheapest (base-load) plants to be kept running and costly peaking plants to be started only for peak hours, minimising generation cost.
  3. Selection of generating units — base load, peak load and standby units are chosen according to the shape of the curve.
  4. Planning maintenance and outages — maintenance is scheduled during low-load periods so that supply is not interrupted.
  5. Tariff (rate) fixing — the load curve gives the load factor and diversity of demand, which are the basis of two-part and time-of-day tariffs.
  6. Forecasting future demand — comparison of successive annual load curves shows the load growth trend for expansion planning.

Importance of the load duration curve

  1. Directly gives the load factor — area under the curve divided by (peak load × total hours) gives the load factor, a key economic index.
  2. Decides the base load and peak load portions — the flat top of the curve represents base load (available almost all the time) and the steep tail represents peak load (available for a few hours only).
  3. Optimum allocation of generation among plants — the curve is divided into blocks; each block is assigned to the plant with the lowest cost for that duration (base-load plants for long-duration blocks, peaking plants for short-duration blocks).
  4. Deciding the size and type of units — the number of hours a load persists tells whether a steam, hydro or gas/diesel unit is suitable.
  5. Estimating energy (units) generated — the area under the curve gives the total energy demanded in kWh, needed for energy accounting and purchase/sale planning.
  6. Reserve and reliability studies — used with the probability of outage to fix spinning reserve and spinning reserve requirements.

In short, the load curve tells when the load occurs and the load duration curve tells for how long it occurs; together they are the basic tools for capacity planning, economic dispatch, tariff design and maintenance scheduling.


Preventive Maintenance of a Transformer

Preventive maintenance is the scheduled, periodic inspection, testing and servicing of a transformer carried out before a fault occurs, to keep it in good operating condition, extend its life and avoid forced outages. It is done in three stages.

1. Daily / routine (online) checks

  • Record oil temperature (top oil) and winding temperature; compare with permissible limits (top oil ≈ 90 °C, winding ≈ 95 °C for ONAN distribution transformers).
  • Check oil level in the conservator against the gauge/magnetic oil level gauge (MOG) marking.
  • Check for oil leakage from the tank, bushings, gaskets, radiator valves and drain valve.
  • Listen for abnormal sound/humming (loose core laminations, DC magnetization, overload).
  • Check silica gel in the breather — colour change from blue to pink means it is saturated and must be replaced or reactivated.
  • Check cooling fans / oil pumps (for forced-cooled units) for correct automatic operation.
  • Record voltage, current and load on all phases; check for unbalanced loading.
  • Inspect bushings for cracks, dirt, chipping and flashover marks; clean them.
  • Check the explosion vent / pressure relief device and the Buchholz relay (oil level, gas accumulation) for oil-filled units above 500 kVA.
  • Verify earthing connections of the tank and neutral.

2. Periodic (shutdown) maintenance — yearly or as per schedule

  • Oil testing: dielectric strength (BDV), acidity, moisture content, sludge and dissolved gas analysis (DGA). Filter or replace oil if BDV is low or moisture exceeds limits; dry out the oil if necessary.
  • Insulation resistance (IR) test of windings with a megger (HV–earth, LV–earth, HV–LV); compare with previous readings and the PI (polarization index) value.
  • Winding resistance and turns ratio test to detect shorted turns or loose connections.
  • Tap changer: clean contacts, check for pitting/burning, lubricate the mechanism, and service the diverter switch oil.
  • Bushings: clean, check tightness of connections, and megger-test.
  • Tightening of all terminal and busbar connections to the specified torque; check for overheating/discoloration.
  • Painting of the tank and radiator to prevent corrosion; cleaning of radiators and cooling ducts.
  • Gaskets and seals — replace hardened or leaking gaskets.
  • Protective devices: test Buchholz relay, pressure relief device, oil temperature indicator, winding temperature indicator and alarm/trip contacts.
  • Earthing resistance measurement and rectification if high.

3. Special / condition-based maintenance

  • DGA at regular intervals to detect incipient faults (arcing, partial discharge, overheating).
  • Thermography of bushings, connections and the tank to locate hot spots.
  • Tan-delta / capacitance test of bushings.
  • Frequency response analysis (FRA) and core insulation test for large or important units.
  • Oil reclamation or replacement when the oil is badly deteriorated.

Records and safety

Every test result, oil sample, and repair must be entered in a transformer history card so that trends can be compared. Before any shutdown work, the transformer must be isolated, earthed, and locked out, and the oil level and oil temperature must be brought to ambient before opening the tank to avoid moisture ingress.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 1st · 5 marks

Write in brief about Smart Meter which is already in operation in certain sector of Nepal Electricity Authority.

Answer

Smart Meter (NEA context) — 5 marks

A smart meter is an advanced electronic electricity meter that records consumption in real time and communicates the data two-way to the utility, unlike a conventional electromechanical or static meter which only measures and displays units consumed.

Main features

  • Interval/Time-of-Day (ToD) metering: records consumption in 15-minute or 30-minute blocks, enabling Time-of-Day and peak/off-peak tariffs.
  • Two-way communication: data is pushed to NEA's MDMS (Meter Data Management System) over GSM/GPRS, RF mesh or PLC, and commands/configuration are sent back to the meter — no manual meter reading.
  • Remote connect/disconnect: supply can be switched on or off from the control centre, useful for defaulters and for load management.
  • Tamper and theft detection: logs events such as cover opening, magnetic influence, reverse current and bypass, and reports them immediately.
  • Power quality and load profile: measures voltage, current, power factor, active/reactive energy and maximum demand; some models support net metering for rooftop solar.
  • Prepaid/Postpaid flexibility: can be configured for prepaid mode with recharge, or postpaid billing.
  • Alarms and outage reporting: last-gasp messages inform NEA of supply interruption and restoration without customer complaint.

Status in NEA

NEA has already rolled out smart meters in selected areas/distribution centres as pilot and phased projects — notably in parts of the Kathmandu Valley (e.g. areas under selected distribution centres), Pokhara and other urban pockets — mainly for high-consumption and ToD customers. The rollout is being extended under NEA's Advanced Metering Infrastructure (AMI) plan, with the aim of reducing commercial losses (non-technical loss), improving billing accuracy and revenue collection, cutting meter-reading cost, and enabling demand-side management. Full nationwide coverage is still in progress, limited by cost, communication coverage and system integration with the existing billing system.

Benefits to NEA: accurate billing, lower commercial loss, remote operation, better load forecasting and improved customer service.

Answer checked by Claude Haiku · Oct 21, 2026

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Questions from NEA Exam Question for Assistant (Assistant Engineer/ Electrical 6th Level · 2nd · 2082) and NEA Question for Assistant Engineer (Assistant Engineer 6th Level · 1st · 2082). Answers are written for this site; check them against your class notes.