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Chapter 2

Electrical Engineering Fundamentals and Circuit Analysis

Lok Sewa Aayog (Public Service Commission) past exam questions

Past questions and answers

12 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 (with load removed), Vth=VocV_{th} = V_{oc}
  • RthR_{th} = equivalent resistance seen from the terminals with all independent sources killed (voltage sources short-circuited, current sources open-circuited).

The load current is then: IL=VthRth+RLI_L = \frac{V_{th}}{R_{th} + R_L}

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 (with load replaced by a short), IN=IscI_N = I_{sc}
  • RNR_N = equivalent resistance seen from the terminals with all independent sources killed, i.e. RN=RthR_N = R_{th}.

The load current is then: IL=IN⋅RNRN+RLI_L = I_N \cdot \frac{R_N}{R_N + R_L}

Relation between the two

Vth=INRN,Rth=RN=VthINV_{th} = I_N R_N, \qquad R_{th} = R_N = \frac{V_{th}}{I_N}

Applications

  1. Circuit simplification — reduces a complicated network to a simple two-element equivalent, so the effect of changing the load (e.g. a variable resistor, a loudspeaker, a sensor) can be found without re-solving the whole circuit each time.
  2. Maximum power transfer — used to determine the condition RL=Rth=RNR_L = R_{th} = R_N for maximum power transfer to the load.
  3. Load analysis in power and electronics circuits — finding voltage, current and power delivered to a particular branch (e.g. a bridge arm, a transistor biasing network, a battery feeding a load).
  4. Source modelling — a practical voltage source is modelled as VthV_{th} in series with RthR_{th}; the same source is modelled as INI_N in parallel with RNR_N (source transformation), useful in network simplification.
  5. Cascaded network analysis — each stage of a multistage network is reduced to its Thevenin/Norton equivalent, simplifying overall analysis.
  6. Sensitivity and fault studies — used to study how terminal voltage and current vary with load, and to locate faults in linear networks.

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 records the highest average power (or current) drawn by an installation over a fixed integration interval, usually 15 or 30 minutes. It works on the thermal (or integrating) principle:

  • A current proportional to the load current is passed through a small heating element (bimetal strip or thermocouple) or through an induction-type/electronic integrating element.
  • The heat produced is proportional to (I^2Rt), i.e. to the energy consumed in the interval, so the deflection of the indicating element is proportional to the average demand over that interval, not the instantaneous peak.
  • A pusher or ratchet mechanism (in thermal type) or a memory register (in electronic type) holds the pointer at the highest value reached. The pointer stays at the maximum value until it is manually reset by the meter reader after billing.
  • In modern electronic MD meters, the microprocessor samples power every few seconds, computes the average over each 15/30-minute block, compares it with the stored maximum, and updates the memory if the new block average is higher. A time-of-day clock also performs the reset (sliding window or fixed block method).

Thus the meter gives the maximum average demand in kW (or kVA) during the billing period, not the total energy (kWh), which is measured separately by the energy meter.

Application

  1. Tariff/billing: Used by utilities for large consumers (industrial, commercial) under two-part or maximum-demand tariff, where the monthly bill = energy charge (Rs/kWh) + demand charge (Rs/kW or Rs/kVA of maximum demand). This penalises consumers who draw heavy loads for short periods.
  2. Load management: Helps the consumer monitor and control peak demand, reschedule heavy loads to off-peak hours, and avoid penalty charges.
  3. System planning: MD records of consumers and feeders give the utility the diversified demand data needed for sizing transformers, feeders and substations, and for load forecasting.
  4. Power factor / kVA demand billing: Combined with a kVA MD meter, it is used to levy power-factor penalties on industrial consumers.
  5. Statistical and regulatory use: Provides data for load-factor calculation, tariff design and energy auditing.

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

Schematic Diagram of a Hydroelectric Power Plant

        Reservoir / Dam
   ┌───────────────────────┐
   │~~~~~~~~~~~~~~~~~~~~~~~│  ← Water surface
   │~~~~~~~~~~~~~~~~~~~~~~~│
   └───────┬───────────────┘
           │  Intake Structure (with trash rack & gate)
           │
           ▼
     ┌───────────┐
     │  Penstock │  (closed conduit, sloping down)
     │  (pipe)   │
     └─────┬─────┘
           │
           ▼
     ┌───────────┐
     │  Surge    │  (surge tank / surge shaft)
     │   Tank    │
     └─────┬─────┘
           │
           ▼
   ┌───────────────┐
   │   Turbine     │  ← Water strikes runner
   │  (in Power    │
   │   House)      │
   └──────┬────────┘
          │  Shaft
          ▼
   ┌───────────────┐
   │  Generator    │
   │  (alternator) │
   └──────┬────────┘
          │
          ▼
   ┌───────────────┐
   │ Transformer   │
   │ (step-up)     │
   └──────┬────────┘
          │
          ▼
     Transmission Line  →  to load centres
          │
          ▼
   Tailrace  →  water returned to river

Main Components and Their Functions

1. Catchment Area and Reservoir

The catchment area is the region that collects rain and snow-melt feeding the plant. A dam across the river stores water in a reservoir, creating the head (height difference) that drives the turbines. The reservoir also stores water to meet demand during dry periods (storage plant) or simply diverts flow (run-of-river plant).

2. Dam

A dam is a barrier built across the river to raise the water level and store water. Types include gravity, buttress, arch and earth-fill dams. The dam:

  • creates the required head,
  • regulates the flow to the turbines,
  • provides storage for lean-season operation.

3. Intake Structure

The intake is the entry point of water from the reservoir into the water-conducting system. It consists of:

  • Trash rack — a screen of steel bars that stops floating debris, leaves and ice from entering the penstock.
  • Intake gate (sluice gate) — controls or shuts off the flow for maintenance.
  • Bell-mouth entry — smooth entry to reduce head loss and turbulence.

4. Headrace / Power Tunnel or Canal

This is the waterway that carries water from the intake to the surge tank or forebay. It may be an open canal (for low heads) or a closed tunnel/pressure shaft (for high heads). It is designed for minimum friction loss, so the cross-section and slope are chosen carefully.

5. Surge Tank

A surge tank is a vertical shaft or chamber connected to the penstock near the power house. Its functions:

  • absorbs the water hammer pressure rise when the turbine gates close suddenly,
  • supplies water to the turbine immediately when the gates open (prevents negative pressure),
  • reduces the length of penstock exposed to water hammer.

6. Penstock

The penstock is a closed pipe (steel, concrete or composite) that conveys water under pressure from the surge tank/forebay to the turbine. It is designed to withstand the internal water pressure plus surge pressure. It may be:

  • above ground on supports, or
  • underground in rock.

7. Turbine

The turbine converts the hydraulic energy of water into mechanical (rotational) energy. Main types:

  • Pelton wheel — for high head, low discharge (impulse turbine),
  • Francis turbine — for medium head, medium discharge (reaction turbine),
  • Kaplan/propeller turbine — for low head, large discharge (reaction turbine).

Water strikes the runner blades and rotates the shaft.

8. Generator (Alternator)

The generator is coupled to the turbine shaft. It converts mechanical energy into electrical energy. The turbine shaft rotates the rotor of the alternator, inducing an emf in the stator windings, producing 3-phase AC at 11 kV or so.

9. Transformer

A step-up transformer raises the generated voltage (e.g., 11 kV) to a high transmission voltage (e.g., 132 kV or 220 kV) to reduce transmission losses over long distances.

10. Switchgear and Protection

Circuit breakers, isolators, relays and control panels protect the generator, transformer and transmission line from faults and allow safe switching.

11. Transmission Line

Carries the electrical power from the power house to load centres (cities, industries).

12. Tailrace

The channel that carries water leaving the turbine back to the river downstream. Its design ensures the water level at the turbine outlet (tailwater level) is correct for efficient operation.

13. Power House

The building that houses the turbine, generator, transformer, switchgear and control equipment. It may be:

  • surface power house (above ground), or
  • underground power house (in rock).

Working Principle (Brief)

Water stored at a high level in the reservoir flows through the intake, headrace and penstock to the turbine. The potential energy of water is converted into kinetic energy, which rotates the turbine runner. The turbine shaft drives the alternator, producing electrical energy. The water then exits through the tailrace to the river. The generated voltage is stepped up by the transformer and sent through the transmission line to consumers.

Answer checked by Claude Haiku · Oct 21, 2026

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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 device that can make, carry and break normal load current as well as fault current (short-circuit current) under abnormal conditions.
  • It automatically trips (opens) on fault, sensed through CTs/relays, and recloses as required.
  • It extinguishes the arc produced during current interruption using a suitable medium — air blast, oil, SF₆ gas or vacuum.
  • It is the main protective and switching device of a power system, capable of interrupting very high fault MVA.

2. Isolator (Disconnecting Switch)

  • It is an off-load (no-load) switching device used to isolate a circuit, equipment or section from the live supply for maintenance and repair.
  • It cannot break or make load current or fault current — it must be operated only when the circuit breaker has already interrupted the current.
  • It provides a visible air gap in the circuit, giving safety assurance to maintenance personnel.
  • It is also used for transferring load from one bus to another and for by-passing a breaker.

3. Earthing Switch

  • It is a switch used to connect a circuit or equipment solidly to earth after isolation, so that any residual/induced charge or accidental energisation is safely discharged to ground.
  • It ensures the section is at earth potential before work begins, protecting personnel from shock.
  • It is not designed to break load or fault current; it is closed only after the isolator is open.
  • It is often mechanically interlocked with the isolator so that earthing can be done only when the isolator is open.

Why Interlocking is Necessary

Interlocking is a safety arrangement that prevents wrong sequence of operation among CB, isolator and earthing switch. It is necessary because:

  1. Isolator cannot break current — if opened on load, a heavy arc will form, damaging the switch and endangering the operator. Hence the isolator must be interlocked so that it can be operated only when the CB is open.
  2. Earthing switch must not be closed on a live circuit — closing it while the line is energised causes a dead short circuit and violent explosion. It must be interlocked to close only after the isolator is open.
  3. CB must not be closed while the earthing switch is closed — otherwise a direct three-phase short circuit occurs. Interlocking prevents this.
  4. It prevents accidental contact with live parts and enforces the correct operating sequence: CB open → Isolator open → Earthing switch closed (and the reverse while restoring supply).
  5. It protects equipment and operating personnel from maloperation and ensures reliability of the supply.

Differential Protection

Principle: It works on Kirchhoff's Current Law — the vector sum of currents entering and leaving the protected zone must be zero in a healthy condition. The current transformers (CTs) on both sides of the protected equipment are connected in a circulating current arrangement to a differential relay.

  • Healthy condition: I₁ = I₂, so the secondary currents circulate and the relay operating current = I₁ − I₂ ≈ 0 → relay does not operate.
  • Internal fault: current flows into the fault, so I₁ ≠ I₂ and the difference current (I₁ − I₂) flows through the relay → relay operates and trips the CB.

The relay is set above the unbalance current caused by CT errors, tap-changing and magnetising inrush, so it remains stable for external faults.

Applications:

  1. Generator protection — stator winding phase and earth faults.
  2. Transformer protection — Merz-Price (circulating current) differential scheme for internal faults.
  3. Busbar protection — high-impedance or biased differential scheme.
  4. Motor protection — large induction/synchronous motors against internal faults.
  5. Feeder and transmission line protection — pilot-wire or carrier-current differential schemes.
  6. Cable protection — for short lengths where both ends are available.
  7. Rectifier/inverter and capacitor bank protection in HVDC and substations.

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

Methods of Testing of Transformer and Principle of Parallel Operation

1. Methods of Testing of Transformer

Transformers are tested to verify their performance, losses, efficiency, temperature rise and dielectric strength. The main tests are:

A. Preliminary / Routine Tests

  1. Measurement of winding resistance – DC is passed through each winding and resistance is measured (by voltmeter–ammeter or Kelvin double bridge). It checks for loose connections, broken strands and contact resistance.
  2. Measurement of insulation resistance – A megger (500 V or 1000 V) is used between windings and between winding and earth. It indicates the health of insulation.
  3. Polarity test – Determines the relative direction of induced emfs in HV and LV windings (additive or subtractive), essential before paralleling.
  4. Ratio test – The turns ratio is checked by applying a known voltage to one winding and measuring the other (voltmeter method or ratio meter).
  5. Dielectric / insulation tests – Power frequency high-voltage test and impulse voltage test are done to check dielectric strength of major and minor insulation.

B. Performance Tests

  1. Open Circuit (No-load) Test – LV winding is excited at rated voltage with HV open. It gives core (iron) loss (constant loss) and no-load current, and hence the shunt branch parameters (R_0) and (X_0).
  2. Short Circuit Test – HV winding is shorted and a small voltage is applied to LV so that rated current flows. It gives copper loss (variable loss) and equivalent resistance and reactance (R_{eq}) and (X_{eq}).
  3. Load test / Back-to-back (Sumpner's) test – Two identical transformers are connected back-to-back to determine efficiency, regulation and temperature rise at full load without actually loading the supply.
  4. Temperature rise (heat run) test – Determines the steady temperature rise of oil and windings at rated load, to check the cooling arrangement.
  5. Dielectric tests on oil – Breakdown voltage test of transformer oil.

From the OC and SC tests, the efficiency and voltage regulation at any load and power factor can be computed.


2. Principle of Parallel Operation of Transformers

Parallel operation means connecting two or more transformers with their primary windings in parallel to the same supply bus and secondary windings in parallel to the same load bus.

Necessity

  • To supply load larger than the capacity of one transformer.
  • To increase reliability — one unit can be shut down for maintenance while the other supplies the load.
  • To reduce spare capacity and improve overall efficiency at light loads.

Principle of Operation

When two transformers are paralleled, they form a closed loop through their secondaries. For no circulating current to flow between them in no-load condition, the induced secondary emfs must be equal in magnitude and in phase. This requires:

  1. Same voltage ratio (turns ratio) – otherwise a potential difference appears across the paralleled secondaries and a circulating current flows even at no load, causing extra copper loss and heating.
  2. Same polarity – the windings must be connected with the same relative polarity; otherwise the emfs add in the loop and a very large circulating current flows (effectively a short circuit).
  3. Same phase sequence and phase angle (vector group) – for three-phase transformers, the phase displacement between primary and secondary must be identical; otherwise a phase difference exists between secondary emfs and circulating current flows.
  4. Equal per-unit (percentage) impedance – the load shared by each transformer is inversely proportional to its per-unit impedance. If per-unit impedances are unequal, the transformer with lower per-unit impedance gets overloaded while the other remains partly loaded. Hence for proper load sharing, percentage impedances should be equal.
  5. Same frequency – since all are fed from the same bus, this is automatically satisfied.

Load Sharing

For two transformers in parallel with equal voltage ratios, the load current shared is:

[ I_1 = I \cdot \frac{Z_2}{Z_1 + Z_2}, \qquad I_2 = I \cdot \frac{Z_1}{Z_1 + Z_2} ]

where (Z_1, Z_2) are the equivalent impedances referred to the same side. Thus each transformer carries a share of load inversely proportional to its impedance, and the total load is the sum of the individual shares.

Advantages

  • Continuity of supply and reliability.
  • Easy maintenance without interrupting the load.
  • Higher efficiency and better load sharing.
  • Future expansion possible by adding units.

Conditions to be Satisfied (Summary)

S.N.Condition
1Equal voltage ratios
2Same polarity
3Same phase sequence and vector group
4Equal per-unit impedance (for equal load sharing)
5Same frequency

If all these conditions are satisfied, the transformers operate satisfactorily in parallel, sharing the load in proportion to their ratings without any circulating current.

Answer checked by Claude Haiku · 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.

Answer

Let
[ S=100\text{ kVA}=100\times10^3\text{ VA} ] Let
[ P_i=\text{iron loss},\qquad P_{cu}=\text{full-load copper loss} ]

1. At half full load, (0.8) p.f. leading

Load fraction (x=0.5), (\cos\phi=0.8)

Output: [ P_{o1}=xS\cos\phi=0.5\times100\times0.8=40\text{ kW} ]

Losses: [ P_{\text{loss1}}=P_{o1}\left(\frac1{\eta_1}-1\right) =40\left(\frac1{0.985}-1\right) =0.60914\text{ kW} ] [ P_{\text{loss1}}=609.14\text{ W} ]

But [ P_{\text{loss1}}=P_i+x^2P_{cu} ] [ P_i+0.25P_{cu}=609.14 \quad (1) ]

2. At full load, unity p.f.

Load fraction (x=1), (\cos\phi=1)

Output: [ P_{o2}=1\times100\times1=100\text{ kW} ]

Losses: [ P_{\text{loss2}}=100\left(\frac1{0.988}-1\right) =1.21457\text{ kW} ] [ P_{\text{loss2}}=1214.57\text{ W} ]

So, [ P_i+P_{cu}=1214.57 \quad (2) ]

3. Solving for losses

Subtract (1) from (2): [ (P_i+P_{cu})-(P_i+0.25P_{cu})=1214.57-609.14 ] [ 0.75P_{cu}=605.43 ] [ P_{cu}=807.25\text{ W} ]

[ P_i=1214.57-807.25=407.32\text{ W} ]

Hence, [ \boxed{P_i=407.32\text{ W}\approx0.407\text{ kW}} ] [ \boxed{P_{cu}=807.25\text{ W}\approx0.807\text{ kW}} ]

4. Maximum efficiency

Maximum efficiency occurs when copper loss equals iron loss: [ x_m^2P_{cu}=P_i ] [ x_m=\sqrt{\frac{P_i}{P_{cu}}} =\sqrt{\frac{407.32}{807.25}} =0.7103 ]

At this load, total loss: [ P_{\text{loss,max}}=2P_i=2\times407.32=814.65\text{ W} ] [ P_{\text{loss,max}}=0.81465\text{ kW} ]

Taking unity p.f. for maximum efficiency: [ P_o=x_mS\cos\phi=0.7103\times100\times1=71.034\text{ kW} ] [ P_{\text{in}}=71.034+0.81465=71.84865\text{ kW} ] [ \eta_{\max}=\frac{71.034}{71.84865}\times100 =98.866% ]

[ \boxed{\eta_{\max}\approx98.87%} ]

Final answers: [ \boxed{P_i=407.32\text{ W},\quad P_{cu}=807.25\text{ W},\quad \eta_{\max}=98.87%} ]

Note: The condition for maximum efficiency is independent of power factor, but its numerical value depends on the load p.f. The value above is at unity p.f.

Answer checked by Claude Haiku · Oct 21, 2026

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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 Testing Devices

Calibration is the process of comparing the reading of a measuring instrument or testing device against a reference standard of known and higher accuracy, and then determining, correcting, or documenting the deviation (error) between the indicated value and the true value. The reference standard used is traceable to a national or international standard (e.g., NIST, or Nepal's NBSM).

In simple terms, calibration establishes the relationship between the output of an instrument and the known values of the quantity being measured. It may involve:

  • Zero adjustment – setting the instrument to read zero when the input is zero.
  • Span/gain adjustment – making the instrument read correctly at full-scale or at a known reference point.
  • Determination of correction factors – recording the error so that readings can be corrected mathematically.

A calibration certificate is issued after calibration, stating the measured deviations, uncertainty, and the conditions under which the test was performed.

Significance in Maintaining Accuracy and Precision

  1. Ensures accuracy (closeness to true value): Regular calibration detects and corrects systematic errors such as drift, aging of components, and zero offset, so that measured values truly represent the actual quantity.

  2. Ensures precision (repeatability/consistency): Calibration verifies that repeated measurements under identical conditions give closely grouped results. It identifies random errors and instrument instability.

  3. Traceability: Calibration links the instrument's readings to national/international standards, making measurements legally and technically valid and comparable across laboratories.

  4. Detects drift and wear: Electrical instruments (ammeters, voltmeters, energy meters, CTs, PTs, multimeters) drift with time, temperature, and usage. Periodic calibration reveals when an instrument is out of tolerance and needs repair or replacement.

  5. Compliance with standards and legal metrology: In Nepal, calibration is mandatory for instruments used in trade (energy meters, weighbridges) under the Nepal Bureau of Standards and Metrology (NBSM) Act. It ensures consumer protection and fair trade.

  6. Reliability of test results: In testing laboratories (e.g., material testing, relay testing, transformer testing), an uncalibrated device gives false pass/fail decisions, leading to unsafe equipment or rejected good products.

  7. Cost and safety benefits: Accurate measurement prevents over-design, material waste, equipment damage, and electrical accidents caused by wrong readings.

  8. Quality assurance (ISO/IEC 17025): Calibration is a mandatory requirement for laboratory accreditation and quality management systems, ensuring consistency and customer confidence.

Conclusion: Calibration is not a one-time activity but a periodic, traceable process. It is the foundation for trustworthy electrical measurements — without it, accuracy and precision cannot be guaranteed, and decisions based on such measurements become unreliable.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd · 5 marks

Briefly explain tariff schemes in Nepal.

Answer

Tariff Schemes in Nepal

Tariff is the rate or schedule of charges at which electrical energy is sold to consumers. In Nepal, the Nepal Electricity Authority (NEA) designs tariff schemes under the Electricity Act, 2049 and the Electricity Rules, 2050, with tariff rates fixed by the independent Electricity Regulatory Commission (ERC), which has replaced the earlier Electricity Tariff Fixation Commission. The main objectives are to recover the cost of generation, transmission and distribution, to meet the required rate of return, to promote efficient and rational use of electricity, and to provide affordable supply to low-income consumers through cross-subsidy.

1. On the basis of consumer category (consumer tariff)

  • Domestic (residential) tariff — for households; charged in declining or telescopic blocks so that small consumers pay a lower rate per unit (kWh) and large consumers pay progressively higher rates. A minimum monthly charge and a demand charge (Rs/kW or Rs/A) also apply.
  • Commercial tariff — for shops, hotels, offices and business houses; higher energy and demand rates than domestic.
  • Industrial tariff — for factories; usually the lowest energy rate to encourage productive use, but with a demand charge based on maximum demand (Rs/kW or Rs/kVA).
  • Agricultural (irrigation) tariff — concessional rate to promote irrigation pumping and food production.
  • Temporary tariff — for construction sites, fairs and temporary connections; highest rate, often with a fixed charge per kW per month.
  • Street light tariff — charged to local bodies/municipalities for public lighting.
  • Water supply, community and institutional (hospitals, schools, religious places) tariff — separate concessional categories.
  • Export/industrial export tariff — special rate for industries exporting their products.

2. On the basis of the form of charge

  • Energy charge (Rs/kWh) — payment for the actual units consumed, measured by the energy meter.
  • Demand charge (Rs/kW or Rs/kVA) — payment for the maximum demand recorded, to recover the fixed cost of capacity kept ready for the consumer.
  • Fixed/minimum charge — a lump sum per month irrespective of consumption, covering meter rent, billing and service connection costs.
  • Reactive energy charge — penalty (or incentive) for poor (or good) power factor, applied to industrial consumers to encourage power-factor improvement.

3. On the basis of rate structure

  • Flat-rate (uniform) tariff — a single rate per unit for all units consumed; simple but does not reflect cost.
  • Block (step) tariff — the rate changes in blocks of consumption; declining block tariff (rate falls as consumption rises) or inverted/ascending block tariff (rate rises as consumption rises). NEA uses inverted block rates for domestic consumers to protect small consumers.
  • Two-part tariff — a fixed/demand charge plus an energy charge; the most common form for NEA consumers.
  • Three-part tariff — fixed charge + demand charge + energy charge; used for large industrial consumers.
  • Time-of-Day (TOD)/peak–off-peak tariff — different rates for peak hours (higher) and off-peak hours (lower) to flatten the load curve; NEA applies TOD to industrial and large consumers.
  • Seasonal tariff — different rates in dry (lean) and wet (monsoon) seasons, since dry-season generation is costlier.
  • Power-factor tariff — as above, linked to reactive power.

4. On the basis of ownership/area

  • NEA tariff — uniform throughout the NEA service area, with separate rates for each consumer category and voltage level (low voltage, medium voltage, high voltage).
  • Community rural electrification tariff — for consumers served by Community Rural Electrification Entities (CREEs), often a single flat rate per unit.
  • Isolated/off-grid tariff — for micro-hydro and solar mini-grids, fixed by the concerned operating company or user committee.

5. Special provisions

  • Demand charge rebate for consumers with high load factor.
  • Power-factor incentive for industrial consumers maintaining power factor above a specified value (e.g., 0.95).
  • Lifeline/minimum consumption block at a very low rate for very small domestic users.
  • Cross-subsidy — commercial and industrial consumers pay relatively more to keep domestic and agricultural tariffs low.

In short, Nepal's tariff schemes are category-based, block-based, two-part or three-part, and increasingly time-differentiated, designed to recover cost, ensure equity, and improve load factor and power factor.

Answer checked by Claude Sonnet · Oct 21, 2026

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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 in a Substation

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

  1. Safety of personnel: If insulation fails, the equipment enclosure becomes live. A low-resistance earth path ensures the enclosure stays at near-earth potential, so a person touching it is not exposed to a dangerous touch voltage.

  2. Return path for fault current: Earthing provides a definite low-impedance path for fault current to flow back to the source, so protective relays and fuses operate quickly and isolate the faulty section.

  3. Limiting overvoltages: Earthing of the neutral and of surge arresters diverts lightning and switching surge currents to earth, limiting overvoltages and protecting transformer and switchgear insulation.

  4. Voltage reference and stability: It fixes the reference potential of the system, keeps phase-to-earth voltages within insulation limits, and improves stability of the neutral point.

  5. Fire and equipment protection: By clearing faults quickly, earthing reduces arcing, fire risk, and thermal/mechanical damage to costly equipment.

  6. Compliance and safety standards: It satisfies statutory safety rules and standards (e.g., touch and step potential limits) mandatory for substation design.

Types used: equipment (body) earthing, neutral earthing (solid, resistance or reactance), and lightning/surge earthing — usually combined in a common grid of earth electrodes, strips and rods to keep earth resistance low (typically ≤ 1 Ω for large substations).


Role of the Cooling System in a Substation and its Significance

Substation equipment — mainly power transformers, reactors, and sometimes cables and busbars — generates heat due to copper loss (I²R) and core loss (eddy and hysteresis). This heat must be removed continuously, otherwise insulation deteriorates and the equipment fails. The cooling system performs this duty.

Role (function)

  1. Heat removal: It transfers the heat generated in windings and core to the surrounding air or water, keeping the temperature within permissible limits.
  2. Maintaining insulation life: It limits the hot-spot temperature of winding insulation (paper/oil), since insulation life roughly halves for every 6–8 °C rise above the rated temperature.
  3. Preserving oil quality: It prevents overheating and oxidation/sludging of transformer oil, thus retaining its dielectric strength and cooling ability.
  4. Enabling higher loading: Efficient cooling allows a transformer of given size to carry more load, or a smaller transformer to do the same duty.

Methods of cooling (transformers)

TypeMethodApplication
ONANOil Natural, Air NaturalSmall distribution transformers
ONAFOil Natural, Air Forced (fans on radiators)Medium/large transformers
OFAF / OFANOil Forced, Air ForcedLarge power transformers
OFWFOil Forced, Water ForcedVery large transformers, hydro stations
ODAN / ODAFOil Directed, Air ForcedEHV transformers

Significance

  1. Reliability of supply: Prevents sudden failure and forced outage of transformers, which would otherwise interrupt bulk power supply.
  2. Safety: Avoids overheating, oil decomposition, gas formation, and possible explosion or fire.
  3. Economy: Extends equipment life, reduces maintenance and replacement cost, and permits economic design (smaller core and tank for the same rating).
  4. System performance: Keeps winding resistance and losses low, maintains voltage regulation and efficiency.
  5. Auxiliary reliability: Cooling fans/pumps are fed from reliable auxiliary supplies with standby (dual) sources and automatic changeover, since loss of cooling can trip the transformer.

Conclusion: Earthing safeguards life and equipment by providing a low-resistance fault and surge path, while the cooling system safeguards insulation by continuous heat removal. Both are essential for safe, reliable, and economical operation of a substation.

Answer checked by Claude Haiku · Oct 21, 2026

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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 span length for which the total annual cost of the transmission line (capital cost of supports + cost of conductors + cost of insulators and fittings + annual cost of energy lost as I2RI^2R loss) is minimum. The main factors governing its choice are:

  1. Cost of supports (towers/poles): Longer spans mean fewer supports per km, reducing the capital cost of towers and their foundations. However, longer spans require taller and stronger towers, which increases the cost per tower. The economic span balances these two opposing effects.

  2. Cost of conductors: For a given sag, a longer span requires greater tension and hence a larger conductor size (or greater sag) to keep stress within limits. This increases conductor cost. Also, longer spans increase the weight of conductor per span to be supported.

  3. Cost of insulators and fittings: Fewer spans per km reduce the number of insulator strings and fittings, lowering cost, but the insulators must be designed for higher mechanical tension.

  4. Right-of-way (wayleave) cost: Longer spans reduce the number of tower sites and the width of right-of-way cleared, lowering land acquisition and clearing costs.

  5. Terrain and topography: In hilly terrain (common in Nepal), longer spans are economical as they reduce the number of tower locations on difficult ground; in plains, shorter spans may be more economical.

  6. Wind and ice loading: Higher wind/ice loads increase the mechanical loading on conductors and supports, tending to reduce the economic span.

  7. Line voltage and conductor material: Higher voltage lines generally use longer economic spans; the conductor material (ACSR, AAAC, copper) and its cost per kg affect the optimum.

  8. Cost of energy losses: Longer spans increase the conductor resistance per span (for a fixed size) and hence I2RI^2R losses; this raises the annual energy cost and pushes the economic span downward.

  9. Labour and maintenance cost: Fewer supports reduce erection and maintenance labour, favouring longer spans, but inspection and repair of long spans may be costlier.

  10. Standardisation and practice: In practice, the economic span is chosen close to standard values used by the utility (e.g., about 200–300 m for 11–33 kV, 300–400 m for 66–132 kV, and 400–500 m for 220 kV and above), so that design, erection and maintenance are simplified.

Mathematically, if CsC_s = cost of one support, CcC_c = cost of conductor per metre, CiC_i = cost of insulators per span, and CeC_e = annual cost of energy loss per metre, the total annual cost per span is minimised by differentiating with respect to span length ll and equating to zero, giving the economic span.


Why a High Voltage Transmission System is Required in Modern Power Systems

Modern power systems require high voltage transmission for the following reasons:

  1. Reduction in transmission losses: Power loss in a line is PL=I2RP_L = I^2 R. For the same power P=VIcos⁡ϕP = VI\cos\phi, increasing voltage VV reduces current II, and since loss varies as the square of current, losses fall sharply. For example, doubling the voltage reduces the current to half and the I2RI^2R loss to one-fourth.

  2. Reduction in conductor material (copper/aluminium) requirement: For a given power and distance, the conductor cross-section is A∝I∝1/VA \propto I \propto 1/V, and for a fixed permissible loss the volume of conductor is inversely proportional to the square of the voltage (Vc∝1/V2V_c \propto 1/V^2). Thus higher voltage greatly reduces conductor cost.

  3. Increase in power transmission capacity: The power transmitted is P=V1V2Xsin⁡δP = \frac{V_1 V_2}{X}\sin\delta for a line. Higher voltage directly increases the power-carrying capability of the line, so more power can be transmitted over the same right-of-way.

  4. Reduction in voltage drop and better voltage regulation: Voltage drop =IR+IX= IR + IX. With higher voltage, current is smaller, so the drop is smaller and regulation improves, keeping the receiving-end voltage close to the sending-end value.

  5. Economical transmission over long distances: Because losses and conductor cost fall and capacity rises, power can be transmitted economically over long distances (e.g., from hydropower stations in the hills to load centres in the Terai), which is essential for a country like Nepal where generation and load centres are far apart.

  6. Interconnection of systems and reliability: High voltage lines (132 kV, 220 kV, 400 kV) allow interconnection of regional grids, enabling pooling of generation, reserve sharing, and improved reliability and stability of supply.

  7. Reduction in the number of circuits and line cost per MW: Fewer high-voltage circuits are needed to carry the same power compared to many low-voltage circuits, reducing total line cost, tower cost and right-of-way requirement.

  8. Meeting growing load demand: As load demand grows rapidly with industrialisation and urbanisation, high voltage transmission is necessary to transmit bulk power efficiently from large generating stations to distant load centres.

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

Answer checked by Claude Sonnet · Oct 21, 2026

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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

Types of Lighting Sources Used in Illumination

Lighting sources are broadly classified into two groups:

1. Incandescent (Filament) Lamps

  • Construction: A tungsten filament sealed in a glass bulb either evacuated or filled with inert gas (argon/nitrogen).
  • Working: Current passes through the filament, heating it to incandescence (about 2500–3000 K); light is emitted as a result of the high temperature.
  • Types: Vacuum lamp, gas-filled lamp, coiled-coil lamp, halogen lamp.
  • Features: Low initial cost, warm colour, simple operation, but low luminous efficacy (10–20 lm/W) and short life (~1000 h).

2. Discharge Lamps

Light is produced by passing current through a gas or vapour, causing excitation and subsequent emission of radiation.

(a) Low-pressure discharge lamps

  • Fluorescent lamp: Mercury vapour at low pressure emits UV radiation, which is converted to visible light by a phosphor coating on the tube. Efficacy ≈ 40–80 lm/W. Requires a choke (ballast) and starter.
  • Sodium vapour lamp (low pressure): Neon gas starts the discharge; sodium vapour gives a monochromatic yellow light. Efficacy ≈ 100–180 lm/W, used for street lighting.

(b) High-pressure discharge lamps

  • High-pressure mercury vapour lamp: Mercury at high pressure gives bluish-white light; efficacy ≈ 40–60 lm/W.
  • High-pressure sodium vapour lamp: Golden-white light, efficacy ≈ 80–120 lm/W, widely used for road lighting.
  • Metal halide lamp: Halides added to mercury vapour improve colour rendering; used in stadiums and studios.

3. Light Emitting Diode (LED)

  • Construction: A p-n junction semiconductor (GaAs, GaP, InGaN) mounted on a heat sink with a plastic lens.
  • Working: Forward bias causes electron-hole recombination, releasing energy as photons.
  • Features: Very high efficacy (100–150 lm/W), long life (>50,000 h), no mercury, instant start; now the preferred modern source.

4. Neon Sign Lamps

  • Construction: Glass tube filled with neon/argon at low pressure with electrodes at both ends.
  • Working: High voltage ionises the gas, producing coloured glow (neon → red, argon → blue). Used for advertisement signs.

5. Arc Lamps (Historical)

  • Construction: Two carbon electrodes separated by a small gap, with a ballast in series.
  • Working: An arc strikes across the gap, producing intense white light. Used earlier for cinema projection and flood lighting.

Thyristor (SCR) — Construction and Working Principle

Construction

A thyristor (Silicon Controlled Rectifier) is a four-layer, three-junction p-n-p-n semiconductor device with three terminals:

TerminalLayerFunction
Anode (A)Outer p-layerMain positive terminal
Cathode (K)Outer n-layerMain negative terminal
Gate (G)Inner p-layerControl terminal

The three junctions are J₁ (p-n near anode), J₂ (middle n-p), and J₃ (p-n near cathode). The device is made of silicon and is mounted on a heat sink for high current ratings. The anode and cathode carry the main current; the gate controls the instant of turn-on.

Working Principle (Two-Transistor Analogy)

The thyristor can be modelled as two interconnected transistors:

  • T₁ (p-n-p): anode, gate, and inner n-layer
  • T₂ (n-p-n): inner n-layer, cathode, and gate

Forward blocking mode (OFF state):

  • Anode is positive w.r.t. cathode, but gate is open.
  • Junctions J₁ and J₃ are forward biased; J₂ is reverse biased.
  • Only a small leakage current flows; the device behaves as an open switch.

Turn-on (Conduction mode):

  • A small positive gate current is applied.
  • This injects electrons into the inner p-layer, increasing the collector current of T₂, which in turn increases the base current of T₁.
  • Regenerative (positive feedback) action rapidly drives both transistors into saturation.
  • The regenerative feedback saturates both transistors, so all three junctions become forward biased and the thyristor latches ON (no J₂ avalanche breakdown is needed; breakover is only the alternative way of turning on at high forward voltage with the gate open).
  • Once ON, the gate loses control; the device continues to conduct even if the gate signal is removed. This is called latching.

Holding current:

  • The minimum anode current required to keep the thyristor in conduction. Below this value, it turns OFF.

Turn-off (Commutation):

  • The anode current must be reduced below the holding current, or the anode supply must be reversed momentarily.
  • The stored charge is swept out and J₂ recovers its blocking ability, restoring the forward blocking state.

V-I Characteristics (summary)

  • Forward blocking region: high voltage, low current.
  • Forward conduction region: low voltage drop (≈1–2 V), high current.
  • Reverse blocking region: small leakage current until reverse breakdown.

Applications

  • Phase-controlled rectifiers, AC voltage controllers, DC motor drives, inverters, cycloconverters, and static switches.

Answer checked by Claude Sonnet · 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 load curve is a graph plotted between load (power demand) on the y-axis and time on the x-axis, showing how the system load varies over a period (daily, monthly or yearly).

Load Duration Curve

A load duration curve is obtained by rearranging the load values of the load curve in descending order of magnitude, so that the x-axis represents the duration (hours) for which a particular load or more is demanded.

Importance

  1. Selection of generating units (plant capacity and type): The load curve shows the base load, peak load and the duration of each. Base load is met by base-load plants (hydro, nuclear, large thermal) which run continuously at high load factor, while peak load is met by quick-starting peak-load plants (gas turbines, pumped storage). The load duration curve directly indicates how many hours a given load level persists, so the most economical mix of base, intermediate and peak plants can be chosen.

  2. Economic loading / load scheduling: The area under the load curve gives the total energy (kWh) consumed in the period. Dividing this by the peak load × time gives the load factor. A high load factor means better utilization of plant and lower cost per kWh. The load duration curve helps decide the order in which units should be committed (merit-order loading) so that total operating cost is minimum.

  3. Plant capacity and reserve determination: The maximum ordinate of the load curve gives the peak demand. The installed capacity must be at least equal to the peak demand plus a reserve margin for maintenance and forced outages. The load duration curve shows the reserve capacity available at different times.

  4. Planning of future expansion: Yearly load curves reveal the trend of load growth. This is the basis for forecasting future demand and planning new generating stations, transmission lines and substations in time.

  5. Maintenance scheduling: By studying the load duration curve, the period of minimum load (light-load period) can be identified, and preventive maintenance of units can be scheduled in that period so that supply is not interrupted and no costly peaking plant has to be run.

  6. Tariff design: Load curves show the peak and off-peak periods. Two-part tariff (maximum demand charge + energy charge) and time-of-day tariff are designed using this information to encourage consumers to shift load from peak to off-peak hours.

  7. Determining spinning reserve and operating cost: The load duration curve gives the load that must be carried for a given number of hours, which fixes the spinning reserve requirement and the fuel/water requirement for the period.


Preventive Maintenance of a Transformer

Preventive maintenance is the periodic, planned 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 unexpected failure.

A. Routine (Frequent) Checks

  1. Oil level: Check the oil level in the main tank and conservator against the gauge; top up with tested insulating oil if low. A falling oil level indicates leakage.
  2. Oil leakage: Inspect joints, gaskets, bushings, valves and the drain plug for leaks and arrest them.
  3. Silica gel breather: Check the colour of silica gel — blue/pink indicates saturation; replace or reactivate by heating. Check and clean the oil seal in the breather cup.
  4. Temperature: Record the oil and winding temperature from the indicators; ensure they are within permissible limits (oil ≈ 90 °C, winding ≈ 95–100 °C). Check the operation of cooling fans/pumps and the alarm and trip contacts.
  5. Buchholz relay: Check the relay for gas collection; drain and test the gas. Ensure the alarm and trip circuits are healthy.
  6. Bushings and insulators: Clean dust, dirt and salt deposits; check for cracks, chipping, tracking marks and tightness of connections.
  7. Earthing: Verify that the tank, neutral and lightning arrester earthing are intact and of low resistance.
  8. Abnormal sound/vibration: Listen for humming, crackling or knocking sounds indicating loose core laminations, poor clamping or partial discharge.
  9. Paint and rust: Touch up damaged paint to prevent corrosion of the tank and radiator.

B. Periodic Tests (Shutdown Maintenance)

  1. Insulation Resistance (IR) test: Measure IR of HV–LV, HV–earth and LV–earth with a megger (500 V/1 kV/2.5 kV as per rating). Compare with previous readings; a sudden fall indicates moisture or insulation deterioration. Also compute the Polarization Index (PI = IR at 10 min / IR at 1 min); PI ≥ 2 is satisfactory.
  2. Oil testing:
    • Dielectric strength (breakdown voltage) test of oil.
    • Acidity test (should be < 0.2 mg KOH/g).
    • Moisture content test.
    • Dissolved Gas Analysis (DGA) to detect incipient faults (arcing, overheating, partial discharge).
    • Furan analysis to assess solid insulation ageing.
  3. Winding resistance measurement: Measure and compare with original values to detect loose connections, shorted turns or contact problems in tap changer.
  4. Turns ratio / voltage ratio test: Verify ratio at all tap positions.
  5. Tap changer maintenance: Clean and lubricate the diverter switch, check contacts for pitting, replace contaminated oil in the diverter chamber, and test the motor drive and limit switches.
  6. Magnetic balance and vector group tests: To detect inter-turn short circuits and verify connection group.
  7. Tan delta / capacitance test of bushings: To detect deterioration of bushing insulation.
  8. Winding temperature indicator and OTI calibration: Verify accuracy of temperature indicators and alarm/trip settings.
  9. Tightening of connections: Re-tighten all busbar, cable and bushing terminal connections to specified torque to avoid hot spots.
  10. Cleaning and painting: Clean the core, coils (if accessible), tank and radiator; repaint the tank.

C. Safety Precautions

  1. Isolate the transformer from both sides, rack out / open the breakers and isolators.
  2. Apply earthing on all terminals and display "Men at Work" tags.
  3. Discharge the winding to earth before touching terminals (to remove trapped charge).
  4. Use proper PPE and calibrated test instruments; maintain a record of all readings for trend comparison.

D. Records and Schedule

Maintain a maintenance log/register giving date, IR values, oil test results, oil level, tap position and remarks. A typical schedule is: routine visual checks monthly, oil and IR tests every 6–12 months, and full shutdown maintenance once a year or as per the manufacturer's recommendation.

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.