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Chapter 2

Electrical Engineering Fundamentals

Lok Sewa Aayog (Public Service Commission) past exam questions

Past questions and answers

5 questions set from this chapter. Most repeated first.

  • 2082 · 2nd Paper · 10 marks

What is PID controller and its significance in power system control? Explain the working principle and applications of such controllers. 3+7=10

Answer

PID Controller and its Significance in Power System Control

1. PID Controller (3 marks)

A PID controller (Proportional–Integral–Derivative controller) is a closed-loop feedback controller that continuously calculates an error signal ( e(t) ) as the difference between a desired setpoint ( r(t) ) and the measured process variable ( y(t) ), and applies a correction based on three separate terms — proportional, integral and derivative.

u(t)=Kp e(t)+Ki∫0te(τ) dτ+Kd de(t)dtu(t) = K_p\,e(t) + K_i \int_0^t e(\tau)\,d\tau + K_d\,\frac{de(t)}{dt}

where:

  • ( u(t) ) = controller output (control signal)
  • ( e(t) = r(t) - y(t) ) = error
  • ( K_p ) = proportional gain, ( K_i ) = integral gain, ( K_d ) = derivative gain

In Laplace (s-domain) form:

Gc(s)=Kp+Kis+Kds=Kp(1+1Tis+Tds)G_c(s) = K_p + \frac{K_i}{s} + K_d s = K_p\left(1 + \frac{1}{T_i s} + T_d s\right)

where ( T_i = K_p/K_i ) is the integral time and ( T_d = K_d/K_p ) is the derivative time.

Significance in power system control:

  • Maintains frequency and voltage at their nominal values despite continuous load changes.
  • Provides fast, accurate and stable response — proportional term gives speed, integral term removes steady-state error, derivative term damps oscillations and improves transient response.
  • Ensures reliability and power quality by keeping deviations within permissible limits (e.g. frequency ( \pm 0.5 ) Hz, voltage ( \pm 5% )).
  • Enables automatic generation control (AGC), excitation control and turbine speed governing without manual intervention.
  • Simple to implement, well understood, and easily tunable, hence widely used in SCADA/EMS-based modern grids.

2. Working Principle (4 marks)

The controller works on the principle of negative feedback: the output is measured, compared with the reference, and the error is driven towards zero.

(a) Proportional (P) action: Output is directly proportional to the instantaneous error, ( u_P = K_p e(t) ). It gives an immediate corrective action — the larger the error, the stronger the correction. However, a pure P-controller leaves a steady-state offset because a finite error is needed to produce a finite output. Increasing ( K_p ) reduces offset but causes overshoot and oscillation, and may lead to instability.

(b) Integral (I) action: Output is proportional to the time integral of the error, ( u_I = K_i \int e,dt ). As long as any error persists, the integral keeps accumulating and the output keeps changing until the error becomes exactly zero. Thus it eliminates steady-state error and ensures the setpoint is finally reached. Too large ( K_i ) causes slow response and oscillation (integral windup); hence anti-windup limiters are used.

(c) Derivative (D) action: Output is proportional to the rate of change of error, ( u_D = K_d \frac{de}{dt} ). It responds to the trend of the error, providing a braking or anticipatory action that reduces overshoot and damps oscillations, improving stability and settling time. Since it acts on slope, it is sensitive to measurement noise; hence a derivative filter is often added. Pure D-action gives no output for a constant error.

Combined action: The three terms are summed to form the control signal. In a power system, this signal drives the governor setpoint (for frequency/load control) or the AVR/excitation system (for voltage control). The controller is tuned (Ziegler–Nichols, Cohen–Coon, or trial-and-error) so that the closed-loop response is fast, well-damped and stable. In practice, PI controllers are most common in AGC (derivative action is avoided due to measurement noise), while PID is used where fast transient response is essential.

3. Applications (3 marks)

In power system control:

  • Automatic Generation Control (AGC) / Load Frequency Control (LFC): PI/PID controllers regulate the turbine-governor setpoint to keep frequency and tie-line power interchange at scheduled values.
  • Automatic Voltage Regulator (AVR): PID control of generator excitation to maintain terminal voltage and improve transient stability.
  • Turbine speed governing: Speed control of hydro/steam turbines during load rejection and islanding.
  • Reactive power and voltage control: SVC, STATCOM and capacitor bank switching.
  • HVDC and FACTS controllers: Current/power flow regulation in converters and damping of inter-area oscillations.
  • Microgrid and distributed generation: Frequency/voltage control of inverter-based sources and energy storage.

Other applications: process control (temperature, pressure, flow, level), motor speed and position control, robotics, aerospace, and industrial automation.

Conclusion: The PID controller is the backbone of automatic control in power systems because it combines speed (P), accuracy (I) and stability (D) in a simple, robust and easily tunable structure, thereby ensuring reliable, quality and economic operation of the grid.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd Paper · 10 marks

Why is a tertiary winding introduced in a star-star (Y-Y) connected power transformer? What are the possible conditions under which tertiary winding can be avoided? Discuss. 6+4=10

Answer

Why a tertiary winding is used in a star-star transformer (6 marks)

A Y-Y transformer is economical for high voltage (graded insulation, earthed neutral), but it has drawbacks that a third, delta-connected tertiary winding on the same core removes.

  1. Third-harmonic magnetising current. The magnetising current needs a third harmonic to give sinusoidal flux. With isolated neutrals, triple-frequency currents of the three phases are in phase and cannot flow in a three-wire star. Flux and phase voltages become distorted with a third-harmonic component. In a closed delta, these in-phase currents circulate and restore sinusoidal flux and voltage.
  2. Zero-sequence and earth-fault path. The delta gives a low-impedance path for zero-sequence currents. This supports earth-fault current for relay operation, reduces the zero-sequence impedance, and limits over-voltage on healthy phases.
  3. Neutral stability under unbalanced load. Unbalanced single-phase loads shift the neutral. The delta carries the unbalance (zero-sequence) ampere-turns and keeps phase voltages nearly balanced.
  4. Auxiliary use. The tertiary can supply station auxiliary load (for example 11 kV or 33 kV), or connect shunt reactors or capacitors for voltage control. It is also used for star-star-delta autotransformers at 220/132 kV and 400/220 kV substations.

Conditions under which the tertiary can be avoided (4 marks)

  1. Five-limb or shell-type core. The in-phase third-harmonic (zero-sequence) flux has an iron return path through the outer limbs, so flux stays sinusoidal. In a three-limb core the flux must return through the air, oil and tank, which causes heating and distortion, so a tertiary is needed there.
  2. Solidly earthed neutral on at least one side, with a delta winding elsewhere in the system or bank, so triple-frequency and zero-sequence currents already have a path.
  3. Balanced three-phase load, so neutral shift and unbalance are negligible.
  4. Low harmonic content where the system is already effectively earthed and the earth-fault level is adequate.
  5. Small distribution ratings, where distortion is tolerable and a Dyn connection can be used instead.
  6. Earth-fault protection and earthing are provided otherwise, for example by a separate earthing transformer (zig-zag).

A tertiary improves waveform, fault capability and neutral stability. It is omitted where the core or the earthing already provides these.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · 2nd Paper · 10 marks

Discuss the criteria/parameters that need to be looked into while selecting the site for developing a hydroelectric power project. 10

Answer

Criteria/Parameters for Selecting a Site for a Hydroelectric Power Project

Site selection is the most critical decision in hydropower development, as it determines technical feasibility, cost, and long-term viability. The following parameters must be examined:

1. Hydrological Parameters

  • Water availability and flow duration curve: Minimum, average and peak discharge data over at least 20–30 years; firm flow (Q₉₀ or Q₉₅) must be adequate to run the turbines.
  • Catchment area and rainfall/snowmelt pattern: Determines runoff volume and seasonal variation.
  • Flood magnitude: Design flood (e.g., 1-in-100 or 1-in-1000 year flood) for spillway and diversion sizing.
  • Sediment load: High sediment reduces reservoir life and erodes turbines; siltation rate must be assessed.

2. Topographic and Geotechnical Parameters

  • Head (gross and net): Greater head means more power per unit of water; a steep gradient between intake and powerhouse is preferred.
  • Storage capacity: For reservoir projects, a suitable valley for impounding water; for run-of-river, adequate drop within a short distance.
  • Geology and rock condition: Stable bedrock for dam, tunnel, powerhouse; absence of faults, landslides, and karst.
  • Seismic zone: Nepal lies in a high seismic zone; fault proximity must be evaluated.

3. Technical and Engineering Parameters

  • Type of scheme feasible: Run-of-river, storage, or pumped storage, based on site conditions.
  • Accessibility: Road access for transporting heavy equipment (turbines, transformers).
  • Transmission distance to load centre: Shorter distance means lower transmission cost and losses.
  • Availability of construction materials: Sand, gravel, stone, and water nearby.

4. Economic and Financial Parameters

  • Cost per MW / per kWh: Capital cost, O&M cost, and levelised cost of energy.
  • Benefit–cost ratio and payback period: Must be financially viable.
  • Power purchase agreement (PPA) prospects: Availability of a buyer (NEA) and tariff rate.
  • Financing availability: Bankable feasibility study, debt–equity structure.

5. Environmental and Social Parameters

  • Environmental Impact Assessment (EIA): Effects on aquatic life, forest, and biodiversity; need for fish ladder or minimum environmental flow.
  • Land acquisition and resettlement: Number of affected households, compensation cost.
  • Public acceptance: Local community support and benefit-sharing (e.g., free electricity, local employment).
  • Protected areas: Avoidance of national parks, conservation areas, and religious sites.

6. Legal, Institutional and Policy Parameters

  • Water rights and licence: Compliance with Water Resources Act, 2049 and Electricity Act, 2049.
  • Hydropower licence from DoED: Survey, generation, and transmission licence.
  • Compliance with national policy: Hydropower Development Policy, 2058; priority to projects with storage and higher capacity.
  • Inter-state/international water treaty issues: For trans-boundary rivers.

7. Other Considerations

  • Reservoir submergence and catchment treatment: Siltation control and watershed management.
  • Minimum environmental flow (e.g., 10% of minimum monthly flow): Legally required.
  • Future expansion potential: Space for additional units or upgrading.

Conclusion: A hydropower site is selected only when hydrological adequacy, favourable topography and geology, technical feasibility, economic viability, environmental and social acceptability, and legal compliance are all satisfied. A multi-criteria decision analysis (MCDA) or weighted scoring is often used to rank alternative sites.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2082 · 2nd Paper · 20 marks

Construction of transmission lines has become a severe challenge for any utility in Nepal due to number of local issues. Failure to complete the transmission line has created problem for power evacuation causing huge financial burden to power plant developers. This has also been the root cause to keep our domestic demand and per capita consumption suppressed. In this context, identify and critically analyze the key local issues faced during construction of transmission line and suggest the remedial measures to resolve them based on the current legal provisions. 12+8=20

Answer

Transmission line construction in Nepal: local issues, analysis and remedies

Transmission lines are the evacuation link for hydropower. Delay in lines such as Marsyangdi Corridor, Dhalkebar-Inaruwa, Solu Corridor and Bheri Corridor has left completed plants unable to sell energy (spilled energy, deemed generation claims), raised interest during construction, and held back domestic consumption. The causes are mainly local rather than technical.

1. Key local issues and critical analysis (12 marks)

(a) Right-of-way (RoW) and land acquisition. Owners resist tower pads and corridors because compensation is seen as low, and the line restricts building and future land use. Valuations differ between district committees, and payment is slow. Squatters and tenants without titles are often left out. This is the largest single cause of delay.

(b) Forest and protected-area clearance. Lines cross community forests, national forests and buffer zones. Tree-felling approval needs EIA/IEE, forest ministry approval, compensatory plantation and consent of community forest user groups. Each step adds months. Clearance is slow because of weak coordination between ministries.

(c) Local obstruction and demands. Local people, political groups and contractors' labour make demands for jobs, local development or higher compensation, and stop work at tower sites. Lack of early consultation and benefit sharing turns this into recurring protest.

(d) Compensation for tree, crop and structures, and tower-base rates. Unlike full acquisition, the line corridor stays with the owner, so many see compensation for only the tower footing as unfair, and demand payment for the full corridor.

(e) Difficult terrain, access and weather. Steep geology, landslides, monsoon floods and lack of roads delay survey and stringing, especially in hill sections.

(f) Weak coordination and institutional capacity. The CDO, local levels, land revenue and forest offices, and NEA work separately. NEA's project teams are short-staffed, procurement is slow, and contractor performance and variation claims add delay.

(g) Funding and contract problems. Delayed payment, financing gaps and weak contractor capacity stall work.

Taken together, these issues are largely a failure of process and trust rather than of engineering. Because transmission is built by NEA alone while generation is built by many private developers, the mismatch in timing is built into the system.

2. Remedial measures under current legal provisions (8 marks)

  1. Use the Land Acquisition Act 2034 and the 2071 resettlement policy properly. Fix compensation transparently at market value through the Compensation Determination Committee, pay on time, and include tenants and landless users in the resettlement plan.
  2. Pay for the corridor, not only the tower base. Provide a standard, notified RoW compensation scheme and recognise loss of land use, as allowed by the Electricity Act 2049 and its rules.
  3. Speed up forest clearance under the Forest Act 2076 and Environment Protection Act 2076 through a single window, with deadlines for decisions and compensatory plantation ready in advance.
  4. Consult early and share benefits. Run public hearings during the EIA, form local coordination committees with local levels, and provide community benefits such as local jobs and small infrastructure.
  5. Use the Local Government Operation Act 2074. Local levels can help in mobilising people and securing approvals; make their role formal in project agreements.
  6. Sequence the work. Complete land and forest clearance before awarding the contract; start the line before or with the power plant.
  7. Strengthen NEA and contractors. Give project units authority, use performance guarantees, liquidated damages and staged payment, and use alternative designs such as underground cable or compact towers in sensitive areas.
  8. Use alternative models. Allow private or public-private investment in transmission under the Electricity Regulatory Commission and the Electricity Act, and make developers and NEA jointly manage evacuation risk.
  9. Support protection. Provide police and administration support for projects of national importance through the CDO, while using mediation first.

Conclusion

Transmission delay in Nepal is a problem of land, forest, community consent and coordination. Fair, timely compensation, a single-window clearance, early local involvement and project sequencing, within the Land Acquisition, Electricity, Forest and Environment laws, can resolve most of it and release the demand that is currently suppressed.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2082 · 2nd Paper · 10 marks

Discuss the Loss of Load Probability (LOLP) and Expected Energy not Served (EENS) while evaluating reliability of a power system. 5+5=10

Answer

Loss of Load Probability (LOLP)

Definition: LOLP is the probability that the system's available generating capacity will be insufficient to meet the system load demand at a given time (or over a given period). It measures the likelihood of a capacity shortfall, not its size.

Evaluation approach:

  1. Build the capacity outage probability table (COPT). For each generating unit, define its forced outage rate (FOR), q=FORq = \text{FOR}, and availability p=1−qp = 1 - q. For nn identical units, the probability that exactly kk units are on outage is

P(k)=(nk)qkp n−kP(k) = \binom{n}{k} q^{k} p^{\,n-k}

so the capacity on outage is O=k×O = k\times (unit size). For units of different sizes and FORs, the table is built by adding one unit at a time (convolution): Pnew(O)=p Pold(O)+q Pold(O−Cu)P_{\text{new}}(O)=p\,P_{\text{old}}(O)+q\,P_{\text{old}}(O-C_u), where CuC_u is the unit's capacity.

  1. Model the load. Use the daily peak load variation curve or the load duration curve (LDC) for the study period.

  2. Compute LOLP. For each day (or hour) jj with peak load LjL_j, find the outage states for which available capacity C−Ok<LjC - O_k < L_j (i.e., reserve is negative). Then:

LOLP=1N∑j=1N∑k∈SjP(Ok)\text{LOLP} = \frac{1}{N}\sum_{j=1}^{N} \sum_{k \in S_j} P(O_k)

where SjS_j is the set of outage states causing loss of load on day jj, and NN is the number of days in the period. Equivalently, LOLP = Σ (probability of capacity deficiency) × (probability of load level).

Interpretation: LOLP is usually expressed in days/year or hours/year. A common planning criterion is 1 day in 10 years (0.1 day/year) — the "1-in-10" standard. It is a dimensionless probability index and does not indicate how severe the shortage is.

Limitations: LOLP treats all shortages as equal — a 1 MW deficit and a 1000 MW deficit give the same LOLP. It also ignores the duration and magnitude of the deficiency. Hence EENS is used as a complementary index.


Expected Energy Not Served (EENS)

Definition: EENS (also called Expected Unserved Energy, EUE) is the expected amount of energy (MWh) that cannot be supplied to consumers over the study period because of generating capacity deficiencies. It quantifies the magnitude/severity of the shortage.

Evaluation approach:

  1. From the COPT, obtain the probability P(Ok)P(O_k) of each capacity outage state.
  2. For each load level LjL_j of duration tjt_j (from the load duration curve), determine the capacity deficiency:

Djk=max⁡(0,  Lj−(C−Ok))D_{jk} = \max\big(0,\; L_j - (C - O_k)\big)

  1. The energy not served for that state and load level is Djk×tjD_{jk} \times t_j.
  2. Sum over all states and load levels, weighted by probabilities:

EENS=∑j∑kP(Ok) [max⁡(0,  Lj−C+Ok)] tj(MWh/period)\text{EENS} = \sum_{j} \sum_{k} P(O_k)\,\big[\max(0,\; L_j - C + O_k)\big]\,t_j \quad \text{(MWh/period)}

Interpretation: EENS gives the expected MWh of unserved demand per year. It is directly related to the economic cost of unreliability through the Value of Lost Load (VOLL):

Expected Interruption Cost=EENS×VOLL (Rs/kWh)\text{Expected Interruption Cost} = \text{EENS} \times \text{VOLL (Rs/kWh)}

Use in planning: EENS is used to size reserve capacity, compare expansion plans, and justify new generation projects on a cost–benefit basis. A plan with lower EENS but higher cost is justified only if the reduction in interruption cost exceeds the additional investment.


Relationship and Combined Use

IndexMeasuresUnitPlanning use
LOLPProbability of shortagedays/yr or hours/yrFrequency of deficiency events
EENSExpected magnitude of shortageMWh/yrSeverity and economic cost

The two indices are complementary: LOLP tells how often load is lost, EENS tells how much energy is lost. A system may have a low LOLP but a high EENS if the few deficiency events are very severe. Therefore, a reliable system should satisfy both criteria — e.g., LOLP ≤ 0.1 day/year and EENS within an acceptable economic limit. Together they allow the planner to trade off reliability against cost and select the least-cost expansion plan that meets the reliability standard.

Answer checked by Claude Sonnet · Oct 21, 2026

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Questions from NEA Question for Assistant Manager 8th (Assistant Manager 8th Level · 1st · 2082) and NEA Loksewa Question for Assistant (Assistant Manager 8th Level · 2nd Paper · 2082). Answers are written for this site; check them against your class notes.