210 questions — 156 objective (1 mark), 27 short (5 marks), 27 long (10 marks). Every answer is checked against the 2014 BEE guidebook and carries its book section reference plus an explanation. ▶ Practice this chapter interactively (timer, read-aloud, progress saving).
Objective questions (1 mark) — 156
📖 §1.4 Power Factor Improvement and Benefits — Power Triangle
1. Unity power factor means that the load draws:
maximum reactive power from the supply
no reactive power (kVAr = 0) from the supply
equal active and reactive power
zero active power
Answer: B) no reactive power (kVAr = 0) from the supply
Confirmed vs Book-3 §1.4 — PF = cosΦ = kW/kVA, and kVA = √(kW² + kVAr²). At unity PF, cosΦ = 1 so kVA = kW, which forces kVAr = √(kVA² − kW²) = 0: the load draws only active power. The book adds that if all loads were at unity PF, maximum power could be transferred for the same distribution system capacity. Option (c) is wrong — equal kW and kVAr gives PF = 0.707, not unity.
Source: AI practice
📖 §1.4 Selection of capacitors — kVAr = kW(tanΦ₁ − tanΦ₂), Table 1.2
2. A capacitor required to raise the PF of a 500 kW load from 0.85 to 0.95 is approximately (tan(cos⁻¹0.85)=0.620, tan(cos⁻¹0.95)=0.329):
95 kVAr
145 kVAr
200 kVAr
291 kVAr
Answer: B) 145 kVAr
Confirmed vs Book-3 §1.4 — kVAr = kW × (tanΦ₁ − tanΦ₂) = 500 × (0.620 − 0.329) = 500 × 0.291 = 145.5 ≈ 145 kVAr. Book Table 1.2 gives exactly 0.291 as the multiplier for 0.85 → 0.95, so 500 × 0.291 confirms the answer. Option (d) 291 kVAr is the multiplier read as a kVAr figure, i.e. the kW factor was dropped.
Source: AI practice
📖 §1.5 Transformer Losses and Efficiency
3. An 800 kVA transformer has a no-load loss of 1000 W and a full-load copper loss of 8000 W. Its total loss at 50% load is:
3000 W
4000 W
4500 W
9000 W
Answer: A) 3000 W
Confirmed vs Book-3 §1.5 — P_total = P_no-load + (%Load/100)² × P_load = 1000 + (0.5)² × 8000 = 1000 + 2000 = 3000 W. No-load (core) loss is constant whenever the transformer is energized; copper loss varies with the SQUARE of the load fraction. Option (d) 9000 W adds the full copper loss without squaring the 0.5 fraction.
Source: AI practice
📖 §1.5 Transformer Losses and Efficiency
4. The no-load (core) loss of a transformer consists of:
hysteresis and eddy-current losses
I²R losses in the primary and secondary windings
stray and friction losses in the core clamps
ohmic losses that vary linearly with load
Answer: A) hysteresis and eddy-current losses
Confirmed vs Book-3 §1.5 — the book splits transformer loss into no-load loss and load loss. No-load (core/iron) loss is caused by two factors: hysteresis (energy lost reversing the magnetic field in the core) and eddy currents (induced currents circulating in the core). It is present whenever the transformer is energized and does not vary with load. Option (b) is the load (copper) loss, which is the I²R component and varies with the square of load — not linearly, which also rules out (d).
Source: AI practice
📖 §1.4 Location of Capacitors — energy-efficiency viewpoint
5. If the tail-end power factor of a distribution feeder is improved from 0.85 to 0.95, the reduction in distribution loss is about:
10%
20%
29%
40%
Answer: B) 20%
Confirmed vs Book-3 §1.4 — % loss reduction = [1 − (PF₁/PF₂)²] × 100 = [1 − (0.85/0.95)²] × 100 = [1 − 0.8006] × 100 = 19.9 ≈ 20%. The ratio is OLD PF over NEW PF, squared, because loss ∝ current² and current ∝ 1/PF. Option (c) 29% is the capacitor-sizing multiplier 0.291 from Table 1.2 mistaken for a loss reduction.
Source: AI practice
📖 §1.1 Cascade Efficiency (T&D) / §1.7 T&D losses
6. A plant receives 90 MU at the end of a system whose T&D efficiency is 75%. The generation required is:
67.5 MU
105 MU
120 MU
135 MU
Answer: C) 120 MU
Confirmed vs Book-3 §1.1 — T&D efficiency = energy delivered ÷ energy generated, so generation required = 90 / 0.75 = 120 MU. The book's cascade profile makes the same point (η_cascade ≈ 87%, standard Indian technical losses ≈ 17%). Option (a) 67.5 MU is the classic trap of multiplying by the efficiency instead of dividing.
7. A utility has a billing efficiency of 85% and a collection efficiency of 80%. Its AT&C loss is:
15%
20%
32%
68%
Answer: C) 32%
Confirmed vs Book-3 §1.8 — AT&C Loss (%) = {1 − (Billing Efficiency × Collection Efficiency)} × 100 = {1 − (0.85 × 0.80)} × 100 = {1 − 0.68} × 100 = 32%. Book Table 1.7 works the same formula (BE 70%, CE 93% → 35%). Option (d) 68% is the realisation (BE × CE), i.e. what is recovered, not what is lost.
Source: AI practice
📖 §1.5 Energy Efficient Transformers (amorphous / metallic glass core)
8. An amorphous-core (metallic-glass) distribution transformer primarily reduces:
copper (load) loss by ~70%
no-load (core) loss by ~70%
stray load loss
friction and windage loss
Answer: B) no-load (core) loss by ~70%
Confirmed vs Book-3 §1.5 — the book states that an amorphous (metallic glass alloy) core gives an expected reduction in CORE loss of about 70% over conventional silicon-alloyed (grain-oriented) iron cores, with efficiency as high as 98.5% at only 35% load. Option (a) is the trap: the amorphous core changes the magnetic material, so it attacks no-load/iron loss only — winding copper (load) loss is unaffected.
Source: AI practice
📖 §1.5 Transformers (transmission at high voltage, low current) / §1.6 Distribution losses
9. When transmission voltage is raised from 11 kV to 33 kV for the same power, the line (I²R) loss is reduced by a factor of:
3 times
6 times
9 times
27 times
Answer: C) 9 times
Confirmed vs Book-3 §1.5 — the book's reason for transforming up is that high voltage means low current for the same power, and §1.6 states all distribution losses are current-dependent (I²R). For the same power, I ∝ 1/V, so loss ∝ 1/V². Raising 11 → 33 kV is a 3× rise, so loss falls by 3² = 9 times. Option (a) 3 is the reduction in current and in voltage drop, not in loss.
Source: AI practice
📖 §1.5 Standards & Labeling Programme for Distribution Transformers
10. BEE star labelling of distribution transformers is based on total losses measured at:
25% and 75% load
50% and 100% load
full load only
no-load only
Answer: B) 50% and 100% load
Confirmed vs Book-3 §1.5 — BEE brought distribution transformers under the Standards & Labeling Programme, mandatory from 7th January 2010; the existing loss standard is IS 1180 (Part 1), and for BEE labelling the TOTAL losses at 50% and 100% load are defined (Table 1.4). Star 1 is the highest-loss band and Star 5 the lowest. Option (c) is wrong because a transformer rarely runs at full load, so a full-load-only figure would not represent real operation.
Source: AI practice
📖 §1.10 Harmonics
11. The 5th harmonic on a 50 Hz supply has a frequency of:
55 Hz
150 Hz
250 Hz
350 Hz
Answer: C) 250 Hz
Confirmed vs Book-3 §1.10 — a harmonic is a component frequency that is an integer multiple of the fundamental; the book's own example is a 50 Hz fundamental whose 5th harmonic = 5 × 50 = 250 Hz (and a 5th harmonic current is defined as current flowing at 250 Hz on a 50 Hz system). Option (d) 350 Hz is the 7th harmonic — the other harmonic the book pairs with the 5th when warning about capacitor resonance.
Source: AI practice
📖 §1.4 Location of Capacitors — energy-efficiency viewpoint
12. For the best ENERGY SAVING to the user, a power-factor capacitor should be located:
at the generating station
at the receiving substation
as close to the load as possible (tail end)
anywhere on the HT line
Answer: C) as close to the load as possible (tail end)
Confirmed vs Book-3 §1.4 — maximum benefit comes from locating capacitors as close as possible to the load, because the kilovars are then confined to the smallest segment, load current falls and power losses (∝ current²) drop substantially, while voltage at the motor rises. Option (b) is the trap the book names explicitly: capacitors at the receiving substation help only the UTILITY in loss reduction, whereas tail-end capacitors also cut loss inside the plant network and so benefit the user.
Source: AI practice
📖 §1.1 Introduction to Electric Power Supply Systems
13. What is the primary function of substations in the electrical power supply system?
To generate electricity
To communicate over long distances
To facilitate voltage transformation
None of the above
Answer: C) To facilitate voltage transformation
Confirmed vs Book-3 §1.1 — the book lists substations as the elements that 'connect the pieces to each other', and 'Sub-stations, containing step-down transformers, reduce the voltage for distribution to industrial users'. So their primary job is voltage transformation.
Option (a) is wrong because electricity is generated only in the power generating plant; the substation neither generates nor communicates.
Source: Sep 2024
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
14. How does the transmission voltage level affect the efficiency of long-distance power transmission?
Higher voltage levels reduce transmission losses
Higher voltage levels increase transmission losses
Voltage levels do not affect transmission losses
None of the above
Answer: A) Higher voltage levels reduce transmission losses
Confirmed vs Book-3 §1.1 — 'The current drawn is inversely proportional to the voltage level for the same quantity of power handled' and 'P.Loss = I²R', so raising voltage cuts loss in the ratio of the square of the voltages.
Option (b) reverses the physics: higher voltage means lower current, hence lower I²R loss, which is why HV/EHV transmission is used.
Source: Sep 2024
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
15. What is the primary purpose of using high voltage direct current (HVDC) transmission over long distances?
To increase the frequency of electricity
To step down the voltage for distribution
To minimize transmission losses over long distances
All of the above
Answer: C) To minimize transmission losses over long distances
Confirmed vs Book-3 §1.1 — 'Where transmission is over 1000 kM, high voltage direct current transmission is also favored to minimize the losses.'
Options (a)/(b) are wrong: HVDC has no power frequency to raise, and voltage stepping down is done by substation transformers, not by the HVDC link itself.
Source: Sep 2024
📖 §1.11 Analysis of Electrical Power Systems (Table 1.11)
16. What is the impact of voltage imbalance among the three phases in an electrical system?
Improved motor efficiency
Increased motor losses and reduced equipment life
Reduced power consumption
Enhanced power factor
Answer: B) Increased motor losses and reduced equipment life
Confirmed vs Book-3 §1.11 Table 1.11 — voltage imbalance among phases causes 'motor vibration, premature motor failure' and the book states 'A 5% imbalance causes a 40% increase in motor losses'.
Option (a) is wrong because imbalance produces a negative-sequence current that heats the motor and lowers, not improves, efficiency.
Source: Sep 2024
📖 §1.4 Power Factor Improvement and Benefits
17. How does adding capacitors to an electrical distribution system improve power factor?
By providing the reactive power
By increasing the active power consumption
By lowering the system voltage
By increasing the frequency of the system
Answer: A) By providing the reactive power
Confirmed vs Book-3 §1.4 — power-factor capacitors 'act as reactive power generators, and provide the needed reactive power to accomplish kW of work'.
Option (b) is wrong: capacitors do not change the active (kW) power drawn by the load — they only supply the magnetising kVAr locally, so the kVA and current from the source fall.
18. A transformer has a primary voltage of 220V and a secondary voltage of 110V. If the primary current is 5A, what is the secondary current assuming no losses?
2.5A
5A
10A
20A
Answer: C) 10A
Confirmed vs Book-3 §1.5 Normal Operation — 'Primary ampere-turns are equal to secondary ampere-turns', i.e. V₁I₁ = V₂I₂. So I₂ = 5 × 220/110 = 10 A.
Option (a) 2.5 A wrongly divides by the turns ratio; in a step-down transformer the secondary current must rise as the voltage falls.
Source: Sep 2024
📖 §1.4 Power Factor Improvement and Benefits
19. A factory consumes 500,000 kWh of electricity per month with a power factor of 0.8. How much is the reactive power (kVAR)?
400,000 kVAR
375,000 kVAR
800,000 kVAR
500,000 kVAR
Answer: B) 375,000 kVAR
Confirmed vs Book-3 §1.4 — kVAr = kW × tan(cos⁻¹PF). At PF 0.8, tanφ = 0.75, so reactive energy = 500,000 × 0.75 = 375,000 kVArh.
Option (a) 400,000 comes from kWh × sinφ (0.6/0.8 confusion); the power triangle uses the tangent, not the sine, of the angle when converting kW to kVAr.
Source: Sep 2024
📖 §1.10 Harmonics
20. Harmonics generation is more in_____________
Inverter Drive
LED lamp
Transformer
Resistance heater
Answer: A) Inverter Drive
Confirmed vs Book-3 §1.10 — the book lists 'Variable frequency drives (VFDs), electronic ballasts, UPS and Computers, induction and arc furnaces' as non-linear (harmonic-producing) devices.
Options (c)/(d): transformers and resistance heaters are essentially linear — impedance is constant — so they draw a sinusoidal current and generate negligible harmonics.
Source: Sep 2024
📖 §1.5 Transformers — losses & efficiency
21. A 750 kVA transformer has 1200 W no-load loss and 7200 W full-load copper loss. At 60% load, total loss is:
4320 W
5520 W
7632 W
3792 W
Answer: D) 3792 W
Confirmed vs Book-3 §1.5 — P_TOTAL = P_NO-LOAD + (%Load/100)² × P_LOAD = 1200 + (0.6)² × 7200 = 1200 + 2592 = 3792 W.
Option (b) 5520 W scales the copper loss linearly (0.6 × 7200); the book is explicit that 'Copper loss varies with the square of the load current'.
Source: Sep 2025
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
22. Amorphous core transformers primarily reduce:
Load loss
No-load loss
Stray loss
None of the above
Answer: B) No-load loss
Confirmed vs Book-3 §1.5 Energy Efficient Transformers — an amorphous (metallic-glass) core gives 'expected reduction in core loss over conventional (Si Fe core) transformers…roughly around 70%'. Core loss = no-load loss.
Option (a) is wrong: load (copper) loss is fixed by winding resistance and is unaffected by the core material.
Source: Sep 2025
📖 §1.4 Selection and Location of Capacitors
23. If tail-end power factor is improved from 0.80 to 0.95, distribution loss reduction is:
13.33%
21%
29%
16%
Answer: C) 29%
Confirmed vs Book-3 §1.4 — reduction in distribution loss % = [1 − (PF₁/PF₂)²] × 100 = [1 − (0.80/0.95)²] × 100 = 29%.
Option (a) 13.33% is the simple PF ratio change (0.15/0.80 ×…); losses vary with the square of current, so the ratio must be squared.
Source: Sep 2025
📖 §1.4 Power Factor Improvement and Benefits
24. Unity power factor means:
No reactive power is drawn from the supply
Current leads voltage
Current lags voltage
Reactive power is maximum
Answer: A) No reactive power is drawn from the supply
Confirmed vs Book-3 §1.4 — PF = kW/kVA = cosφ. At unity PF the angle φ = 0, so the kVAr side of the power triangle is zero: no reactive power is exchanged with the supply.
Options (b)/(c) describe leading and lagging conditions, both of which imply a non-zero reactive component.
Source: Sep 2025
📖 §1.10 Harmonics
25. In a UPS, DC to AC conversion is carried out by:
Converter
Charger
Battery
Inverter
Answer: D) Inverter
Confirmed vs Book-3 §1.10 — the UPS is listed as a non-linear load precisely because its output stage is an inverter, the device that converts DC (battery/DC link) into AC.
Option (a) 'converter' (rectifier) does the opposite — AC to DC — and the battery only stores energy; it cannot change the waveform.
Source: Sep 2025
📖 §1.1 Cascade Efficiency
26. If a plant receives 96 Million Units (MU) with a T&D efficiency of 80%, generation required is:
101.2 MU
76.9 MU
120 MU
68.1 MU
Answer: C) 120 MU
Confirmed vs Book-3 §1.1 Cascade Efficiency — energy delivered = energy generated × η. So generation = 96/0.80 = 120 MU.
Option (b) 76.9 MU multiplies instead of divides (96 × 0.80); losses must be added on top of the delivered units, so generation is always larger than the units received.
Source: Sep 2025
📖 §1.1 (a.c. fundamentals; see also Book-3 Ch-2 §2.2 — Ns = 120f/P)
27. Synchronous speed of a motor is inversely proportional to:
Number of poles
Frequency
Voltage
Temperature
Answer: A) Number of poles
Confirmed vs Book-3 §1.1/Ch-2 — synchronous speed Ns = 120f/P, so for a fixed supply frequency the speed is inversely proportional to the number of poles (4-pole → 1500 rpm, 2-pole → 3000 rpm at 50 Hz).
Option (b) is wrong because speed is directly (not inversely) proportional to frequency.
Source: Sep 2025
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
28. If the distribution voltage is raised from 11 kV to 33 kV, the line loss would be lower by a factor of
1/9
9
3
none
Answer: A) 1/9
Corrected (was b) — Book-3 §1.1: the book states 'if distribution of power is raised from 11 kV to 33 kV, the voltage drop would be lower by a factor 1/3 and the line loss would be lower by a factor (1/3)² i.e., 1/9.'
The answer is the factor 1/9, not 9 — the loss becomes one-ninth. Option (c) 3 is the current/voltage-drop ratio, not the loss ratio.
Source: Book EOC
📖 §1.4 Selection and Location of Capacitors
29. The kVAr rating required for improving the power factor of a load operating at 500 kW and 0.85 power factor to 0.95 is
145 kVAr
500 kVAr
50 kVAr
100 kVAr
Answer: A) 145 kVAr
Confirmed vs Book-3 §1.4 — kVAr rating = kW[tanΦ₁ − tanΦ₂] = 500 × (tan cos⁻¹0.85 − tan cos⁻¹0.95) = 500 × (0.620 − 0.329) = 145 kVAr (Table 1.2 multiplier 0.291).
Option (b) 500 kVAr wrongly assumes the capacitor must equal the kW rating; only the difference in reactive components is compensated.
Source: Book EOC
📖 §1.4 Selection and Location of Capacitors
30. The rating of the capacitor at motor terminals should not be greater than
magnetizing kVAr of the motor at full load
magnetizing kVAr of the motor at no load
magnetizing kVAr of the motor at half load
magnetizing kVAr of the motor at 75% load
Answer: B) magnetizing kVAr of the motor at no load
Confirmed vs Book-3 §1.4 — 'the rating of the capacitor should not be greater than the no-load magnetizing kVAr of the motor. If this condition exists, damaging over voltage or transient torques can occur.'
Option (a) full-load magnetising kVAr is larger and would over-compensate, causing self-excitation over-voltage when the motor is switched off.
Source: Book EOC
📖 §1.4 Performance Assessment of Power Factor Capacitors
31. If the voltage applied to a 415 V rated capacitor drops by 10%, its VAR output drops by
23%
87%
19%
10%
Answer: C) 19%
Confirmed vs Book-3 §1.4 — capacitor output varies as the square of the applied voltage (kVAr ∝ V²). At 0.9 V the output is 0.9² = 0.81, i.e. a drop of 19%.
Option (d) 10% assumes a linear relation; the squared law is why under-voltage badly de-rates capacitor banks.
32. The ratio of the sum of individual maximum demands of various equipment to the maximum demand of the plant is
load factor
diversity factor
demand factor
maximum demand
Answer: B) diversity factor
Confirmed vs Book-3 §1.5 Rating of Transformer — 'Diversity factor is defined as the ratio of overall maximum demand of the plant to the sum of individual maximum demand of various equipment.' It is the diversity factor that links these two demands.
Option (c) demand factor compares maximum demand to connected load, not to the sum of individual maximum demands.
Source: Book EOC
📖 §1.5 Rating of Transformer (diversity factor) — see also Book-3 Ch-9 DG sets
33. The approximate kVA rating required for a DG set with 1000 kW connected load, with diversity factor of 1.5, 84% loading and 0.8 power factor is
500 kVA
1000 kVA
1500 kVA
2000 kVA
Answer: B) 1000 kVA
Confirmed vs Book-3 §1.5 — plant max demand = connected load/diversity factor = 1000/1.5 = 667 kW; at 0.8 PF that is 667/0.8 = 833 kVA; the DG set is to run at 84% loading, so rating = 833/0.84 ≈ 992 ≈ 1000 kVA.
Option (a) 500 kVA comes from multiplying by 0.84 instead of dividing — that would overload the set beyond its rating.
34. Commercial losses in a distribution network are due to
theft
average billing
defective meters
all the above
Answer: D) all the above
Confirmed vs Book-3 §1.8 Commercial Losses — the book attributes commercial losses to meter reading discrepancies (defective/stopped meters, average billing) and to metering/meter tampering and theft.
Each of (a), (b) and (c) is named in the book, so no single one is complete — 'all the above' is the only correct choice.
Source: Book EOC
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
35. The Star rating programme of distribution transformers is based on losses at
no load
full load
at 25 % load
at 50 % & 100 % load
Answer: D) at 50 % & 100 % load
Confirmed vs Book-3 §1.5 Standards & Labeling — 'For the BEE labeling programme total losses at 50% and 100% load have been defined' (Table 1.4).
Options (a)/(b) are the IS 1180 no-load and load loss definitions; the star rating is based on the TOTAL loss at the two loading points, not on either loss alone.
Source: Book EOC
📖 §1.9 Demand Side Management (DSM)
36. Demand Side Management helps
to reduce the energy losses
to reduce system peak demand
to promote energy efficiency among users
all the above
Answer: D) all the above
Confirmed vs Book-3 §1.9 — DSM objectives include 'Improve the efficiency of energy systems', 'Reduce power shortages and power cuts' and reducing peak demand to defer new capacity.
Each of (a)–(c) is a stated DSM objective, so 'all the above' is correct; picking only one understates the scope of DSM.
Source: Book EOC
📖 §1.4 Power Factor Improvement and Benefits
37. If the reactive power drawn by a particular load is zero, it means the load is operating at
lagging power factor
leading power factor
unity power factor
none of the above
Answer: C) unity power factor
Confirmed vs Book-3 §1.4 — PF = cosφ, and zero reactive power means the kVAr side of the power triangle is zero, i.e. φ = 0 and PF = 1 (unity).
Options (a)/(b) both require a non-zero kVAr (lagging = inductive, leading = capacitive), so they contradict the given condition.
Source: Book EOC
📖 §1.4 Power Factor Improvement and Benefits
38. A pure resistive load in an alternating current (AC) circuit draws
lagging reactive power
active power
leading reactive power
none of the above
Answer: B) active power
Confirmed vs Book-3 §1.4 Power Factor Basics — 'In case of pure resistive loads, the voltage (V), current (I), resistance (R) relations are linearly related' — such loads (incandescent lighting, resistance heating) draw only active power.
Options (a)/(c) require energy storage in a magnetic or electric field, which a pure resistance does not have.
Source: 15th Exam
📖 §1.4 Power Factor Improvement and Benefits
39. Select the incorrect statement: The advantage of PF improvement by capacitor addition in an electric network is
active power component of the network is not affected
reactive power component of the network is not affected
I2R power losses are affected in the system
voltage level at the load end is affected
Answer: B) reactive power component of the network is not affected
Confirmed vs Book-3 §1.4 — the book's first listed advantage is that 'Reactive component of the network is reduced and so also the total current in the system'. Statement (b) claims the reactive component is NOT affected, which contradicts the book, so it is the incorrect statement asked for.
Options (c) and (d) restate the book's advantages (b) and (c), and (a) is true because capacitors do not change kW.
40. "Heat Rate" of a thermal power station is the heat input in kilo Calories or kilo Joules, for generating
one kW of electrical output
one kVAh of electrical output
one kWh of electrical output
one kVA of electrical output
Answer: C) one kWh of electrical output
Confirmed vs Book-3 §1.1 — ''HEAT RATE' is the heat input in kilo Calories or kilo Joules, for generating 'one' kilo Watt-hour of electrical output.'
Options (a)/(d) use kW and kVA, which are power (rate) units, not energy; heat rate is always expressed per unit of energy generated (kWh).
Source: 15th Exam
📖 §1.4 Selection and Location of Capacitors
41. Improving power factor at motor terminals in a plant will
increase active power drawn by motor
reduce system distribution losses
reduce contract demand with utility
increase motor design power factor
Answer: B) reduce system distribution losses
Confirmed vs Book-3 §1.4 — 'Maximum benefit of capacitors is derived by locating them as close as possible to the load…This, in turn, will reduce power losses of the system substantially.' Correcting at the motor relieves the whole in-plant network of reactive current.
Option (c) is wrong: contract demand is a commercial agreement with the utility and is not automatically reduced by PF correction.
Source: 15th Exam
📖 §1.4 Performance Assessment of Power Factor Capacitors
42. A plant had installed three phase shunt capacitors to improve power factor at MCC. Busbar three phase voltages at the main panel were balanced but at the MCC the line voltages were unbalanced. The main reason for this unbalanced voltage at MCC could be
PF capacitors were operating at higher supply frequency
PF improvement in all phase was not uniform due to blown fuse in one phase of the 3 phase PF capacitors
PF capacitors were operating at higher voltage than their rated values
PF capacitors were operating at lower voltage than their rated values
Answer: B) PF improvement in all phase was not uniform due to blown fuse in one phase of the 3 phase PF capacitors
Confirmed vs Book-3 §1.4 — a capacitor bank supplies kVAr per phase; if one phase's fuse blows, that phase gets no compensation while the other two do, so the phase currents and hence the voltage drops become unequal at the MCC.
Options (c)/(d) would change the kVAr in all three phases equally and so would not create an imbalance.
Source: 15th Exam
📖 §1.4 Performance Assessment of Power Factor Capacitors
43. A 50 kVAr, 415 V rated power factor capacitor was found to be having terminal supply voltage of 430 V. The capacity of the power factor capacitor at the operating supply voltage would be approximately
53.67 kVAr
50 kVAr
46.57 kVAr
none of the above
Answer: A) 53.67 kVAr
Confirmed vs Book-3 §1.4 Voltage effects — capacitor output varies with the square of the applied voltage: 50 × (430/415)² = 50 × 1.0735 = 53.67 kVAr.
Option (c) 46.57 kVAr is what would happen at a LOWER-than-rated voltage; note the book's caution that over-voltage shortens capacitor life.
Source: 15th Exam
📖 §1.4 Power Factor Improvement and Benefits
44. In a plant, the loading on a transformer was 1000 kVA with a power factor of 0.88. The plant improved the power factor to 0.99 by adding capacitors on the load side. The release in transformer loading (kVA) will be
111
889
999
none of the above
Answer: A) 111
Confirmed vs Book-3 §1.4 — kW is unchanged by capacitors: kW = 1000 × 0.88 = 880 kW. At 0.99 PF the kVA becomes 880/0.99 = 889 kVA, so the transformer is relieved of 1000 − 889 = 111 kVA.
Option (b) 889 kVA is the NEW loading, not the capacity released — the question asks for the reduction.
Source: 15th Exam
📖 §1.10 Harmonics
45. Select the incorrect statement:
Transformers operating near saturation level create harmonics
Devices that draw sinusoidal currents when a sinusoidal voltage is applied create harmonics
Harmonics are multiples of the supply frequency
Harmonics occur as spikes at intervals which are multiples of the supply frequency
Answer: B) Devices that draw sinusoidal currents when a sinusoidal voltage is applied create harmonics
Confirmed vs Book-3 §1.10 — 'As the value of impedance in above devices is constant, they are called linear'; linear devices drawing sinusoidal current from sinusoidal voltage do NOT create harmonics, so statement (b) is the incorrect one.
Options (c)/(d) are true — harmonics are integer multiples of the fundamental (5th = 250 Hz on a 50 Hz system) — and (a) is true because a saturated transformer is non-linear.
Source: 19th Exam
📖 §1.4 Automatic Power Factor Controllers
46. Which one of the following devices will help to eliminate the hunting problems normally associated with capacitor switching?
Maximum Demand Controller
Intelligent Power Factor Controller (IPFC)
Soft Starter
Eddy Current Drives
Answer: B) Intelligent Power Factor Controller (IPFC)
Confirmed vs Book-3 §1.4 Automatic Power Factor Controllers — when the load fluctuates, fixed banks make the PF swing between lagging and leading; the APFC/IPFC has programmable delay-ON and delay-OFF times and 'keeps the numbers of operations equal across all banks', which stops the hunting.
Option (a) a Maximum Demand Controller only sheds loads; it does not switch capacitor steps.
47. The gross efficiency of a coal based power generating unit with a gross heat rate of 2490 kcal/kWh is
40%
34.5 %
33.3%
45.2%
Answer: B) 34.5 %
Confirmed vs Book-3 §1.1 — 1 kWh ≡ 860 kCal, and heat rate is inversely proportional to efficiency, so η = 860/2490 = 34.5%.
Option (c) 33.3% would correspond to a heat rate of about 2580 kCal/kWh; always divide 860 by the given heat rate.
Source: 14th Exam
📖 §1.1 Cascade Efficiency
48. The efficiencies of a power plant and transmission system are 40% and 97% respectively. The distribution system loss is 23%. The cascade efficiency of generation, transmission and distribution is
8.92 %
29.87%
40 %
23%
Answer: B) 29.87%
Confirmed vs Book-3 §1.1 Cascade Efficiency — multiply the stage efficiencies: 0.40 × 0.97 × (1 − 0.23) = 0.40 × 0.97 × 0.77 = 0.2987 = 29.87%.
Option (a) 8.92% comes from multiplying by the 23% loss instead of by the 77% efficiency; cascade efficiency always uses efficiencies, never losses.
Source: 14th Exam
📖 §1.4 Selection and Location of Capacitors
49. The rating of power factor correction capacitors at induction motor terminals should be
90% of no load magnetizing kVAr of induction motor
100 % of no load magnetizing kVAr of induction motor
80% of no load magnetizing kVAr of induction motor
none of the above
Answer: A) 90% of no load magnetizing kVAr of induction motor
Confirmed vs Book-3 §1.4 — the book's rule is that 'the rating of the capacitor should not be greater than the no-load magnetizing kVAr of the motor'; standard practice therefore sizes terminal capacitors at about 90% of that value to keep a safety margin against self-excitation.
Option (b) 100% sits exactly on the limit and risks the 'damaging over voltage or transient torques' the book warns about.
Source: 14th Exam
📖 §1.4 Power Factor Improvement and Benefits
50. Select the correct statement: Power factor
is the ratio of active and reactive power
is the ratio of reactive and apparent power
is the ratio of active and apparent power
of a pure inductive and capacitive load is unity
Answer: C) is the ratio of active and apparent power
Confirmed vs Book-3 §1.4 — 'The ratio of kW to kVA is called the power factor', i.e. active power ÷ apparent power (PF = cosΦ), always ≤ 1.
Option (d) is wrong: a purely inductive or capacitive load draws only reactive power, so its power factor is zero, not unity.
Source: 14th Exam
📖 §1.4 Power Factor Improvement and Benefits
51. If the maximum demand is 3500 kVA at 0.88 p.f., the maximum demand will reduce by ______ kVA if PF is improved to 0.98:
3143
357
3897
maximum demand will not reduce
Answer: B) 357
Confirmed vs Book-3 §1.4 — capacitors do not change kW: kW = 3500 × 0.88 = 3080 kW. New kVA = 3080/0.98 = 3143 kVA, so the demand falls by 3500 − 3143 = 357 kVA.
Option (a) 3143 is the new maximum demand itself, not the reduction the question asks for.
Source: 14th Exam
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
52. If the apparent power drawn over a recording cycle of 30 minutes is 5000 kVA for 0.5 minutes, 3400 kVA for 20 minutes and 1800 for 9.5 minutes, the MD recorder will compute MD as
5000 kVA
3400 kVA
2920 kVA
1800 kVA
Answer: C) 2920 kVA
Confirmed vs Book-3 §1.2 — maximum demand is the time-integrated demand over the recording cycle: [(5000×0.5)+(3400×20)+(1800×9.5)]/30 = 87,600/30 = 2920 kVA.
Option (a) 5000 kVA is the instantaneous peak; the book stresses MD 'is not the instantaneous demand drawn, as is often misunderstood'.
Source: 14th Exam
📖 §1.4 Selection and Location of Capacitors
53. The rating of the PF correction capacitors at motor terminals for a 37 kW, 2-pole induction motor will be ______ in comparison to the same sized 6-pole induction motor
more
less
same
sometime less or more
Answer: B) less
Confirmed vs Book-3 §1.4 — the terminal capacitor is sized on the motor's no-load magnetising kVAr, which falls as the number of poles falls (higher-speed machines have a smaller magnetising current for the same kW).
So a 2-pole 37 kW motor needs LESS kVAr than a 6-pole 37 kW motor; option (a) 'more' reverses the relationship.
Source: 14th Exam
📖 §1.2 Electricity Billing — tariff structure
54. If the metered kWh is 95, kVAh is 100 and kVARh is 31, the power factor will be:
0.95
0.61
0.69
unity
Answer: A) 0.95
Confirmed vs Book-3 §1.2 — the trivector meter records kWh, kVArh and kVAh; average power factor = kWh/kVAh = 95/100 = 0.95.
Option (b)/(c) come from using kVArh in the ratio; the kVArh value (31) is only a cross-check (√(95²+31²) ≈ 100).
Source: 14th Exam
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
55. Energy efficient distribution transformer core is made up of ______.
silicon alloyed iron (grain oriented)
copper
amorphous core - metallic glass alloy
none of the above
Answer: C) amorphous core - metallic glass alloy
Confirmed vs Book-3 §1.5 — 'the latest technology is to use amorphous material — a metallic glass alloy for the core', giving about 70% lower core loss and 98.5% efficiency even at 35% load.
Option (a) grain-oriented silicon iron is the CONVENTIONAL core the book contrasts with; copper is a winding material, not a core material.
56. In a transformer on load, if the secondary voltage is one-fourth the primary voltage, then the secondary current will be
four times the primary current
sixteen times the primary current
one-fourth the primary current
two times the primary current
Answer: A) four times the primary current
Confirmed vs Book-3 §1.5 — 'Primary ampere-turns are equal to secondary ampere-turns', so V₁I₁ = V₂I₂. With V₂ = V₁/4, the secondary current must be four times the primary current.
Option (c) is the trap: current and voltage move in OPPOSITE directions in a transformer, so the low-voltage side carries the larger current.
Source: Jul 2022
📖 §1.4 Performance Assessment of Power Factor Capacitors
57. If V1 is the actual supply voltage and V2 is the rated voltage of a capacitor, the reactive kVAr produced would be in the ratio of
V2²/V1²
V1²/V2²
1 - V2²/V1²
1 + V2²/V1²
Answer: B) V1²/V2²
Confirmed vs Book-3 §1.4 Voltage effects — capacitor output varies as the square of the applied voltage, so kVAr_actual/kVAr_rated = V₁²/V₂² (V₁ = actual supply voltage, V₂ = rated voltage).
Option (a) inverts the ratio, which would wrongly predict MORE kVAr at reduced voltage.
Source: 14th Exam
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
58. In a 22 kV feeder, if the voltage is raised from 22 kV to 66 kV for the same loading conditions, the voltage drop in the same feeder system would be lowered to
1/2
1/3
1/9
unpredictable value
Answer: B) 1/3
Confirmed vs Book-3 §1.1 — 'Higher voltage transmission and distribution thus would help to minimize line voltage drop in the ratio of voltages'. Raising 22 kV to 66 kV is a 3:1 rise, so the voltage drop falls to 1/3.
Option (c) 1/9 is the reduction in line LOSS (ratio of the squares), not in voltage drop — the two must not be interchanged.
Source: Set-A
📖 §1.5 Transformers — losses & efficiency
59. Normally, the efficiency of a distribution transformer at full load varies anywhere between
96 to 99 %
80 to 85 %
60 to 70%
50%- 60%
Answer: A) 96 to 99 %
Confirmed vs Book-3 §1.5 Transformer Losses and Efficiency — 'The efficiency varies anywhere between 96 to 99 percent.'
Options (b)–(d) are far too low; a transformer is a static machine with no friction or windage losses, which is why it is the most efficient item of electrical plant.
Source: Set-A
📖 §1.8 Estimation of Technical Losses in Distribution System
60. What is the reduction in distribution loss if the current flowing through the distribution line is reduced by 10%?
10%
81%
19%
None of the above
Answer: C) 19%
Confirmed vs Book-3 §1.8 — line loss ∝ I², so at 0.9 I the loss becomes 0.81 of the original, a reduction of 19%.
Option (a) 10% assumes loss varies directly with current; option (b) 81% is the remaining loss, not the reduction.
Source: 18th Exam
📖 §1.4 Power Factor Improvement and Benefits
61. Power factor is the ratio of
kVAr/kW
(kW²+kVAr²)^0.5/kW
kW/(kW²+kVAr²)^0.5
kVAr/(kW²+kVAr²)^0.5
Answer: C) kW/(kW²+kVAr²)^0.5
Confirmed vs Book-3 §1.4 — from the power triangle (Figure 1.9), kVA = √(kW² + kVAr²) and PF = kW/kVA, i.e. kW/√(kW²+kVAr²).
Option (b) inverts the ratio (that is 1/PF); option (a) kVAr/kW is tanΦ, which is used for capacitor sizing, not for power factor.
Source: Set-A
📖 §1.4 Selection and Location of Capacitors
62. The electricity bill shows an average power factor of 0.72 with an average kW demand of 627. How much kVAr is required to improve the power factor to 0.95? (tanφ1=0.964, tanφ2=0.329)
398
144
95
627
Answer: A) 398
Confirmed vs Book-3 §1.4 worked example — kVAr = P(tanΦ₁ − tanΦ₂) = 627 × (0.964 − 0.329) = 398 kVAr; the Table 1.2 multiplier route (0.635 × 627) gives the same 398 kVAr, and the book advises installing 400 kVAr.
Option (d) 627 is the kW itself; the capacitor compensates only the difference in reactive components.
Source: Set-A
📖 §1.5 Transformers — losses & efficiency
63. Where transformer loading is known, the actual transformer loss at a given load can be computed as:
No Load Loss + (Actual kVA/rated kVA) x Load Loss
No Load Loss + (Actual kVA/rated kVA)² x Load Loss
No Load Loss + (Actual kVA/rated kVA) x Load Loss²
[No Load Loss + {(Actual kVA/rated kVA) x Load Loss}]²
Answer: B) No Load Loss + (Actual kVA/rated kVA)² x Load Loss
Confirmed vs Book-3 §1.5 — 'the actual transformer loss at given load = No load loss + (load kVA/Rated kVA)² × (Full load loss)', because copper loss varies with the square of the load current.
Option (a) scales the copper loss linearly, which would badly over-estimate the loss at part load.
Source: Set-A
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
64. In BEE Star labeled distribution transformers, which of the following losses are defined?
total loss at 50% and 100% loading
total loss at 75% loading
total loss at 75% and 100% loading
total loss at 100% loading
Answer: A) total loss at 50% and 100% loading
Confirmed vs Book-3 §1.5 Standards & Labeling — 'For the BEE labeling programme total losses at 50% and 100% load have been defined' (Table 1.4 lists max losses at 50% and 100% for 1–5 star).
Option (c) 75% and 100% is the common distractor; 75% is not a defined labelling point in IS 1180/BEE star rating.
Source: Set-A
📖 §1.2 Electricity Billing — tariff structure
65. A 2500 MW super thermal power station generated 15786 million units in the year 2011-12. Its Plant Load Factor (PLF) is:
60%
65%
72%
79%
Answer: C) 72%
Confirmed vs Book-3 §1.2 — PLF = actual generation ÷ maximum possible generation = 15,786 MU ÷ (2500 MW × 8760 h = 21,900 MU) = 72%.
Option (d) 79% would need about 17,300 MU; the key step is converting the installed capacity into annual units using 8760 hours.
Source: Set-A
📖 §1.3 Electrical Load Management and Maximum Demand Control
66. Which of the following statements is not true of a maximum demand controller
switches off non-essential loads in logical sequence
alarm is sounded when demand approaches a preset value
voltage level is closely regulated
plant equipment selected for load management can be programmed
Answer: C) voltage level is closely regulated
Confirmed vs Book-3 §1.3 — a maximum demand controller gives 'Accurate prediction of demand', 'Visual and audible alarm', 'Automatic load shedding in a predetermined sequence' and programmable load selection.
Voltage regulation is not among its functions — that is done by transformer tap changers (§1.5), so statement (c) is the untrue one.
67. Which of the following is wrong with reference to heat rate of a coal fired thermal power plant?
Heat rate indicates the overall energy efficiency of a power plant
When calculating plant heat rate, the energy input to the system is GCV of the fuel
Lower the heat rate the better
860 kCal per kWh is practically achievable
Answer: D) 860 kCal per kWh is practically achievable
Confirmed vs Book-3 §1.1 — 1 kWh ≡ 860 kCal is the THEORETICAL heat equivalent (100% efficiency); real Indian coal plants run at 28–35% efficiency, i.e. about 2450–3070 kCal/kWh, so 860 kCal/kWh is not practically achievable.
Options (a)–(c) are all correct book statements, including 'lower the heat rate, higher is the generation efficiency'.
Source: 17th Sep-2016
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
68. In electrical power system, transmission efficiency increases as:
both voltage and power factor increase
both voltage and power factor decrease
voltage increases but power factor decreases
voltage decreases but power factor increases
Answer: A) both voltage and power factor increase
Confirmed vs Book-3 §1.1 — for the same power, current falls as voltage rises AND as power factor rises (I = P/(√3·V·cosφ)); since loss = I²R, efficiency improves when both increase.
Option (c) is wrong because a falling power factor raises the current for the same kW and therefore raises the I²R loss.
Source: 17th Sep-2016
📖 §1.4 Power Factor Improvement and Benefits
69. A 10 MVA generator has power factor 0.86 lagging. The reactive power produced will be:
10 MVAr
8 MVAr
5 MVAr
1.34 MVAr
Answer: C) 5 MVAr
Confirmed vs Book-3 §1.4 power triangle — kVAr = kVA × sinφ = 10 × sin(cos⁻¹0.86) = 10 × 0.51 = 5.1 ≈ 5 MVAr.
Option (b) 8 MVAr is the ACTIVE power (10 × 0.86 = 8.6 MW); the reactive side of the triangle uses the sine, not the cosine.
Source: 17th Sep-2016
📖 §1.5 Transformers — losses & efficiency
70. The no-load loss and copper loss of a 500 kVA transformer is 900 watts and 6400 watts respectively. What is the total loss at 50% of transformer loading?
4100 watts
6850 watts
2500 watts
3650 watts
Answer: C) 2500 watts
Confirmed vs Book-3 §1.5 — total loss = no-load loss + (%load)² × full-load copper loss = 900 + (0.5)² × 6400 = 900 + 1600 = 2500 W.
Option (d) 3650 W would come from halving the copper loss (900 + 3200); the book's formula squares the load fraction.
Source: 17th Sep-2016
📖 §1.4 Selection and Location of Capacitors
71. The percentage reduction in distribution losses when tail end power factor is raised from 0.8 to 0.95 is:
29%
15.8%
71%
84%
Answer: A) 29%
Confirmed vs Book-3 §1.4 — reduction in distribution loss % = [1 − (PF₁/PF₂)²] × 100 = [1 − (0.8/0.95)²] × 100 = [1 − 0.709] × 100 = 29%.
Option (c) 71% is the RESIDUAL loss fraction (0.709), not the reduction — a very common slip.
72. In a Three Phase Transformer, the secondary side line current is 139.1A, and secondary voltage is 415V. The rating of the transformer would be:
50 kVA
150 kVA
100 kVA
63 kVA
Answer: C) 100 kVA
Confirmed vs Book-3 §1.5 — for a 3-phase transformer, kVA = √3 × V × I = 1.732 × 415 × 139.1 = 99,980 VA ≈ 100 kVA.
Option (d) 63 kVA would result from omitting the √3 factor and using single-phase VA (415 × 139.1 ≈ 57.7 kVA).
Source: 17th Sep-2016
📖 §1.4 Performance Assessment of Power Factor Capacitors
73. Shunt capacitors connection is normally adopted for:
Distribution Voltage improvement
Power factor improvement
Both a and b
None of these
Answer: B) Power factor improvement
Corrected (was c) — Book-3 §1.4 Performance Assessment of PF Capacitors: 'Shunt capacitor connections are adopted for almost all industry/end user applications, while series capacitors are adopted for voltage boosting in distribution networks.'
So the purpose for which shunt capacitors are normally adopted is power factor improvement; voltage boosting is the role of SERIES capacitors, which is why (a) and hence (c) 'both' are ruled out. (The load-end voltage does rise slightly as a by-product — book advantage (c) — but that is not what shunt banks are installed for.)
Source: 17th Sep-2016
📖 §1.4 Performance Assessment of Power Factor Capacitors
74. A company installed a new 100 kVAr, 415Volt capacitor but the power analyzer indicates that it is operating at 93 kVAr. The reason could be:
Operation is at low load
Higher Voltage at terminals
Lower voltage at terminals
None of the above
Answer: C) Lower voltage at terminals
Confirmed vs Book-3 §1.4 — capacitor output ∝ V², so 93/100 = (V/415)² gives V = 415 × √0.93 ≈ 400 V — the terminal voltage is BELOW the rated 415 V.
Option (b) higher voltage would give MORE than 100 kVAr (and shorten capacitor life), so it cannot explain a 93 kVAr reading.
Source: 17th Sep-2016
📖 §1.4 Power Factor Improvement and Benefits
75. The kVA reduction by improving the power factor of a plant operating at 400 kW load from 0.85 to 0.95 is:
40
49
72
None of the above
Answer: B) 49
Confirmed vs Book-3 §1.4 — kVA₁ = 400/0.85 = 470.6 and kVA₂ = 400/0.95 = 421.1, so the kVA reduction is 470.6 − 421.1 ≈ 49 kVA.
Option (c) 72 is roughly the capacitor kVAr needed (400 × (0.620 − 0.329) = 116) confusion; the question asks for the drop in apparent power, not the capacitor size.
Source: 17th Sep-2016
📖 §1.8 Estimation of Technical Losses in Distribution System
76. For a supply end Voltage of 10.6 kV and receiving end Voltage of 9.8 kV, the percentage regulation works out to:
0.80
8.16
7.55
None of these
Answer: B) 8.16
Confirmed vs Book-3 §1.8 Voltage Regulation — 'Percentage regulation = 100 (Es − Er)/Er' = 100 × (10.6 − 9.8)/9.8 = 8.16%.
Option (c) 7.55% divides by the SENDING end voltage (10.6); the book's formula uses the receiving-end voltage as the reference.
77. Which of the following can be attributed to commercial loss in electrical distribution system?
lengthy low voltage lines
low load side power factor
faulty consumer service meters
undersize conductors
Answer: C) faulty consumer service meters
Confirmed vs Book-3 §1.8 — commercial losses arise from 'Meter Reading', 'Metering' (tampering, stopped/defective meters) and collection efficiency, i.e. energy consumed but not correctly metered, billed or collected.
Options (a), (b) and (d) — long LV lines, poor PF and undersized conductors — are all listed under TECHNICAL losses (I²R), not commercial losses.
Source: 16th Exam (alt set)
📖 §1.5 Transformers — losses & efficiency
78. Which Loss in a Distribution Transformer is predominant if the transformer is loaded to 75% of its rated capacity?
core loss
copper loss
hysteresis loss
magnetic field loss
Answer: B) copper loss
Confirmed vs Book-3 §1.5 — core loss is constant while copper loss varies as (%load)². At 75% load a typical 500 kVA unit (Table 1.3) has copper loss 0.75² × 6450 = 3630 W against a fixed core loss of only 900 W, so copper loss dominates.
Option (a) core loss dominates only at light loading (below roughly 35–40% load); (c) hysteresis is a COMPONENT of core loss, not a separate answer.
Source: 17th Sep-2016
📖 §1.4 Automatic Power Factor Controllers
79. Which of the following cannot be controlled by automatic power factor controllers?
KW
voltage
Power factor
KiloVAr
Answer: A) KW
Confirmed vs Book-3 §1.4 — an APFC 'monitors the displacement power factor' and switches capacitor banks in or out via relay outputs, so it controls kVAr, the resulting power factor, and indirectly the voltage.
Active power (kW) is set by the process load; capacitors supply no kW, so the controller cannot control it.
Source: 17th Sep-2016
📖 §1.5 Transformers — losses & efficiency
80. Which loss in a distribution transformer is dominating; if the transformer is loaded at 68% of its rated capacity?
core loss
copper loss
hysteresis loss
magnetic field loss
Answer: B) copper loss
Confirmed vs Book-3 §1.5 — core loss is constant but copper loss varies as (%load)²; at 68% loading a typical 500 kVA unit (Table 1.3) has copper loss 0.68² × 6450 = 2982 W against a fixed core loss of only 900 W.
Option (a) core loss dominates only at light loading (roughly below 35–40%); (c) hysteresis loss is a component OF the core loss, not a separate answer.
Source: 16th Exam (alt set)
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
81. Trivector meter measures three vectors representing:
active, reactive and maximum demand
active, power factor and apparent power
active, harmonics and maximum demand
active, reactive and apparent power
Answer: D) active, reactive and apparent power
Confirmed vs Book-3 §1.2 — the trivector meter's outputs are 'Active energy in kWh…Reactive energy in kVArh…and Apparent energy in kVAh' (plus the recorded maximum demand). Those three energy vectors are what give it its name.
Option (a) lists maximum demand, which is a derived DEMAND reading, not one of the three measured vectors.
Source: 16th Exam (alt set)
📖 §1.2 Electricity Billing — tariff structure
82. Time of the Day metering (TOD) is a way to:
reduce the peak demand of the distribution company
increase the revenue of the distribution company
increase the peak demand
increase the maximum demand in an industry
Answer: A) reduce the peak demand of the distribution company
Confirmed vs Book-3 §1.2/§1.3 — 'Time of Day (TOD) rates like peak and non-peak hours…' are used by utilities to 'influence end user in better load management', so that consumers shift load away from the utility's peak.
Option (c)/(d) are the opposite of the intent — the higher peak-hour tariff is designed to discourage, not encourage, drawl at peak.
Source: 16th Exam
📖 §1.4 Selection and Location of Capacitors
83. The percentage reduction in distribution losses when tail end power factor raised from 0.85 to 0.95 is:
10.1%
19.9%
71%
84%
Answer: B) 19.9%
Confirmed vs Book-3 §1.4 — reduction in distribution loss % = [1 − (PF₁/PF₂)²] × 100 = [1 − (0.85/0.95)²] × 100 = [1 − 0.8006] × 100 = 19.9%.
Option (c) 71% is a residual-loss figure from the 0.8→0.95 version of this question; always square the PF ratio and subtract from 1.
Source: 16th Exam
📖 §1.2 Electricity Billing — tariff structure
84. The components of two part tariff structure for HT & EHT category consumers are:
one part for capacity (or demand) drawn and second part for actual energy drawn
one part for actual Power Factor and second part for actual energy drawn
one part for capacity (or demand) drawn and second part for actual reactive energy drawn
one part for actual apparent energy drawn and second part for actual reactive energy drawn
Answer: A) one part for capacity (or demand) drawn and second part for actual energy drawn
Confirmed vs Book-3 §1.2 — 'the electricity billing…in High Tension (HT) category, is often done on two-part tariff structure, i.e. one part for capacity (or demand) drawn and the second part for actual energy drawn during the billing cycle.'
Option (c) is wrong because reactive energy is charged separately as a PF penalty/bonus, not as the second part of the two-part tariff.
Source: 16th Exam
📖 §1.5 Transformers — losses & efficiency
85. The actual measured load of 1000 kVA transformer is 400 kVA. Find out the total transformer loss corresponding to this load if no load loss is 1500 Watts and full load Copper Loss is 12,000 Watts.
1920 watts
1500 watts
3420 watt
13500 watts
Answer: C) 3420 watt
Confirmed vs Book-3 §1.5 — total loss = no-load loss + (load kVA/rated kVA)² × full-load copper loss = 1500 + (400/1000)² × 12,000 = 1500 + 1920 = 3420 W.
Option (a) 1920 W is only the copper component; the constant no-load (core) loss is present whenever the transformer is energised and must be added.
Source: 16th Exam (alt set)
📖 §1.11 Analysis of Electrical Power Systems (Table 1.11)
86. The percentage imbalance when line-line voltages are 415 V, 418 V and 408 V is:
1.047%
0.32%
1.44%
none of the above
Answer: A) 1.047%
Confirmed vs Book-3 §1.11 Table 1.11 — voltage imbalance is expressed as the deviation of a line voltage from the average line voltage: average = (415+418+408)/3 = 413.7 V, and the +4.3 V deviation of the 418 V line gives 4.3/413.7 = 1.047%.
Option (b) 0.32% ignores the average-voltage reference. (Strict NEMA practice takes the LARGEST deviation, i.e. 408 V → 5.7/413.7 ≈ 1.37%; either way the imbalance is ~1%, far below the 5% that the book says raises motor losses by 40%.)
Source: 16th Exam (alt set)
📖 §1.4 Power Factor Improvement and Benefits
87. Power factor Improvement will result in:
reduction in active power
reduction in active current
reduction in reactive power
all the above
Answer: C) reduction in reactive power
Confirmed vs Book-3 §1.4 — capacitors 'act as reactive power generators', so PF improvement cuts the REACTIVE power (and hence the total current and kVA) drawn from the source.
Option (a) is wrong because kW is fixed by the work done, and (b) is wrong because the ACTIVE component of current (I·cosφ) is unchanged — it is the reactive component that disappears, so (d) cannot be right.
Source: 16th Exam (alt set)
📖 §1.4 Power Factor (see also Book-3 Ch-9 DG Sets)
88. Lower power factor of a DG set demands __________.
lower excitation currents
no change in excitation currents
higher excitation currents
none of the above
Answer: C) higher excitation currents
Confirmed vs Book-3 §1.4/Ch-9 — an alternator supplies the load's reactive kVAr from its field; the lower the power factor, the larger the kVAr for the same kW, so the field (excitation) current must be increased.
Option (a) is the reverse: only at high/unity power factor can the machine run with reduced excitation, which is why DG sets are de-rated at low PF.
Source: Mar 2023
📖 §1.3 Load management (see also Book-3 Ch-5 Fans and Blowers)
89. Flow control by damper operation in fan system will ___________
increase energy consumption
reduce energy consumption
increase system resistance
none of the above
Answer: B) reduce energy consumption
Confirmed vs Book-3 Ch-5 — closing a damper moves the operating point up the fan curve to a lower flow, and the fan's absorbed power falls, so energy consumption does reduce compared with full flow.
Option (c) increased system resistance is the MECHANISM, not the energy outcome the question asks about; note that damper control saves far less than speed (VFD) control, which follows the cube law.
Source: Mar 2021 (Set B)
📖 §1.10 Harmonics
90. Find the Total Harmonic Distortion (THD) for current for the following readings: fundamental (50 Hz) = 250 A, third harmonic = 50 A, fifth harmonic = 35 A.
58 %
48 %
24%
34 %
Answer: C) 24%
Confirmed vs Book-3 §1.10 — THD_current = √(ΣIₙ²)/I₁ × 100 = √(50² + 35²)/250 × 100 = 61.03/250 × 100 = 24.4% ≈ 24%.
Option (d) 34% comes from adding the harmonics arithmetically ((50+35)/250 = 34%); harmonic components must be combined as a root-sum-square.
Source: 16th Exam
📖 §1.4 Performance Assessment of Power Factor Capacitors
91. A company installed a 130 kVAr, 600 Volt capacitor but the power meter indicates that it is only operating at 119 kVAr. The reason out of the following could be:
operating at low load
high voltage
low voltage
low current
Answer: C) low voltage
Confirmed vs Book-3 §1.4 Voltage effects — kVAr ∝ V², so 119/130 = (V/600)² gives V = 600 × √0.9154 ≈ 574 V, i.e. the terminal voltage is BELOW rating.
Option (b) a high voltage would give MORE than the rated 130 kVAr (and shorten capacitor life); capacitor output does not depend on the plant load, ruling out (a).
Source: 16th Exam (alt set)
📖 §1.3 Electrical Load Management and Maximum Demand Control
92. An Industrial Consumer has a load pattern of 2000 kW, 0.8 lag for 12 hrs and 1000 kW unity power factor for 12 hrs. The load factor is:
0.5
0.75
0.6
0.2
Answer: B) 0.75
Confirmed vs Book-3 §1.3 Load Curve — load factor = average load ÷ maximum demand = [(2000×12)+(1000×12)]/24 ÷ 2000 = 1500/2000 = 0.75.
Option (a) 0.5 is the ratio of the two load levels (1000/2000); the load factor must use the time-weighted AVERAGE load, and power factor plays no part in it.
Source: 18th Exam
📖 §1.10 Harmonics
93. Which of the following is not likely to create harmonics in an electrical system?
soft starters
variable frequency drives
uninterrupted power supply source (UPS)
electric heater
Answer: D) electric heater
Confirmed vs Book-3 §1.10 — the book classes heaters as LINEAR loads ('Incandescent lamps, heaters and, to a great extent, motors are linear systems') because their impedance is constant, so they draw a sinusoidal current and create no harmonics.
Options (a)–(c) are all power-electronic (non-linear) devices — soft starters, VFDs and UPS — which the book lists as harmonic sources.
Source: Jul 2022
📖 §1.10 Harmonics
94. The source of maximum harmonics among the following, in a plant power system is
100 CFL lamps of 11 W to 25 W
500 kW, 3 Phase, 415 V, 50 Hz resistance furnace
5 kVA UPS for computer system
variable frequency drive for 225 kW motive load
Answer: D) variable frequency drive for 225 kW motive load
Confirmed vs Book-3 §1.10 — VFDs head the book's list of non-linear loads, and the harmonic current injected scales with the load, so a 225 kW drive is by far the biggest source here.
Option (b) is the trap: a 500 kW resistance furnace is large but LINEAR (constant impedance), so it injects no harmonics; the 5 kVA UPS is non-linear but negligible in size.
Source: 18th Exam
📖 §1.4 Power Factor Improvement and Benefits
95. The combined power factor of a set of incandescent bulbs totaling 20 kW and two motors, each of 20 kW with power factor of 0.80 is
0.89
0.90
0.80
none of the above
Answer: A) 0.89
Confirmed vs Book-3 §1.4 power triangle — incandescent bulbs are resistive (20 kW, 0 kVAr); each 20 kW motor at 0.8 PF draws 20 × tan(cos⁻¹0.8) = 15 kVAr, so totals are 60 kW and 30 kVAr. kVA = √(60² + 30²) = 67.1, giving PF = 60/67.1 = 0.89.
Option (c) 0.80 wrongly assumes the motors' PF applies to the whole plant — the unity-PF lighting load pulls the combined PF up. [Option (a) repaired from '0.88' to the computed 0.89.]
Source: 18th Exam
📖 §1.2 Electricity Billing — tariff structure
96. The daily average power factor is 0.95 and the energy consumption is 2200 kWh. The average kVARh drawn is ______
1900
2315
722.5
None of the above
Answer: C) 722.5
Confirmed vs Book-3 §1.4 — kVArh = kWh × tan(cos⁻¹PF) = 2200 × tan(cos⁻¹0.95) = 2200 × 0.3287 = 723 ≈ 722.5 kVArh.
Option (a) 1900 is 2200 × 0.95/1.1-type confusion; the conversion from kWh to kVArh always uses the TANGENT of the phase angle.
Source: 18th Exam
📖 §1.8 Estimation of Technical Losses in Distribution System
97. HVDS (High Voltage Distribution System) is preferred to
reduce technical loss in distribution system
improve voltage regulation
comply with regulatory mandate
reduce energy bill for the end consumer
Answer: A) reduce technical loss in distribution system
Confirmed vs Book-3 §1.8 Measures to reduce technical losses — HVDS replaces long 415 V LT lines with 11 kV lines feeding small (16/25 kVA) pole-mounted transformers at the load centres; at higher voltage the current, and hence the I²R loss, is far lower.
Option (b) voltage regulation does improve, but it is a by-product; the book's stated purpose of HVDS is reduction of technical distribution losses.
Source: 18th Exam
📖 §1.4 Power Factor Improvement and Benefits
98. Improving power factor at motor terminals in a factory will
increase active power
release distribution transformer capacity
reduce contract demand
increase motor efficiency
Answer: B) release distribution transformer capacity
Confirmed vs Book-3 §1.4 — advantage (d): 'kVA loading on the source generators as also on the transformers and lines up to the capacitors reduces giving capacity relief. A high power factor can help in utilizing the full capacity of the electrical system.'
Option (c) is wrong — contract demand is a contractual figure with the utility, which must be renegotiated; PF correction only lowers the recorded kVA.
Source: 18th Exam
📖 §1.4 Power Factor Improvement and Benefits
99. In a rolling mill, the loading on the transformer was 1200 kVA at power factor 0.86. The plant improved the power factor to 0.98 by adding capacitors. What is the reduction in kVA?
144
147
171
163.3
Answer: B) 147
Corrected (was d) — Book-3 §1.4: capacitors do not change kW, so kW = 1200 × 0.86 = 1032 kW. At 0.98 PF the loading becomes 1032/0.98 = 1053 kVA, so the reduction is 1200 − 1053 = 147 kVA.
The printed option 163.3 does not follow from the data; option (a) 144 comes from rounding kW to 1030. Always work kW first, then divide by the new PF.
Source: 19th Exam
📖 §1.10 Harmonics
100. In a solar PV system the conversion from DC to AC is carried out by
Converter
Charger
Battery
Inverter
Answer: D) Inverter
Confirmed vs Book-3 §1.10 — the inverter is the stage that converts DC into AC (solar array/DC link → AC grid), and the book lists inverter-fed drives and UPS among non-linear, harmonic-producing loads for exactly this reason.
Option (a) 'converter' (rectifier) performs the reverse AC→DC conversion; a battery only stores DC.
Source: 19th Exam
📖 §1.10 Harmonics
101. The 5th and 7th harmonic in a 50 Hz power supply system will have:
Voltage and current distortions with 55 Hz & 57 Hz
Voltage and current distortions with 500 Hz & 700 Hz
Voltage and current distortions with 250 Hz & 350 Hz
No voltage and current distortion at all
Answer: C) Voltage and current distortions with 250 Hz & 350 Hz
Confirmed vs Book-3 §1.10 — 'the fundamental frequency is 50 Hz, and then the 5th harmonic is five times that frequency, or 250 Hz'; likewise the 7th is 7 × 50 = 350 Hz.
Option (a) 55 Hz/57 Hz adds the harmonic number instead of multiplying by it — harmonics are integer MULTIPLES of the fundamental.
Source: 19th Exam
📖 §1.4 Performance Assessment of Power Factor Capacitors
102. A 5 kVAr, 415 V rated power factor capacitor was found to be having 5.5 kVAr operating capacity. The operating supply voltage at the same supply frequency would be approximately.
400 V
415 V
435 V
None of the above
Answer: C) 435 V
Confirmed vs Book-3 §1.4 — kVAr ∝ V², so V = 415 × √(5.5/5) = 415 × 1.0488 ≈ 435 V.
Option (a) 400 V would REDUCE the output below 5 kVAr; a higher-than-rated output always means a higher-than-rated terminal voltage (which shortens capacitor life).
103. Aggregate Technical & Commercial loss in distribution system covers
I2R losses of all transformers
Transmission & distribution loss
Only transmission losses
Energy and monetary loss
Answer: D) Energy and monetary loss
Confirmed vs Book-3 §1.8 — AT&C loss = {1 − (Billing Efficiency × Collection Efficiency)} × 100; it combines the energy not billed (technical + theft) with the money billed but not collected, i.e. both an energy and a monetary loss.
Option (b) is only the technical part; AT&C deliberately goes beyond T&D loss by adding the commercial/collection dimension.
Source: 19th Exam
📖 §1.1 Cascade Efficiency
104. A medium voltage end consumer receives 83 million units with a transmission and distribution cascade efficiency of 82%. The million units generated will be ____________.
101.2
68.1
83
None of the above
Answer: A) 101.2
Confirmed vs Book-3 §1.1 Cascade Efficiency — units generated = units received ÷ cascade efficiency = 83/0.82 = 101.2 MU.
Option (b) 68.1 MU multiplies by 0.82 instead of dividing; generation must always exceed the units delivered to the consumer.
Source: Sep 2019
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
105. In an electrical power system, transmission efficiency increases as _______________.
both voltage and power factor increases
both voltage and power factor decreases
voltage increases but power factor decreases
Voltage decreases but power factor increases
Answer: A) both voltage and power factor increases
Confirmed vs Book-3 §1.1 — I = P/(√3·V·cosφ), and loss = I²R; raising either the voltage or the power factor lowers the current and hence the I²R loss, so transmission efficiency improves when both increase.
Option (d) is wrong because a lower voltage raises the current, wiping out any gain from a better power factor.
106. In electrical distribution system, commercial loss covers discrepancies due to ________________.
Meter Reading
Metering
Collection Efficiency
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §1.8 Commercial Losses — the book attributes commercial losses to discrepancies in 'Meter Reading', 'Metering' and 'Collection efficiency'.
Each option names one of the three listed heads, so only 'all of the above' covers the book's definition.
Source: Sep 2019
📖 §1.10 Harmonics
107. Harmonics generation will be more in _________________.
Inverter drives
LED Lamps
Transformers
Resistance heaters
Answer: A) Inverter drives
Confirmed vs Book-3 §1.10 Causes of Harmonics — 'Common non-linear loads include variable speed drives (AC as well as DC), induction furnaces, LED based and CFL lamps, certain types of UPS & computer power supplies.' Inverter drives are the largest such source in industry.
Transformers and resistance heaters are linear (constant impedance) and generate negligible harmonics; LED lamps do generate harmonics but at a far smaller magnitude than a drive.
Source: Sep 2019
📖 §1.5 Transformers — losses & efficiency
108. Among the following, _____________ has highest design efficiency.
High tension motors
Power transformers
Alternators
Electric melting furnaces
Answer: B) Power transformers
Confirmed vs Book-3 §1.5 — 'The efficiency [of transformers] varies anywhere between 96 to 99 percent' — the highest of any electrical machine, because a transformer is static (no friction or windage loss).
Options (a)/(c): rotating machines lose energy to friction, windage and slip, so their design efficiency is lower; furnaces are far lower still.
Source: Sep 2019
📖 §1.8 Estimation of Technical Losses in Distribution System
109. Technical loss in a distribution system can be reduced by _________________.
Maintaining low HT/LT ratio
Accurate meter reading
High voltage supply to consumers
Improving Collection Efficiency
Answer: C) High voltage supply to consumers
Confirmed vs Book-3 §1.8 Measures to reduce technical losses — the book recommends the High Voltage Distribution System (HVDS), replacing long LT lines with 11 kV lines and pole-mounted transformers at the load centres, because loss = I²R and higher voltage means lower current.
Options (b)/(d) — accurate meter reading and collection efficiency — reduce COMMERCIAL losses, not technical losses.
Source: Sep 2019
📖 §1.5 Transformers — losses & efficiency
110. A 500-kVA transformer is designed for No load loss of 750 watts and load loss of 5700 Watts. The calculated total transformer loss is 1662 watts. What will be the percentage loading of the transformer?
54.8 %
29 %
40 %
25.7 %
Answer: C) 40 %
Confirmed vs Book-3 §1.5 — total loss = no-load + (%load)² × load loss ⇒ 1662 = 750 + x² × 5700 ⇒ x² = 912/5700 = 0.16 ⇒ x = 0.40, i.e. 40% loading.
Option (b) 29% comes from taking 912/5700 = 16% without the square root; the square root step is essential because copper loss varies as the square of the load.
Source: Sep 2019
📖 §1.10 Harmonics
111. Find the Total Harmonic Distortion (THD) for current for the following current readings. Current at 50 Hz fundamental = 250 A, Third harmonic current = 50 A, fifth harmonic current = 35 A.
58 %
48 %
24%
34 %
Answer: C) 24%
Confirmed vs Book-3 §1.10 — THD_current = √(ΣIₙ²)/I₁ × 100 = √(50² + 35²)/250 × 100 = 61.03/250 × 100 = 24.4% ≈ 24%.
Option (d) 34% comes from adding the harmonic currents arithmetically ((50+35)/250); harmonic components must be combined as a root-sum-square.
Source: 16th Exam (alt set)
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
112. In the city electrical distribution scheme, a proposal is being prepared to upgrade 33 kV network to 66 kV. The distribution loss, corresponding to the same quantum of load in the proposed upgraded system will be
less by 25%
less by 33%
less by 75%
none of the above
Answer: C) less by 75%
Confirmed vs Book-3 §1.1 — line loss varies as the inverse square of the voltage: (33/66)² = 1/4, so the loss falls to 25% of its old value, i.e. it is LESS by 75%.
Option (a) 'less by 25%' quotes the remaining loss instead of the reduction — read the wording carefully.
Source: 9th Dec-2009
📖 §1.4 Power Factor Improvement and Benefits
113. Select the correct statement:
the advantage of PF improvement by capacitor addition in an electric network is that active power component of the network is reduced
the power factor indicated in the monthly electricity bill is the lowest power factor recorded at any time during the billing month
PF capacitors operating at lower voltage than their rated values have higher operating kVArs than their rated values
the power factor of an induction motor decreases with decrease in percentage motor loading
Answer: D) the power factor of an induction motor decreases with decrease in percentage motor loading
Confirmed vs Book-3 §1.4 — an induction motor's magnetising (reactive) current is nearly constant while its kW falls with load, so kW/kVA — the power factor — falls as loading falls; this is why lightly loaded motors are a major cause of poor plant PF.
Option (a) is wrong (capacitors do not reduce kW), (b) is wrong (the bill shows the AVERAGE PF, kWh/kVAh) and (c) is wrong (below rated voltage a capacitor gives LESS kVAr, since kVAr ∝ V²).
Source: 9th Dec-2009
📖 §1.1 Cascade Efficiency
114. If the efficiencies of a power plant, transmission and distribution systems are 30%, 95% & 85% respectively, the cascade efficiency of power generation, and transmission system is given by
24.23%
28.5%
80.75%
95%
Answer: B) 28.5%
Confirmed vs Book-3 §1.1 Cascade Efficiency — the question asks only for generation AND transmission: 0.30 × 0.95 = 0.285 = 28.5%.
Option (a) 24.23% is the full cascade INCLUDING the 85% distribution stage (0.30 × 0.95 × 0.85) — that stage was not asked for.
Source: 9th Dec-2009
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
115. If the apparent power drawn over a recording cycle of 30 minutes is 3000 kVA for 10 minutes, 2400 kVA for 15 minutes and 2900 kVA for 5 minutes, the MD recorder will compute MD as
3000 kVA
2400 kVA
2683 kVA
none of the above
Answer: C) 2683 kVA
Confirmed vs Book-3 §1.2 — MD = time-integrated demand over the cycle = [(3000×10) + (2400×15) + (2900×5)]/30 = 80,500/30 = 2683 kVA.
Option (a) 3000 kVA is the highest instantaneous block; the book stresses the meter registers the AVERAGE over the whole 30-minute recording cycle.
Source: 9th Dec-2009
📖 §1.4 Automatic Power Factor Controllers
116. Automatic power factor controller using kVAr control, requires sensing of
current
voltage
capacitance
both a and b
Answer: D) both a and b
Confirmed vs Book-3 §1.4 — an APFC computes the reactive power/displacement power factor, which requires both the current and the voltage (and the angle between them) to be sensed before it can switch capacitor steps.
Option (a) alone gives no angle information, so no kVAr or PF can be derived from current sensing by itself.
Source: 9th Dec-2009
📖 §1.3 Electrical Load Management and Maximum Demand Control
117. Maximum demand controller is used to
switch off non-essential loads in a logical sequence
switch off essential loads in a logical sequence
controls the reactive power of the plant
all the above.
Answer: A) switch off non-essential loads in a logical sequence
Confirmed vs Book-3 §1.3 Shedding of Non-Essential Loads — the MD controller provides 'Automatic load shedding in a predetermined sequence' and 'Automatic restoration of load' when the demand approaches a preset limit.
Option (b) is wrong: only NON-essential loads are shed; option (c) describes an APFC, which handles reactive power.
Source: 9th Dec-2009
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
118. For the same quantity of power handled by a distribution line, the lower the voltage
the higher the current drawn and higher the distribution loss
the lower the current drawn and lower the distribution loss
the lower the voltage drop and lower the distribution loss
the higher the voltage drop and lower the distribution loss
Answer: A) the higher the current drawn and higher the distribution loss
Confirmed vs Book-3 §1.1 — 'For the same quantity of power handled, lower the voltage, higher the current drawn and higher the voltage drop', and since P.Loss = I²R the distribution loss rises too.
Option (b) reverses this; the whole rationale for HV transmission is that a lower current means a lower I²R loss.
119. The ratio of overall maximum demand of the plant to the sum of individual maximum demand of various equipment is ______.
load factor
diversity factor
demand factor
maximum demand
Answer: B) diversity factor
Confirmed vs Book-3 §1.5 Rating of Transformer — 'Diversity factor is defined as the ratio of overall maximum demand of the plant to the sum of individual maximum demand of various equipment…Diversity factor will always be less than one.'
Option (a) load factor is average load ÷ maximum demand, and (c) demand factor is maximum demand ÷ connected load — different ratios.
Source: 10th Jul-2010
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
120. Maximum demand charges for a billing cycle is calculated by the Utility based on:
the instantaneous demand drawn
the time integrated demand over the predefined recording cycle
the kVARh drawn per demand cycle
the kWh drawn during peak load period
Answer: B) the time integrated demand over the predefined recording cycle
Confirmed vs Book-3 §1.2 — 'while maximum demand is recorded, it is not the instantaneous demand drawn, as is often misunderstood, but the time integrated demand over the predefined recording cycle' (typically 30 minutes).
Option (a) is exactly the misconception the book warns against; a momentary current spike does not by itself set the billed demand.
Source: 10th Jul-2010
📖 §1.10 Harmonics
121. The total amount of harmonics present in the system is expressed using ___.
Total Harmonic Factor
Total Harmonic Ratio
Total Harmonic Distortion
Crest Factor
Answer: C) Total Harmonic Distortion
Confirmed vs Book-3 §1.10 — 'Total Harmonic Distortion (THD) expresses the amount of harmonics', computed as the root-sum-square of the harmonic components as a percentage of the fundamental.
Option (d) Crest Factor is the peak-to-RMS ratio of a waveform — related to distortion but not a measure of total harmonic content.
Source: 10th Jul-2010
📖 §1.1 Industrial End User — 'ONE Unit saved = TWO Units Generated'
122. Efficiency of power generation in a power plant is 30%, T&D losses are 23%, distribution loss within the factory is 6%, and equipment end use efficiency is 65%. The overall cascade system efficiency from fuel input to end-use will be
2.69%
14.11%
4.21
none of the above
Answer: B) 14.11%
Confirmed vs Book-3 §1.1 — multiply the stage efficiencies: 0.30 × (1 − 0.23) × (1 − 0.06) × 0.65 = 0.30 × 0.77 × 0.94 × 0.65 = 0.1411 = 14.11%.
Option (a) 2.69% comes from multiplying by the loss fractions (0.23, 0.06) instead of by the efficiencies (0.77, 0.94).
Source: 10th Jul-2010
📖 §1.4 Power Factor Improvement and Benefits
123. Select the correct Statement: The advantage of PF improvement by capacitor addition in an electric network is
apparent power component of the network is reduced
active power component of the network is reduced
I2R power losses are reduced in the system from the point of installation to the load end
voltage level at the load end is not improved
Answer: A) apparent power component of the network is reduced
Confirmed vs Book-3 §1.4 — capacitors reduce the reactive component and hence the total current, so the APPARENT power (kVA) drawn from the source falls; the book's advantage (d) notes the resulting kVA/capacity relief.
Option (b) is false because kW is set by the work done; (c) is false because losses fall only UPSTREAM of the capacitor, and (d) is false because the load-end voltage does improve.
Source: 10th Jul-2010
📖 §1.4 Performance Assessment of Power Factor Capacitors
124. A power factor capacitor designed for 10 kVAr at 415 V was found to be operating at 405 V. The effective capacity of the capacitor would be
9.75 kVAr
10 kVAr
9.52 kVAr
none of the above
Answer: C) 9.52 kVAr
Confirmed vs Book-3 §1.4 Voltage effects — output kVAr ∝ V²: 10 × (405/415)² = 10 × 0.9524 = 9.52 kVAr.
Option (a) 9.75 kVAr assumes a linear voltage relation (405/415 × 10); the squared law is why even a small under-voltage noticeably de-rates a bank.
Source: 10th Jul-2010
📖 §1.5 Transformers — losses & efficiency
125. Among the following electrical equipment _______ has the highest design efficiency
synchronous motor
DC shunt motor
induction motor
transformer
Answer: D) transformer
Confirmed vs Book-3 §1.5 — the transformer is a static machine with no friction or windage loss, and the book gives its efficiency as 96–99%, higher than any rotating machine.
Options (a)–(c) are all rotating machines whose efficiency is reduced by friction, windage and (for induction motors) slip losses.
Source: 10th Jul-2010
📖 §1.10 Harmonics
126. The 5th and 7th harmonic in a 50 Hz power environment will have:
voltage and current distortions with 55 Hz & 57 Hz
voltage and current distortions with 500 Hz & 700 Hz
voltage and current distortions with 250 Hz & 350 Hz
no voltage and current distortion at all
Answer: C) voltage and current distortions with 250 Hz & 350 Hz
Confirmed vs Book-3 §1.10 — the harmonic order multiplies the fundamental: 5 × 50 = 250 Hz and 7 × 50 = 350 Hz.
Option (b) 500/700 Hz would be the 10th and 14th harmonics; option (a) adds instead of multiplying.
Source: 10th Jul-2010
📖 §1.10 Harmonics
127. The source of maximum harmonics among the following in a plant power system could be:
100 CFL lamps of 11 W to 25 W
500 kW, 3 Phase, 415 V, 50 Hz resistance furnace
5 kVA UPS for computer system
Variable Frequency Drive for 225 kW motive load
Answer: D) Variable Frequency Drive for 225 kW motive load
Confirmed vs Book-3 §1.10 — variable frequency drives head the book's list of non-linear loads, and the harmonic current injected scales with the load size, so a 225 kW VFD dwarfs the other options.
Option (b) a resistance furnace is a LINEAR load (constant impedance) and generates no harmonics despite its 500 kW rating; the 5 kVA UPS is non-linear but tiny.
128. Which of the following are emerging technological solutions in electric power distribution system?
Intelligent meters for improved system operation and customer relationship management
Intelligent or smart meters to replace older systems to allow customers a clear picture of their energy use profile
SCADA system to control and data acquisition for complete T&D system
all the above
Answer: D) all the above
Confirmed vs Book-3 §1.8 Measures to Reduce Commercial Losses — the book calls for 'Installation of Electronic meters with (TOD, tamper proof, data and remote reading facility)', accurate metering plans and energy audit as a tool, all of which smart meters and SCADA deliver.
Each option describes one of these emerging solutions, so only 'all the above' is complete.
Source: 10th Jul-2010
📖 §1.4 Automatic Power Factor Controllers
129. Capacitors with automatic power factor controller when installed in a plant:
reduces apparent power drawn from grid
reduces the voltage of the plant
reduces the reactive power drawn from grid
increases the load current of the plant
Answer: C) reduces the reactive power drawn from grid
Confirmed vs Book-3 §1.4 — capacitors 'act as reactive power generators', supplying the magnetising kVAr locally so it need not be drawn from the grid; the APFC keeps that compensation matched to a fluctuating load.
Option (d) is the opposite of what happens — the line current falls; note that the apparent power (a) also falls, but the direct, primary action of a capacitor is on the REACTIVE power.
Source: Jul 2022
📖 §1.4 Selection and Location of Capacitors
130. If power factor is improved from PF1 to PF2 then the reduction in distribution losses in an electric network is proportional to:
ratio of PF1 to PF2
square root of (PF1/PF2)
square of (PF1/PF2)
none of the above
Answer: C) square of (PF1/PF2)
Confirmed vs Book-3 §1.4 — the book gives the reduction in distribution loss % as [1 − (PF₁/PF₂)²] × 100, so the loss after correction is the square of the power-factor ratio times the loss before: loss₂/loss₁ = (PF₁/PF₂)².
The governing term is therefore the SQUARE of (PF₁/PF₂); options (a) and (b) use the plain ratio and its square root, which under-state the benefit. Memorise the full form [1 − (PF₁/PF₂)²] × 100 for the numericals.
Source: 11th Feb-2011
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
131. In the BEE labeling programme for distribution transformers, the total transformer losses at
50% and 100% loading have been defined.
only 50% loading have been defined.
only 100% loading have been defined.
25%, 50% and 100% loading have been defined.
Answer: A) 50% and 100% loading have been defined.
Confirmed vs Book-3 §1.5 Standards & Labeling — 'For the BEE labeling programme total losses at 50% and 100% load have been defined', with 1 star the highest-loss and 5 star the lowest-loss segment (Table 1.4).
Option (d) adds 25% loading, which belongs to no BEE/IS 1180 labelling point.
Source: 11th Feb-2011
📖 §1.4 Performance Assessment of Power Factor Capacitors
132. A 5 kVAr, 415 V rated power factor capacitor was found to be having 5.5 kVAr operating capacity. The operating supply voltage at the same supply frequency would approximately be
400 V
415 V
435 V
none of the above
Answer: C) 435 V
Confirmed vs Book-3 §1.4 — kVAr ∝ V², so V = 415 × √(5.5/5) = 415 × 1.049 ≈ 435 V.
Option (b) 415 V would give exactly the rated 5 kVAr; an output ABOVE rating can only come from an over-voltage, which the book warns shortens capacitor life.
Source: 11th Feb-2011
📖 §1.4 Performance Assessment of Power Factor Capacitors
133. Busbar Voltages at the main electrical panel were balanced but at the following Motor Control Circuit (MCC), fitted with PF Correction capacitors, the voltages were unbalanced by about 3%. The possible reason for this could be
motors connected to MCC were operating at partial loads
motors connected to MCC were overloaded
excessive kVAr of Capacitors than required at MCC
blown fuse in one phase of the 3 phase capacitor bank connected to the MCC.
Answer: D) blown fuse in one phase of the 3 phase capacitor bank connected to the MCC.
Confirmed vs Book-3 §1.4 — a three-phase bank compensates each phase separately; a blown fuse in one phase removes that phase's kVAr only, so the three phase currents (and hence the volt-drops down the feeder) become unequal, producing the 3% imbalance at the MCC.
Options (a)/(b) alter all three phases symmetrically and (c) over-compensation would raise the voltage in all three phases equally — none of these creates an imbalance.
Source: 11th Feb-2011
📖 §1.5 Transformers — losses & efficiency
134. The iron losses in a transformer are proportional to:
kVA load
square of kVA load
cube of kVA load
none of the above
Answer: D) none of the above
Confirmed vs Book-3 §1.5 — 'Core loss occurs whenever the transformer is energized; core loss does not vary with load.' Iron loss is therefore independent of the kVA loading, so none of the stated proportionalities applies.
Option (b) 'square of kVA load' describes the COPPER loss (P = I²R), which is the classic confusion in this question.
Source: 11th Feb-2011
📖 §1.5 Transformers — losses & efficiency
135. The total loss for a transformer loading at 60% and with no load and full load losses of 3 kW and 25 kW respectively, is
3 kW
12 kW
18 kW
25 kW
Answer: B) 12 kW
Confirmed vs Book-3 §1.5 — total loss = no-load loss + (%load)² × full-load copper loss = 3 + (0.6)² × 25 = 3 + 9 = 12 kW.
Option (c) 18 kW comes from scaling the copper loss linearly (0.6 × 25 = 15); the load fraction must be squared.
Source: 11th Feb-2011
📖 §1.10 Harmonics
136. Harmonics are generated by __________.
HT motors
transformers
LT motors
variable frequency drives
Answer: D) variable frequency drives
Confirmed vs Book-3 §1.10 — 'Harmonic voltages and currents in an electric power system are a result of non-linear electric loads', and the book's list of non-linear loads is headed by variable frequency drives.
Options (a)–(c): motors and transformers have essentially constant impedance and are treated as linear (a transformer only creates harmonics when driven into saturation).
Source: Mar 2023
📖 §1.4 Power Factor Improvement and Benefits
137. The power factor of an electrical system having an active power of 100 kW and reactive power of 80 kVAr will be ___________
0.81
0.88
0.78
cannot be determined
Answer: C) 0.78
Confirmed vs Book-3 §1.4 power triangle — kVA = √(kW² + kVAr²) = √(100² + 80²) = 128.06 kVA, so PF = 100/128.06 = 0.78.
Option (b) 0.88 would come from 100/(100+80)×… ; always build the vector sum first, never add kW and kVAr arithmetically.
138. Aggregate Technical & Commercial losses in distribution system covers ________.
only PR losses of all transformers
only transmission & distribution loss
only transmission losses
energy and monetary loss
Answer: D) energy and monetary loss
Confirmed vs Book-3 §1.8 — AT&C loss = {1 − (Billing Efficiency × Collection Efficiency)} × 100, so it captures both the units lost (technical + theft/un-metered) and the revenue billed but never collected.
Option (b) covers only the technical T&D component and ignores the entire commercial/collection side.
Source: Mar 2021
📖 §1.2 Electricity Billing — tariff structure
139. The two-part tariff structure for HT category consumers are
one part for capacity drawn and second part for actual energy drawn
one part for actual power and second part for actual reactive power drawn
one part for capacity drawn and second part for actual reactive power drawn
one part for actual apparent energy drawn and second part for actual reactive energy drawn
Answer: A) one part for capacity drawn and second part for actual energy drawn
Confirmed vs Book-3 §1.2 — 'the electricity billing…in High Tension (HT) category, is often done on two-part tariff structure, i.e. one part for capacity (or demand) drawn and the second part for actual energy drawn during the billing cycle.'
Option (c) is wrong: reactive energy is billed separately as a PF penalty/bonus, not as the second part of the two-part tariff.
140. Technical & Commercial losses in distribution system covers ___________
only I^2R losses of all transformers
only transmission & distribution loss
only transmission losses
energy and monetary loss
Answer: D) energy and monetary loss
Confirmed vs Book-3 §1.8 — AT&C loss = {1 − (Billing Efficiency × Collection Efficiency)} × 100, combining unbilled energy with uncollected revenue — an energy loss and a monetary loss.
Option (a) I²R transformer losses are only one technical component; AT&C is a much wider utility performance measure.
Source: Mar 2021 (Set B)
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
141. In BEE Star labelled distribution transformers, which of following losses are defined?
total loss at 50% and 100% loading
total loss at 75 % loading
total loss at 75% and 100% loading
total loss at 100% loading
Answer: A) total loss at 50% and 100% loading
Confirmed vs Book-3 §1.5 Standards & Labeling — 'For the BEE labeling programme total losses at 50% and 100% load have been defined' (Table 1.4).
Option (c) 75% and 100% is the standard distractor; 75% loading is not a BEE star-rating reference point.
Source: Jul 2022
📖 §1.4 Performance Assessment of Power Factor Capacitors
142. If V1 is actual supply voltage and V2 is the rated voltage of a capacitor, the reactive kVAr produced would be in the ratio of
V1²/V2²
V1²/V2
1 - V1²/V2²
1 + V1²/V2²
Answer: A) V1²/V2²
Confirmed vs Book-3 §1.4 Voltage effects — capacitor output varies with the square of the applied voltage, so the kVAr actually produced is in the ratio V₁²/V₂² of the rated value (V₁ = actual, V₂ = rated).
Option (c) 1 − V₁²/V₂² gives the FRACTIONAL DROP in output, not the output itself — that form is used when a question asks 'by how much does the VAr output drop'.
Source: Jul 2022
📖 §1.4 Selection and Location of Capacitors
143. The ratings of the PF correction capacitors at motor terminals for a 37 kW induction motor at 3000 rpm synchronous speed will be __________ in comparison to the same sized induction motor at 1500 rpm synchronous speed
more
less
same
dependent on the connected load
Answer: B) less
Confirmed vs Book-3 §1.4 — 3000 rpm synchronous speed is a 2-pole machine and 1500 rpm is a 4-pole machine; the no-load magnetising kVAr (on which the terminal capacitor is sized) is smaller for the higher-speed, lower-pole machine.
So the 3000 rpm motor needs LESS kVAr; option (a) reverses the relationship between pole number and magnetising current.
Source: Jul 2022
📖 §1.5 Transformers — losses & efficiency
144. The total loss for a transformer loading at 60% with no load and full load losses of 3 kW and 25 kW respectively, would be
3 kW
12 kW
18 kW
25 kW
Answer: B) 12 kW
Confirmed vs Book-3 §1.5 — total loss = no-load loss + (%load)² × full-load copper loss = 3 + (0.6)² × 25 = 3 + 9 = 12 kW.
Option (c) 18 kW scales the copper loss linearly with load; the book's formula squares the load fraction because P = I²R.
Source: Jul 2022
📖 §1.11 (lighting terminology; see Book-3 Ch-8 §8.2 Lighting System)
145. A device that distributes filters or transforms the light emitted from one or more lamps is
Control gear
Luminaire
Lamp
Starter
Answer: B) Luminaire
Confirmed vs Book-3 Ch-8 §8.2 — a luminaire is the complete lighting unit that 'distributes, filters or transforms the light emitted from one or more lamps', including the housing, reflector and the parts that hold and protect the lamp.
Option (c) the lamp is only the light SOURCE inside the luminaire; control gear (a) is the ballast/driver circuit.
Source: Jul 2022
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
146. If distribution line has losses of 100kW and voltage is raised from 33KV to 66KV, then line losses will be __________.
50KW
100KW
200KW
25KW
Answer: D) 25KW
Confirmed vs Book-3 §1.1 — line loss varies as the inverse square of the voltage: doubling 33 kV to 66 kV divides the loss by (66/33)² = 4, so 100 kW becomes 25 kW.
Option (a) 50 kW assumes an inverse LINEAR relation; the book's rule is 'the line power loss in the ratio of square of voltages'.
Source: Mar 2023
📖 §1.5 Transformers — losses & efficiency
147. A 2000KVA Transformer has full load losses of 200KW. If the transformer is running at 50% loading, then what will be the load losses?
10KW
50 KW
20KW
None of the above
Answer: B) 50 KW
Confirmed vs Book-3 §1.5 — load (copper) loss varies with the square of the load: at 50% loading the load loss = 200 × (0.5)² = 50 kW.
Option (c) 20 kW assumes the loss falls to one-tenth and option (a) halves it linearly; the squared-load rule (P = I²R) is the whole point of the question. [Option repaired: the printed '5 KW' is arithmetically impossible for a 200 kW full-load loss at 50% load.]
Source: Mar 2023
📖 §1.4 Power Factor Improvement and Benefits
148. What will be the Power Factor of the system having Active Power as 812 KW and Reactive Power as 418 KVAR.
0.89
0.51
0.5
1
Answer: A) 0.89
Confirmed vs Book-3 §1.4 (Figure 1.11) — kVA = √(kW² + kVAr²) = √(812² + 418²) = 913 kVA, so PF = 812/913 = 0.89. These are exactly the book's post-correction figures in the chemical-industry worked example.
Option (b) 0.51 is kVAr/kW-type confusion; power factor is always the ACTIVE power divided by the apparent power.
149. If the Collection Efficiency is 90% and Billing Efficiency is 95% then AT&C losses are __________.
85.5%
14.5%
15%
None of the above
Answer: B) 14.5%
Confirmed vs Book-3 §1.8 — AT&C Losses = {1 − (Billing Efficiency × Collection Efficiency)} × 100 = {1 − (0.95 × 0.90)} × 100 = {1 − 0.855} × 100 = 14.5%.
Option (a) 85.5% is the PRODUCT itself (the efficiency retained), not the loss — remember to subtract from 1.
Source: Mar 2023
📖 §1.9 DSM (building energy performance; see Book-3 Ch-10 ECBC)
150. The unit of Energy Performance Index (EPI) for rating the building is __________.
kWh/sq mtr/yr
kWh/sq mtr/hr
Wh/sq mtr/hr
sq mtr/Wh/yr
Answer: A) kWh/sq mtr/yr
Confirmed vs Book-3 Ch-10/ECBC — the Energy Performance Index (EPI) of a building is its annual energy consumption per unit of built-up area, expressed in kWh/m²/year.
Option (c) Wh/m²/hr is an instantaneous intensity, not the annual index used for star-rating buildings under the BEE programme.
Answer: B) aggregate technical and commercial losses
Confirmed vs Book-3 §1.8 — 'The above losses are collectively categorized as AT & C (Aggregate Technical & Commercial) losses', combining the technical (I²R, transformation) losses with the commercial (theft, metering, collection) losses.
Option (a) misreads 'technical' as 'transmission' and 'commercial' as 'current' — the acronym is fixed by the book.
Source: Mar 2023
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
152. The nearest kVAr compensation required for improving the power factor of a 1000 kW load from 0.95 leading power factor to unity power factor is __________.
328 kVAr
750 kVAr
1000 kVAr
none of the above
Answer: A) 328 kVAr
Confirmed vs Book-3 §1.11 Solved Example — kVAr = kW[tan(cos⁻¹PF₁) − tan(cos⁻¹PF₂)] = 1000 × (0.329 − 0) = 328.7 ≈ 328 kVAr, exactly the book's 0.95→unity route.
Option (c) 1000 kVAr wrongly equates the capacitor size with the kW; only the reactive component (kW × tanΦ₁) has to be cancelled.
Source: Mar 2023
📖 §1.5 Transformers — losses & efficiency
153. The no-load loss and copper loss of a 500 kVA transformer are 1600 Watts and 6400 Watts respectively. What is the total loss at 50% of transformer loading?
4100 Watts
6850 Watts
2500 Watts
3200 Watts
Answer: D) 3200 Watts
Confirmed vs Book-3 §1.5 — total loss = no-load loss + (%load)² × full-load copper loss = 1600 + (0.5)² × 6400 = 1600 + 1600 = 3200 W.
Option (c) 2500 W is the answer for a 900 W no-load loss (the sister question from the 17th exam); option (b) 6850 W adds the FULL copper loss without squaring the load fraction. [Option repaired: the printed '3650 Watts' does not follow from the stated 1600 W / 6400 W data.]
lower the heat rate of a power generating unit, higher is the generation efficiency
one kilo Watt hour of electrical energy being equivalent to 3600 kilo Joules of thermal energy
'Heat Rate' is directly proportional to the efficiency of power generation
design 'Heat Rate' of a 210 MW thermal generating unit is lower than that of a 110 MW thermal generating unit
Answer: C) 'Heat Rate' is directly proportional to the efficiency of power generation
Confirmed vs Book-3 §1.1 — 'The 'HEAT RATE' is inversely proportional to efficiency of power generation i.e., lower the heat rate, higher is the generation efficiency.' Statement (c) says DIRECTLY proportional, which flatly contradicts the book, so it is the wrong statement.
Statements (a) and (b) are book facts (1 kWh ≡ 3600 kJ ≡ 860 kCal), and (d) is true as repaired: larger units are more efficient, so a 210 MW set has a LOWER design heat rate than a 110 MW set. [Option (d) repaired so that exactly one statement is wrong.]
Source: 11th Feb-2011
📖 §1.10 Harmonics
155. Select the incorrect statement:
harmonics occur as spikes at intervals which are multiples of the supply frequency
harmonics are multiples of the fundamental frequency
induction motors are the major sources of harmonics
transformers operating near saturation level create harmonics
Answer: C) induction motors are the major sources of harmonics
Confirmed vs Book-3 §1.10 — induction motors are given by the book as examples of LINEAR systems ('to a great extent, motors are linear systems'); the major harmonic sources listed are VFDs, electronic ballasts, UPS/computers and induction/arc furnaces. So (c) is the incorrect statement.
Statements (a), (b) and (d) are all true book content — harmonics are integer multiples of the fundamental, and a saturated transformer is non-linear. [Options re-lettered a)–d); they were printed without labels.]
Source: 13th Sep-2012
📖 §1.10 Harmonics
156. Select the incorrect statement:
harmonics occur as spikes at intervals which are multiples of the supply frequency
harmonics are not multiples of the fundamental frequency
induction motors are not the major sources of harmonics
transformers operating near saturation level create harmonics
Answer: B) harmonics are not multiples of the fundamental frequency
Confirmed vs Book-3 §1.10 — 'A harmonic is a component frequency of the signal that is an integer multiple of the fundamental frequency' (5th harmonic on 50 Hz = 250 Hz). Statement (b) says harmonics are NOT multiples of the fundamental, which contradicts the definition, so it is the incorrect statement.
Statement (c) is true because the book classes motors as largely linear, and (d) is true because a saturated transformer behaves non-linearly. [Options re-lettered a)–d); they were printed without labels.]
Source: 15th Aug-2014
Short questions (5 marks) — 27
📖 §1.4 Selection of capacitors — kVAr = kW(tanΦ₁ − tanΦ₂), Table 1.2
1. A plant draws an average load of 800 kW at a power factor of 0.75. Calculate the capacitor rating (kVAr) needed to raise the PF to 0.95. (tan(cos⁻¹0.75)=0.882, tan(cos⁻¹0.95)=0.329)
Model answer: Formula (Book-3 §1.4): kVAr rating = kW × (tanΦ₁ − tanΦ₂), where Φ₁ = cos⁻¹(present PF) and Φ₂ = cos⁻¹(desired PF).
Step 1: Φ₁ = cos⁻¹ 0.75 → tanΦ₁ = 0.882.
Step 2: Φ₂ = cos⁻¹ 0.95 → tanΦ₂ = 0.329.
Step 3: kVAr = 800 × (0.882 − 0.329) = 800 × 0.553 = 442.4 kVAr.
Cross-check with Book Table 1.2: the multiplier for 0.75 → 0.95 is 0.553, and 800 × 0.553 = 442.4 kVAr.
Answer: install the next standard bank size, i.e. about 450 kVAr (the book rounds its own 398 kVAr example up to a 400 kVAr installation).
Confirmed vs Book-3 §1.4 — method: take tan of the present and desired power-factor angles, subtract, multiply by the average kW drawn, then round UP to the next standard capacitor size. Note the kW is unchanged by PF correction; only the kVAr (and hence kVA) is reduced.
Source: AI practice
📖 §1.8 AT&C Losses (Table 1.7 method)
2. A discom has units input 80 MU, units billed 64 MU, amount billed ₹700 million, and revenue collected ₹560 million. Find the billing efficiency, collection efficiency and AT&C loss.
Model answer: Book-3 §1.8 definitions:
Billing Efficiency (BE) = (Total units billed ÷ Total units input) × 100 = (64 ÷ 80) × 100 = 80%.
Collection Efficiency (CE) = (Revenue collected ÷ Amount billed) × 100 = (560 ÷ 700) × 100 = 80%.
AT&C Loss (%) = {1 − (BE × CE)} × 100 = {1 − (0.80 × 0.80)} × 100 = {1 − 0.64} × 100 = 36%.
So 36% of the energy input is lost technically plus commercially (unbilled units plus uncollected revenue).
Confirmed vs Book-3 §1.8 — the same three-step method as book Table 1.7 (BE 70%, CE 93% → AT&C 35%). Keep BE and CE as FRACTIONS inside the bracket; a common error is to add the two loss percentages (20% + 20% = 40%), which double-counts.
Source: AI practice
📖 §1.4 Advantages and Cost benefits of PF improvement by capacitor addition
3. List any four (4) benefits of improving the power factor of an industrial installation.
Model answer: Book-3 §1.4 lists (any four):
1) Reduced kVA / maximum demand charges in the utility bill — for the same kW, a higher PF means lower kVA.
2) Reduced I²R distribution losses (kWh) within the plant network, because the reactive component of current is cancelled locally.
3) Better/raised voltage at the load end — improved motor terminal voltage and hence motor performance.
4) Released (freed) capacity of transformers, cables and switchgear up to the capacitor point, allowing future load addition without new investment.
5) Elimination of the utility's power-factor penalty (and often earning a PF rebate).
Confirmed vs Book-3 §1.4 — the book's worked example shows the mechanism: adding 410 kVAr took a plant from 1160 kVA at 0.70 PF down to 913 kVA at 0.89 PF for the same 812 kW, dropping transformer loading from about 78% to 60% and avoiding the penalty.
Source: AI practice
📖 §1.4 Selection and Location of Capacitors
4. During April-2003 a plant recorded a maximum demand of 600 kVA and an average PF of 0.82 lag. The utility requires a minimum average PF of 0.92 lag, and every 1% dip in PF attracts a penalty of Rs 10,000 per month. (a) Calculate the improvement in PF for May-2003 by installing 100 kVAr capacitors. (b) Calculate the penalty to be paid, if any, during May-2003.
Model answer: At 600 kVA and 0.82 PF: kW = 600 x 0.82 = 492 kW; kVAr = 600 x sin(cos^-1 0.82) = 600 x 0.5724 = 343.4 kVAr. After adding 100 kVAr capacitor, new kVAr = 243.4; new kVA = sqrt(492^2 + 243.4^2) = sqrt(242064 + 59243) = 548.9 kVA; new PF = 492/548.9 = 0.896 (about 0.90). (a) PF improves from 0.82 to ~0.90 (an improvement of about 0.08). (b) Required PF is 0.92, so PF still falls short by about 2% (0.92 - 0.90); penalty for May-2003 is approximately 2 x Rs 10,000 = Rs 20,000.
Resolve load into kW and kVAr, subtract the capacitor kVAr, recompute kVA and PF; compare with the 0.92 target to find the shortfall and penalty.
Source: Book EOC
📖 §1.4 Selection and Location of Capacitors (Location of Capacitors, Figure 1.13)
5. Which is the best location for capacitor banks for power factor improvement from an energy-conservation point of view? Why?
Model answer: The best location is as close to the load (at the individual motor/equipment terminals) as possible. Locating capacitors near the load relieves the entire upstream system (cables, transformers, switchgear and the distribution network) of reactive current, so I^2R losses are reduced throughout, voltage profile improves and released capacity becomes available. Central/group compensation at the substation improves the utility-side PF but does not reduce losses in the internal distribution network.
Compensation nearest the load minimizes reactive current flow over the longest path and gives the maximum loss reduction and capacity release.
Source: Book EOC
📖 §1.10 Harmonics — Effects of Harmonics
6. List any five problems that can arise due to harmonics in a system.
Model answer: 1) Overheating of transformers, motors and neutral conductors. 2) Increased I^2R and eddy-current/iron losses lowering efficiency. 3) Nuisance tripping of circuit breakers and blowing of capacitor fuses. 4) Resonance with power-factor capacitors leading to capacitor failure. 5) Malfunction/measurement errors in electronic equipment, meters and protective relays (plus reduced power factor and equipment derating).
Harmonics cause extra heating, losses, resonance, nuisance tripping and interference with electronic/metering equipment.
Source: Book EOC
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
7. Explain why power is generated at lower voltage and transmitted at higher voltages.
Model answer: Generators are built for moderate voltages (around 11 kV) because insulating the windings for very high voltage inside a machine is impractical and uneconomic. For transmission, power is stepped up to high voltage so that, for a given power, the line current is reduced; since line loss is proportional to I^2R, raising the voltage greatly lowers transmission losses and voltage drop, and allows smaller-cross-section conductors. The voltage is then stepped down for distribution and utilisation.
Low generation voltage suits machine insulation; high transmission voltage lowers current and hence I^2R losses for a given power.
Source: Book EOC
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
8. A trivector-meter installed in a mini steel plant with a 15-minute cycle records the following during the maximum demand period: 1000 kVA for 10 min, 2000 kVA for 5 min, 750 kVA for 10 min, 1500 kVA for 5 min. What is the maximum demand during the 15-minute interval?
Model answer: Maximum demand is the TIME-INTEGRATED demand over the meter's 15-minute cycle (Book-3 Sec.1.2), not the instantaneous peak, so evaluate each 15-minute window and take the highest:
Window 1 (1000 kVA for 10 min + 2000 kVA for 5 min) = [(1000 x 10) + (2000 x 5)] / 15 = 20,000/15 = 1333 kVA
Window 2 (750 kVA for 10 min + 1500 kVA for 5 min) = [(750 x 10) + (1500 x 5)] / 15 = 15,000/15 = 1000 kVA
Maximum demand registered = 1333 kVA. Note that the momentary 2000 kVA block is NOT the billed demand - the meter averages it over the full 15-minute cycle.
MD = energy (kVAh) accumulated in the worst 15-minute window divided by the 0.25 h interval; the 1000 kVA (10 min) + 2000 kVA (5 min) window averages 1333 kVA.
Source: Book EOC
📖 §1.4 Selection and Location of Capacitors
9. A process plant consumes 12,500 kWh per month at 0.9 power factor. What is the percentage reduction in distribution losses per month if the PF is improved to 0.96 at the load end? Assume existing distribution loss is 4% of the plant energy consumption.
Model answer: Distribution loss is proportional to I^2, hence to 1/PF^2. Reduction factor = 1 - (PF_old/PF_new)^2 = 1 - (0.9/0.96)^2 = 1 - 0.8789 = 0.1211, i.e. about 12.1% reduction in losses. In energy terms, existing loss = 4% of 12,500 = 500 kWh/month; new loss = 500 x (0.9/0.96)^2 = 439.5 kWh; saving ≈ 60.5 kWh/month.
Losses scale as 1/PF^2; improving PF from 0.9 to 0.96 cuts losses by 1 - (0.9/0.96)^2 ≈ 12.1%.
Source: Book EOC
📖 §1.4 Power Factor Improvement and Benefits
10. A 37 kW, 3 phase, 415 V induction motor draws 56 A and 33 kW power at 410 V. What is the apparent and reactive power drawn by the motor at the operating load?
Model answer: Apparent power = 1.7321 x 0.410 x 56 = 39.769 kVA. Active power = 33 kW. Reactive power = √(39.769² - 33²) = √(1581.57 - 1089) = 22.19 kVAr.
Apparent power S = √3 x V x I; reactive power Q = √(S² - P²).
11. Compute AT&C (Aggregate Technical and Commercial) Losses for the given data: Input Energy Ei=20 MU, Energy Billed Metered E1=16 MU, Un-metered E2=1 MU, Total Billed Eb=17 MU, Amount Billed Ab=Rs.800 lakhs, Gross Amount Collected AG=Rs.820 lakhs, Arrears Collected Ar=Rs.40 lakhs.
Model answer: Amount collected without arrears Ac = AG - Ar = 820 - 40 = 780. Billing Efficiency BE = Eb/Ei = 17/20 = 85%. Collection Efficiency CE = Ac/Ab = 780/800 = 97.5%. AT&C Loss = [1 - (BE x CE)] x 100 = [1 - (0.85 x 0.975)] x 100 = 17.12%.
AT&C Loss = 1 - (Billing Efficiency x Collection Efficiency).
Source: 15th Exam
📖 §1.10 Harmonics
12. Harmonic measurements gave: current at 50 Hz = 300 A, at 150 Hz = 42 A, at 250 Hz = 33 A. Calculate the Total Harmonic Distortion in current.
Model answer: THD(I) = √[(42/300)² + (33/300)²] x 100 = √(0.0196 + 0.0121) x 100 = √0.0317 x 100 = 17.8%.
THD = √(sum of squares of harmonic currents)/fundamental x 100.
Source: 14th Exam
📖 §1.2 Electricity Billing — tariff structure (Time of Day tariff) + §1.3
13. How do the Time-Of-Day (TOD) metering and billing benefit the utilities as well as consumers?
Model answer: A ToD meter records demand, time and energy, and ToD tariff sets higher rates at peak periods and lower rates at off-peak. Consumers benefit by shifting load to off-peak hours at the lowest tariff; the higher peak tariff discourages peak drawl. This reduces the utility's maximum demand, saves peak-time power procurement at high rates, and maximises load factor for better utility financials.
ToD tariff aligns consumer behaviour with utility load management benefits.
14. Compute AT&C Losses: Input Energy Ei=11 MU, Energy Billed Metered=7 MU, Un-metered=1 MU, Total Billed=8 MU, Amount Billed=Rs.450 lakhs, Gross Collected=Rs.460 lakhs, Arrears Collected=Rs.40 lakhs.
Model answer: Ac (amount collected without arrears) = AG - Ar = 460 - 40 = Rs.420 lakhs.
Billing Efficiency BE = Eb/Ei x 100 = 8/11 x 100 = 72.7%.
Collection Efficiency CE = Ac/Ab x 100 = 420/450 x 100 = 93.3%.
AT&C Loss = [1 - (BE x CE)] x 100 = [1 - (0.727 x 0.933)] x 100 = [1 - 0.6786] x 100 = 32.1%.
(Book-3 Sec.1.8, Table 1.7 method: note that arrears must be stripped out of the gross collection before computing collection efficiency.)
AT&C Loss = 1 - (Billing Efficiency x Collection Efficiency).
Source: Set-A
📖 §1.1 Industrial End User — 'ONE Unit saved = TWO Units Generated'
15. One unit of electricity in end-use application is equivalent to about two units of electricity generated. Substantiate with the cascade efficiency from generating plant ex-bus to end-use. Assume: Generator yard substation efficiency 98%; T&D loss = 20%; End-use application efficiency = 65%.
Model answer: Cascade efficiency = 0.98 × (1 − 0.20) × 0.65 = 0.5096. Therefore one unit at end use = 1/0.5096 = 1.96 ≈ 2 units at ex-generator bus.
16. A single line diagram shows a 100 kW heater load and a 200 kW motor (200 m from the 415 V LT bus). Main incoming line PF is 0.85 lag. Calculate the rating of capacitors to improve the main incoming line PF to 0.9 lag.
Model answer: The capacitor formula of Book-3 Sec.1.4 is kVAr = kW[tanPhi1 - tanPhi2], where kW is the TOTAL active power flowing at the point whose power factor is being corrected - here the main incoming line, which carries 100 kW (heater) + 200 kW (motor) = 300 kW.
tan(cos^-1 0.85) = 0.6197; tan(cos^-1 0.90) = 0.4843.
kVAr = 300 x (0.6197 - 0.4843) = 300 x 0.1354 = 40.6 kVAr, say a 40-45 kVAr bank.
Check: at 0.85 PF, kVAr drawn = 300 x 0.6197 = 186 kVAr; after correction 300 x 0.4843 = 145 kVAr; difference = 41 kVAr.
The reactive power comes only from the motor (the heater is resistive), but the SIZING must use the total 300 kW measured at the incoming line - using only the 200 kW motor load under-sizes the bank.
Compensate only inductive (motor) load; kVAr = kW[tanφ1 − tanφ2].
17. Fill in the blanks: (a) Voltage levels can be varied without isolating the connected load to the transformer using ___. (b) Use of ___ starter is appropriate for a high number of motor starts/stops per hour. (c) Operating a highly under-loaded motor in star mode reduces voltage by a factor of ___. (d) ___ is the ratio of dissolved solids in circulating water to dissolved solids in makeup water. (e) In SI units, ___ is the measure of light output of a lamp.
Model answer: (a) On-load tap changer (OLTC); (b) Soft starter; (c) √3 (square root of three); (d) Cycles of Concentration (COC); (e) Lumens.
Definitional fill-in-the-blanks across transformers, motor starters, COC and photometry.
Source: 18th Exam
📖 §1.4 Power Factor Improvement and Benefits
18. S-5: List any five benefits of power factor improvement in an industrial power distribution system.
Model answer: 1. Reduced kVA demand and lower maximum-demand charges. 2. Reduced line/transformer current → lower I²R distribution losses. 3. Released capacity in transformers, cables and switchgear for additional load. 4. Improved voltage regulation (less voltage drop) at the load. 5. Avoidance of low-PF penalties and possible PF-incentive rebates from the utility. (Also: longer equipment life due to reduced heating.)
Standard PF-improvement benefits from Book-3 Chapter 1 (the paper directs 'Refer Guide Book No 3, Chapter 1, Page No 11'); model answer supplied from chapter notes.
Source: 19th Exam
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
19. A foundry draws 2500 kW. Demand during furnace operation: 5 min 2940 kVA, 7 min 2550 kVA, 3 min 2777 kVA. Billing meter monitors demand every 15 minutes. Calculate the maximum demand registered and the average PF during the interval.
Model answer: Maximum demand is the time-integrated kVA over the 15-minute cycle (Book-3 Sec.1.2):
MD = [(2940 x 5) + (2550 x 7) + (2777 x 3)] / 15 = [14,700 + 17,850 + 8331] / 15 = 40,881/15 = 2725.4 kVA.
Average power factor over the interval = kW / kVA = 2500 / 2725.4 = 0.917 (about 0.92 lag).
Note the demand billed is the 2725 kVA average, not the 2940 kVA instantaneous peak.
20. S-8: A 100 kW heater load and a 200 kW motor (200 m from the 415V LT bus). Main incoming line PF is 0.85 lag. Calculate the rating of capacitors to improve the PF of the main incoming line to 0.9 lag.
Model answer: Use kVAr = kW[tanPhi1 - tanPhi2] with the TOTAL active power at the point being corrected (Book-3 Sec.1.4). The main incoming line carries 100 kW heater + 200 kW motor = 300 kW at 0.85 lag.
tan(cos^-1 0.85) = 0.6197; tan(cos^-1 0.90) = 0.4843.
kVAr required = 300 x (0.6197 - 0.4843) = 40.6 kVAr, i.e. a 40-45 kVAr capacitor bank.
(The heater draws no kVAr, but it does carry active power through the incoming line, so it must be included in the kW used for sizing.)
📖 §1.4 Performance Assessment of Power Factor Capacitors
21. a) A 10 kVAr, 415 V rated power factor capacitor was found to be having terminal supply voltage of 440 V. Calculate the capacity of the power factor capacitor at the operating supply voltage. b) What would be the nearest kVAr compensation required for changing the power factor of a 500 kW load from 0.9 lead to unity power factor?
Model answer: a) Capacitor output varies as the square of the applied voltage (Book-3 Sec.1.4, Voltage effects): kVAr = 10 x (440/415)^2 = 10 x 1.124 = 11.24 kVAr. The bank delivers more than its rating, but running above rated voltage shortens capacitor life.
b) At 0.9 LEADING the load is already over-compensated - the current leads the voltage - so NO further capacitive kVAr is required. To reach unity, capacitance must instead be REMOVED: excess kVAr = 500 x tan(cos^-1 0.9) = 500 x 0.4843 = 242 kVAr of the existing bank should be switched out.
a) Capacitor kVAr varies with square of voltage ratio. b) Load is already at leading PF, so additional capacitive compensation is not needed.
23. S-2: Match the following load-shape objectives of any Demand Side Management (DSM) programme of a utility. (i) Peak Clipping, (ii) Valley filling, (iii) Load shifting, (iv) Conservation, (v) Load building - with the corresponding load-shape diagrams a-e.
Model answer: i - c; ii - d; iii - b; iv - e; v - a
Standard DSM load-shape matching as given in the official key.
24. Fill in the blanks (1 mark each): a) Heat rate of a thermal power plant is expressed in ___; b) The ___ loss is independent of load in a transformer; c) ___ is used to reduce the dew point in a compressed air system; d) The speed of an energy efficient motor will be more than the standard motor of same capacity because ___ decreases.
Model answer: a) kCal/kWh (or kJ/kWh) - the heat input per kWh generated; 1 kWh = 860 kCal = 3600 kJ.
b) Core loss (iron loss / no-load loss) - the book states core loss 'occurs whenever the transformer is energized; core loss does not vary with load'.
c) An air dryer (refrigerant or desiccant type) is used to lower the dew point of compressed air.
d) SLIP - an energy-efficient motor has lower rotor losses and therefore lower slip, so for the same number of poles it runs slightly faster than a standard motor.
Standard BEE fill-in answers: thermal heat rate in kCal/kWh, transformer iron/core loss is load-independent, air dryer lowers dew point, EE motor has higher speed because slip decreases.
Source: Mar 2021 (Set B)
📖 §1.4 Power Factor Improvement and Benefits + §1.5 — multi-chapter True/False
25. State True or False (1 Mark each): 1. In an industrial electrical system operating at unity power factor, addition of further capacitors will reduce the maximum demand (kVA). 2. In a step-down transformer for a given load the current in the primary will be much lower than the current in the secondary. 3. For the same no of poles and kVA rating, the RPM of an energy efficient motor is higher than that of a standard motor. 4. The advantage of evaporative cooling is that it is possible to obtain water temperatures below the wet bulb economically. 5. A fluid coupling changes the speed of the driven equipment without changing the speed of the motor.
Model answer: 1. FALSE - at unity power factor the reactive component is already zero; adding more capacitors makes the current LEAD and the kVA (and hence maximum demand) rises again.
2. TRUE - primary ampere-turns = secondary ampere-turns, so V1I1 = V2I2. In a step-down transformer the primary is the high-voltage side and therefore carries the LOWER current.
3. TRUE - an energy-efficient motor has lower rotor I^2R loss and hence lower slip, so for the same number of poles and rating it runs at a slightly HIGHER rpm than a standard motor.
4. FALSE - evaporative cooling can approach but never economically go below the wet bulb temperature; the difference (cold water temp - WBT) is the 'approach'.
5. TRUE - a fluid coupling varies the output (driven) speed by varying the oil fill while the motor continues to run at its own constant speed.
At unity PF capacitors over-correct and can increase kVA; primary current of step-down transformer is lower than secondary; EE motor runs at nearly same/slightly higher speed but not 'higher RPM' as stated; evaporative cooling cannot go below wet bulb; fluid coupling varies output speed while motor runs constant.
26. Fill in the blanks for the following: 1. The main input energy used for refrigeration in vapor absorption refrigeration plants is _____. 2. One ton of refrigeration is equivalent to _____ kW. 3. Stray losses in induction motor generally are proportional to the square of the _____ current. 4. The unit of Ah*EPI is _____. 5. If the pump impeller diameter is reduced by 10% then head reduces by _____%. 6. A 4 pole 50Hz motor operating with slip of 3% will have a shaft speed of _____ RPM. 7. Effective Aperture Glazing (EA) = VLT × _____. 8. In an amorphous core distribution transformer, no-load loss is _____ than a conventional transformer. 9. As the condensing temperature increases, kW/TR of refrigeration system will _____. 10. The extent of drying compressed air is expressed by the term _____.
Model answer: 1. Thermal energy (steam / waste heat / hot water / fuel gas) - the vapour absorption machine is heat-driven, not compressor-driven.
2. 3.51 kW (1 TR = 3024 kCal/h = 3.51 kW).
3. ROTOR current - stray load losses in an induction motor vary as the square of the rotor current.
4. EPI (Energy Performance Index) is expressed in kWh/sq.m/year.
5. 19% - head varies as the square of impeller diameter, so 1 - 0.9^2 = 0.19.
6. 1455 rpm - Ns = 120 x 50/4 = 1500 rpm; at 3% slip, N = 1500 x 0.97 = 1455 rpm.
7. Window-to-Wall Ratio (WWR): Effective Aperture = VLT x WWR.
8. LESS (about 70% lower core loss than a conventional CRGO silicon-iron core - Book-3 Sec.1.5).
9. INCREASE - a higher condensing temperature raises the compression ratio and hence the specific power kW/TR.
10. Dew point (atmospheric / pressure dew point).
Fill-in answers from BEE Book-3 fundamentals; 1 TR=3.517 kW; head∝D² so 10% dia drop → ~19% head drop; N=120f/p×(1-s)=1500×0.97=1455 rpm.
Source: Jul 2022
📖 §1.4 Power Factor Improvement and Benefits
27. A textile industry had installed a 2 MVA transformer. The initial demand of the plant was 1500 kVA with power factor of 0.75. Industry has installed 450 kVA capacitor at the motor end. Calculate the following: 1. Reduction in apparent power (kVA); 2. Improved power factor; 3. Revised % loading of transformer after installing the capacitor.
Model answer: Real power = 1500×0.75 = 1125 kW. Old reactive power = √(1500²−1125²) = 992 kVAr. After 450 kVAr capacitor, revised kVAr = 992−450 = 542 kVAr. Revised apparent power = √(1125²+542²) = 1248 kVA. Reduction in apparent power = 1500−1248 = 252 kVA. Improved PF = 1125/1248 = 0.90. Revised % loading = 1248/2000 = 62.4%.
Resolve into real and reactive components, subtract capacitor kVAr, recompute apparent power, pf and transformer loading (kVA/rating).
1. A private power distribution company has implemented new digital metering and billing systems to improve efficiency in a residential zone. After six months of operation, the following data was recorded:
• Input energy to the system = 75 MU
• Metered billed energy = 56 MU
• Unmetered average billing = 4 MU
• Amount billed = ₹680 million
• Total amount received = ₹600 million
• Arrears collected = ₹90 million
• Purchased energy cost = ₹8.50 per kWh
i) Estimate the Aggregate Technical and Commercial (AT&C) loss (%) and the revenue realized per kWh (4 Marks)
ii) Calculate the revenue loss per kWh to the company due to AT&C loss (1 Mark)
Model answer: Input Energy = 75 MU = 75,000,000 kWh
Metered Billed Energy = 56 MU
Unmetered Average Billing = 4 MU
Total Energy Billed = 56 + 4 = 60 MU
Amount Billed = ₹680 million; Arrears Collected = ₹90 million; Amount Received = ₹600 million; Purchased Energy Cost = ₹8.50/kWh
i) Billing Efficiency = (60 / 75) × 100 = 80.0%
Collection Efficiency = ((600 - 90) / 680) × 100 = (510 / 680) × 100 = 75.0%
AT&C Loss (%) = 1 - (Billing Efficiency × Collection Efficiency) = 1 - (0.80 × 0.75) = 1 - 0.60 = 40.0%
Revenue Realized per kWh = (600 - 90) / 75 = 510 / 75 = ₹6.80/kWh
ii) Revenue Loss = ₹8.50 - ₹6.80 = ₹1.70/kWh
Final Answers: i) AT&C Loss = 40.0%, Revenue Realized = ₹6.80/kWh; ii) Revenue Loss per kWh = ₹1.70/kWh
AT&C loss = 1 − (Billing Efficiency × Collection Efficiency); revenue realized = net amount received / input energy.
Source: Sep 2025
📖 §1.4 Selection and Location of Capacitors
2. A steel manufacturing facility is powered by a 3-phase, 6.6 kV, 50 Hz supply and operates the following electrical loads: An electric arc furnace consumes 1.2 MW at a lagging power factor of 0.65, a bank of induction motors for rolling operations consumes 800 kW at a 0.80 lagging power factor, and the lighting and instrumentation systems consume 100 kW at unity power factor. Due to utility regulations, the overall plant power factor must be improved to 0.95 lagging. A capacitor bank will be installed for compensation. As an energy auditor evaluate the following:
a. Total active power consumption. (1 Mark)
b. Total initial apparent power drawn by the facility. (1 Mark)
c. The operating power factor. (1 Mark)
d. Determine the total reactive power required to achieve the desired power factor. (2 Mark)
Model answer: Load-wise breakdown (kVAr = kW x tan(cos^-1 PF)):
Arc furnace: 1200 kW, PF 0.65 -> kVA = 1200/0.65 = 1846 kVA, kVAr = 1846 x sin(49.5deg) = 1403 kVAr
Induction motors: 800 kW, PF 0.80 -> kVA = 1000 kVA, kVAr = 600 kVAr
Lighting & instrumentation: 100 kW, PF 1.0 -> kVA = 100 kVA, kVAr = 0
a) Total active power = 1200 + 800 + 100 = 2100 kW
b) kVA must be added VECTORIALLY, not arithmetically: total kVAr = 1403 + 600 + 0 = 2003 kVAr, so total kVA = sqrt(2100^2 + 2003^2) = 2902 kVA
c) Operating power factor = kW/kVA = 2100/2902 = 0.724 lag
d) kVAr required = kW[tan(cos^-1 0.724) - tan(cos^-1 0.95)] = 2100 x (0.9538 - 0.3287) = 1313 kVAr; equivalently 2003 - (2100 x 0.3287) = 1313 kVAr.
(Capacitor bank of about 1300-1350 kVAr, preferably switched in steps through an APFC because the arc-furnace PF swings over the melting cycle - Book-3 Sec.1.4.)
Sum kW and kVAr per load; operating PF = ΣkW/ΣkVA; compensation kVAr = kW(tanφ1 − tanφ2).
Source: Sep 2025
📖 §1.9 Demand Side Management (DSM)
3. A distribution company (DISCOM) plans to implement a comprehensive Demand Side Management (DSM) initiative to reduce its peak load and overall energy procurement cost. The program targets Residential consumers (LED replacement) and Industrial consumers (load shifting). The DISCOM supplies electricity to 10,000 households, each using 4 CFL bulbs (30 W each), used for 5 hours per day during evening peak hours. The DISCOM replaces each CFL bulb with a 9 W LED bulb. Procurement cost of each LED bulb is ₹100, provided to consumers at a subsidized rate of ₹70/bulb. Administrative cost per household is ₹10. DISCOM also serves 50 industrial consumers, each with a shiftable evening load of 100 kW, used from 6 PM to 10 PM. DISCOM incentivizes them to shift load from 10 PM to 2 AM by offering ₹2 per kWh shifted. Eighty percent of industrial consumers agree. Power purchase cost: Evening Peak (5–10 PM) ₹7/kWh; Late Night (10 PM–6 AM) ₹3/kWh. Calculate:
a) For residential consumers: i) total daily energy savings (kWh/day) from LED replacement (2 Marks); ii) daily cost savings for DISCOM (1 Mark); iii) total one-time cost to DISCOM including subsidies and administrative costs (2 Marks); iv) simple payback period in days (1 Mark).
b) For industrial consumers: i) total energy shifted in kWh/day (1 Mark); ii) net daily savings for DISCOM considering power cost reduction and incentive payout (1 Mark).
c) Estimate the carbon emission avoidance due to above two DSM activity, if emission factor of grid electricity is 0.716 tCO2/MWh (2 Marks).
Model answer: a) Residential (LED replacement)
i) Daily energy saving = 10,000 households x 4 lamps x 5 h x (30 - 9) W / 1000 = 4200 kWh/day
ii) Daily cost saving to DISCOM (all in the 5-10 PM peak block at Rs.7/kWh) = 4200 x 7 = Rs.29,400/day
iii) One-time cost = subsidy 10,000 x 4 x (100 - 70) = Rs.12,00,000 + administrative 10,000 x 10 = Rs.1,00,000 -> Rs.13,00,000
iv) Simple payback = 13,00,000 / 29,400 = 44.2 days (about 44 days)
b) Industrial (load shifting)
i) Energy shifted = 50 consumers x 100 kW x 4 h x 0.80 = 16,000 kWh/day
ii) Net daily saving = 16,000 x (7 - 3 - 2) = Rs.32,000/day (Rs.4/kWh avoided cost less Rs.2/kWh incentive)
c) Carbon avoidance: only the LED programme saves ENERGY (load shifting merely re-times it, so it avoids peak cost but not kWh). CO2 avoided = 4200 kWh/day x 0.716 tCO2/MWh / 1000 = 3.0 tCO2/day, i.e. about 1098 tCO2/year.
This is peak clipping + conservation in Book-3 Sec.1.9 terms - the classic DSM combination of an end-use efficiency measure with a load-shifting tariff incentive.
LED savings = households×bulbs×hrs×ΔW; payback = one-time cost / annual saving; load-shift saving = energy×(peak − offpeak − incentive); CO2 = energy saved × emission factor.
4. A 20 MW co-generation plant operates at a daily load factor of 85% and 8% auxiliary power consumption. The power is generated at 11 kV. Out of the total energy generated, 45% is exported to the grid through a 15 MVA transformer with 99% efficiency. Additionally, 35% of the generated energy is supplied to mill motors at 600 Volts through an 8 MVA step-down transformer with 98.5% efficiency. The remaining energy is used for other LT loads and auxiliaries at 415 Volts through a 4 MVA transformer with 98.2% efficiency. Calculate the following:
1. Daily energy generation in MWh.
2. Daily energy exported in MWh to the grid at 33 kV.
3. Daily mill motors consumption in MWh at 600 V.
4. Daily LT loads and auxiliary consumption in MWh at 415 V.
5. Daily transformer losses in kWh and % transformer losses.
Model answer: 1. Gross generation = 20 MW x 0.85 load factor x 24 h = 408 MWh/day; net (after 8% auxiliary consumption) = 408 x 0.92 = 375.36 MWh/day
2. Export = 375.36 x 0.45 = 168.91 MWh at the 11 kV bus; delivered through the 15 MVA transformer at 99% = 167.22 MWh/day
3. Mill motors = 375.36 x 0.35 = 131.38 MWh; delivered through the 8 MVA transformer at 98.5% = 129.41 MWh/day
4. LT loads & auxiliaries = 375.36 x 0.20 = 75.07 MWh; delivered through the 4 MVA transformer at 98.2% = 73.72 MWh/day
5. Transformer losses = (168.91-167.22) + (131.38-129.41) + (75.07-73.72) = 1.69 + 1.97 + 1.35 = 5.01 MWh = about 5010 kWh/day.
As a percentage of the 375.36 MWh actually passing through the transformers, loss = 5010/375,360 = 1.33% (1.23% if expressed on gross generation of 408 MWh).
This is the cascade-efficiency idea of Book-3 Sec.1.1 applied inside a plant: each transformation stage multiplies its own efficiency onto the energy delivered.
Gross gen = MW×LF×24; net = gross×(1−aux); split by % and apply transformer efficiencies; losses = before − after.
5. The maximum demand approved by a utility is 5500 kVA and the tariff provides for a minimum billing demand of 80% of approved. Records of the past 12 months show the monthly maximum demand recorded is around 4200 kVA. Will there be any benefit in surrendering part of the contract demand? If so, what kVA do you recommend surrendering? Give the cost saving, if the unit rate for kVA demand is Rs 200.
Model answer: Minimum billing demand at present = 80% of 5500 = 4400 kVA, which is higher than the actual recorded demand of ~4200 kVA, so the plant pays for 4400 kVA. By surrendering demand, the contract demand should be reduced so that 80% of the new contract demand just covers the actual maximum demand of 4200 kVA: new contract demand = 4200/0.8 = 5250 kVA. Recommend surrendering 5500 - 5250 = 250 kVA. New billing demand = 4200 kVA, saving 4400 - 4200 = 200 kVA per month. Monthly saving = 200 x Rs 200 = Rs 40,000, i.e. about Rs 4,80,000 per year.
Lower the contract demand until 80% of it equals the actual peak (4200 kVA); billing demand falls from 4400 to 4200 kVA, saving 200 kVA x Rs 200.
Source: Book EOC
📖 §1.9 Demand Side Management (DSM)
6. Explain Demand Side Management (DSM) and the various techniques used for DSM. What are the benefits of DSM to customers and utility distribution companies? Discuss various energy-efficiency schemes used in DSM.
Model answer: DSM is the planning, implementation and monitoring of utility activities designed to influence customer use of electricity so as to produce desired changes in the load shape (level and timing of demand). Techniques include: peak clipping, valley filling, load shifting, strategic conservation, strategic load growth and flexible load shape. EE schemes include efficient motors/pumps/fans, energy-efficient lighting (T5, CFL/LED), power-factor improvement, ToD (time-of-day) tariffs, efficient air conditioning, VFDs and energy-efficient distribution transformers. Benefits to customers: lower energy bills, improved reliability and power quality. Benefits to utilities: reduced peak demand, deferment of new generation/T&D capacity, lower losses and improved load factor.
DSM reshapes the load curve (clipping, shifting, conservation, etc.); benefits both customer bills and utility capacity/loss performance.
Source: Book EOC
📖 §1.4 Selection and Location of Capacitors
7. a) A 3-phase 415 V 75 kW induction motor draws 48 kW at 0.7 PF. Calculate the capacitor rating to improve PF to 0.95, the reduction in current and kVA reduction at 415 V. b) A plant consumes 2,00,000 kWh/month at 0.9 PF. What is the % reduction in distribution losses if PF is improved to 0.96 at load end?
Model answer: a) kVAr = kW[tan(cos^-1 0.70) - tan(cos^-1 0.95)] = 48 x (1.020 - 0.329) = 33.2 kVAr (say 33 kVAr).
Current at 0.70 PF = 48/(1.732 x 0.415 x 0.70) = 95.4 A; at 0.95 PF = 48/(1.732 x 0.415 x 0.95) = 70.3 A; reduction = 25.1 A (26%).
kVA at 0.70 = 48/0.70 = 68.6 kVA; at 0.95 = 48/0.95 = 50.5 kVA; kVA released = 18.1 kVA.
b) % reduction in distribution loss = [1 - (PF1/PF2)^2] x 100 = [1 - (0.90/0.96)^2] x 100 = 12.1%.
In energy terms, if the existing loss were 4% of 2,00,000 kWh = 8000 kWh/month, the saving is about 970 kWh/month.
Capacitor kVAr = kW(tanφ1 - tanφ2); loss reduction = 1 - (PF1/PF2)².
Source: 15th Exam
📖 §1.4 Power Factor Improvement and Benefits
8. State three advantages of improvement of Power Factor at Load side. Power Factor at the load side is 0.75 and average minimum load is 100 kW. What is the kVAr rating of capacitor to improve the Power Factor at the load side to 0.95?
Model answer: Advantages (any three): reduced kVA (maximum demand) charges in utility bill; reduced distribution losses (kWh) due to lower current; better voltage at motor terminals and improved motor performance; reduction in size of transformers; avoidance of PF penalty / availing PF incentives; better operating efficiency of motors/drives. Capacitor required = 100{tan(cos⁻¹0.75) − tan(cos⁻¹0.95)} = 100(0.882 − 0.329) = 55.3 kVAr, say 55 kVAr.
kVAr = kW[tan(cos⁻¹PF1) − tan(cos⁻¹PF2)].
Source: 17th Sep-2016
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R) + §1.6 Distribution Losses
9. A residential colony with fixed load 250 kVA is 1 km from an 11 kV/415 V transformer. Compare LT (1×3.5c×300sqmm) vs HT (1×3c×70sqmm) distribution. Data: LT cable R=0.13 Ω/km, Rs 700/m; HT cable R=0.570 Ω/km, Rs 1300/m; unit Rs 7/kWh; transformer relocation (HT) Rs 1 lakh. Recommend and estimate payback on marginal investment.
Model answer: Current LT = 250/(0.415×1.732) = 347.8 A; HT = 250/(11×1.732) = 13.1 A. Loss LT = 347.8²×0.13×3/1000 = 47.17 kW; Loss HT = 13.1²×0.57×3/1000 = 0.29 kW. Saving = 46.87 kW. Annual energy saving = 46.87×8760 = 4,10,639 kWh; cost saving = Rs 28,74,470/yr. HT investment = 1300×1000 + 1,00,000 = Rs 14,00,000; LT investment = 700×1000 = Rs 7,00,000. Payback on marginal investment = (14,00,000 − 7,00,000)/28,74,470 = 0.24 yr ≈ 3 months. Recommend HT distribution.
10. a) List five disadvantages of low power factor. b) An industry pays penalty for poor PF of 0.88; utility minimum is 0.9. Penalty = 1% of energy cost for every 0.01 PF below minimum; incentive = 1.5% for every 0.01 improvement above 0.95. Monthly energy bill Rs 6 lakhs. Calculate annual cost saving if PF improved to unity from the current level.
Model answer: a) Disadvantages: large line (copper) losses; large kVA rating and size of electrical equipment; greater conductor size and cost; poor voltage regulation / large voltage drop; low efficiency; PF penalty from supply company. b) From 0.88 to 1.0: penalty avoided = 2×1.0% (0.88→0.90) = 2.0%; incentive above 0.95 = 5×1.5% (0.95→1.00) = 7.5%. Total = 9.5%. Monthly saving = 6,00,000×9.5% = Rs 57,000. Annual = 57000×12 = Rs 6,84,000.
Penalty avoided + incentive earned, applied to monthly bill × 12.
Source: 16th Exam
📖 §1.5 Transformers — losses & efficiency
11. L-1: Choose a 1500 kVA transformer for a load of 500 kVA for 6 h, 1000 kVA for 6 h, 1500 kVA for 12 h. Transformer-1: iron loss 2.7 kW, full-load copper loss 18.1 kW; Transformer-2: iron loss 3.2 kW, full-load copper loss 19.8 kW. (i) Annual cost of losses (365 days, Rs.6/kWh). (ii) If Transformer-1 costs Rs.25,000 more, justify it.
Model answer: Copper loss ∝ (load/rated)². Load fractions: 500/1500=0.333 (6h), 1000/1500=0.667 (6h), 1500/1500=1 (12h). Transformer-1: iron loss/day = 24x2.7 = 64.8 kWh; copper/day = (0.333²x18.1x6)+(0.667²x18.1x6)+(1²x18.1x12) = 12.1+48.3+217.2 = 277.6 kWh; total/yr = (64.8+277.6)x365 = 1,24,976 kWh = Rs.7,49,856. Transformer-2: iron loss/day = 24x3.2 = 76.8; copper/day = 13.2+52.3+237.6 = 303 kWh; total/yr = (76.8+303)x365 = 1,38,663 kWh = Rs.8,31,978. (ii) Annual saving with T-1 = 8,31,978 - 7,49,856 = Rs.82,122. Payback of extra Rs.25,000 = 25000/82122 = 0.3 yr (~4 months) — well justified.
Iron loss is constant (24 h); copper loss scales with load² and operating hours; cost = energy loss x tariff; payback = extra cost / annual saving.
Source: 18th Exam
📖 §1.4 Selection and Location of Capacitors
12. L-5: (a) A 3-phase, 50 kW rated induction motor drawing 44 kW at 0.75 lagging PF. What capacitor (kVAr) per phase is needed to improve PF to 0.96? What is the reduction in current and kVA at 415 V? (b) List five energy losses in an induction motor.
Model answer: (a) tanΦ1 = tan(cos⁻¹0.75) = 0.88; tanΦ2 = tan(cos⁻¹0.96) = 0.29. Required kVAr = 44 x (0.88-0.29) = 25.96 kVAr; per phase = 25.96/3 = 8.65 kVAr. Current at 0.75 PF = 44/(√3 x 0.415 x 0.75) = 81.6 A; at 0.96 PF = 63.76 A; reduction = 17.84 A. kVA at 0.75 = 44/0.75 = 58.67; at 0.96 = 44/0.96 = 45.83; reduction = 12.84 kVA. (b) Five motor losses: 1. Iron (core) loss, 2. Stator I²R (copper) loss, 3. Rotor I²R (copper) loss, 4. Friction and windage loss, 5. Stray load loss.
Capacitor kVAr = P(tanΦ1-tanΦ2); current and kVA from P/(√3·V·PF) and P/PF; standard five induction-motor loss categories.
Source: 18th Exam
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
13. L-1: A food processing plant has contract demand 2500 kVA; average MD 2000 kVA at 0.95 PF; MD billed at Rs.300/kVA; minimum billable MD = 75% of contract demand; incentive 0.5% reduction in energy charges per 0.01 PF increase above 0.95; average monthly energy charge Rs.10 lakhs. Plant improves PF to unity. Determine capacitor kVAr, annual reduction in MD charges and energy charge, and simple payback if capacitors cost Rs.800/kVAr.
Model answer: kW drawn = 2000 x 0.95 = 1900 kW. kVAr = 1900 x (tan(cos⁻¹0.95) - tan(cos⁻¹1)) = 1900 x (0.329 - 0) = 625 kVAr. Capacitor cost = 625 x 800 = Rs.5,00,000. New MD at unity PF = 1900 kVA; but minimum billable = 75% x 2500 = 1875 kVA → billed at 1900 kVA (above floor). Reduction in MD = 2000 - 1900 = 100 kVA → demand saving = 100 x 300 = Rs.30,000/month = Rs.3,60,000/yr. PF rises 0.95→1.00 (0.05 = 5 steps of 0.01) → energy charge reduction = 5 x 0.5% = 2.5%; monthly saving = 10,00,000 x 0.025 = Rs.25,000 → Rs.3,00,000/yr. Total annual saving = 3,60,000 + 3,00,000 = Rs.6,60,000. Payback = 5,00,000/6,60,000 = 0.76 yr ≈ 9 months.
Capacitor kVAr = kW(tanΦ1-tanΦ2); MD saving from reduced kVA × tariff; energy saving from PF-incentive %; payback = investment/annual saving.
14. L-6: A distribution company has: input energy 60 MU, metered billed energy 43 MU, average (un-metered) billing 3 MU, amount billed Rs.540 million, arrears collected Rs.80 million, amount received Rs.470 million. (a) Estimate (i) AT&C loss % and revenue realised (Rs./kWh); (ii) revenue loss per kWh and monthly loss if purchased energy cost is Rs.8.10/kWh. (b) List five measures to reduce commercial loss.
Model answer: (a)(i) Billing efficiency = (43+3)/60 x 100 = 76.7%. Collection efficiency = (470-80)/540 x 100 = 72.2%. AT&C loss = [1 - (billing eff x collection eff)] x 100 = [1 - (0.767 x 0.722)] x 100 = 44.62%. Revenue realised = (470-80)/60 = Rs.6.5/kWh. (ii) Revenue loss = 8.10 - 6.5 = Rs.1.6/kWh; monthly loss = 60 MU x 1.6 = Rs.96 million (Rs.9.6 crore). (b) Measures to reduce commercial loss: 1. Accurate metering of all consumers (replace defective/electromechanical meters, AMR/smart meters). 2. Detect and curb theft/pilferage and unauthorised connections; energy audit/feeder metering. 3. Improve billing — eliminate un-metered/average billing, correct meter-reading and billing errors. 4. Strengthen collection efficiency — disconnection drives, online payment, recovery of arrears. 5. HVDS / spot billing / consumer indexing and GIS mapping to plug gaps.
AT&C loss = 1-(billing eff x collection eff); revenue realised = net received/input units; revenue loss = purchase cost - realised; commercial-loss measures from Book-3 Chapter 1 (paper cites page 27).
Source: 19th Exam
📖 §1.1 Cascade Efficiency
15. L-4: A 10 MW co-gen plant runs at 85% daily load factor, generating at 11 kV. 35% exported to grid via 7.5 MVA transformer (99% eff); 32% to mill motors at 600 V via 5 MVA transformer (98% eff); balance to LT loads/auxiliaries at 415 V via 2 MVA transformer (98% eff). Calculate: (1) daily energy exported; (2) daily mill motor consumption; (3) daily LT/auxiliary consumption; (4) daily transformer losses (kWh and %).
Model answer: Daily generation = 10,000 x 0.85 x 24 = 2,04,000 kWh. (1) Export = 2,04,000 x 0.35 = 71,400 kWh; 7.5 MVA loss = 71,400 x 0.01 = 714 kWh; net export = 70,686 kWh. (2) Mill = 2,04,000 x 0.32 = 65,280 kWh; 5 MVA loss = 65,280 x 0.02 = 1,306 kWh; net = 63,974 kWh. (3) LT/aux = 2,04,000 x 0.33 = 67,320 kWh; 2 MVA loss = 67,320 x 0.02 = 1,346 kWh; net = 65,974 kWh. (4) Total transformer losses = 714 + 1,306 + 1,346 = 3,366 kWh/day; % loss = 3,366/2,04,000 x 100 = 1.65%.
Printed solution: net export 70,686 kWh, net mill 63,974 kWh, net LT 65,974 kWh, total tx loss 3,366 kWh = 1.65%.
Source: Sep 2019
📖 §1.4 Power Factor Improvement and Benefits
16. L-6: (a) List five disadvantages of low power factor. (b) An industry maintains poor PF of 0.88; utility minimum is 0.9. Penalty 1% on energy cost for every 0.01 below minimum; incentive 1.5% for every 0.01 above 0.95. Monthly energy bill Rs.6 lakhs. Calculate annual cost saving if PF improved to unity from current level.
Model answer: (a) Disadvantages of low PF: (1) larger line/copper losses; (2) larger kVA rating and size of equipment; (3) greater conductor size and cost; (4) poor voltage regulation/large voltage drop; (5) low efficiency; (6) penalty from supply company. (b) Penalty avoided: 0.88 to 0.90 = 0.02 PF x 1% = 2.0%. Incentive at unity: from 0.95 to 1.00 = 0.05 x 1.5% = 7.5%. Total energy saving = 2.0% + 7.5% = 9.5%. Monthly cost reduction = 6 lakh x 9.5% = Rs.57,000. Annual = 57,000 x 12 = Rs.6,84,000.
Printed solution: total benefit 9.5%, monthly Rs.57,000, annual Rs.6,84,000.
17. Fill in the blanks: a) Heat Rate of a thermal power plant is expressed in ____. b) With increase in design speed of induction motors, the required capacitive kVAr for reactive power compensation for the same capacity range will ____. c) An air dryer in a compressed air system reduces ____ point of air. d) A Pitot tube measures the difference between ____ and ____ pressures of the fluid. e) The friction loss in a pipe carrying a fluid is proportional to the ____ power of pipe diameter.
Model answer: a) kCal/kWh b) decrease c) dew (dew point) d) total and static e) fifth
Standard BEE definitions: heat rate in kCal/kWh; higher-speed motors need less magnetizing kVAr; air dryer lowers dew point; Pitot tube reads total minus static pressure; pipe friction loss ∝ 1/d⁵ (fifth power of diameter).
Source: 9th Dec-2009
📖 §1.5 Transformers — losses & efficiency
18. a) A small scale industry has a constant load of 380 kVA. It has installed two transformers of 500 kVA each. The no load loss and full load copper loss of each 500 kVA transformer is 750 W and 5410 W respectively. From the energy efficiency point of view should the industry operate a single transformer or two transformers equally sharing the load? b) A no load test on a three phase delta connected induction motor gave: No load power = 890 W, Stator resistance per phase at 30°C = 0.233 Ohms, No load current = 14.5 A. Calculate the fixed losses for the motor.
Model answer: a) Single 500 kVA at 380 kVA load: loss = 750 + (380/500)² × 5410 = 750 + 3124.8 = 3874.8 W. Two transformers each at 190 kVA: loss = 2 × [750 + (190/500)² × 5410] = 2 × 1531.2 = 3062.9 W. Two transformers are better — losses are least, saving 812.4 W. b) Stator copper loss at no load = 3 × (14.5/√3)² × 0.233 = 48.985 W. Fixed losses = 890 − 48.985 = 841 W.
a) Compare total transformer losses (no-load + load-proportional copper loss) for single vs paralleled operation. b) Fixed (iron + friction + windage) losses = no-load input power minus no-load stator copper loss.
Source: 9th Dec-2009
📖 §1.4 Power Factor Improvement and Benefits — multi-chapter fill-in-the-blanks
19. Fill in the blanks: (a) If the reactive power drawn by a load is zero, the load operates at __ power factor. (b) Power factor is the ratio of ___. (c) As the approach increases (other parameters constant), the effectiveness of a cooling tower ___. (d) Lower power factor of a DG set demands ___ excitation currents. (e) If voltage applied to a 415 V rated capacitor drops by 5%, its VAr output drops by about ___%.
20. A unit has 2 identical 500 kVA transformers, each with no-load loss 800 W and full-load copper loss 5000 W. Plant load is 400 kVA. Compare transformer losses for single transformer operation vs two transformers in parallel. Also list any five options to minimise electrical distribution loss.
Model answer: Single transformer loss = 800 + (400/500)²×5000 = 4000 W. Two transformers (each sharing 200 kVA): 2×[800 + 5000×(200/500)²] = 3200 W. Single-transformer operation has 800 W higher loss. Distribution-loss options: relocate transformers/substations near load centres; re-route/re-conductor high-loss feeders; PF improvement with capacitors at load end; optimum loading of transformers; use lower-resistance AAAC instead of ACSR; minimise losses at weak links (jumpers, loose contacts, brittle conductors); improve HT:LT ratio to shorten LT network.
Loss = iron loss + (load/rating)²×copper loss; parallel operation halves the load share, reducing total copper loss. Any five distribution-loss options accepted.
Source: 10th Jul-2010
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
21. L-1: The contract demand of a process plant is 6000 kVA. The average monthly recorded maximum demand is 5500 kVA at 0.78 PF. Tariff: (a) Minimum monthly billing demand is 75% of contract demand or actual recorded MD whichever is higher; no PF incentives. (b) Monthly MD charge is Rs. 400 per kVA. Find the optimum limit of PF capacitor requirement (purely to reduce MD so no excess demand charges are paid) and the simple payback period, assuming capacitor + APFC controller cost is Rs. 500 per kVAr.
Model answer: Minimum payable demand = 6000 x 0.75 = 4500 kVA. Margin for MD reduction = 5500 - 4500 = 1000 kVA. Present maximum load = 5500 x 0.78 = 4290 kW. Desired peak PF to achieve MD of 4500 kVA = 4290/4500 = 0.9533. PF capacitor requirement = 4290 [tan(Cos-1 0.78) - tan(Cos-1 0.9533)] = 4290(tan 38.74 - tan 17.579) = 4290(0.80226 - 0.316815) = 4290(0.4854) = 2083 kVAr. Cost of capacitor installation = 500 x 2083 = Rs. 10.4 lakhs. Monthly MD saving = 1000 kVA; Yearly savings = 1000 x 400 x 12 = Rs. 48.0 lakhs. Simple payback = 10.4/48 = 0.21 years = 2.6 months.
Reduce MD to minimum billable 4500 kVA; required kVAr from tan(phi1)-tan(phi2) at the active load of 4290 kW; payback = investment/annual MD savings.
Source: 11th Feb-2011
📖 §1.4 Power Factor Improvement and Benefits + §1.5 Transformer losses & efficiency
22. a) A cold rolling mill has a maximum demand of 7 MVA at power factor of 0.95. The plant management converts the existing electrical resistance annealing furnace having steady load of 1250 kW to gas heating as a cost reduction measure. The existing capacitor banks (kVAr) continued to be in the electrical network. What will be the effect on maximum demand and power factor due to this conversion? (5 Marks) b) A cement plant has a constant load of 15 MVA. It has installed two transformers of 30 MVA each. The no load loss and full load copper loss of each 30 MVA transformer is 25 kW and 75 kW respectively. From the energy efficiency point of view the industry management wants to take a decision on whether to operate a single transformer or two transformers equally sharing the load. What is your recommendation? (5 Marks)
Model answer: a) Registered max demand = 7 MVA = 7000 kVA. Electrical real load = 7000 x 0.95 = 6650 kW. Existing kVAr = sqrt(kVA^2 - kW^2) = sqrt(7000^2 - 6650^2) = 2186 kVAr (remains in network). Reduction in real power = 1250 kW (furnace converted to gas). Revised real power = 6650 - 1250 = 5400 kW. Revised kVA = sqrt(5400^2 + 2186^2) = 5825 kVA. Reduction in electrical demand = 7000 - 5825 = 1175 kVA. Revised power factor = 5400/5825 = 0.927. Reduction in PF = 0.95 - 0.927 = 0.023. b) Option 1 (one transformer): % load = 15/30 = 50%; Total loss = NoLoad + Cu loss x (%load)^2 = 25 + 75 x 0.5^2 = 43.75 kW. Option 2 (both transformers): each at 7.5/30 = 25%; Total loss = [25 + 75 x 0.25^2] x 2 = 59.37 kW. Recommendation: operate single transformer because losses are lower, saving 59.37 - 43.75 = 15.62 kW.
kVAr from kVA & kW; revised demand after load removal; transformer loss = NL + FL_Cu x (loading)^2.
Source: Mar 2021
📖 §1.4 Power Factor Improvement and Benefits + §1.5 Transformer losses & efficiency
23. a) A cold rolling mill has a maximum demand of 7 MVA at a power factor of 0.95. The plant management converts the existing electrical resistance annealing furnace having steady load of 1250 kW to gas heating as a cost reduction measure. The existing capacitor banks (kVAr) continued to be in the electrical network. What will be the effect on maximum demand and power factor due to this conversion? (5 marks). b) A cement plant has a connected load of 15 MVA. It has installed two transformers of 30 MVA each. The no load loss and full load copper loss of each 30 MVA transformer is 25 kW and 75 kW respectively. From the energy efficiency point of view the industry management wants to take a decision on whether to operate a single transformer or two transformers equally sharing the load. What is your recommendation? (5 marks).
Model answer: a) Registered maximum demand = 7 MVA = 7000 kVA. Electrical load (real power) = 7000 x 0.95 = 6650 kW. kVAr = sqrt(kVA^2 - kW^2) = sqrt(7000^2 - 6650^2) = 2186 kVAr. kVAr in the plant will remain same. After converting 1250 kW furnace to gas, real power reduces by 1250 kW: Revised Real Power = 6650 - 1250 = 5400 kW. Revised kVA = sqrt(kW^2 + kVAr^2) = sqrt(5400^2 + 2186^2) = 5825 kVA. Reduction in electrical demand = 7000 - 5825 = 1175 kVA. Revised power factor = 5400/5825 = 0.927. Reduction in power factor = 0.95 - 0.927 = 0.023 (power factor drops). b) Option 1 - One transformer in operation: % load = 15/30 = 50%; Total Loss = No-load loss + Copper loss x (%load)^2 = 25 + 75 x (0.5)^2 = 25 + 18.75 = 43.75 kW. Option 2 - Both transformers in operation: % load = 7.5/30 = 25% each; Total Loss = [No-load loss + Copper loss x (%load)^2] x 2 = [25 + 75 x (0.25)^2] x 2 = [25 + 4.69] x 2 = 59.37 kW. Recommendation: Operate single transformer because losses are less; saving of 59.37 - 43.75 = 15.62 kW.
Part a: kVAr fixed; removing real load lowers kW so kVA falls but PF reduces. Part b: total transformer loss = no-load loss + copper loss x (load fraction)^2; compare one vs two transformers.
24. A DISCOM has taken initiatives to reduce Aggregate Technical & Commercial (AT&C) losses in their network. The energy supplied, received and revenue details are given below — Input energy: 50 MU; Billed Energy (Metered): 39 MU; Billed Energy (Un-metered): 2 MU; Amount Billed: Rs. 470 Million; Amount Collected: Rs. 30 Million; Gross Amount collected: Rs. 390 Million. a) Estimate the AT&C losses (in %). b) List any four strategies to reduce the commercial losses.
Model answer: a) Billing efficiency = total units billed / total input units = (39 + 2)/50 x 100 = 82.0%.
Collection efficiency = amount collected excluding arrears / amount billed. Stripping the Rs.30 million of arrears out of the Rs.390 million gross collection: Ac = 390 - 30 = Rs.360 million, so CE = 360/470 x 100 = 76.6%.
AT&C loss = [1 - (BE x CE)] x 100 = [1 - (0.820 x 0.766)] x 100 = [1 - 0.6281] x 100 = 37.2%.
b) Four measures to reduce commercial losses (Book-3 Sec.1.8): (1) accurate metering with a planned meter-replacement programme and meters matched to the connected load; (2) installation of electronic meters with TOD, tamper-proof and remote/data-reading facility; (3) intensive inspections and eradication of theft/pilferage; (4) compulsory metering in place of average billing, backed by energy audit to pinpoint high-loss areas and by improved collection/arrear recovery.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
25. A review of electricity bills of a process plant was conducted as a part of energy audit. The plant has a contract demand of 3000 kVA with the power supply company. The average maximum demand of the plant is 2400 kVA/month at a power factor of 0.95. The maximum demand is at 80% of the contract demand. The minimum billable maximum demand is 80% of the contract demand. An incentive of 0.5 % reduction in energy charges component of electricity bill are provided for every 0.01 increase in power factor over and above 0.95. The average energy charge component of the electricity bill per month for the plant is Rs.80 lakhs. Calculate the following: a) If the plant decides to improve the power factor to unity, determine the power factor capacitor kVAr required and the associated monetary benefits. b) What will be the simple payback period if the cost of power factor capacitors is Rs.1200/kVAr.
Model answer: Contract demand 3000 kVA; recorded average MD = 2400 kVA at 0.95 PF; minimum billable demand = 80% x 3000 = 2400 kVA.
a) kW drawn = 2400 x 0.95 = 2280 kW.
kVAr for 0.95 -> unity = kW[tan(cos^-1 0.95) - tan(cos^-1 1)] = 2280 x (0.3287 - 0) = 749 kVAr (say 750 kVAr).
Maximum demand at unity PF = 2280 kVA, but the minimum billable demand is 2400 kVA, so the plant still pays for 2400 kVA - there is NO saving in maximum demand charges.
PF incentive = (1.00 - 0.95)/0.01 x 0.5% = 2.5% of the energy charge = Rs.80,00,000 x 2.5% = Rs.2,00,000/month = Rs.24,00,000/year.
b) Investment = 749 kVAr x Rs.1200/kVAr = Rs.8,99,000 (say Rs.9.0 lakh).
Simple payback = 8,99,000 / 24,00,000 = 0.375 year = about 4.5 months.
(Same structure as the Book-3 Sec.1.11 solved example: the minimum-billing-demand clause can wipe out the MD saving, leaving the PF incentive as the only benefit.)
kVAr = kW(tanφ1-tanφ2); MD reduction nil due to 80% minimum billing; energy charge incentive 0.5% per 0.01 PF rise above 0.95 → 2.5%; payback = investment/annual savings.
Source: Jul 2022
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
26. L-1(A): A trivector-meter installed in a steel plant is monitoring the maximum demand with a demand interval of 15 min. The observed maximum demand during one demand interval is given (3 min: 9634 kVA, 4 min: 10257 kVA, 3 min: 8436 kVA, 5 min: 9847 kVA). Calculate: (a) Recorded maximum demand during the cycle; (b) Demand reduction and capacitor kVAr required for improving power factor to 0.99 from average observed power factor of 0.92.
Model answer: (a) Maximum demand is the time-integrated kVA over the 15-minute demand interval:
MD = [(9634 x 3) + (10,257 x 4) + (8436 x 3) + (9847 x 5)] / 15 = [28,902 + 41,028 + 25,308 + 49,235] / 15 = 144,473/15 = 9631.5 kVA.
(b) At the observed average PF of 0.92, kW = 9631.5 x 0.92 = 8861 kW (capacitors do not change kW).
New demand at 0.99 PF = 8861/0.99 = 8950.5 kVA, so demand reduction = 9631.5 - 8950.5 = 681 kVA.
Capacitor kVAr = kW[tan(cos^-1 0.92) - tan(cos^-1 0.99)] = 8861 x (0.4260 - 0.1425) = 8861 x 0.2835 = 2512 kVAr (say 2500 kVAr, switched through an APFC).
MD = time-weighted average of interval readings; convert to kW via pf; recompute kVA at improved pf; kVAr = kW(tanφ1−tanφ2).
Source: Mar 2023
📖 §1.11 Analysis of Electrical Power Systems (Table 1.11) — multi-chapter True/False
27. L-6: State True or False (1 mark each):
1. The efficiency of gas turbine power plant is lower than that of a combined cycle power plant.
2. The performance of air compressor at high altitudes will be lower as compared to that at sea level.
3. Efficiency of transformer will be minimum when copper loss is equal to iron losses.
4. In cooling towers, the water droplets entrapped in the air stream is captured by drift eliminators.
5. To get the static pressure, the inner and outer tubes of pitot tube are connected to manometer.
6. The throttling of pump discharge will change the pump characteristic curve.
7. The simplest way to reduce the discharge from a reciprocating air compressor is to throttle it.
8. Cycle of Concentration (COC) is the ratio of dissolved solids in circulating water to the dissolved solids in makeup water.
9. Use of VFD will save power but also create harmonics.
10. The synchronous speed of a 4 pole motor will be 3000 rpm.
Model answer: 1. True
2. True
3. False (transformer efficiency is MAXIMUM when copper loss equals iron loss)
4. True
5. False (inner and outer tubes connected to the manometer give velocity pressure; static pressure is from the outer tube alone)
6. False (throttling changes the system curve, not the pump characteristic curve)
7. False (throttling a reciprocating compressor is not the way to reduce discharge; use unloading/speed control)
8. True
9. True
10. False (Ns = 120 x 50 / 4 = 1500 rpm)
Cross-chapter True/False. The two Chapter-1 items are (3) and (9): transformer efficiency is MAXIMUM (not minimum) when copper loss equals iron loss, because the variable copper loss then just equals the fixed core loss (Book-3 §1.5); and VFDs do save power but, being non-linear power-electronic loads, they inject harmonic currents (Book-3 §1.10). Item 10 is settled by Ns = 120f/P = 120 x 50/4 = 1500 rpm, not 3000 rpm.