161 questions — 127 objective (1 mark), 21 short (5 marks), 13 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).
1. The synchronous speed of a 6-pole, 50 Hz induction motor is:
1500 rpm
1000 rpm
750 rpm
3000 rpm
Answer: B) 1000 rpm
Confirmed vs Book-3 §2.3 — Ns = 120 × f / P = 120 × 50 / 6 = 1000 rpm. The book's 50 Hz table lists 2 poles = 3000, 4 = 1500, 6 = 1000, 8 = 750 rpm. Option (a) 1500 rpm is the 4-pole answer — the usual slip of the pen when the pole count is misread.
Source: AI practice
📖 §2.3 Motor Characteristics — Slip
2. A 4-pole, 50 Hz induction motor runs at 1440 rpm. Its slip is:
2%
4%
6%
0%
Answer: B) 4%
Confirmed vs Book-3 §2.3 — first Ns = 120 × 50 / 4 = 1500 rpm, then Slip (%) = [(Ns − Full load speed) / Ns] × 100 = [(1500 − 1440)/1500] × 100 = 60/1500 = 4%. Option (a) 2% would correspond to 1470 rpm; the slip must always be taken as a fraction OF THE SYNCHRONOUS speed, not of the running speed.
Source: AI practice
📖 §2.4 Motor Efficiency / §2.7 Motor Loading
3. The nameplate kW rating of an induction motor represents its:
input (electrical) power
output (shaft) power
reactive power
apparent power
Answer: B) output (shaft) power
Confirmed vs Book-3 §2.4 — efficiency is defined as the mechanical energy delivered at the rotating shaft divided by the electrical energy input at the terminals, so the nameplate kW is the rated OUTPUT (shaft) power. The book's loading formula confirms it: rated input kW = nameplate full-load kW ÷ nameplate full-load efficiency. Option (a) is wrong because a 75 kW, 90% motor draws 75/0.9 = 83.3 kW at the terminals.
Source: AI practice
📖 §2.4 Motor Efficiency
4. A 90% efficient motor with a nameplate rating of 45 kW, running at full load, draws an input power of about:
40.5 kW
45 kW
50 kW
54 kW
Answer: C) 50 kW
Confirmed vs Book-3 §2.4 — η = P_out / P_in, so P_in = P_out / η = 45 / 0.90 = 50 kW. The nameplate 45 kW is shaft output, so the input must always be LARGER than the nameplate rating. Option (a) 40.5 kW multiplies output by efficiency instead of dividing, which would imply the motor generates power.
Source: AI practice
📖 §2.6 Energy Efficient Motors — loss breakdown
5. The largest single component of loss in a typical induction motor is the:
iron (core) loss, 20–25%
stator + rotor I²R (copper) loss, 55–60%
friction and windage loss, 8–12%
stray load loss, 4–5%
Answer: B) stator + rotor I²R (copper) loss, 55–60%
Confirmed vs Book-3 §2.6 — the book's loss table gives stator + rotor I²R = 55–60%, core (iron) = 20–25%, friction & windage = 8–12%, stray load = 4–5% of total losses. Copper loss is therefore the dominant component and it is load-variable (∝ current²), which is why part-load running hurts efficiency. Option (a) iron loss is the largest FIXED loss but only about a quarter of the total.
Source: AI practice
📖 §2.6 Design Improvements in EEMs / core loss
6. Eddy-current loss in a motor's core is reduced primarily by:
using thinner laminations
adding more copper
a larger cooling fan
a smaller air gap
Answer: A) using thinner laminations
Confirmed vs Book-3 §2.6 — core loss is hysteresis plus eddy current; the book states that eddy currents are reduced by THINNER LAMINATIONS, while hysteresis is reduced by lower-loss silicon steel and lower flux density. Option (d) a smaller air gap is indeed an EEM design feature, but the book credits it with reducing the magnetizing current (and hence I²R), not the eddy-current loss.
Source: AI practice
📖 §2.7 Improving Motor Loading by Operating in Star Mode
7. For motors that consistently run below 40% of rated load, an energy-saving measure is to:
run them in delta connection
re-wire them to permanent star connection
increase the supply voltage
add a soft starter
Answer: B) re-wire them to permanent star connection
Confirmed vs Book-3 §2.7 — for motors consistently operating below 40% of rated capacity the book recommends changing from standard delta to PERMANENT STAR by re-configuring the terminal box wiring and resetting the over-current relay. Star reduces the winding voltage by a factor of √3 and electrically downsizes the motor to 1/3, so it then runs near full load with higher efficiency and PF. Option (d) is wrong because a soft starter only controls the START; it does nothing for a continuously under-loaded running motor.
Source: AI practice
📖 §2.7 Improving Motor Loading by Operating in Star Mode
8. A 24 kW (delta-connected) motor that is permanently re-wired to star mode is effectively derated to about:
24 kW
14 kW
8 kW
12 kW
Answer: C) 8 kW
Confirmed vs Book-3 §2.7 — in star mode the motor is electrically downsized to 1/3 of its delta rating, so 24 ÷ 3 = 8 kW. The book's own example is a 15 kW delta motor derated to 5 kW in star, with the warning that the load must never cross the derated capacity. Option (b) 14 kW is the trap of dividing by √3 (the VOLTAGE factor) instead of by 3 (the POWER factor).
Source: AI practice
📖 §2.9 Speed Control of Motors — Soft Starter
9. A soft starter is correctly described as a device that:
improves the steady-state running efficiency of the motor
reduces starting inrush current and mechanical stress on the drive train
varies the running speed of the motor by changing the supply frequency
corrects the motor power factor by switching capacitor banks
Answer: B) reduces starting inrush current and mechanical stress on the drive train
Confirmed vs Book-3 §2.9 — the book explains that rapid acceleration draws inrush currents of about +600% of normal run current and transfers excess starting torque into chains, belts, gears and seals; a soft starter gives a controlled, stepless release of power, so its listed advantages are less mechanical stress, improved power factor, lower maximum demand and less mechanical maintenance. It acts only during starting, so option (a) is wrong — running efficiency is unchanged; option (c) describes a VFD, not a soft starter.
Source: AI practice
📖 §2.7 Power Supply Quality — Table 2.3 (voltage variation)
10. Motor torque is proportional to the square of the supply voltage. A 10% drop in voltage reduces the torque by about:
10%
19%
21%
5%
Answer: B) 19%
Confirmed vs Book-3 §2.7 — Table 2.3 lists starting and maximum running torque as a function of (voltage)², and gives −19% at 90% voltage. Working: T ∝ V², so at 0.9 p.u. the torque is (0.9)² = 0.81 p.u., i.e. a fall of about 19%. Option (c) 21% is the mirror-image error of using 1.1² − 1 (the +21% figure the same table gives for 110% voltage).
Source: AI practice
📖 §2.7 Power Supply Quality — BIS voltage/frequency tolerance
11. Per BIS, an induction motor should deliver its rated output within a supply voltage and frequency tolerance of:
±10% V, ±5% f
±6% V, ±3% f
±3% V, ±6% f
±5% V, ±5% f
Answer: B) ±6% V, ±3% f
Confirmed vs Book-3 §2.7 — the book states the BIS standard: a motor shall deliver its rated output with a voltage variation of ±6% and a frequency variation of ±3%, and notes that Indian supply fluctuations commonly exceed this. Do not confuse this with the ±10% supply-voltage limits used elsewhere; the pairing to memorise is 6 for volts, 3 for hertz.
Source: AI practice
📖 §2.4 Field Tests — Stator I²R losses, temperature correction of resistance
12. The stator resistance of a motor is 0.30 Ω at 30°C. Corrected to 120°C (copper constant = 235), it becomes:
0.30 Ω
0.34 Ω
0.40 Ω
0.45 Ω
Answer: C) 0.40 Ω
Confirmed vs Book-3 §2.4 — the book gives R₂/R₁ = (235 + t₂)/(235 + t₁) for copper. Working: (235 + 120)/(235 + 30) = 355/265 = 1.34, so R₂ = 0.30 × 1.34 = 0.402 ≈ 0.40 Ω. The book notes modern motors run at 100–120 °C, so correcting only to 75 °C understates the resistance. Option (a) 0.30 Ω ignores the correction entirely and would under-state the I²R loss by about a third.
Source: AI practice
📖 §2.2 Motor Types
13. Which of the following best describes an induction motor's operation?
It uses direct current to create mechanical energy.
It generates a rotating magnetic flux that induces current in the rotor.
It operates synchronously with the AC supply frequency.
It requires external excitation to operate.
Answer: B) It generates a rotating magnetic flux that induces current in the rotor.
Confirmed vs Book-3 §2.2 Motor Types — A 3-phase stator winding sets up a flux of constant magnitude rotating at synchronous speed; as it sweeps the stationary shorted rotor bars it induces an e.m.f (Faraday) and hence rotor current (Lenz), which produces torque. Option (c) is the tempting wrong choice: the rotor never catches the stator field, it always runs below synchronous speed with slip > 0 — only the synchronous motor runs at supply speed, and only it needs external DC excitation (d).
Source: Sep 2024
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
14. How does operating a motor in star mode affect its performance?
It reduces the voltage and derates the motor capacity.
It increases motor speed.
It improves the power factor at high loads.
It eliminates the need for external capacitors.
Answer: A) It reduces the voltage and derates the motor capacity.
Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — Changing from delta to permanent star reduces the winding voltage by a factor of √3 and electrically downsizes the motor to about 1/3 of its delta rating (a 15 kW delta motor becomes 5 kW in star), so a chronically <40 % loaded motor then operates near full load with better efficiency and power factor. (b) is wrong — speed actually drops slightly in star mode, which is why the book warns against it where output depends on motor speed.
Source: Sep 2024
📖 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD
15. In a VFD-controlled motor, what happens when the supply frequency is reduced while maintaining the same voltage?
The motor speed increases.
The motor efficiency improves.
The motor draws higher current and may overheat.
The motor torque decreases.
Answer: C) The motor draws higher current and may overheat.
Confirmed vs Book-3 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD — The book states that when supply frequency is reduced the equivalent circuit impedance falls, so the motor draws higher current and the flux rises towards saturation; that is why V and f must be varied together at a constant ratio (V/f control). (d) is the trap — with correct V/f the torque stays almost constant up to base speed; torque only falls above base speed due to field weakening.
Source: Sep 2024
📖 §2.9 Soft Starter
16. Which of the following describes the function of a soft starter in a motor system?
It increases the motor's full-load speed.
It converts AC power to DC power.
It reduces the inrush current during motor start-up.
It improves the motor's efficiency at low speeds.
Answer: C) It reduces the inrush current during motor start-up.
Confirmed vs Book-3 §2.9 Soft Starter — A soft starter delivers a controlled release of power giving smooth, stepless acceleration, so it limits the direct-on-line inrush that can reach about +600 % of normal run current. (d) is wrong — it is a starting device, not a speed controller, and gives no running-efficiency gain at low speed.
Source: Sep 2024
📖 §2.4 Motor Efficiency
17. A motor operates at 75% load with an efficiency of 88%. If the motor's rated power is 20 kW, what is the actual output power?
15 kW
13.2 kW
17.6 kW
14.4 kW
Answer: A) 15 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Nameplate kW is the rated shaft output, so at 75 % loading the output is 0.75 × 20 = 15 kW. (b) 13.2 kW is the tempting error — it multiplies the output again by the 88 % efficiency, but efficiency relates output to electrical input, it does not reduce the shaft output.
Source: Sep 2024
📖 §2.4 Motor Efficiency
18. What is the efficiency of motor with the following nameplate details 22 kW, 415V, 42 A, 0.8 p.f, 1475 rpm?
94.5%
91%
89.9%
None of the above
Answer: B) 91%
Confirmed vs Book-3 §2.4 Motor Efficiency — Rated input = √3 × 415 × 42 × 0.8 = 24.15 kW, so rated efficiency = 22 / 24.15 = 91 %. (c) 89.9 % and (a) 94.5 % do not follow from the nameplate data; the common slip is to forget the √3 or the power factor in the input calculation.
Source: Sep 2024
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
19. Iron losses in an electric motor can be reduced by using
More copper and large conductors
Use of thinner gauge lower loss core steel
Use of low loss fan design
Optimised design and strict quality control
Answer: B) Use of thinner gauge lower loss core steel
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — Table 2.2 attributes iron-loss reduction to thinner gauge, lower-loss core steel (less eddy current) plus a longer core to lower flux density. (a) more copper / larger conductors reduces stator I²R loss, not iron loss, and (c) low-loss fan design attacks friction & windage — both are the tempting confusions.
Source: Sep 2025
📖 §2.5 Motor Selection
20. Energy savings by motor replacement can be worked out by:
KW output (ηold – ηnew)
KW output (ηnew – ηold)
KW output (1/ηold – 1/ηnew)
KW output (1/ηnew – 1/ηold)
Answer: C) KW output (1/ηold – 1/ηnew)
Confirmed vs Book-3 §2.5 Motor Selection — The book gives kW savings = kW output × [1/ηold − 1/ηnew]; savings arise from the difference in the INPUT power drawn for the same shaft output. (a)/(b), the plain difference of efficiencies, is the tempting wrong form — it has the wrong units and grossly under-states savings.
Source: Sep 2025
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
21. Stray losses in a motor are mainly caused by:
Leakage flux induced by load currents
Hysteresis and eddy currents
Frictional losses
Copper winding losses
Answer: A) Leakage flux induced by load currents
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — Stray load losses are caused by leakage flux induced by the load currents in the laminations, vary as the square of load current and account for 4–5 % of total losses. (b) hysteresis and eddy currents are the core (iron) losses, a different, load-independent category — that is the usual confusion.
Source: Sep 2025
📖 §2.7 Voltage Unbalance
22. Voltage unbalance in motors:
Reduces motor temperature
Increases motor slip
Causes excessive heating and reduces life
Improves torque
Answer: C) Causes excessive heating and reduces life
Confirmed vs Book-3 §2.7 Voltage Unbalance — Voltage unbalance produces a current unbalance 6–10 times as large and an additional temperature rise of 2 × (% unbalance)²; insulation life halves for every 10 °C rise, so it is a leading cause of premature motor failure. (b) is the trap — the harm is overheating and derating, not a useful change in slip; the book recommends unbalance at motor terminals not exceed 1 %.
Source: Sep 2025
📖 §2.9 Soft Starter
23. Soft starters are used to:
Increase motor speed
Reduce inrush current
Convert AC to DC
Improve efficiency
Answer: B) Reduce inrush current
Confirmed vs Book-3 §2.9 Soft Starter — Soft starters ramp the voltage to give smooth acceleration, cutting the DOL inrush of about +600 % of run current and the associated mechanical stress. (d) is the tempting answer — the listed advantages are less mechanical stress, improved power factor, lower maximum demand and less maintenance, not improved running efficiency.
Source: Sep 2025
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
24. An air compressor is driven by an IE3 premium efficiency motor. Compared to an IE2 motor of the same rating, which of the following statements is most accurate?
The IE3 motor will always consume less power under all load conditions.
The IE3 motor achieves higher efficiency mainly by reducing copper and iron losses.
The IE3 motor has lower inrush current during starting compared to IE2.
The IE3 motor achieves efficiency by increasing slip.
Answer: B) The IE3 motor achieves higher efficiency mainly by reducing copper and iron losses.
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — Design improvements in energy-efficient motors target the dominant loss groups: stator + rotor I²R (55–60 % of losses, reduced by more/larger copper conductors) and core losses (20–25 %, reduced by thinner low-loss silicon steel and a longer core). (c) is wrong — EEMs do not have lower inrush; (d) is wrong because EEMs have LOWER slip (about 1 % faster than standard motors).
Source: Sep 2025
📖 §2.3 Motor Characteristics
25. A 4-pole, 50 Hz induction motor runs at 1470 rpm. Slip is:
0.02
0.20
0.25
0.30
Answer: A) 0.02
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 4 = 1500 rpm; slip = (1500 − 1470)/1500 = 0.02 (2 %). (b) 0.20 is the tempting decimal-point error — an induction motor running that far below synchronous speed would be grossly overloaded.
Source: Sep 2025
📖 §1.1 (a.c. fundamentals; see also Book-3 Ch-2 §2.2 — Ns = 120f/P)
26. 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
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
27. With a decrease in speed of the motor, the required capacitive kVAr
increases
decreases
does not change
none of the above
Answer: A) increases
Corrected (was b) — Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings: the book states that the required capacitive kVAr INCREASES with a decrease in speed of the motor, because the magnetizing current of a low-speed (many-pole) motor is larger than that of a high-speed motor of the same HP. Table 2.5 shows this directly — for a 50 HP motor the capacitor rating rises from 10 kVAr at 3000 rpm to 22 kVAr at 500 rpm. (b) 'decreases' is exactly the reversed reading of the table.
Source: Book EOC
📖 §2.7 Power Supply Quality — Table 2.3 Effect of Voltage & Frequency Variation
28. Reduction in supply voltage by 10% will change the torque of the motor by
38%
19%
9.5%
no change
Answer: B) 19%
Confirmed vs Book-3 §2.7 Power Supply Quality — Table 2.3 Effect of Voltage & Frequency Variation — Torque varies as (voltage)², so at 90 % voltage the torque is 0.9² = 0.81 of rated, i.e. a 19 % reduction — the figure printed in Table 2.3 for starting and maximum running torque at 90 % voltage. (c) 9.5 % is the trap of assuming torque falls linearly with voltage instead of with its square.
Source: Book EOC
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
29. One low-investment measure to improve the efficiency of a squirrel cage induction motor which operates consistently below 40% of its rated capacity is by
operating it in star mode
replacing it with a correctly sized motor
operating in delta mode
none of the above
Answer: A) operating it in star mode
Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — For motors consistently operating below 40 % of rated capacity the book recommends permanent star operation — an inexpensive measure needing only re-wiring at the terminal box and resetting the overload relay. (b) replacing with a correctly sized motor also works but is a HIGH-investment option, which is why it is not the answer to a 'low investment' question.
Source: Book EOC
📖 §2.2 Motor Types
30. In an induction motor, the magnetic field is established in
stator winding only
rotor winding only
stator and rotor windings
none of the above
Answer: C) stator and rotor windings
Confirmed vs Book-3 §2.2 Motor Types — The stator sets up the rotating field and the induced rotor current sets up its own alternating field in the rotor, so a magnetic field exists in both stator and rotor windings; the interaction of the two produces torque. (a) 'stator only' is the tempting answer, but without rotor flux there would be no torque at all.
Source: Book EOC
📖 §2.7 Motor Loading — Measuring Load
31. A 7.5 kW, 415 V, 14.5 A, 1460 RPM, 3-phase rated induction motor with full-load efficiency of 88% draws 10.1 A and 5.1 kW of input power. The percentage loading of the motor is about
60 %
70 %
50%
none of the above
Answer: A) 60 %
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — % loading = input kW / (nameplate kW ÷ full-load efficiency) = 5.1 / (7.5/0.88) = 5.1 / 8.523 = 59.8 % ≈ 60 %. Do not use the current ratio 10.1/14.5 = 70 % — the book expressly warns that loading must not be estimated as a ratio of currents, which is what option (b) rewards.
Source: Book EOC
📖 §2.4 Motor Efficiency
32. An induction motor rated for 75 kW and 90% efficiency, operating at full load, will
deliver 83.3 kW
deliver 75 kW
draw 75 kW
draw 67.5 kW
Answer: B) deliver 75 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — The nameplate kW is the rated OUTPUT, so at full load the motor delivers 75 kW and draws 75/0.9 = 83.3 kW. (a) 'deliver 83.3 kW' is the classic inversion — 83.3 kW is the input, not the output.
Source: Book EOC
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
33. Stator phase resistance at 30 degC is 0.264 ohms. At 120 degC its value will be
0.264 ohms
0.354 ohms
0.237 ohms
none of the above
Answer: B) 0.354 ohms
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — R₂/R₁ = (235 + t₂)/(235 + t₁) = (235+120)/(235+30) = 355/265 = 1.34, so R₁₂₀ = 0.264 × 1.34 = 0.354 Ω — exactly the worked example in the book. (c) 0.237 Ω comes from inverting the ratio; winding resistance must RISE with temperature.
Source: Book EOC
📖 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives
34. Hydrodynamic principle for speed control is used in
DC drives
fluid coupling
pulse width modulation
eddy current drive
Answer: B) fluid coupling
Confirmed vs Book-3 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives — Fluid couplings work on the hydrodynamic principle — an impeller and a runner transmit power through low-viscosity mineral oil across an air gap, with slip inherent to the transfer. (d) eddy current drive is the tempting alternative but works on DC-excited eddy-current clutch action, not hydrodynamics.
Source: Book EOC
📖 §2.3 Motor Characteristics
35. A four-pole induction motor operating at 50 Hz with 1% slip will run at an actual speed of
1500 RPM
1515 RPM
1485 RPM
none of the above
Answer: C) 1485 RPM
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 4 = 1500 rpm; with 1 % slip the actual speed = 1500 × (1 − 0.01) = 1485 rpm. (b) 1515 rpm is the trap of adding the slip — an induction motor always runs BELOW synchronous speed.
Source: Book EOC
📖 §2.4 Motor Efficiency
36. The kW rating indicated on the name plate of an induction motor indicates
rated input of the motor
rated output of the motor
maximum input power which the motor can draw
maximum instantaneous input power of the motor
Answer: B) rated output of the motor
Confirmed vs Book-3 §2.4 Motor Efficiency — The nameplate kW of an induction motor is the rated mechanical OUTPUT available at the shaft; the input is that output divided by the efficiency. (a) 'rated input' is the standard misconception — it would make a 75 kW, 90 % motor deliver only 67.5 kW.
Source: Book EOC
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
37. Which loss is considered the most unreliable or complicated to measure in electric motor efficiency testing?
stator Cu loss
rotor Cu loss
stator Iron loss
stray loss
Answer: D) stray loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — Stray load losses are the hardest to determine — IEEE 112 gives a complicated method rarely used on the shop floor, and IS/IEC simply assume a fixed 0.5 % of input while the actual value is likely to be 1–3 %. (a)/(b) stator and rotor I²R losses are directly computable from measured resistance, current and slip, so they are not the unreliable ones.
Source: Aug 2014
📖 §2.3 Motor Characteristics
38. For a 6 pole induction motor operating at 49.5 Hz, the percentage slip at a shaft speed of 950 RPM will be
4.0 %
5.0 %
0.04 %
none of the above
Answer: A) 4.0 %
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 49.5 / 6 = 990 rpm; slip = (990 − 950)/990 × 100 = 4.04 % ≈ 4 %. (b) 5 % is what you get by wrongly using 1000 rpm (i.e. assuming 50 Hz) — the book's slip formula must use the synchronous speed at the ACTUAL supply frequency.
Source: Aug 2014
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
39. A 75 kW squirrel cage induction motor with static PF correction capacitors across the motor terminals got damaged along with capacitors once supply was switched off due to power failure. The possible reason for the motor burn out could be
motor was oversized
motor was undersized
charging current of the capacitors was more than the magnetizing current of the motor
charging current of the capacitor was only 85% of the motor magnetizing current
Answer: C) charging current of the capacitors was more than the magnetizing current of the motor
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — The book limits terminal capacitors to not more than 90 % of the motor's no-load kVAr, warning that higher capacitors could result in over-voltages and motor burn-outs; on supply failure an over-sized capacitor bank self-excites the still-spinning motor and the resulting over-voltage destroys motor and capacitors. (a)/(b) sizing of the motor itself has nothing to do with a burn-out that occurs at the instant of switch-off.
Source: Aug 2014
📖 §2.4 Motor Efficiency
40. An induction motor rated for 75 kW and 94 % efficiency, operating at full load, will
deliver 70.5 kW
deliver 75 kW
draw 75 kW
deliver 79.78 kW
Answer: B) deliver 75 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Nameplate kW is the rated shaft output, so a 75 kW motor at full load delivers 75 kW and draws 75/0.94 = 79.8 kW. (d) 79.78 kW is the deliberate trap — it is the INPUT power, not the delivered output.
Source: Aug 2014
📖 §2.8 Rewinding Effects on Energy Efficiency
41. The performance of rewinding of an induction motor can be assessed by which of the following factors?
no load current
stator resistance per phase
load current
no load current and stator resistance per phase
Answer: D) no load current and stator resistance per phase
Confirmed vs Book-3 §2.8 Rewinding Effects on Energy Efficiency — The book says comparison of no-load current AND stator resistance per phase of the rewound motor with the original values at the same voltage is the indicator of rewind efficacy — no-load current reveals core damage, resistance per phase reveals thinner wire / more turns. (a) or (b) alone is incomplete; load current (c) depends on the driven load, not the rewind.
Source: Mar 2021 (Set B)
📖 §2.3 Motor Characteristics
42. For a synchronous speed of 1500 rpm at a mains frequency of 50 Hz, the induction motor will have _________ number of poles.
8
6
4
2
Answer: C) 4
Confirmed vs Book-3 §2.3 Motor Characteristics — From Ns = 120 f / P, P = 120 × 50 / 1500 = 4 poles — consistent with the book's list of Indian synchronous speeds 3000/1500/1000/750 rpm for 2/4/6/8 poles at 50 Hz. (b) 6 poles would give 1000 rpm, which is the common mis-pick.
Source: Aug 2013
📖 §2.7 Motor Loading — Measuring Load
43. A 7.5 kW, 415 V, 14.5 A, 1460 RPM rated 3-phase induction motor with full-load efficiency 90% draws 9.1 A and 4.6 kW of input power. The percentage loading of the motor is about
55.2 %
61.3 %
67.5 %
none of the above
Answer: A) 55.2 %
Corrected (was b) — Book-3 §2.7 Motor Loading — Measuring Load: % loading = input kW / (rated kW ÷ full-load efficiency) = 4.6 / (7.5/0.90) = 4.6 / 8.333 = 55.2 %, which is option (a). The recorded 61.3 % does not follow from the book formula; 9.1/14.5 = 62.8 % is the current-ratio method that the book expressly forbids ('loading should not be estimated as the ratio of currents').
Source: Aug 2013
📖 §2.4 Motor Efficiency
44. The power input to the rotor of a three phase induction motor is 42.3 kW. If the motor operates at a slip of 1.30%, the total mechanical power developed will be:
42.3 kW
41.75 kW
5.48 kW
47.79 kW
Answer: B) 41.75 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Mechanical power developed = (1 − s) × rotor input = (1 − 0.013) × 42.3 = 41.75 kW; the rotor I²R loss is slip × rotor input = 0.55 kW. (c) 5.48 kW is the trap of computing something other than (1−s)×Pr; (a) 42.3 kW ignores the rotor copper loss altogether.
Source: Aug 2013
📖 §2.7 Motor Loading — Measuring Load
45. Which parameters need to be measured to assess the percentage loading of a motor by the slip method neglecting voltage correction?
motor speed
synchronous speed
operating motor speed and frequency
operating current
Answer: C) operating motor speed and frequency
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — The slip method needs the operating (actual) motor speed AND the supply frequency, because the synchronous speed used in slip % = (Ns − Nr)/Ns must be computed as 120 f / P at the ACTUAL frequency. (a) speed alone is insufficient — at 49.5 Hz a 4-pole motor's Ns is 1485 rpm, not 1500 rpm, and the slip changes materially.
Source: Aug 2013
📖 §2.2 Motor Types
46. Direct current motors are used in special applications where
high torque starting or where smooth acceleration over a broad speed range is required
low torque starting or where steady acceleration over a narrow speed range is required
normal torque starting or where high acceleration over a broad speed range is required
low torque starting or where smooth acceleration over a broad speed range is required
Answer: A) high torque starting or where smooth acceleration over a broad speed range is required
Confirmed vs Book-3 §2.2 Motor Types — The book states DC motors are used in special applications where high starting torque or smooth acceleration over a broad speed range is required, speed being directly proportional to armature voltage. The other options simply invert 'high' to 'low' or 'broad' to 'narrow' — they describe applications for which an ordinary squirrel cage motor would suffice.
Source: Year not recorded
📖 §2.3 Motor Characteristics
47. A 3-phase, 415 volts, 50 Hz, 100 kW, 6 pole squirrel cage induction motor with a rated slip of 2% will have a full load rotor speed of
1470 rpm
980 rpm
1020 rpm
none of the above
Answer: B) 980 rpm
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 6 = 1000 rpm; with 2 % slip the rotor speed = 1000 × 0.98 = 980 rpm. (a) 1470 rpm is the trap of using 4 poles; (c) 1020 rpm adds the slip instead of subtracting it, which is impossible for an induction motor.
Source: Year not recorded
📖 §2.4 Motor Efficiency
48. In an induction motor the loss which is independent of motor load is
I²R loss of stator
I²R loss of rotor
friction and windage loss
all of the above
Answer: C) friction and windage loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Fixed losses — magnetic core losses and friction & windage losses — are independent of motor load; friction & windage arise from bearing friction and the ventilation fan and account for 8–12 % of total losses. (a)/(b) stator and rotor I²R are the variable losses, proportional to the square of load current, so (d) 'all of the above' is wrong.
Source: Year not recorded
📖 §2.8 Rewinding Effects on Energy Efficiency
49. Rewinding can affect which of the following factors that contribute to deterioration in motor efficiency:
winding and slot design and winding material selection
heat applied to strip windings, damaging insulation between laminations, increasing eddy current losses
change in the air gap may affect power factor and output torque
all the above
Answer: D) all the above
Confirmed vs Book-3 §2.8 Rewinding Effects on Energy Efficiency — The book lists all three effects: winding and slot design and winding material, heat used to strip old windings damaging inter-lamination insulation and raising eddy-current losses, and a changed air gap affecting power factor and output torque. Choosing any single option is the tempting error — the book explicitly groups them together.
Source: Year not recorded
📖 §2.3 Motor Characteristics
50. If the measured line current of a 3-phase delta-connected induction motor is 25.98 A, what will be the phase current?
15 A
45 A
8.96 A
30 A
Answer: A) 15 A
Confirmed vs Book-3 §2.3 Motor Characteristics — In a delta connection the phase current = line current / √3 = 25.98 / 1.732 = 15 A. (b) 45 A multiplies instead of divides; the √3 relation is the same one used in the book's field-test example, where stator copper loss is computed as 3 × (I_line/√3)² × R.
Source: Year not recorded
📖 §2.3 Motor Characteristics
51. The power factor of a squirrel cage induction motor
decreases at low motor loading
decreases at high motor loading
remains constant and is independent of load
cannot be predicted
Answer: A) decreases at low motor loading
Confirmed vs Book-3 §2.3 Motor Characteristics — As load falls the active current falls but the magnetizing current does not (it is set by the supply voltage), so the power factor drops sharply at part load — Figure 2.2 shows exactly this. (c) 'independent of load' is the misconception the book is written against: under-loaded induction motors are the main cause of low plant power factor.
Source: Year not recorded
📖 §2.2 Motor Types
52. The slip of a synchronous motor will be
more than the induction motor
less than the induction motor
zero
load dependent
Answer: C) zero
Confirmed vs Book-3 §2.2 Motor Types — A synchronous motor's DC-excited rotor field locks onto the stator rotating field, so it runs exactly at synchronous speed and the slip is zero — the slip energy being supplied by the DC excitation. (d) 'load dependent' describes an induction motor, whose slip grows with load.
Source: Year not recorded
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
53. The effect of increasing the air gap in an induction motor will increase:
power factor
speed
capacity
magnetizing current
Answer: D) magnetizing current
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — A larger air gap needs more magnetizing current to drive flux across it; that is why the book's loss-reduction advice is the opposite — 'lowering the operating flux density and possible shortening of air gap' reduces the magnetizing component of current. (a) power factor is the trap: more magnetizing current makes power factor WORSE, not better.
Source: 17th Sep-2016
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
54. In no load test of a poly-phase induction motor, the measured power by the wattmeter consists of:
core loss
copper loss
core loss, windage & friction loss
stator copper loss, iron loss, windage & friction loss
Answer: D) stator copper loss, iron loss, windage & friction loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — The no-load wattmeter reading is the total no-load input, i.e. stator I²R at no-load current PLUS core (iron) loss PLUS friction & windage; the book subtracts (no-load current)² × stator resistance from it to leave core + F&W. (c) is the tempting answer because it names what is LEFT after that subtraction, not what the wattmeter actually reads.
Source: 17th Sep-2016
📖 §2.2 Motor Types
55. The power factor of a synchronous motor:
Improves with increase in excitation and may even become leading at high excitations
Decreases with increase in excitation
Is independent of its excitation
None of the above
Answer: A) Improves with increase in excitation and may even become leading at high excitations
Confirmed vs Book-3 §2.2 Motor Types — A synchronous motor's power factor is set by its DC field excitation: raising excitation improves the power factor and at over-excitation the machine draws leading current, which is why synchronous motors are used for plant PF correction. (c) 'independent of excitation' describes the induction motor, whose PF cannot be controlled this way.
Source: 17th Sep-2016
📖 §2.3 Motor Characteristics
56. A 4 pole 50 Hz induction motor is running at 1470 rpm. What is the slip value?
20%
2%
30%
40%
Answer: B) 2%
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 4 = 1500 rpm; slip = (1500 − 1470)/1500 × 100 = 2 %. (a) 20 % is a decimal slip mis-read; normal full-load slip of a squirrel cage motor is only 1–3 %.
Source: Jul 2022
📖 §2.4 Motor Efficiency
57. An Induction motor rated 15 kW and 90% efficiency, at full load will:
Draw 15 kW
Draw 13.5 kW
Deliver 16.66 kW
Deliver 15 kW
Answer: D) Deliver 15 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Nameplate kW is the rated output, so a 15 kW motor at full load DELIVERS 15 kW and draws 15/0.9 = 16.67 kW. (c) 'deliver 16.66 kW' is the standard inversion trap — 16.66 kW is the input drawn, not the shaft output.
Source: 17th Sep-2016
📖 §2.7 Motor Loading — Measuring Load
58. A 50 hp motor with a full load efficiency of 90 percent was found to be operating at 25 kW input. The percent Motor Load is:
75%
67%
60%
25%
Answer: C) 60%
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — 50 hp = 37.3 kW output; rated input = 37.3/0.90 = 41.4 kW; % load = 25 / 41.4 = 60 %. (a) 75 % comes from dividing 25 kW by the hp figure without converting to kW and without dividing by efficiency — the book's formula always compares measured input kW with rated kW ÷ rated efficiency.
Source: 17th Sep-2016
📖 §2.7 Voltage Unbalance
59. The voltage unbalance in three phase supply is 1.5%. If the motor is operating at 100°C, the additional temperature rise in °C due to voltage unbalance is:
4.5
9
0
none of the above
Answer: A) 4.5
Confirmed vs Book-3 §2.7 Voltage Unbalance — Additional temperature rise = 2 × (% voltage unbalance)² = 2 × 1.5² = 4.5 °C. (b) 9 °C is the trap of multiplying by the unbalance instead of squaring it (the book's own example gives 8 °C for 2 % unbalance, confirming the square law).
Source: 17th Sep-2016
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
60. The inexpensive way to improving energy efficiency of a motor which operates consistently at below 40% of rated capacity is by _____.
operating in star mode
replacing with correct sized motor
operating in delta mode
none
Answer: A) operating in star mode
Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — For motors consistently below 40 % of rated capacity, star operation is the inexpensive fix — only terminal-box re-wiring and an overload-relay reset are required. (b) replacing with a correctly sized motor gives the same benefit but is the capital-intensive option, so it fails the 'inexpensive' test.
Source: Sep 2015
📖 §2.9 Soft Starter
61. Star – delta starter of an induction motor:
reduces voltage by inserting resistance in rotor circuit
reduces voltage by inserting resistance in stator circuit
reduces the voltage applied to motor windings at start
inserts capacitance in the stator
Answer: C) reduces the voltage applied to motor windings at start
Confirmed vs Book-3 §2.9 Soft Starter — A star-delta starter connects the windings in star at start, reducing the voltage across each winding by √3 and hence the starting current and torque, before switching to delta. (a)/(b) inserting resistance is how a slip-ring motor is started; the book notes star-delta gives only a partial solution because peaks can recur at transition.
Source: Sep 2015
📖 §2.9 Speed Control of Motors
62. Slip power recovery system is applicable in case of:
squirrel cage induction motor
wound rotor motor
synchronous motor
DC shunt motor
Answer: B) wound rotor motor
Confirmed vs Book-3 §2.9 Speed Control of Motors — Slip power recovery collects the excess rotor power from the SLIP RINGS and returns it to the shaft or the supply, so it is applicable only to wound-rotor (slip-ring) motors. (a) a squirrel cage rotor has no external electrical connection at all, so its slip power cannot be tapped.
Source: Sep 2015
📖 §2.2 Motor Types
63. Rotating magnetic field is produced in a___________
single-phase induction motor
three-phase induction motor
DC series motor
all of the above
Answer: B) three-phase induction motor
Confirmed vs Book-3 §2.2 Motor Types — A 3-phase supply fed to the stator windings sets up a flux of constant magnitude rotating at synchronous speed — this is the defining statement of the 3-phase induction motor. A single-phase motor produces a pulsating field needing an auxiliary phase to start, and a DC series motor has a stationary field, so (d) is wrong.
Source: Sep 2015
📖 §2.6 Energy Efficient Motors
64. Motor efficiency will be improved by:
reducing the slip
increasing the slip
reducing the diameter of the motor
decreasing the length of the motor
Answer: A) reducing the slip
Confirmed vs Book-3 §2.6 Energy Efficient Motors — The book states that in polyphase induction motors slip is a measure of motor winding losses — the lower the slip the higher the efficiency; energy-efficient motors run about 1 % faster than standard ones for this reason. (b) increasing the slip does the opposite: rotor I²R loss = slip × rotor input.
Source: Sep 2015
📖 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD
65. Installation of Variable frequency drives (VFD) allows the motor to be operated with:
constant current
lower start-up current
higher voltage
none of the above
Answer: B) lower start-up current
Confirmed vs Book-3 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD — By selecting the proper V/f ratio the VFD keeps the starting current well under control, avoiding supply sag and motor heating, and it also provides overcurrent protection. (c) 'higher voltage' is wrong — the VFD varies voltage in proportion to frequency and cannot exceed rated voltage above base speed.
Source: Sep 2015
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
66. In a no load test of a poly-phase induction motor, the measured power by the wattmeter consists of:
core loss only
copper loss only
core loss, windage & friction loss
full load copper loss
Answer: C) core loss, windage & friction loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — With no shaft load the input covers the core (iron) loss plus friction & windage plus a small stator I²R at no-load current; of the choices offered, core loss with windage & friction is the correct grouping. (a) core loss only is incomplete — the book separates the two by repeating the test at variable voltage and taking the intercept as F&W.
Source: Sep 2015
📖 §2.7 Motor Loading — Measuring Load
67. A 50 hp motor with a full load efficiency rating of 90 percent was metered and found to be operating at 25 kW. The percent motor load is:
75%
50%
60%
25%
Answer: C) 60%
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — 50 hp = 37.3 kW; rated input = 37.3/0.90 = 41.4 kW; % load = 25/41.4 = 60 %. (b) 50 % is the trap of comparing 25 kW with the 50 hp figure treated loosely as 50 kW — hp must first be converted to kW and then divided by the efficiency.
Source: Sep 2015
📖 §2.4 Motor Efficiency
68. A 22 kW, 415 V, 45A, 0.8 PF, 1475 RPM, 4 pole 3 phase induction motor operating at 420 V, 40 A and 0.8 PF. What will be the rated efficiency?
85.0%
94.5%
89.9%
88.2%
Answer: A) 85.0%
Corrected (was c) — Book-3 §2.4 Motor Efficiency: rated efficiency = rated output / rated input, and rated input = √3 × 415 × 45 × 0.8 = 25.88 kW, giving 22 / 25.88 = 85.0 % — option (a). The recorded 89.9 % does not follow from the nameplate figures; the sister question with a 42 A nameplate gives 91 %, which shows how sensitive the result is to the nameplate current used.
Source: Sep 2015
📖 §2.3 Motor Characteristics
69. A 10 HP/7.5 kW, 415 V, 14.5 A, 1460 RPM, 3 phase rated induction motor, after decoupling from the driven equipment, was found to be drawing 3 A at no load. The current drawn by the motor at no load is high because of
faulty ammeter reading
very high supply frequency
loose motor terminal connections
poor power factor as the load is almost reactive
Answer: D) poor power factor as the load is almost reactive
Confirmed vs Book-3 §2.3 Motor Characteristics — At no load the current is almost entirely magnetizing (reactive), so a 3 A no-load draw on a 14.5 A motor is normal and simply reflects a very poor no-load power factor. (b)/(c) a frequency error or loose connection would show up as speed or heating faults, not as a routine high no-load current — the book advises keeping a record of no-load input power and current precisely because it is expected to be high.
Source: Sep 2017
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
70. In a no load test of a 3-phase induction motor, the measured power by the wattmeter consists of:
core loss
copper loss
core loss, windage & friction loss
stator copper loss, iron loss, windage & friction loss
Answer: D) stator copper loss, iron loss, windage & friction loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — The no-load wattmeter reads the whole no-load input: stator copper loss at no-load current plus iron loss plus windage & friction. The book's very next step — 'F&W and core losses = No load power − (No load current)² × stator resistance' — proves the stator I²R term is included in the reading; (c) omits it.
Source: Sep 2017
📖 §2.4 Motor Efficiency
71. A 22 kW, 415 V, 45 A, 0.8 pf, 1475 rpm, 4 pole 3-phase induction motor operating at 420 V, 40 A and 0.8 pf. What will be the motor efficiency?
85.0 %
94.5 %
89.9 %
None of the above
Answer: A) 85.0 %
Corrected (was d) — Book-3 §2.4 Motor Efficiency: the efficiency follows directly from the nameplate: rated input = √3 × 415 × 45 × 0.8 = 25.88 kW, so η = 22 / 25.88 = 85.0 %, which IS listed as option (a) — therefore 'none of the above' cannot be correct. At the measured 420 V / 40 A / 0.8 pf the input is 23.3 kW, i.e. about 90 % loading, where the book's own solved example assumes efficiency is unchanged.
Source: Sep 2018
📖 §2.9 Soft Starter
72. Use of soft starters for induction motors results in
lower mechanical stress
lower power factor
higher maximum demand
All the above
Answer: A) lower mechanical stress
Confirmed vs Book-3 §2.9 Soft Starter — The book's listed advantages of soft start are less mechanical stress, improved power factor, lower maximum demand and less mechanical maintenance. (b) and (c) state the exact opposite of two of those advantages, so (d) 'all the above' cannot hold.
Source: Jul 2022
📖 §2.7 Motor Loading — Measuring Load
73. A 7.5 kW, 415 V, 15 A, 970 RPM, 3 phase rated induction motor with full load efficiency of 86 % draws 7.5 A and 3.23 kW of input power. The percentage loading of the motor is about
37 %
43 %
50 %
None of the above
Answer: A) 37 %
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — % loading = 3.23 / (7.5/0.86) = 3.23 / 8.721 = 37 %. (c) 50 % would follow from the current ratio 7.5/15, which the book forbids ('loading should not be estimated as the ratio of currents') because power factor collapses at part load.
Source: Sep 2018
📖 §2.3 Motor Characteristics
74. A two pole induction motor operating at 50 Hz, with 1 % slip will run at an actual speed of
3000 RPM
3030 RPM
2970 RPM
None of the above
Answer: C) 2970 RPM
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 2 = 3000 rpm; at 1 % slip the actual speed = 3000 × 0.99 = 2970 rpm. (b) 3030 rpm adds the slip — an induction motor can never exceed synchronous speed while motoring.
Source: Sep 2018
📖 §2.9 Soft Starter
75. _____________ is not used for speed control.
Variable Frequency drive
Soft starter
Hydraulic coupling
Eddy current drives
Answer: B) Soft starter
Confirmed vs Book-3 §2.9 Soft Starter — A soft starter only ramps the voltage during starting and stopping; once at full voltage it gives no speed control. VFDs, hydraulic (fluid) couplings and eddy-current drives are all listed by the book as speed-control methods, so (b) is the odd one out.
Source: Sep 2019
📖 §2.6 Energy Efficient Motors
76. When compared to standard motors, energy efficient motors will have ____________.
Higher slip
Higher starting torque
Lower No load current
All the above
Answer: C) Lower No load current
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Energy-efficient designs lower the operating flux density and shorten the air gap to cut the magnetizing component of current, so the no-load current is lower than in a standard motor. (b) is the classic trap — the book warns that 'starting torque for efficient motors may be LOWER than for standard motors'; (a) is also wrong because EEMs have lower slip.
Source: Sep 2019
📖 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives
77. A DC excitation is used to vary the speed of _____________.
Eddy Current Coupling
fluid coupling
variable frequency drive
None of the above
Answer: A) Eddy Current Coupling
Confirmed vs Book-3 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives — In an eddy-current drive the freely revolving secondary member is separately excited by a DC field winding, and varying that DC excitation varies the output speed; its drawback is poor efficiency at low speeds. (b) a fluid coupling is purely hydrodynamic — speed is varied by the oil fill, not by excitation.
Source: Sep 2019
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
78. Rating of PF correction capacitors for Induction Motors terminal should be
100 % kVAr of the induction motor
20 % of Motor Rating
25 % of Motor rating
90 % of the no-load kVAr induction motor
Answer: D) 90 % of the no-load kVAr induction motor
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — The book fixes the size of a terminal-connected capacitor from the motor's no-load reactive kVAr, selecting it 'to not exceed 90 % of the no-load kVAR of the motor' because higher capacitors could cause over-voltages and burn-outs. (a) 100 % is exactly the over-correction the book warns against; percentages of motor kW rating (b)/(c) are not the book's basis.
Source: Sep 2019
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
79. Rating of power factor correction capacitors at induction motor terminals should be
100% of no load magnetizing kVAr of induction motor
90% of no load magnetizing kVAr of induction motor
120% of no load magnetizing kVAr of induction motor
none of the above
Answer: B) 90% of no load magnetizing kVAr of induction motor
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — Same rule: the capacitor is selected not to exceed 90 % of the motor's no-load (magnetizing) kVAr, which can only be established by a no-load test. (c) 120 % would leave the machine critically or over-corrected, risking self-excitation over-voltage on supply interruption.
Source: 9th Dec-2009
📖 §2.3 Motor Characteristics
80. What is the % slip of a 4 pole induction motor if the shaft speed at 49.5 Hz supply frequency is 1460 rpm?
1.68
2.66
1.71
none of the above
Answer: A) 1.68
Corrected (was c) — Book-3 §2.3 Motor Characteristics: the slip formula uses the SYNCHRONOUS speed in the denominator. Ns = 120 × 49.5 / 4 = 1485 rpm, so slip = (1485 − 1460)/1485 × 100 = 1.68 % — option (a). The recorded 1.71 % comes from dividing by the rotor speed 1460 instead of by Ns; (b) 2.66 % comes from wrongly assuming Ns = 1500 rpm at 50 Hz.
Source: 9th Dec-2009
📖 §2.2 Motor Types
81. During induction motor operation, magnetic field is established in
stator winding only
rotor winding only
stator and rotor windings
at carbon brushes
Answer: C) stator and rotor windings
Confirmed vs Book-3 §2.2 Motor Types — The stator carries the rotating flux and the induced rotor current creates its own alternating field, so a magnetic field exists in both stator and rotor windings; their interaction produces torque. (d) 'at carbon brushes' applies to a slip-ring or DC machine and has nothing to do with field production.
Source: 9th Dec-2009
📖 §2.7 Motor Loading — Measuring Load
82. An induction motor rated for 7.5 kW and 90 % efficiency at full load, was drawing 5 kW. The percentage loading on the motor is
60 %
66.66%
74%
none of the above
Answer: A) 60 %
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — % loading = measured input / (rated kW ÷ rated efficiency) = 5 / (7.5/0.90) = 5 / 8.333 = 60 %. (b) 66.66 % is the trap of dividing 5 kW by the 7.5 kW OUTPUT rating, forgetting to convert the rating to its equivalent input.
Source: 9th Dec-2009
📖 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD
83. The largest potential for electricity savings with variable speed drives is generally for:
variable torque applications
constant torque loads
constant power load
combination of above
Answer: A) variable torque applications
Confirmed vs Book-3 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD — The book states the largest potential for electricity savings with VSDs is generally in variable torque applications such as centrifugal pumps and fans, where power changes as the cube of speed — a 20 % speed cut gives almost 50 % input power reduction. (b) constant torque loads are also suitable for VSDs but their power falls only linearly with speed, so the savings are smaller.
Source: 9th Dec-2009
📖 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives
84. In a fluid coupling, connecting an induction motor and a fan
motor speed can be changed by the fluid coupling
fan speed can be changed by the fluid coupling
both motor and fan speed can be changed by the fluid coupling
none of the above is possible
Answer: B) fan speed can be changed by the fluid coupling
Confirmed vs Book-3 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives — A fluid coupling varies the speed of the DRIVEN equipment without changing the speed of the motor — the impeller runs at motor speed and slip between impeller and runner sets the output speed. (a)/(c) are wrong precisely because the motor stays on a fixed-frequency supply and keeps its own speed.
Source: 9th Dec-2009
📖 §2.7 Motor Loading — Measuring Load
85. A 3 phase, 7.5 kW, 415 V, 15 A, 1480 RPM rated induction motor with full load efficiency of 90% draws 5 A at rated voltage and 0.5 power factor. The percentage loading of the motor is about
21.56%
23.96%
33.33%
none of the above
Answer: A) 21.56%
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — Input = √3 × 415 × 5 × 0.5 = 1.797 kW; rated input = 7.5/0.90 = 8.333 kW; % loading = 1.797 / 8.333 = 21.56 %. (b) 23.96 % is the trap of dividing the input by the 7.5 kW OUTPUT rating instead of by the equivalent rated input.
Source: 10th Jul-2010
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
86. An induction motor installed with static PF correction capacitors across the motor terminals got damaged along with capacitors once it was disconnected from the supply. The possible reason among the following was
charging current of the capacitor was only 80% of the motor magnetising current
motor PF was over corrected or critically corrected (unity power factor)
motor was oversized
motor was undersized
Answer: B) motor PF was over corrected or critically corrected (unity power factor)
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — Over-correction (capacitor kVAr at or above the motor's no-load kVAr) makes the disconnected but still-spinning motor self-excite; the resulting over-voltage destroys both motor and capacitors — the book therefore caps the capacitor at 90 % of no-load kVAr. (a) under-correction at 80 % is exactly the safe condition the book recommends, so it cannot be the cause.
Source: 10th Jul-2010
📖 §2.3 Motor Characteristics
87. A 4-pole squirrel cage induction motor operates with 1% slip at full load. What is the approximate full load RPM at a grid frequency of 49.5 Hz?
1485
1470
1500
none of the above
Answer: B) 1470
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 49.5 / 4 = 1485 rpm; at 1 % slip the speed = 1485 × 0.99 = 1470 rpm. (a) 1485 rpm ignores the slip and (c) 1500 rpm ignores the reduced grid frequency — both are the standard traps in this question.
Source: 10th Jul-2010
📖 §2.3 Motor Characteristics
88. The power factor of an induction motor
increases with increase in motor loading
decreases with increase in motor loading
is independent of motor loading
increases with decrease in motor loading
Answer: A) increases with increase in motor loading
Confirmed vs Book-3 §2.3 Motor Characteristics — As loading rises the active component of current grows while the magnetizing current stays essentially fixed, so power factor improves with load; Figure 2.2 shows power factor dropping sharply at part load. (d) states the reverse and is the misconception the book's section on under-loading is written to correct.
Source: 10th Jul-2010
📖 §2.4 Motor Efficiency
89. An induction motor rated for 15 kW and 93% efficiency, operating at full load at the rated parameters, will
deliver 15 kW
deliver 16.12 kW
draw 15 kW
draw 13.95 kW
Answer: A) deliver 15 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Nameplate kW is the rated shaft output, so the motor delivers 15 kW and draws 15/0.93 = 16.12 kW. (b) 'deliver 16.12 kW' is the trap — 16.12 kW is the electrical INPUT, and no motor can deliver more than its rating at full load.
Source: 10th Jul-2010
📖 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives
90. In a variable speed drive using hydraulic coupling
motor speed changes
driven equipment speed changes
both a & b
neither a nor b
Answer: B) driven equipment speed changes
Confirmed vs Book-3 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives — A hydraulic (fluid) coupling changes the speed of the driven equipment while the motor itself continues to run at its fixed supply-determined speed. (c) 'both' is the tempting answer, but only a VFD can change the MOTOR's speed, by changing supply frequency.
Source: 10th Jul-2010
📖 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD
91. Energy savings potential of variable torque applications compared to constant torque application is:
higher
lower
equal
none of the above
Answer: A) higher
Confirmed vs Book-3 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD — For variable torque loads (centrifugal fans and pumps) power varies as the cube of speed, so a small speed reduction gives a large power saving; the book calls this the largest potential for VSD savings. Constant torque loads are also suitable but their power falls only in proportion to speed, so their savings potential is lower, making (b)/(c) wrong.
Source: 10th Jul-2010
📖 §2.4 Motor Efficiency
92. Slip ring induction motors, in general, have a …… design efficiency in comparison with the squirrel cage induction motors for similar ratings
lower
higher
same
none of the above
Answer: A) lower
Confirmed vs Book-3 §2.4 Motor Efficiency — The book states plainly that squirrel cage motors are normally more efficient than slip-ring motors, so a slip-ring machine has a lower design efficiency for the same rating — the slip rings, brushes and rotor windings add loss. (b) 'higher' is the reversed reading of the same sentence.
Source: 10th Jul-2010
📖 §2.3 Motor Characteristics
93. The synchronous speed (rpm) of a 2 pole induction motor at 49.5 Hz supply frequency is:
3000
2970
1500
none of the above
Answer: B) 2970
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 49.5 / 2 = 2970 rpm. (a) 3000 rpm is the trap of using the nominal 50 Hz instead of the actual 49.5 Hz supply frequency — synchronous speed is directly proportional to frequency.
Source: 11th Feb-2011
📖 §2.4 Motor Efficiency
94. kW rating indicated on the name plate of an induction motor indicates
rated input of the motor
maximum input power which the motor can draw
rated output of the motor
maximum instantaneous input power of the motor
Answer: C) rated output of the motor
Confirmed vs Book-3 §2.4 Motor Efficiency — The kW on an induction motor nameplate is the rated mechanical OUTPUT at the shaft; the input is that value divided by efficiency. (a) 'rated input' is the standard misconception — it would make a 75 kW, 90 % efficient motor deliver only 67.5 kW.
Source: 11th Feb-2011
📖 §2.3 Motor Characteristics
95. A 7.5 kW, 415 V, 14.0 A, 1480 RPM, three phase rated squirrel cage induction motor, after decoupling from the driven equipment, was found to be drawing 3.5 A at no load. The current drawn by the motor at no load is high because of
very high supply frequency at the time of no load test
faulty ammeter reading
very poor power factor as the load is almost inductive
loose motor terminal connections
Answer: C) very poor power factor as the load is almost inductive
Confirmed vs Book-3 §2.3 Motor Characteristics — At no load the current is almost purely magnetizing (inductive), giving a very poor no-load power factor, so 3.5 A on a 14 A motor is normal — the book recommends recording no-load current for exactly this reason. (b)/(d) an instrument fault or loose connection would not produce a consistent, expected no-load reading.
Source: 11th Feb-2011
📖 §2.3 Motor Characteristics
96. A six pole induction motor operating at 49.6 Hz, with 980 RPM actual speed, will have operating % slip of
1.21%
2%
0%
none of the above
Answer: A) 1.21%
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 49.6 / 6 = 992 rpm; slip = (992 − 980)/992 × 100 = 1.21 %. (b) 2 % results from assuming Ns = 1000 rpm at 50 Hz — the slip must always be referred to the synchronous speed at the measured frequency.
Source: 11th Feb-2011
📖 §2.10 Star Labeling of Energy Efficient Induction Motors
97. Eff1 (as per IS 12615:2004) induction motor is
endorsed by BEE as high efficiency label
having same efficiency as of Eff2
having less efficiency than Eff 2 motor
not covered in BEE labeling scheme for motors
Answer: A) endorsed by BEE as high efficiency label
Confirmed vs Book-3 §2.10 Star Labeling of Energy Efficient Induction Motors — Eff1 was the high-efficiency class under IS 12615:2004 and is the class endorsed by BEE under its labelling scheme for energy-efficient motors (now carried forward as IE2/IE3 classes under IS 12615:2011). (c) reverses the classes — Eff1 is more efficient than Eff2, not less.
Source: 11th Feb-2011
📖 §2.6 Energy Efficient Motors
98. Select the feature which does not apply to energy efficient motors by design:
energy efficient motors last longer
starting torque for efficient motors may be lower than for standard motors
energy efficient motors have high slips which results in speeds about 1% lower than standard motors
energy efficient motors have low slips which results in speeds about 1% higher than standard motors
Answer: C) energy efficient motors have high slips which results in speeds about 1% lower than standard motors
Confirmed vs Book-3 §2.6 Energy Efficient Motors — The book states that less slippage in energy-efficient motors results in speeds about 1 % FASTER than standard counterparts, so statement (c) — high slip and 1 % lower speed — is the one that does not apply. (b) is a genuine EEM feature the book warns about ('starting torque for efficient motors may be lower'), so it cannot be the answer to a 'does not apply' question.
Source: 11th Feb-2011
📖 Book-3 Ch-2 Electric Motors — speed control / Ch-1 power factor (cross-chapter item filed under Ch-3)
99. Which of following is not used for speed control ?
fluid coupling
eddy current
soft starter
variable frequency drive
Answer: C) soft starter
Confirmed vs Book-3 Ch-2 Electric Motors — Fluid couplings, eddy-current couplings and variable frequency drives all vary the output speed of a drive.
A soft starter only ramps the applied voltage during starting to limit inrush current and starting torque; once running, the motor returns to full speed, so it is not a speed-control device. Note this is a drives item filed within the compressed-air set.
Source: Mar 2021
📖 §2.6 Energy Efficient Motors
100. Which of the following is true for energy efficient motors ?
starting torque is higher than standard motors
slip is higher than standard motors
no-load current is higher than standard motors
speed is about 1 % higher than standard motors
Answer: D) speed is about 1 % higher than standard motors
Corrected (was a) — Book-3 §2.6 Energy Efficient Motors: the book says 'starting torque for efficient motors may be LOWER than for standard motors' and that 'less slippage in energy efficient motors results in speeds about 1 % faster than in standard counterparts'. So the true statement is the higher speed, not a higher starting torque. The printed options also repeated the same 'starting torque is higher' text twice, so they have been reconstructed from the book so that exactly one option is correct.
Source: Mar 2021
📖 §2.8 Rewinding Effects on Energy Efficiency
101. The performance of rewinding of an induction motor can be assessed by which of the following ?
no load current
stator resistance per phase
load current
both no load current and stator resistance per phase
Answer: D) both no load current and stator resistance per phase
Confirmed vs Book-3 §2.8 Rewinding Effects on Energy Efficiency — Rewind efficacy is judged by comparing BOTH the no-load current and the stator resistance per phase against the original values at the same voltage — no-load current exposes core damage from stripping heat, resistance per phase exposes thinner wire or altered turns. (c) load current reflects the driven load, not the quality of the rewind.
Source: Mar 2021
📖 §2.3 Motor Characteristics
102. The theoretical synchronous speed of 4 pole motor operating at 50 Hz will be ___________
1500 rpm
3000 rpm
200 rpm
1450 rpm
Answer: A) 1500 rpm
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 4 = 1500 rpm — one of the standard Indian synchronous speeds listed in the book (3000/1500/1000/750 rpm for 2/4/6/8 poles). (d) 1450 rpm is a typical full-load RUNNING speed, not the synchronous speed; (b) 3000 rpm is the 2-pole value.
Source: Mar 2021 (Set B)
📖 §2.4 Motor Efficiency
103. The output of a 900 kW rated motor operating with 90% efficiency is ________.
900 kW
1000 kW
810 kW
none of the above
Answer: A) 900 kW
Corrected (was b) — Book-3 §2.4 Motor Efficiency: the kW rating on the nameplate is the rated OUTPUT of the motor, so the output of a 900 kW rated motor at full load is 900 kW; 1000 kW (= 900/0.9) is the INPUT it draws. The book's own end-of-chapter question makes this explicit — a 75 kW, 90 % motor at full load 'delivers 75 kW'.
Source: Mar 2021
📖 §2.6 Energy Efficient Motors
104. Which of the following is not true of energy efficient motors?
starting torque is higher than standard motors
starting torque is lower than standard motors
slip is lower than standard motors
speed is higher than standard motors
Answer: A) starting torque is higher than standard motors
Confirmed vs Book-3 §2.6 Energy Efficient Motors — The book warns that 'starting torque for efficient motors may be lower than for standard motors', so (a) — higher starting torque — is NOT true of energy-efficient motors. (c) lower slip and (d) about 1 % higher speed are both stated by the book as genuine EEM characteristics, so they cannot answer a 'not true' question.
Source: Mar 2021 (Set B)
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
105. The inexpensive way to improve energy efficiency of a motor which operates consistently at below 40% of rated capacity is by ___
Operating in Star mode
Replacing with correct sized motor
Operating in delta mode
Operating in VFD mode
Answer: A) Operating in Star mode
Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — For motors consistently loaded below 40 % of rating, permanent star operation is the inexpensive fix — re-wiring the terminal box and resetting the overload relay, with no new equipment. (d) VFD operation would also cut losses but requires substantial capital, failing the 'inexpensive' criterion.
Source: Jul 2022
📖 §2.8 Rewinding Effects on Energy Efficiency
106. The performance of winding of an induction motor can be assessed by which of the following factors?
load current
stator resistance
no load current
both b and c
Answer: D) both b and c
Confirmed vs Book-3 §2.8 Rewinding Effects on Energy Efficiency — The book names two indicators of rewind quality: the no-load current and the stator resistance per phase, both compared with the original values at the same voltage. (a) load current is set by the driven machine and tells nothing about the winding work, so only the combination of (b) and (c) is correct.
Source: Jul 2022
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
107. In an Energy Efficient Motor, the efficiency is increased by increasing __________.
stator winding cross sectional area
fan losses
conductor resistance of rotor
stator winding resistance
Answer: A) stator winding cross sectional area
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — Table 2.2 states that 'use of more copper and larger conductors increases cross-sectional area of stator windings. This lowers resistance (R) of the windings and reduces losses due to current flow' — so increasing the stator winding cross-section raises efficiency. (d) increasing stator winding RESISTANCE is the exact opposite, and (b) increasing fan losses contradicts the low-loss fan design the book recommends.
Source: Mar 2023
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
108. By locating the capacitor near the motor terminal __________.
motor power factor increases
motor energy consumption decreases
system power factor decreases
line losses increases
Answer: B) motor energy consumption decreases
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — A capacitor improves power factor only from its point of installation back towards the generating side; placing it at the motor terminals therefore maximises the length of cable that carries reduced current, cutting I²R losses and voltage drop and so reducing the energy drawn from the system. (a) is the tempting wrong option — the book states explicitly that a capacitor at the starter terminals 'won't improve the operating PF of the motor', only the PF upstream.
Source: Mar 2023
📖 §2.3 Motor Characteristics
109. Slip% of Induction Motor is calculated as (Ns= Synchronous Speed, Nr= Full Load Rated Speed) __________.
(Ns-Nr)x100/Ns
(Ns-Nr)x100/Nr
(Nr-Ns)x100/Ns
(Ns+Nr)x100/Ns
Answer: A) (Ns-Nr)x100/Ns
Confirmed vs Book-3 §2.3 Motor Characteristics — The book's formula is Slip (%) = (Synchronous Speed − Full Load Rated Speed) / Synchronous Speed × 100, i.e. the difference is always referred to Ns. Dividing by Nr (option b) is the standard error — it is what turns a correct 1.68 % slip into a wrong 1.71 %. The printed options repeated the same expression three times, so the distractors have been reconstructed from the book formula.
Source: Mar 2023
📖 §2.3 Motor Characteristics
110. Synchronous speed of motor is directly proportional to __________.
No. of Poles
Frequency
Terminal Voltage
All of the above
Answer: B) Frequency
Confirmed vs Book-3 §2.3 Motor Characteristics — From Ns = 120 f / P, synchronous speed is directly proportional to the supply frequency and INVERSELY proportional to the number of poles. (a) is therefore the trap — more poles give a lower speed (2/4/6/8 poles give 3000/1500/1000/750 rpm at 50 Hz); terminal voltage does not enter the relation at all.
Source: Mar 2023
📖 §2.4 Motor Efficiency
111. Select the incorrect statement:
slip ring induction motors are normally less efficient than squirrel cage induction motors
high speed squirrel cage induction motors are normally less efficient than low speed squirrel cage induction motors
the capacitor requirement for PF improvement at induction motor terminal increases with decrease in rated speed of the induction motor
induction motor efficiency increases with increase in its rated capacity
Answer: B) high speed squirrel cage induction motors are normally less efficient than low speed squirrel cage induction motors
Confirmed vs Book-3 §2.4 Motor Efficiency — The book states that 'higher-speed motors are normally more efficient than lower-speed motors', so statement (b) reverses the book and is the incorrect one. (a), (c) and (d) are all book-supported — squirrel cage beats slip-ring on efficiency, capacitor kVAr rises as speed falls (Table 2.5), and motor efficiency increases with rated capacity.
Source: 9th Dec-2009
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
112. Select the incorrect statement:
required PF capacitor kVAr at induction motor terminal increases with decrease in speed of the motor
PF capacitor improves power factor from the point of installation back to the load side
induction motor efficiency increases with increase in its rated capacity
the largest potential for electricity savings with variable speed drives is generally in variable torque applications
Answer: B) PF capacitor improves power factor from the point of installation back to the load side
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — The book states that a PF capacitor 'improves power factor from the point of installation back to the generating side' — i.e. upstream, not towards the load — so statement (b) is the incorrect one. (a), (c) and (d) all restate book facts: capacitor kVAr rises as motor speed falls (Table 2.5), efficiency rises with rated capacity, and variable torque applications offer the largest VSD savings.
Source: 11th Feb-2011
📖 §2.8 Rewinding effects on energy efficiency
113. The performance of rewinding of an induction motor can be assessed by which of the following factors?
no load current
stator resistance per phase
load current
both no load current and stator resistance per phase
Answer: D) both no load current and stator resistance per phase
The book's test for rewind quality is a BEFORE-and-AFTER comparison of two measurable quantities: no-load current / no-load loss (which rises if the core laminations were heat-damaged during burn-out, i.e. higher iron loss) and stator resistance per phase (which rises if thinner conductor was used, i.e. higher copper loss).
Because the two indicators point at two different loss mechanisms - core loss and copper loss - you need both, so 'both' is the only complete answer.
Book figure: a careless rewind typically costs 1-5% efficiency, and the damage compounds with each successive rewind.
Source: Aug 2013
📖 §2.3 Motor characteristics (synchronous speed and slip)
114. For a synchronous speed of 1500 rpm, at a given mains frequency of 50 Hz, the induction motor will have _________ number of poles.
8
6
4
2
Answer: C) 4 poles (P = 120 x 50 / 1500 = 4)
Ns = 120f/P, so P = 120f/Ns = (120 x 50)/1500 = 4 poles.
Learn the 50 Hz ladder by heart - 2 poles = 3000, 4 = 1500, 6 = 1000, 8 = 750 rpm - and remember these are SYNCHRONOUS speeds; the nameplate full-load speed is always a little lower (e.g. 1440-1480 rpm on a 4-pole).
The common slip is writing Ns = 120P/f, which inverts the formula.
Source: Aug 2013
📖 §2.4 Motor efficiency (percentage loading by input-power method)
115. A 7.5 kW, 415 V, 14.5 A, 1460 RPM rated 3 phase induction motor with full load efficiency of 90%, draws 9.1 A and 4.6 kW of input power. The percentage loading of the motor is about
Loading by the kW method compares measured input with RATED INPUT, not with rated output: rated input = rated output/full-load efficiency = 7.5/0.90 = 8.33 kW.
% load = 4.6/8.33 x 100 = 55.2%.
Dividing 4.6 by the 7.5 kW output gives 61.3% - that is exactly the distractor at option (b). Always convert the shaft rating to an input rating first by dividing by the efficiency. The 9.1 A reading is not needed here.
Source: Aug 2013
📖 §2.4 Motor efficiency (slip, rotor input and rotor copper loss)
116. The power input to a rotor of three phase induction motor is 42.3 kW. If the induction motor is operating at a slip of 1.30 % the total mechanical power developed will be :
42.3 kW
41.75 kW
5.48 kW
47.79 kW
Answer: B) 41.75 kW (mechanical power = (1 - s) x rotor input = 0.987 x 42.3 = 41.75 kW)
The rotor power split: rotor copper loss = s x rotor input, and mechanical power developed = (1 - s) x rotor input.
Working: (1 - 0.013) x 42.3 = 0.987 x 42.3 = 41.75 kW; the 0.55 kW difference is the rotor I^2R loss.
Marks are lost by using s as 1.30 instead of 0.0130 - convert the percentage to a fraction first. Memory hook: slip is the fraction of rotor input burnt in the rotor bars, so low slip = efficient rotor.
Source: Aug 2013
📖 §2.4 Motor efficiency (slip method of load estimation)
117. Which parameters need to be measured to assess the percentage loading of a motor by slip method neglecting voltage correction?
motor speed
synchronous speed
operating motor speed and frequency
operating current
Answer: C) operating motor speed and frequency
Slip method: % load = (Ns - N)/(Ns - N_full-load) x 100, where Ns = 120f/P. To evaluate it you need the OPERATING speed (a tachometer or stroboscope reading) and the supply FREQUENCY, because Ns itself depends on f.
Supply frequency matters: a fall of even 1 Hz shifts Ns by 30 rpm on a 4-pole motor and can badly distort the estimated load.
Synchronous speed alone (option b) is not measured, it is calculated - which is precisely why frequency has to be measured.
Source: Aug 2013
📖 §2.7 Starting methods (star-delta starter)
118. Star - delta starter of an induction motor
reduces voltage by inserting resistance in rotor circuit
reduces voltage by inserting resistance in stator circuit
reduces voltage through a transformer
reduces the supply voltage due to change in connection configuration
Answer: D) reduces the supply voltage due to change in connection configuration - In star connection each winding receives V/sqrt(3) (58%) of line voltage, so starting current and torque fall to about one-third. No resistance or transformer is involved (those describe rotor-resistance, stator-resistance and auto-transformer starters respectively).
In star each winding sees V_line/sqrt(3) = 58% of the line voltage, so the starting current and starting torque both fall to about 1/3 of their direct-on-line values. Nothing is inserted in the circuit - the reduction comes purely from re-configuring the winding connection.
The three wrong options each describe a different starter: rotor-resistance (slip-ring motors), stator-resistance, and auto-transformer starting.
Remember why torque falls as the SQUARE of voltage: T is proportional to V^2, so 0.58^2 = 1/3.
Source: Sep 2015
📖 §2.9 Speed control of motors (slip power recovery)
119. Slip power recovery system is applicable in case of
squirrel cage induction motor
wound rotor motor
synchronous motor
DC shunt motor
Answer: B) wound rotor motor - Slip power recovery (static Scherbius / Kramer drive) taps the rotor slip power through slip rings and feeds it back to the supply. This requires access to the rotor winding, which only a wound rotor (slip ring) induction motor provides.
Slip power recovery (static Scherbius / Kramer drive) extracts the slip power from the rotor circuit through slip rings and returns it to the supply instead of wasting it as heat in a rotor resistor.
That requires physical ACCESS to the rotor winding, which only a wound-rotor (slip-ring) induction motor provides - a squirrel cage rotor is short-circuited internally with no terminals to tap.
Efficiency point: with a plain rotor rheostat the slip power (s x rotor input) is burnt as heat; slip power recovery makes sub-synchronous speed control efficient.
Source: Sep 2015
📖 §2.2 Motor types (rotating magnetic field)
120. Rotating magnetic field is produced in a ___________
single-phase induction motor
three-phase induction motor
DC series motor
all of the above
Answer: B) three-phase induction motor - Three balanced currents displaced 120 electrical degrees in three space-displaced windings produce a constant-magnitude rotating field at synchronous speed Ns = 120f/P. A single-phase winding alone produces only a pulsating field (needing an auxiliary winding to start), and a DC motor has a stationary field.
Three balanced phase currents, displaced 120 electrical degrees in time and fed to three windings displaced 120 degrees in space, produce a constant-magnitude field that rotates at Ns = 120f/P. This is the whole basis of the induction motor.
A single-phase winding alone produces only a PULSATING field - it cannot self-start and needs an auxiliary/capacitor winding to create the second phase. A DC machine has a stationary field with a rotating armature.
Memory hook: three phases in space + three phases in time = rotation.
Source: Sep 2015
📖 §2.4 Motor efficiency (slip and rotor losses) / §2.6 Energy efficient motors
121. Motor efficiency will be improved by
reducing the slip
increasing the slip
reducing the diameter of the motor
decreasing the length of the motor
Answer: A) reducing the slip - Rotor copper loss = slip x air-gap power, and motor efficiency is approximately (1 - s) neglecting other losses. Lower slip means lower rotor I^2R loss and higher efficiency; that is why energy-efficient motors run at slightly higher speed (lower slip) than standard motors.
Rotor copper loss = s x air-gap (rotor input) power, so the rotor's own efficiency is (1 - s). Cutting slip directly cuts the rotor I^2R loss and lifts overall efficiency.
That is why an energy-efficient motor runs slightly FASTER (lower slip) than the standard motor it replaces - a useful clue in the exam. Caution for centrifugal loads: that small speed rise can increase the load's power draw as speed^3, partly eating the saving.
Motor diameter and length are design parameters, not operating variables, so (c) and (d) are irrelevant.
Source: Sep 2015
📖 §2.9 Speed control of motors (variable frequency drives)
122. Installation of Variable frequency drives (VFD) allows the motor to be operated with
constant current
lower start-up current
higher voltage
none of the above
Answer: B) lower start-up current - A VFD starts the motor at low frequency and low voltage (constant V/f), so the motor develops rated torque while drawing near-rated (typically 100-150%) current instead of the 6-7 times DOL inrush. It also gives soft start/stop and speed control.
A VFD starts the motor at low frequency and low voltage on a constant V/f ratio, so full torque is available while the current stays near rated - typically 100-150% - instead of the 6-7 times rated DOL inrush.
Extra benefits: soft start/stop reducing mechanical shock, and speed control which on centrifugal fans and pumps saves power as speed^3 (the affinity laws).
A VFD never raises the supply voltage, and the current it draws varies with load, so (a) and (c) are wrong.
Source: Sep 2015
📖 §2.4 Motor efficiency (no-load test)
123. In a no load test of a poly-phase induction motor, the measured power by the wattmeter consists of:
core loss
copper loss
core loss, windage & friction loss
stator copper loss, iron loss, windage & friction loss
Answer: D) stator copper loss, iron loss, windage & friction loss - At no load the motor still draws the magnetising current, so the wattmeter reading = stator I^2R (copper) loss + core/iron loss + friction and windage loss. Rotor copper loss is negligible because slip is almost zero. Option (c) is incomplete since it omits the stator copper loss that the wattmeter actually measures.
At no load the motor still draws its magnetising current, so the wattmeter reads stator I^2R loss + iron (core) loss + friction & windage loss. Rotor copper loss is negligible because slip is almost zero.
The standard exam step follows straight from this: iron + F&W = no-load input MINUS the no-load stator copper loss (3 x I_phase^2 x R_phase), computed at the measured winding temperature.
Option (c) is the trap - it is right in spirit but incomplete, because it drops the stator copper loss the wattmeter genuinely measures.
Source: Sep 2015
📖 §2.4 Motor efficiency (percentage loading by input-power method)
124. A 50 hp motor with a full load efficiency rating of 90 percent was metered and found to be operating at 25 kW. The percent motor load is
75%
50%
60%
25%
Answer: C) 60% - Rated shaft output = 50 hp x 0.746 = 37.3 kW. Rated input at full load = 37.3/0.90 = 41.44 kW. % load = measured input / rated full-load input x 100 = 25/41.44 x 100 = 60.3% ~ 60%.
Convert horsepower first: 1 hp = 0.746 kW, so rated output = 50 x 0.746 = 37.3 kW, and rated INPUT = 37.3/0.90 = 41.44 kW.
% load = measured input / rated input = 25/41.44 x 100 = 60.3% ~ 60%.
Two classic errors: comparing 25 kW with the 37.3 kW output (gives 67%, wrong) and forgetting the 0.746 conversion altogether. Rule: measured INPUT must always be compared against rated INPUT.
Source: Sep 2015
📖 §2.4 Motor efficiency (efficiency from nameplate data)
125. A 22 kW, 415 kV, 45 A, 0.8 PF, 1475 RPM, 4 pole 3 phase induction motor operating at 420 V, 40 A and 0.8 PF. What will be the rated efficiency
85.0%
94.5%
89.9%
88.2%
Answer: A) 85.0% - Rated efficiency uses the NAMEPLATE values (the 420 V/40 A figures describe the present operating point and are a distractor). Rated input = 1.732 x 415 x 45 x 0.8 = 25,875.7 W = 25.88 kW. Rated output = 22 kW. Efficiency = 22/25.88 x 100 = 85.02% ~ 85.0%. (The nameplate '415 kV' is an obvious misprint for 415 V.)
RATED efficiency uses nameplate values only - the 420 V / 40 A figures describe the present operating point and are pure distraction.
Rated input = sqrt(3) x V x I x PF = 1.732 x 415 x 45 x 0.8 = 25,876 W = 25.88 kW; efficiency = 22/25.88 x 100 = 85.0%.
Traps: dropping the sqrt(3) (gives 147%, obviously absurd - a sanity check worth doing), and reading '415 kV' literally, which is a misprint for 415 V.
Source: Sep 2015
📖 §2.3 Motor characteristics (load torque characteristics) / §2.9 Speed control
126. Which of the following is an example of variable torque equipment ?
centrifugal pump
reciprocating compressor
screw compressor
roots blower
Answer: A) centrifugal pump
Variable-torque loads are the centrifugal machines - pumps, fans, blowers - where torque varies as speed^2 and power as speed^3. Constant-torque loads are the positive-displacement machines - reciprocating and screw compressors, roots blowers, conveyors - where torque is roughly independent of speed.
Only the centrifugal pump is centrifugal here, so it is the variable-torque equipment.
Why it matters: VFDs pay back fastest on variable-torque loads, because a 20% speed cut saves about 50% of the power (0.8^3 = 0.51).
Source: Sep 2017
📖 §2.4 Motor efficiency (efficiency from nameplate data)
127. A 22 kW, 415 V, 45 A, 0.8 pf, 1475 rpm, 4 pole 3 phase induction motor operating at 420 V, 40 A and 0.8 pf. What will be the motor efficiency?
85.0 %
94.5 %
89.9 %
None of the above
Answer: A) 85.0 %
Efficiency = output/input, and the input must be computed from the NAMEPLATE ratings: sqrt(3) x 415 x 45 x 0.8 = 25.88 kW against 22 kW output = 85.0%.
The 420 V and 40 A are the present operating point, deliberately supplied to tempt you into 1.732 x 420 x 40 x 0.8 = 23.28 kW, which would give 94.5% - option (b), the planted wrong answer.
Sanity rule: a standard 22 kW motor sits around 88-91% efficiency, so any answer above 94% on nameplate data should make you re-read the question.
Source: Sep 2018
Short questions (5 marks) — 21
📖 §2.3 Synchronous Speed / §2.4 Motor Efficiency
1. A 4-pole, 50 Hz induction motor delivers a shaft output of 30 kW. A measurement gives an input power of 33.0 kW. Determine (a) the synchronous speed, and (b) the operating efficiency.
Model answer: (a) Synchronous speed (Book-3 §2.3): Ns = 120 × f / P = 120 × 50 / 4 = 6000/4 = 1500 rpm.
(b) Operating efficiency (Book-3 §2.4): η = P_out / P_in, where P_out is the mechanical power at the shaft and P_in the electrical power at the terminals.
η = 30.0 / 33.0 = 0.909 = 90.9%.
Equivalently, total losses = 33.0 − 30.0 = 3.0 kW, and η = 1 − (P_loss/P_in) = 1 − 3.0/33.0 = 90.9% — the book's alternative form.
Confirmed vs Book-3 §2.3–§2.4 — the shaft/nameplate kW is always the OUTPUT, so it goes on top of the efficiency ratio; the measured 33.0 kW is the input. The actual running speed would be slightly below 1500 rpm because of slip, but synchronous speed itself depends only on f and P.
Source: AI practice
📖 §2.4 Energy Savings by Motor Replacement
2. A motor delivers a shaft output of 20 kW. It is to be replaced by an energy-efficient motor. The old efficiency is 0.88 and the new is 0.93. For 6000 operating hours per year at ₹7/unit, find the annual energy cost saving.
Model answer: Formula (Book-3 §2.4): kW saving = kW output × [(1/η_old) − (1/η_new)].
Step 1: 1/η_old = 1/0.88 = 1.1364; 1/η_new = 1/0.93 = 1.0753.
Step 2: kW saving = 20 × (1.1364 − 1.0753) = 20 × 0.0611 = 1.222 kW.
Step 3: Annual energy saving = 1.222 × 6000 = 7332 kWh/year.
Step 4: Annual cost saving = 7332 × ₹7 = ₹51,324 ≈ ₹51,300 per year.
(The cost benefit is then judged by comparing this against the price premium of the energy-efficient motor.)
Confirmed vs Book-3 §2.4 — the saving arises because the SAME shaft output is produced from a smaller input: input falls from 20/0.88 = 22.73 kW to 20/0.93 = 21.51 kW. Always use the actual SHAFT OUTPUT kW in the formula, not the nameplate rating, otherwise an under-loaded motor's saving is badly over-stated.
Source: AI practice
📖 §2.6 Energy Efficient Motors — loss breakdown; §2.4 Intrinsic losses
3. State the four categories of losses in an induction motor and the approximate percentage share of each in the total losses.
Model answer: Book-3 groups motor losses into FIXED (independent of load) and VARIABLE (load dependent), in four categories:
1) Stator + rotor resistance (I²R / copper) losses — 55–60% of total losses; VARIABLE, ∝ resistance × current². Reduced by more copper and larger conductor cross-section (copper rotor bars in place of aluminium).
2) Magnetic core (iron) losses — hysteresis + eddy current — 20–25%; FIXED. Reduced by low-loss silicon steel, a longer core (lower flux density) and thinner laminations.
3) Friction and windage losses — 8–12%; FIXED. Reduced by superior bearings and a low-loss/smaller fan.
4) Stray load losses — 4–5%; VARIABLE, ∝ (load current)². Reduced by optimised slot/tooth geometry, air gap and strict quality control.
Confirmed vs Book-3 §2.4 and §2.6 — the split to memorise is 55–60 / 20–25 / 8–12 / 4–5. Because the fixed losses stay constant while output falls, efficiency and power factor both collapse at part load — the reason the book pushes correct sizing and star-mode operation below 40% loading.
Source: AI practice
📖 §2.7 Motor Loading — Measuring Load
4. A 3 phase Induction motor has the following details:
Name plate details: 55 kW, 415 V, 95 A, 0.9 p.f, 50 Hz
Running load details: 410 V, 75 A, 0.80 p.f, 48 Hz
Calculate the following:
a) loading percentage,
b) Rated efficiency,
Model answer: Actual power drawn by the motor = 1.732 × 410 × 75 × 0.80 / 1000 = 42.6 kW
Rated input power = 1.732 × 415 × 95 × 0.90 / 1000 = 61.5 kW
Percentage loading of motor = 42.6 / 61.5 = 69.3 %
Rated efficiency of motor = (55 / 61.5) × 100 = 89.4%
5. Calculate the % voltage unbalance if the measured voltages are V_RY = 425, V_YB = 418, V_BR = 423.
Model answer: Average voltage = (425 + 418 + 423)/3 = 1266/3 = 422 V. Maximum deviation from average = |418 - 422| = 4 V (the largest of |425-422|=3, |418-422|=4, |423-422|=1). % voltage unbalance = (max deviation / average) x 100 = (4/422) x 100 = 0.95%.
% unbalance = (maximum deviation of any line voltage from the average) / average x 100 = 4/422 x 100 ≈ 0.95%.
Source: Book EOC
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
6. Why is it beneficial to operate motors in star mode for under-loaded motors?
Model answer: For a normally delta-connected motor that is consistently lightly loaded, reconnecting in star reduces the phase voltage to 1/sqrt(3) of line voltage. This lowers the flux density and hence the iron (core) and magnetizing losses, reduces the no-load current and improves the power factor and operating efficiency at light load. It is a low-cost measure, but is only suitable where the load never exceeds about 30-40% of rating, since star operation lowers the available torque.
Star connection cuts winding voltage, reducing iron/magnetizing losses and improving light-load efficiency and PF.
Source: Book EOC
📖 §2.9 Soft Starter
7. Explain the working of a soft starter and its advantage over other conventional starters.
Model answer: A soft starter uses thyristors (SCRs) connected in the supply lines whose firing angle is gradually advanced to ramp the applied voltage up from a low value to full voltage over a set time. This gives a smooth, gradual increase in motor torque and limits the starting current and mechanical/electrical shock. Compared with DOL or star-delta starters, it limits inrush current and torque transients, reduces voltage dips and mechanical stress on couplings/belts, allows controlled acceleration/deceleration, and reduces maintenance. (Note: it reduces starting current, not running energy.)
Soft starter ramps voltage via SCRs, limiting starting current and torque shock smoothly versus DOL/star-delta.
Source: Book EOC
📖 §2.8 Rewinding Effects on Energy Efficiency
8. How does efficiency loss occur in a rewound motor? How do you check the efficacy of a rewound motor?
Model answer: During rewinding, the old windings are often removed by burning out in an oven; excessive temperature can damage the core lamination insulation, increasing eddy-current and hysteresis (iron) losses. Poor-quality wire of smaller cross-section, changes in the number of turns, slot-fill or winding configuration increase copper (I^2R) losses. The cumulative effect typically lowers efficiency by 1-2% (or more) per rewind. Efficacy can be checked by a no-load test before and after rewinding to compare no-load (iron + friction) losses, by measuring stator resistance, and by comparing the no-load current and input power of the rewound motor with the original/standard values; a core-loss test on the stator core also indicates lamination damage.
Burn-out heat raises iron losses and poorer winding raises copper losses; verify via no-load test and core-loss/stator-resistance comparison.
Source: Book EOC
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
9. List the typical losses in induction motors.
Model answer: 1) Stator copper (I^2R) losses in the stator winding. 2) Rotor copper (I^2R) losses in the rotor bars. 3) Iron/core losses (hysteresis and eddy current) in the stator and rotor laminations. 4) Friction and windage (mechanical) losses in bearings and due to air movement. 5) Stray load losses. Fixed losses (iron + friction and windage) are roughly constant, while copper and stray losses vary with load.
Five categories: stator copper, rotor copper, iron/core, friction and windage, and stray load losses.
Source: Book EOC
📖 §2.3 Motor Characteristics
10. A 15 kW, 415 V, 4 pole, 50 Hz, 3-phase squirrel cage induction motor has full-load efficiency 92% and PF 0.89. Find at full load: a) input power in kW, b) current drawn, c) RPM at full-load slip of 0.8%.
Model answer: a) Pin = 15/0.92 = 16.304 kW. b) I = 16,304/(√3 x 415 x 0.89) = 25.48 A. c) Ns = 120 x 50/4 = 1500 rpm; N = 1500(1-0.008) = 1488 rpm.
Input = output/efficiency; I = Pin/(√3 V cosφ); N = Ns(1-slip).
Source: Aug 2013
📖 §2.4 Motor Efficiency
11. A 415 V, 15 kW, 3-ph, 50 Hz induction motor at full load: 88% efficiency, 0.85 PF lagging. a) Find current drawn. b) If replaced by a 92.5% efficient motor with 0.92 PF, what are the power savings in kW and kVA?
Model answer: a) Input power = 15/0.88 = 17.05 kW. Line current = 17.05×1000/(√3×415×0.85) = 27.91 A. kVA = 17.05/0.85 = 20.06 kVA. b) New input = 15/0.925 = 16.216 kW; new kVA = 16.216/0.92 = 17.62 kVA. Savings = 17.05 − 16.216 = 0.834 kW and 20.06 − 17.62 = 2.44 kVA.
Input=output/η; I=kW/(√3·V·PF); kVA=kW/PF.
Source: 17th Sep-2016
📖 §2.7 Voltage Unbalance
12. A 75 kW, 415 V, 140 A, 4 pole, 50 Hz, 3-phase squirrel-cage induction motor (full-load efficiency 87.6%) has measured terminal voltages 415, 418, 420 V and currents 137, 132, 137 A. Estimate the additional temperature rise of the motor due to unbalanced voltage supply.
Model answer: Mean voltage = (415 + 418 + 420)/3 = 417.67 V.
Maximum deviation from the mean = |415 − 417.67| = 2.67 V (larger than the 0.33 V and 2.33 V deviations).
% voltage unbalance (NEMA) = 2.67 / 417.67 × 100 = 0.64 %.
Additional temperature rise = 2 × (% voltage unbalance)² = 2 × (0.64)² = 0.82 °C.
This is within the book's recommended limit of 1 % unbalance at motor terminals, so no derating is called for; note that current unbalance can still be 6–10 times the voltage unbalance.
Voltage unbalance = max deviation from mean ÷ mean x 100; additional heating ≈ 2 x (% unbalance)².
Source: Sep 2017
📖 §2.8 Rewinding Effects on Energy Efficiency
13. S-7: How does a motor lose its efficiency upon rewinding? What two parameters will indicate the efficacy of the rewinding?
Model answer: Efficiency loss on rewinding: the burnout/stripping of the old winding (especially using high temperature or mechanical force) can damage the stator core lamination insulation, increasing iron (core) losses; changes in winding wire gauge, number of turns, or winding tightness alter the copper resistance and increase I²R losses; poor slot fill and handling raise stray and friction losses. A typical rewound motor can lose 1-2% (or more) efficiency. The two parameters that indicate the efficacy/quality of the rewind are: (1) the no-load (core) loss / no-load current and (2) the stator winding resistance (copper loss / I²R) — both measured before and after to confirm losses have not increased.
Book-3 Chapter 2 content (paper says 'Refer Guide Book No 3, Chapter 2, Page No 61'); model answer supplied from chapter notes — core-loss test and winding-resistance test indicate rewind quality.
Source: Sep 2018
📖 §2.7 Motor Loading — Measuring Load
14. A 15 kW, 415 V, 26 A, 4 pole, 50 Hz, 3 phase squirrel cage induction motor has full load efficiency and power factor of 90% and 0.89. An energy auditor measures: Supply voltage 408 V, Current 15 A, PF 0.81, Supply frequency 49.9 Hz, RPM 1488. Find at the operating conditions: 1) Power input in kW, 2) % motor loading, 3) % slip.
Model answer: 1) Power input = √3 × V × I × PF = 1.732 × 408 × 15 × 0.81 = 8,586 W = 8.59 kW.
2) Rated input = rated output / full-load efficiency = 15 / 0.90 = 16.67 kW; % motor loading = 8.59 / 16.67 × 100 = 51.5 %. (Using the nameplate √3 × 415 × 26 × 0.89 = 16.63 kW gives the same 51.6 %. The current ratio 15/26 = 58 % must NOT be used — the book forbids estimating loading from currents.)
3) Synchronous speed at 49.9 Hz = 120 × 49.9 / 4 = 1,497 rpm; % slip = (1,497 − 1,488)/1,497 × 100 = 0.60 %.
Power input from √3×V×I×PF; loading vs rated input (rated kW/efficiency); slip from synchronous speed at measured frequency.
Source: 9th Dec-2009
📖 §2.3 Motor Characteristics
15. A 7.5 kW, 415 V, 2 pole, 50 Hz, 3 phase squirrel cage induction motor has full load efficiency 90% and power factor 0.88. At full load rated values find: (a) input power in kW; (b) current drawn; (c) RPM at full load slip of 1%.
Model answer: (a) Pin = 8.333 kW; (b) I = 13.17 A; (c) N = 2970 RPM
(a) Pin = 7.5/0.90 = 8.333 kW. (b) I = 8333/(√3×415×0.88) = 13.17 A. (c) Ns = 120×50/2 = 3000 RPM; N = 3000×(1−0.01) = 2970 RPM.
Source: 10th Jul-2010
📖 §2.4 Motor Efficiency
16. S-3: The power input to a three phase induction motor is 52 kW. If the induction motor is operating at a slip of 1.9% and with total stator losses of 1.30 kW, find the total mechanical power developed.
Model answer: Stator input = 52 kW; Stator losses = 1.30 kW; Stator output = 52 - 1.30 = 50.7 kW = Rotor input; Slip = 1.9%; Mechanical Power Output = (1 - s) x Rotor Input = (1 - 0.019) x 50.7 = 0.981 x 50.7 = 49.737 kW.
Rotor input = stator input - stator losses; mechanical power = (1-slip) x rotor input.
Source: 11th Feb-2011
📖 §2.7 Motor Loading — Measuring Load
17. A three-phase induction motor has the following details: Name plate details: 55 kW, 415V, 95A, 0.90 PF, 50 Hz. Running load details: 410V, 75A, 0.80 PF, 48 Hz. Calculate the loading percentage and rated efficiency of the motor.
Model answer: Actual power drawn = 1.732×410×75×0.80/1000 = 42.6 kW. Rated input power = 1.732×415×95×0.90/1000 = 61.5 kW. Loading percentage = 42.6/61.5 = 69.3%. Rated efficiency = 55/61.5 = 89.4%.
Input power = √3×V×I×pf; loading = actual input/rated input; rated efficiency = rated output/rated input.
Source: Mar 2023
📖 §2.3 Motor characteristics + §2.4 Motor efficiency (input power, current, speed)
18. A 15 kW, 415 V, 4 pole, 50 Hz, 3 Phase squirrel cage induction motor has a full load efficiency of 92% and power factor of 0.89. Find the following if the motor operates at full load rated values. a) input power in kW b) current drawn by the motor c) RPM at a full load slip of 0.8%
Model answer: a) Pin (Input power) = 15 / 0.92 = 16.304 kW
b) I (Input current) = 16.304 / (1.732 x 0.415 x 0.89) = 25.48 A
c) Ns = 120 x f / p = 120 x 50 / 4 = 1500 RPM
N = Ns (1 - S) = 1500 (1 - 0.008) = 1488 RPM
Three formulas, applied in order: input kW = output/efficiency = 15/0.92 = 16.304 kW; I = kW/(sqrt(3) x kV x PF) = 16.304/(1.732 x 0.415 x 0.89) = 25.48 A; Ns = 120f/P = 120 x 50/4 = 1500 rpm, so N = Ns(1 - s) = 1500 x (1 - 0.008) = 1488 rpm.
Note that the current must be computed from the INPUT kW, not the 15 kW shaft output - using 15 kW gives 23.4 A and loses the mark.
Other regulars: dropping the sqrt(3), and leaving voltage in volts instead of kV so the answer comes out 1000x wrong.
19. Fill in the blanks for the following: a) Voltage levels can be varied without isolating the connected load to the transformer using ______________ b) Use of ________ starter is appropriate in case of high number of motor starts and 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 the 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) sqrt(3) (i.e. square root of three)
d) Cycles of Concentration (COC)
e) Lumens
......5 marks (each one carries one mark)
Blank (a): an ON-LOAD TAP CHANGER (OLTC) shifts taps without de-energising - an off-load tap changer requires isolating the load, which is exactly the distinction being tested.
Blank (b): soft starters suit frequent starts/stops because they limit both inrush current and mechanical shock. Blank (c): re-connecting an under-loaded motor from delta to star drops the winding voltage by a factor of sqrt(3) (to 58%), cutting the iron loss - only safe below about 30-40% load.
Blanks (d) and (e): Cycles of Concentration (COC) = TDS in circulating water / TDS in make-up water; the SI measure of a lamp's light output is the lumen (luminous flux), while lux is illuminance on a surface - do not swap the two.
Source: Sep 2017
📖 §2.7 Factors affecting motor efficiency (voltage unbalance)
20. A 75 kW, 415 V, 140 Amp, 4 pole, 50 Hz, 3-phase squirrel cage induction motor has a full load efficiency of 87.6%. The measured operating motor terminal voltages in a 3-phase supply are 415 V, 418 V & 420 V. The current drawn in 3-phase supply are 137 Amp, 132 Amp & 137 Amp. Estimate the additional temperature rise of motor, due to unbalanced voltage supply.
Model answer: Additional temperature rise:
Phase R: V = 415, deviation from mean = -2.67
Phase Y: V = 418, deviation from mean = 0.33
Phase B: V = 420, deviation from mean = 2.33
Mean = 417.67 V
Voltage unbalance = Maximum deviation from mean / mean voltage
= 2.67 x 100 / 417.67 = 0.639% ......3 Marks
Additional temperature rise = 2 x (% voltage unbalance)^2
= 2 x (0.639)^2 = 0.8166% ......2 Marks
Two formulas, both examinable: % voltage unbalance = (maximum deviation from the mean voltage / mean voltage) x 100, and additional temperature rise (%) = 2 x (% voltage unbalance)^2.
Working: mean = (415 + 418 + 420)/3 = 417.67 V; deviations -2.67, +0.33, +2.33; take the LARGEST magnitude, 2.67; unbalance = 2.67/417.67 x 100 = 0.639%; extra temperature rise = 2 x 0.639^2 = 0.82%.
Mistakes that cost marks: averaging the deviations instead of taking the maximum, using the current readings (they are a distractor here), and forgetting to square the unbalance. Book limit: keep unbalance below 1% at the motor terminals, else derate.
Source: Sep 2017
📖 §2.8 Rewinding effects on energy efficiency
21. How does a motor lose its efficiency upon rewinding? (2.5 Marks) What two parameters will indicate the efficacy of the rewinding? (2.5 Marks)
Model answer: (The paper prints only "Refer Guide Book No 3, Chapter 2, Page No 61". Model answer from the 2014 BEE Book-3, Chapter 2:)
Loss of efficiency on rewinding: It is generally observed that rewound motors have a lower efficiency than the original, typically a drop of 1% to 5% (an average of about 1-2% for good practice, up to 5% for poor practice). The main causes are:
- Excessive heat applied while stripping the old winding (burn-out ovens / blow torch) damages the inter-laminar insulation of the stator core, increasing eddy current and hysteresis (iron) losses.
- Use of a smaller conductor cross-section / poorer winding-space utilisation than the original, which increases stator I2R (copper) loss.
- Change of winding configuration, number of turns, coil pitch or winding pattern from the original design.
- Mechanical damage to the stator slots and the air gap, and poor bearing/assembly practice, increasing friction and stray load losses.
Each subsequent rewind compounds the loss, so repeated rewinding of small motors is often uneconomical compared to replacement with an energy-efficient motor.
Two parameters that indicate the efficacy of the rewinding:
1. No-load current (and no-load loss / no-load input power) of the rewound motor, compared with the original value - a higher no-load current indicates increased core losses / air-gap damage.
2. Stator winding resistance per phase (and the resulting load or full-load current / I2R loss), compared with the original value - a higher resistance indicates a reduced conductor size and higher copper loss.
Three loss mechanisms explain the drop: burn-out heat damages the inter-laminar insulation of the stator core (higher eddy-current and hysteresis loss); a smaller conductor cross-section or poorer slot fill raises stator I^2R loss; and any change in turns, coil pitch or winding pattern alters the design flux and stray losses.
Book figure to quote: rewound motors typically lose 1-5% efficiency, and the loss compounds with each successive rewind - which is why replacing a small, repeatedly rewound motor with an energy-efficient motor usually wins.
The two efficacy indicators are the no-load current / no-load loss (flags core damage) and the stator resistance per phase (flags reduced conductor size) - each compared against the pre-rewind value, which is why keeping those records matters.
Source: Sep 2018
Long questions (10 marks) — 13
📖 §2.7 Motor Loading — Measuring Load
1. A 7.5 kW, 415 V, 14.5 A, 4-pole, 50 Hz, 3-phase squirrel cage induction motor has full-load efficiency and PF of 89% and 0.88. Measured operating data: supply voltage 410 V, current 9.5 A, PF 0.8, frequency 49.8 Hz, RPM 1480. Find (1) power input in kW, (2) % motor loading, (3) % slip.
Model answer: (1) Power input = √3 × V × I × PF = 1.732 × 410 × 9.5 × 0.8 = 5,396 W = 5.40 kW.
(2) Rated input = rated output / full-load efficiency = 7.5 / 0.89 = 8.43 kW; % motor loading = measured input / rated input = 5.40 / 8.43 × 100 = 64.1 % (the book warns the loading must NOT be taken as the current ratio 9.5/14.5 = 65.5 %).
(3) Synchronous speed at the measured 49.8 Hz = 120 × 49.8 / 4 = 1,494 rpm; % slip = (1,494 − 1,480) / 1,494 × 100 = 0.94 %.
Input from sqrt(3)VI cos(phi); loading vs rated input 7.5/0.89; slip from Ns at 49.8 Hz (1494 RPM) versus 1480 RPM.
Source: Book EOC
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
2. A 50 HP/415 V, 60 A, 1475 rpm, 3-phase delta-connected squirrel cage motor gave no-load test data: 415 V, 18 A, 50 Hz, stator resistance/phase 0.27 ohm, no-load power 1080 W, ambient 35 degC. Calculate (i) iron + friction and windage losses, (ii) stator resistance at 120 degC, (iii) stator copper loss at 120 degC, (iv) full-load slip and rotor input (rotor losses = slip x rotor input), (v) motor input (stray losses = 0.5% of rated power), (vi) full-load efficiency and PF.
Model answer: (i) No-load stator copper loss = 3 × (I/√3)² × R = 3 × (18/√3)² × 0.27 = 87.5 W; iron + friction & windage loss = 1,080 − 87.5 = 992.5 W.
(ii) R₁₂₀ = R₃₅ × (235 + 120)/(235 + 35) = 0.27 × 355/270 = 0.355 Ω per phase.
(iii) Stator copper loss at full load at 120 °C = 3 × (60/√3)² × 0.355 = 1,278 W.
(iv) Ns = 1,500 rpm, so full-load slip = (1,500 − 1,475)/1,500 = 0.0167 (1.67 %); rated output = 50 HP = 37.3 kW; rotor input = output/(1 − s) = 37,300 / 0.9833 = 37,932 W.
(v) Motor input = rotor input + stator Cu loss + (iron + F&W) + stray = 37,932 + 1,278 + 992.5 + (0.005 × 37,300 = 186.5) = 40,389 W.
(vi) Full-load efficiency = 37,300 / 40,389 × 100 = 92.4 %; full-load PF = 40,389 / (√3 × 415 × 60) = 0.94.
Separate no-load losses into Cu and (iron+F&W); correct stator R to 120 degC; sum stator/rotor Cu, fixed and stray losses to get input, efficiency and PF.
Source: Book EOC
📖 §2.4 Motor Efficiency
3. A 3-phase induction motor: rated 37 kW, 415 V, 66 A, 0.88 pf; operating 410 V, 49 A, 0.76 pf. Efficiency constant 50-100% load. Plant runs 7000 h/yr at Rs.6/unit. Proposed replacement: 30 kW EE motor at 92% efficiency costing Rs.75,000 (salvage of old Rs.10,000). a) Rated efficiency and loading of existing motor. b) Loading with EE motor. c) Payback period.
Model answer: Rated input = 1.732 x 0.415 x 66 x 0.88 = 41.746 kW; rated efficiency = 37/41.746 = 88.63%. Actual input = 1.732 x 0.410 x 49 x 0.76 = 26.44 kW; loading = 26.44/41.746 = 63.3%; shaft output = 37 x 0.633 = 23.44 kW. EE motor (30 kW): loading = 23.44/30 = 78%. Annual savings = 23.44 x (1/0.8863 - 1/0.92) x 7000 x 6 = Rs.40,740. Payback = (75,000 - 10,000)/40,740 = 1.59 years.
Rated input via √3VIcosφ; loading = actual/rated input; savings from efficiency difference applied to shaft load x hours x tariff.
Source: Aug 2014
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
4. 4-pole delta-connected 37 kW, 415 V, 63 A, 1475 rpm squirrel cage motor. No-load test: V=415 V, I=17 A, f=50 Hz, stator resistance/phase=0.260 Ω at 30°C, no-load power=1152 W. Calculate: i) iron+friction+windage loss, ii) stator resistance at 120°C, iii) stator copper loss at full load at 120°C, iv) full-load slip and rotor input, v) motor input (stray=0.5% of rated output), vi) full-load efficiency.
Model answer: i) Stator Cu loss at 30°C = 3 x (17/√3)² x 0.260 = 75.13 W; Pi+fw = 1152 - 75.13 = 1076.87 W. ii) R120 = 0.260 x (235+120)/(235+30) = 0.3483 Ω/phase. iii) Stator Cu loss at full load 120°C = 3 x (63/√3)² x 0.3483 = 1382.3 W. iv) Ns=1500; slip = (1500-1475)/1500 = 1.66%; rotor input = 37000/(1-0.01666) = 37,626.86 W. v) Input = 37,626.86 + 1382.3 + 1076.87 + (0.005 x 37000) = 40,271.03 W. vi) Efficiency = 37000/40,271.03 x 100 = 91.87%.
No-load test separates losses; resistance temperature-corrected; efficiency = output/input.
Source: Aug 2013
📖 §2.3 Motor Characteristics
5. Fill in the blanks: 1) A motor operable at both lagging and leading PF is the ___ motor. 2) A 50 Hz 3-phase IM with full-load speed 1440 rpm has ___ poles. 3) In a centrifugal pump velocity energy is converted to pressure energy by ___. 4) If pumped liquid density is twice water, HP required is ___ times. 5) Friction loss in a pipe is proportional to the fifth power of ___. 6) A 10 MVA generator at 0.866 PF lagging produces ___ MVAR. 7) TEFC motors are ___ efficient than SPDP motors. 8) Low-speed squirrel cage motors are ___ efficient than high-speed ones. 9) Harmonics are multiples of the ___ frequency. 10) For same rating, slip ring induction motors are ___ efficient than squirrel cage.
Model answer: 1) Synchronous. 2) 4. 3) volute or diffuser. 4) 2. 5) pipe diameter. 6) 5. 7) more. 8) less. 9) fundamental. 10) less.
Standard motor/pump fill-in answers from BEE guidebook.
Source: Year not recorded
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
6. a) How do you calculate the velocity of air/gas in a duct using the average differential pressure and density? b) No-load test on a delta-connected 37 kW IM: nameplate 3-phase, 415 V, 50 Hz, 55 A; no-load V=415 V, I=17 A, f=50 Hz, stator phase resistance at 30°C=0.24 Ω, no-load power=955 W. Find: i) iron+friction+windage loss, ii) stator copper loss at full load at 120°C, iii) no-load power factor.
Model answer: a) Velocity V (m/s) = Cp x √(2 x 9.81 x Δp x γ)/γ, where Cp = pitot constant (≈0.85), Δp = average differential pressure (mmWC) over the cross-section, γ = air/gas density at test condition. b) Stator Cu loss at 30°C = 3 x (17/√3)² x 0.24 = 69.36 W; Pi+fw = 955-69.36 = 885.64 W. Stator resistance at 120°C = 0.24 x (120+235)/(30+235) = 0.322 Ω. Stator Cu loss at full load = 3 x (55/√3)² x 0.322 = 974.05 W. No-load PF = 955/(√3 x 415 x 17) = 0.078.
Pitot velocity formula; no-load test loss separation and resistance temperature correction.
Source: Year not recorded
📖 §2.5 Motor Selection
7. L-3: Replace 30 motors with energy efficient motors. Data: 7.5 kW, 75% load, old eff 86%, new eff 89%, 12 nos; 11.5 kW, 85% load, old 88%, new 91%, 7 nos; 15 kW, 70% load, old 89%, new 92%, 11 nos. Motor loading same in both cases; calculate annual energy savings for 4000 hours/year.
Model answer: Savings per motor = rated kW × % loading × (1/ηold − 1/ηnew), the book's motor-replacement relation.
7.5 kW, 75 % load, 12 nos: old input = 7.5 × 0.75 / 0.86 = 6.541 kW; new = 7.5 × 0.75 / 0.89 = 6.320 kW; saving = 0.221 kW × 4,000 h × 12 = 10,584 kWh.
11.5 kW, 85 % load, 7 nos: old = 11.5 × 0.85 / 0.88 = 11.108 kW; new = 11.5 × 0.85 / 0.91 = 10.742 kW; saving = 0.366 kW × 4,000 h × 7 = 10,254 kWh.
15 kW, 70 % load, 11 nos: old = 15 × 0.70 / 0.89 = 11.798 kW; new = 15 × 0.70 / 0.92 = 11.413 kW; saving = 0.385 kW × 4,000 h × 11 = 16,931 kWh.
Total annual energy savings ≈ 37,769 kWh (≈ 37,800 kWh/year).
Printed solution: total 37,640 kWh/year (10,560 + 10,360 + 16,720).
Source: Sep 2019
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
8. Efficiency assessment of a 37 kW, 415 V, 62 A, 1480 rpm, 3-phase delta SCIM. No-load test: 415 V, 18 A, 50 Hz, stator resistance 0.275 Ω/phase at 30°C, no-load power 1164 W. Calculate: (i) iron+friction&windage loss; (ii) stator resistance at 120°C; (iii) stator copper loss at full load at 120°C; (iv) full load slip and rotor input; (v) motor input (stray loss 0.5% of rated output); (vi) full load efficiency and power factor.
(i) Pst30 = 3×(18/√3)²×0.275 = 89.1 W; Pi+fw = 1164−89.1 = 1074.9 W. (ii) R120 = 0.275×(235+120)/(235+30) = 0.368 Ω. (iii) Pst120 = 3×(62/√3)²×0.368 = 1414.5 W. (iv) S = (1500−1480)/1500 = 0.01333; Pr = 37000/(1−0.01333) = 37499.87 W. (v) Pin = 37499.87+1414.5+1074.9+0.005×37000 = 40174.27 W. (vi) η = 37000/40174.27 = 92.1%; PF = 40174.27/(√3×415×62) = 0.9014.
Source: 10th Jul-2010
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
9. a) List five losses in electrical motors and discuss about the measures taken by the motor manufacturers to make it energy efficient motor. (5 Marks) b) List four energy conservation opportunities in pumping system. (5 Marks)
Model answer: a) Book-3 §2.6 (Table 2.2 and the loss break-up). The five losses in an induction motor are: 1. Stator I²R (copper) loss and 2. Rotor I²R loss — together 55–60 % of total losses; 3. Core / iron loss (hysteresis + eddy current) — 20–25 %; 4. Friction & windage loss — 8–12 %; 5. Stray load loss — 4–5 %.
Manufacturers' measures in energy-efficient motors: more copper and larger-cross-section conductors and larger rotor bars (lower R, lower I²R); thinner, lower-loss silicon steel laminations and a longer core to lower flux density (lower eddy-current and hysteresis loss); smaller air gap and lower flux density to cut magnetizing current; low-loss fan design and superior bearings (lower friction & windage); optimised slot/tooth geometry and strict quality control (lower stray load loss). Result: full-load efficiency 3–7 percentage points higher, with mounting dimensions kept to IS 1231.
b) Pumping-system opportunities (Book-3 Ch. 6): avoid oversizing / re-size the pump to the actual duty; use a VFD instead of throttle-valve or bypass control; trim or replace the impeller to match the required head; and replace worn or low-efficiency pumps and eliminate unnecessary head losses (oversized valves, fittings, clogged strainers).
Motor losses and EE measures, and pumping ECOs from Guidebook-3.
Source: Mar 2021
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
10. a) List five losses in electrical motors and discuss about the measures taken by the motor manufacturers to reduce the losses in energy efficient motor (5 marks). b) List five energy conservation opportunities in pumping system (5 marks).
Model answer: a) Book-3 §2.6 (Table 2.2 and the loss break-up, p. 51). Five motor losses: stator I²R (copper) loss, rotor I²R loss (together 55–60 % of losses), core/iron loss (20–25 %), friction & windage loss (8–12 %) and stray load loss (4–5 %).
Measures taken in energy-efficient motors: use of more copper and larger conductors, and larger rotor bars, to lower winding resistance and hence I²R losses; thinner gauge, low-loss silicon steel laminations plus a longer core to reduce flux density and so eddy-current and hysteresis losses; a smaller air gap to reduce magnetizing current; low-loss fan design and superior bearings to cut friction & windage; optimised slot/tooth geometry and strict quality control to minimise stray load loss. These give 3–7 percentage points higher full-load efficiency and lower operating temperature.
b) Five pumping-system energy conservation opportunities (Book-3 Ch. 6): install variable speed drives in place of throttling; correctly size / de-stage or trim the impeller to the actual duty; replace inefficient or oversized pumps with high-efficiency units; reduce system resistance (pipe sizing, valve and fitting losses, clean strainers); and stop-start or sequence multiple pumps to match demand instead of running on bypass/recirculation.
Standard motor loss categories with EE design measures and recognized pump system energy conservation measures per Guidebook-3.
Source: Mar 2021 (Set B)
📖 §2.3 Motor Characteristics
11. The input parameter measured for a 15 kW, 3 phase, 415 V induction motor are 25 A and 12 kW at 410 V. Calculate the following: a) Apparent power drawn by the motor at the operating load (3 Marks); b) Reactive Power drawn by the motor at the operating load (1 Mark); c) Operating power factor (1 Mark).
Model answer: a) Apparent power = √3 × 0.410 × 25 = 17.75 kVA. b) Reactive power = sqrt(apparent power² - active power²) = sqrt(17.75² - 12²) = 13.07 kVAr. c) Operating power factor = Active power/Apparent power = 12/17.75 = 0.676.
S=√3VI, Q=√(S²-P²), PF=P/S.
Source: Jul 2022
📖 §2.5 Motor Selection
12. L-1(B): A spinning unit proposed to replace 50 no's of 22kW IE2 motors with IE3 motors under National Motor Replacement Programme. Present operating details: Motor Loading 74%, IE2 Motor Efficiency 90%, IE3 Motor Efficiency 93%, Annual operating hours 7000 Hrs, Energy Cost Rs.10/kWh, CO2 Emission from electricity 0.85 Kg/kWh. Calculate annual energy savings and emission reduction in Tons of CO2.
Model answer: Motor rating = 22 kW; loading = 74 %, so shaft load per motor = 22 × 0.74 = 16.28 kW.
Input with IE2 motor = 16.28 / 0.90 = 18.09 kW; input with IE3 motor = 16.28 / 0.93 = 17.51 kW.
Power saving per motor = 18.09 − 17.51 = 0.584 kW (book relation: kW saving = kW output × [1/ηold − 1/ηnew]).
Annual energy saving per motor = 0.584 × 7,000 = 4,088 kWh; for 50 motors = 204,400 kWh/year.
Annual cost saving at Rs. 10/kWh = Rs. 20,44,000 (≈ Rs. 20.4 lakh).
CO₂ reduction at 0.85 kg CO₂/kWh = 204,400 × 0.85 = 173,740 kg ≈ 173.7 tonnes CO₂ per year.
Input power = output/efficiency for each class; per-motor saving × hours × 50 gives kWh; cost = kWh×rate; CO2 = kWh×0.85.
Source: Mar 2023
📖 §2.4 Motor efficiency (no-load test and full-load efficiency assessment)
13. An efficiency assessment test was carried out for a standard 4 pole squirrel cage induction motor in a chemical plant. The motor specifications are as under: Motor rated specification: 3 phase delta connected, 37 kW, 415 Volt, 63 Amps, 1475 rpm. The following data was collected during the no-load test on the motor: Voltage = 415 Volts; Current = 17 Amps; Frequency = 50 Hz; Stator resistance per phase = 0.260 Ohms at 30 °C; No load power = 1152 Watts. Calculate the following: (i) Iron plus friction and windage losses. (ii) Stator resistance at 120 °C. (iii) Stator copper loss at full load at operating temperature of 120 °C. (iv) Full load slip and rotor input assuming rotor losses are slip times rotor input. (v) Motor input assuming that stray losses are 0.5% of the motor rated output power. (vi) Motor full load efficiency
Model answer: (i) Iron plus friction and windage loss, Pi+fw
No load power, Pnl = 1152 Watts
Stator copper loss at 30 °C, Pst.cu = 3 x (17/√3)² x 0.260 = 75.13 Watts
Pi+fw = Pnl - Pst.cu = 1152 - 75.13 = 1076.87 W
(ii) Stator resistance at 120 °C
R(120 °C) = 0.260 x (120 + 235)/(30 + 235) = 0.3483 ohms per phase
(iii) Stator copper losses at full load at 120 °C
Pst.cu(120 °C) = 3 x (63/√3)² x 0.3483 = 1382.3 Watts
(iv) Full load slip
S = (1500 - 1475)/1500 = 0.01666 or 1.66%
Rotor input, Pr = Poutput/(1 - S) = 37000/(1 - 0.01666) = 37000/0.98334 = 37626.86 Watts
(v) Motor full load input power
Pinput = Pr + Pst.cu(120 °C) + (Pi+fw) + Pstray
= 37626.86 + 1382.3 + 1076.87 + (0.005 x 37000)
= 40271.03 Watts
(where stray losses = 0.5% of rated output, assumed)
(vi) Motor efficiency at full load = (Poutput/Pinput) x 100
= (37000/40271.03) x 100 = 91.87 %
The five standing rules for this classic 10-marker: (1) for a DELTA-connected motor the phase current is I_line/sqrt(3), so stator copper loss = 3 x (I/sqrt(3))^2 x R_phase; (2) resistance rises with temperature as R2 = R1 x (235 + t2)/(235 + t1) for copper - 235 is the copper constant, and using 273 here is a standard mark-loser;
(3) iron + friction & windage = no-load input minus the no-load stator copper loss (1152 - 75.13 = 1076.87 W); (4) rotor input = output/(1 - s) with s = (1500 - 1475)/1500 = 0.0167; (5) stray loss is taken as 0.5% of rated output.
Input = 37,626.86 + 1382.3 + 1076.87 + 185 = 40,271 W, so efficiency = 37,000/40,271 = 91.87%. Note the copper loss is computed twice - once cold at 30 degC for the no-load split, once hot at 120 degC for the full-load loss.
Source: Aug 2013
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