ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersMotors and power supply on farms and in processing plants: choosing, starting, protecting and paying for electric drives.
In the exams: ICAR AIEEA-PG Unit II (explicit; not in the GATE AG syllabus); NCEES PE electric motors; rural electrification in PRC ABE
ICAR AIEEA-PG — Agricultural Engineering & Technology (Unit II)
Quick recall and one-step numericals on electric motors and power distribution: synchronous speed and slip, rotor frequency, torque–slip relations and the effect of voltage and rotor resistance, star-delta and autotransformer starting ratios, single-phase motor types and their starting methods, DC motor EMF, back EMF and speed, transformer connections and all-day efficiency, and power-factor correction.
How our questions for ICAR AIEEA-PG spread across the units of Electrical Machines & Power Utilisation. Difficulty: 41 easy, 28 medium, 3 hard.
| Three-phase induction motors: slip and torque-speed curve | 10 | |
| Single-phase motors, including capacitor-start | 9 | |
| Wiring and protection: fuses, MCB, ELCB and earthing | 9 | |
| DC machines | 8 | |
| Motor starters | 7 | |
| Variable-frequency drives | 7 | |
| Rural power distribution and tariffs | 7 | |
| Transformers in practice | 6 | |
| Power factor correction | 5 | |
| Selecting motors for pumps, threshers and processing machines | 4 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Which DC motor gives the highest starting torque per ampere and so suits traction-type loads?
C. Series motor
In a series motor flux rises with armature current, so torque rises roughly as Ia² below saturation, giving the highest starting torque per ampere. Shunt, separately excited and permanent-magnet motors have nearly fixed flux, so torque rises only in proportion to current; weakening the field lowers torque per ampere further.
What is the full-load low-voltage line current of a three-phase 63 kVA, 11 kV/433 V distribution transformer feeding an irrigation cluster?
A. 84.0 A
I = S/(√3 V) = 63000/(1.732 × 433) = 84.0 A. 145.5 A forgets the √3 of a three-phase supply. 3.31 A is the HV-side current. 48.5 A divides by 3 instead of √3.
What must a three-phase motor have for it to be started with a star-delta starter?
D. Six winding ends brought out and a delta rating at line voltage
The starter connects the windings first in star and then in delta, so both ends of each phase (six terminals) must be accessible, and the motor must run in delta at the supply line voltage. Slip rings are for rotor-resistance starting; centrifugal switches belong to single-phase motors.
At roughly what speed does the centrifugal switch of a split-phase or capacitor-start motor disconnect the starting winding?
B. About 70–80 % of synchronous speed
The switch opens at about 70–80 % of synchronous speed, after the motor has enough running torque on the main winding. At 20–30 % the motor would lack torque; induction motors never reach or exceed synchronous speed in motoring.
Under IEC 60034-1, what does duty type S1 mean?
A. Continuous running at constant load until thermal equilibrium
S1 is continuous duty: constant load long enough to reach thermal equilibrium, as for irrigation pumps. S2 is short-time duty, S3 intermittent periodic duty, and higher S-numbers cover starting, braking and speed changes.
A 400 V, 50 Hz motor is fed by a VFD at 35 Hz under constant V/f control. What voltage does the drive apply?
B. 280 V
V/f = 400/50 = 8 V/Hz; at 35 Hz V = 8 × 35 = 280 V. Keeping 400 V would over-flux and saturate the motor. 196 V scales by (35/50)². 571 V inverts the ratio.
A centrifugal pump working against friction only is slowed to 80 % speed. What happens to its head?
C. Falls to 64 % of the original
Head ∝ N², so 0.8² = 0.64. 80 % is the flow ratio, 51.2 % the power ratio. Head does not stay constant when speed changes.
What is the main purpose of converting a rural low-voltage network to a high-voltage distribution system (HVDS)?
C. Shorten low-voltage lines by placing small transformers near each few consumers
HVDS takes the medium-voltage line close to loads and uses many small transformers, so LT lines are short — cutting losses, improving voltage and making illegal hooking harder. It does not change consumer voltage, replace MV lines with LT cable, or convert the network to SWER.
A 4-pole alternator in a farm standby generator set must supply 50 Hz. At what speed must the engine drive it?
B. 1500 rev/min
N = 120f/P = 120 × 50/4 = 1500 rev/min (check: f = PN/120 = 4 × 1500/120 = 50 Hz). 3000 rev/min is for a 2-pole machine; 750 rev/min for 8 poles; 1800 rev/min gives 60 Hz.
Which device protects a pump motor against running on two lines when one phase of the rural supply fails?
D. Single-phasing preventer or phase-failure relay
A phase-failure relay (single-phasing preventer) detects loss or severe imbalance of a phase and trips the contactor; a correctly set overload relay also helps. Surge arresters handle lightning, HRC fuses clear short circuits, and RCDs detect earth leakage — none detects single-phasing reliably.
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