Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust
Question Statement:
A single phase 230 V electric motor while running at 1400 rpm develops a torque of 3.1 Nm. If the phase angle between the voltage and current is 38° and the power efficiency of the motor is 80%, the amount of electric current drawn by the electric motor is
Options:
(A) 2.470 A
(B) 3.135 A
(C) 4.810 A
(D) 5.512 A
Official Answer Key:
A, B
Step-by-Step Solution & Explanation
Output (shaft) power: P_out = Tω = 3.1 × (2π×1400/60) = 3.1 × 146.6 = 454.5 W. Input electrical power: P_in = P_out/η = 454.5/0.80 = 568.1 W. For single-phase AC: P_in = V·I·cos φ ⟹ I = P_in/(V cos φ) = 568.1/(230×cos38°) = 568.1/(230×0.788) = 3.135 A.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.