QuestionAn electric motor is connected to a 12 V supply and draws a current of 2.5 A. It lifts a 3.0 kg mass at a steady speed through a height of 1.5 m in 4.0 s. (g = 9.81 N kg−1)(a) Calculate the electrical power input to the motor.
(b) Calculate the charge that flows through the motor in 4.0 s.
(c) Calculate the efficiency of the motor.
(d) Suggest why the efficiency is less than 100%.(a)P = VI = 12 V × 2.5 A = 30 W (b)Q = It = 2.5 A × 4.0 s = 10 C The electrical energy input is therefore W = QV = 10 C × 12 V = 120 J, which agrees with Pt = 30 W × 4.0 s.(c) The useful output is the gain in gravitational potential energy:ΔEgrav = mgΔh = 3.0 × 9.81 × 1.5 = 44 J efficiency = useful energy output / total energy input = 44 J / 120 J = 0.37 (37%) (d) The current heats the resistance of the motor's coils (power I2R), and energy is also dissipated by friction in the bearings and as sound. This energy is wasted, so the useful output is less than the input.Exam tip: efficiency has no unit. Give it as a decimal or a percentage, and keep full values until the last step: 44.1 J / 120 J = 0.368, which rounds to 0.37.
(b) Calculate the charge that flows through the motor in 4.0 s.
(c) Calculate the efficiency of the motor.
(d) Suggest why the efficiency is less than 100%.(a)