An electric motor operating at a voltage of 380 V and consuming a current of 4 A

An electric motor operating at a voltage of 380 V and consuming a current of 4 A lifts a 150 kg load to a height of 12 m in 20 s. Efficiency of the electric motor ..

Let’s start solving the problem by determining the efficiency. Efficiency is the ratio of useful work to complete work, as a percentage (efficiency = (A useful / Acomplete) * 100%)). Under the conditions of this problem, we call the power of the electric motor (or electric power) full work. As you know, it will be equal to the product of the voltage (in this case, the electric motor) by the current that it consumes and will be equal to: P = U * I = 380 * 4 = 1520 V (this power expresses the full “potential” of this motor – the limit value , which he cannot exceed). This engine is adapted for lifting loads. Lifting the load (mechanical work) in this case will be his useful work. The load rises to a height of 12 meters evenly and in a straight line in 20 seconds (that is, for any equal intervals of time it travels equal distances) – we find its speed: v = s / t = 12/20 = 0.6 m / s. The formula of mechanical power for a uniformly moving body: N = m * g * v = 150 * 10 * 0.6 = 900 J (where m is the mass of the load; g is the gravitational acceleration = 9.8 (for simplicity of calculations, we will round up to the value equal to 10); v-speed of the load). Now we can calculate the efficiency = (900/1520) * 100% = 59%. Therefore, the correct answer in this problem is at number 3.