What will be the capacity of the current used for the supply, which operates on single 240 V AC phase, a motor with 75% efficiency of 5 HP and 0.8 power factor.

Concept:1 HP = 735.5 Watt.Efficiency is the ratio of OUTPUT power to the motor to input power to the motor.Electrical input power P = V × I × cos ϕWhere, P = PowerV = Supply VoltageI = Input Currentcos ϕ = Power Factor (ϕ = ANGLE between supply voltage and input current) Note:- If rating of any machine is given that means that is its output power rating.Calculation:Given: Output power rating of SINGLE PHASE Induction motor = 5 HP = 5 × 735.5 W = 3677.5 W.Supply voltage (V) = 240 VEfficiency (η) = 75 % = 0.75Power factor (cos ϕ) = 0.8 \(\eta = \frac{{Output\;power}}{{Input\;power}}\)\(Input\;Power\;\left( P \right) = \;\frac{{Output\;power}}{\eta } = \;\frac{{3677.5}}{{0.75}} = 4903.33\;W\)Now P = V × I × cos ϕ4903.33 = 240 × I × 0.8I = 25.53 A