For a single-phase motor of 2 HP rating, supply voltage is 240 V ac. If the efficiency is 70% and power factor is 0.8, find the input current.

Concept:

1 HP = 746 Watt.

Efficiency is the ratio of output power to the motor to input power to the motor.

Electrical input power P = V × I × cos ϕ

Where, P = Power

V = Supply Voltage

I = Input Current

cos 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 = 2 HP = 2 × 746 W = 1492 W.

Supply voltage (V) = 240 V

Efficiency (η) = 70 % = 0.7

Power factor (cos ϕ) = 0.8  

\(\eta = \frac{{Output\;power}}{{Input\;power}}\)

\(Input\;Power\;\left( P \right) = \;\frac{{Output\;power}}{\eta } = \;\frac{{1492}}{{0.7}} = 2131.43\;W\)

Now P = V × I × cos ϕ

2131.43 = 240 × I × 0.8

I = 11.1 A

So our closest possible answer is 10.95 A.