Question

An electric motor is rated at 2 HP – 250 V. Calculate:

1. Current flowing through it.
2. Energy consumed by it in one second.
3. Time in which it will consume 90 kWh of energy.

[1 HP = 750 W]

Answer 1

P = 2 HP = 2 × 750 = 1500 W; V = 250 V

(a) Current flowing through it:

\[I = \frac{P}{V}\]

\[= \frac{1500}{250}\]

= 6 A

(b) Energy consumed by it in one second:

E = P × t

= 1500 W × 1 s

= 1500 J

(c) Time in which it will consume 90 kWh of energy:

E = 90 kWh; P = 1500 W = 1.5 kW

\[t = \frac{E}{P}\]

\[= \frac{90}{1.5}\]

= 60 h