Question

An electric iron is rated at 230 V, 750 W. Calculate; (i) the maximum current, and; (ii) the number of units of electricity it would use in 30 minutes.

Answer 1

Answer:

(i) Maximum current drawn = 3.36 A

(ii) Number of units of electricity used = 0.375 unit

Explanation:

As per the question,

Given data for an electric iron:

Voltage = 230 V

Power = 750 W

Time = 30 minutes = 0.5 hours

Formula used:

Power = Voltage × Current

Energy = Power × Time

(i) For the maximum current drawn:

Power = Voltage × Current

⇒ 750 = 230 × I

⇒ I = 3.26 Ampere

∴ Maximum current drawn = 3.36 A

(ii) For the number of units of electricity:

Energy = Power × Time

⇒ Energy = 750 × 0.5

⇒ Energy = 0.375 kWh

As we know that 1 kWh = 1 unit

Therefore, the number of units of electricity used by an electric iron = 0.375 unit

Answer 2

V = 230V

P = 750W

T = 30 minutes = 0.5 hours

I = ?

P = V × I

750 = 230×I

I=750÷230

I=3.26 A

Therefore, the maximum current would be 3.26A

Now,

Energy=?

E=P×T

E=750×0.5

E=0.375kW

Therefore, the number of units of electricity it would use in 30 minutes is 0.375kWh