Question

An electric iron consumes energy at a rate of 840 W when heating
is at the maximum rate and 360 W when the heating is at the
minimum. The voltage is 220 V. What are the current and the
resistance in each case?​

Answer 1

Answer:

When heating at max rate,

power, P = 840W

V = 220V

We know P = VI

⇒ 840 = 220 × I

⇒ I = 840/220

⇒ I = 4 A

R = V/I = 220/4 = 55 Ω

So current = 4A and resistance = 55Ω

When heating at minimum rate,

power, P = 360W

V = 220V

We know P = VI

⇒ 360 = 220 × I

⇒ I = 360/220

⇒ I = 1.636 A

R = V/I = 220/1.636 = 134.45 Ω

So current = 1.636A and resistance = 134.45Ω

Answer 2

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