Question

an electric iron consumes energy at a rate of 840w ..when heating is at the maximum .the ..the voltage .is .220v ..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

Explanation: