Question

An electric iron consumes energy at a rate of 840w when heating is at the max rate and 360w when heating is at minimum.the voltage is 220v,what are the current and resistance in each case

Answer 1

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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