1.

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

Answer»

When heating at max rate,power, P = 840WV = 220V

We know P = VI⇒ 840 = 220× I⇒ I = 840/220⇒ I = 4A

R = V/I = 220/4 = 55Ω

So current = 4A and resistance = 55Ω

When heating at minimum rate,power, P = 360WV = 220V

We know P = VI⇒ 360 = 220× I⇒ I = 360/220⇒ I = 1.636A

R = V/I = 220/1.636 = 134.45 Ω

Socurrent = 1.636A and resistance = 134.45Ω

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