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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? |
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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 Like my answer if you find it useful! |
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