solve this numerical . an electric iron consumes energy at the 840volt .When heating at the maximun rate at 360v . when heating up is at the maximum rate voltage is 220volt . what is the current and the resistance in each case
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Power input (P)=IV
Heating at Maximum rate (I)=840/220=3.82A
Resistance is = V/I = 220/3.82 = 57.79Ω
Heating at minimum rate= 360/220 =1.64A
Resistance is= 220/1.64 = 134.15Ω
Heating at Maximum rate (I)=840/220=3.82A
Resistance is = V/I = 220/3.82 = 57.79Ω
Heating at minimum rate= 360/220 =1.64A
Resistance is= 220/1.64 = 134.15Ω
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