A electric iron consumes energy at the rate of
(a) `840 W` when heating is at the maximum rate and
(b) `360 W` when the heating is at the minimum. The voltage is `220 V` What are the current and the resistance in each case ?

Answer

(a) `3.82 A, 57.60 Omega`
(b) `1.64 A, 134.15 Omega`.

Explanation

(a) `P = VI, I = (P)/(V) = (840 W)/(220 V) = 3.82 A`
`R = (V)/(I) = (220 V)/(3.82 A) = 57.60 Omega`
(b) `P' = Vi', I' = (P')/(V) = (360 W)/(220 V) = 1.64 A`
`R' = (V')/(I') = (220 V)/(1064 A) = 134.15 Omega`.