An electric iron consumes energy at a rate of $840\, W$ when heating is at the maximum rate and $360\, W$ when the heating is at the minimum. The voltage is $220\, V$. What are the current and the resistance in each case?
(N/A) We know that the power input is $P = V \times I$.
Thus,the current is $I = P / V$.
$(a)$ When heating is at the maximum rate $(P = 840\, W)$:
$I = 840 / 220 = 3.82\, A$.
The resistance is $R = V / I = 220 / 3.82 = 57.60\, \Omega$.
$(b)$ When heating is at the minimum rate $(P = 360\, W)$:
$I = 360 / 220 = 1.64\, A$.
The resistance is $R = V / I = 220 / 1.64 = 134.15\, \Omega$.
Thus,the current is $I = P / V$.
$(a)$ When heating is at the maximum rate $(P = 840\, W)$:
$I = 840 / 220 = 3.82\, A$.
The resistance is $R = V / I = 220 / 3.82 = 57.60\, \Omega$.
$(b)$ When heating is at the minimum rate $(P = 360\, W)$:
$I = 360 / 220 = 1.64\, A$.
The resistance is $R = V / I = 220 / 1.64 = 134.15\, \Omega$.
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