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

Explanation

From equation of electric power, we know that the power input is P = VI.
Thus, the current `I = P/V`
( a ) When heating is at the maximum rate,
`I = (840 W)/(220V) = 3.82 A ,`
and the resistance of the electric iron is
`R =V/I = (220V)/(3.82A) =57.60 Omega.`
(b )When heating is at the minimum rate,
`I =(360W)/(220V) = 1.64A `,
and the resistance of the electric iron is
`R = V/I = (220W)/(1.64A) = 134.15 Omega`
Note: It is clear that when the heating rate of an electric iron is different, resistance of its heating coil is different eventhough the source voltage remains the same.