An electric iron consumes energyat 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?

An electric iron consumes energyat a rate of 840 W, when heating is at

1.	An electric iron consumes energyat 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?
Answer» Solution :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.