Question

An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case?

Answer 1

Case: 1
Power (P) = 420W
Applied Voltage (V) = 220V
Current \(\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{420}{220}=1.9 \mathrm{A}\)

Case: 2
Power (P) = 180 W
Applied Voltage (V) = 2.20 V
Current \(\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{180}{220}=0.8 \mathrm{A}\)