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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?
Solution
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 \(\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}=\frac{180}{220}=0.8 \mathrm{A}\)
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