1.

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 220V. What is the current in each case.

Answer»

Solution :Case : 1
POWER (P) = 420W
Applied Voltage (V) = 220V
CURRENT I `=(P)/(V)=(420)/(220)=1.9A`
Case : 2
Power (P) = 180W
Applied Voltage (V) = 2.20V
Current I `=(P)/(V)=(180)/(220)=0.8A`


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