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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. |
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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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