An electric iron is rated 1.5 kW at 100 V. Calculate the capacity of the fuse that should be used for the electric iron.
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Answer:
Power of the electric iron P=2 kW
Time for which the iron is operated daily t=2 h
Daily energy consumption E
daily
=Pt=2 kW×2 h=4 kWh
Total energy consumed in one week E
week
=7×E
daily
=7×4=28 kWh
Total cost =Rs.4.25×E
week
=Rs.4.25×28=Rs.119
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