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A 1 kQ heater is meant to operate at 200 V. (a) what is its resistance ? (b) How much power will it consume if the line voltage drops to 100 V? (c) how many units of electrical energy will it consume in a month (of 30 days) if it operates 10 hr daily at the specified voltage? |
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Answer» The resistance of an elecric appliance is given by `R=(V_(S)^(2))/(W) so, R=((200)^(2))/(1000)=40Omega` (b) The actual power consumed by an electric appliance is given by `P=((V_(A))/(V_(S)))^(2)xxW` so, `P=((100)/(200))^(2)xx1000=250W` (c) The total electrical energy consumed by an electric appliance in a specified time is given by `E=(sumW_(1)h_(1))/(1000)kWh` so, `E=(1000xx(10xx30))/(1000)=300kWh` |
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