physics problem on current electricity (2 parts)?

C

Cindy

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A typical household circuit is capable of carrying 30.0 Amps of current at 120 V before the circuit breaker will trip. How many 1500-W hair dryers can run off one such circuit?

In the rush to get ready for lecture, a physics professor leaves the hair dryer described in the previous problem running and does not turn it off until he gets home 8.1 hours later. How much will this add, in dollars, to his next electric bill (assume electricity costs $0.078 per kilowatt-hr)?

please show all work studding for a test.
 
Let n be the number of hair dryers. Then the total watt will become 1500 n W. The maximum power that can be provided by the supply is P =VI = 120*30 = 3600 W.
So by equating 1500 n = 3600, we can get the value of n as 2.4. So n should not exceed 2.4. This means only 2 hair dryers can be used.
So in one second electrical energy consumed will be 1500 J. For 8.1 hour there will be 8.1*3600 seconds. So daily the energy consumed will be 1500*8.1*3600 J.
In kilo watt hour, it will be arrived by dividing this by 3.6*10^6 J. This comes to 12.15 kW hr.
So the additional cost will come to $0.9477
 
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