Question

You are cooking breakfast for yourself and a friend using a 1100 W waffle iron and a 490 W coffeepot. Usually, you operate these appliances from a 110 V outlet for 0.500 h each day.

(a) At 12 cents per kWh, how much do you spend to cook breakfast during a 30.0 day period?
$  

(b) You find yourself addicted to waffles and would like to upgrade to a 2200 W waffle iron that will enable you to cook twice as many waffles during a half-hour period, but you know that the circuit breaker in your kitchen is a 20 A breaker. Can you do the upgrade?
yes or no

Give the current that this new waffle iron (along with the coffee pot) draws to support your answer.
I =  A

Answer 1

a) The energy used by the waffle iron and the coffeepot for one day is
E = P x t
E = (1200 W x 0.500 + 490 x 0.500 )
E = 845 Wh
The energy used per month in kWh
E = 0.845 x 30 =  25.35 kWh
The money spent = 25.35 kWh x 12 cents/ kWh
=  304.2 cents
= 3.042 dollars

b) No,
The circuit breaker allows a maximum current draw of 20 A. If the 2200 W waffle iron is used, the current drawn is
I = P / V
I = 2200 W / 110 V
I = 20 A
and the coffeepot draws a current of magnitude
I = 490 / 110 V
I =  4.45 A
Thus the total current drawn from the circuit is
I = 24.45 A
Which is higher than the circuit breaker permits. Thus the waffle iron can not be used along with the coffee pot