A radio operates by means of a 9.0-V battery that supplies it with a 50.0-mA current

Question

A radio operates by means of a 9.0-V battery that supplies it with a 50.0-mA current

a)If the cost of the battery is $2.49 and it lasts for 300.0 h, what is the cost per kWh to operate the radio in this manner?

b)The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays $0.12 per kWh. What does it now cost to operate the radio for 300.0 h?

Please show work

Thank you

Answer 1

a) The power used by the radio is the voltage times the current so (9)(0.050) = 0.45 Watts. Converting watts to kilowatts means dividing by 1000 so 0.45/1000 = 4.5 x 10^-4 kW. To determine the kW-hours, just multiply the kilowatts and hours so (4.5 x 10^-4 kW)(300 hours) = 0.135 kWh. The cost per kWh is $2.49/(0.135 kWh) = $18.44/kWh

b) Assuming that the converter is 100% efficient, the power used is still 0.45 W or 4.5 x 10^-4 kW and for 300 hours, that is still 0.135 kWh. If the cost is $0.12 per kWh, then the cost to operate for 300 hours is (0.135 kWh)•($0.12/kWh) = $0.0162 or about 2 cents