Transistor radio/kWh problem

[wakefreak90]

**Transistor radio/kWh problem**

A transistor radio operates by means of a 10.0 V battery that supplies it with a 48 mA current.

(a) If the cost of the battery is $0.55 and it lasts for 300 h, what is the cost per kWh to operate the radio in this manner?
3.819 dollars/kWh


(b) The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays 6¢ per kWh. What does it now cost to operate the radio for 300 h?
? cents

I got Question A but I have tried everything on Question B and I do not understand what I am doing wrong. For some reason, I think it is easier than what I'm trying to do.

Please help!

 

[Dadface]

Work out the number of kilowatt hours and multiply by 6.Note that the question gives the current in mA and not A and note also the the power must be expressed in kW and not W.

 

[wakefreak90]

So if one kWh = 3.6E6 J, and the total energy is 518,400 J..

(3.6E6)/(518400) = 6.9444

--> (6.9444)(6) = 41.666

Is 41.666 cents the answer? I only have two more tries left and I don't want to get it wrong. I'm not sure if those are the correct unit conversions

 

[Dadface]

W=10*48/1000
kW=above/1000
kWh=above*300
cost=above*6