What Is the Cost to Operate a Transistor Radio for 300 Hours on Household Power?

  • Thread starter Thread starter wakefreak90
  • Start date Start date
  • Tags Tags
    Transistor
Click For Summary

Homework Help Overview

The discussion revolves around calculating the cost to operate a transistor radio using both a battery and household power over a specified duration. The subject area includes electrical power calculations and cost analysis related to energy consumption.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the conversion of current from milliamperes to amperes and the calculation of power in kilowatts. There are attempts to calculate the total energy consumed and its cost based on the household electricity rate.

Discussion Status

Some participants have provided calculations and suggested methods for determining the cost of operating the radio on household power. However, there is uncertainty regarding unit conversions and the correctness of the calculations, particularly in the context of the second question.

Contextual Notes

Participants note the importance of correctly converting units from milliamperes to amperes and watts to kilowatts, as well as the implications of the different power sources on cost calculations. There is also a mention of limited attempts remaining for the original poster to solve the problem.

wakefreak90
Messages
4
Reaction score
0
**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!
 
Physics news on Phys.org


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.
 


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
 


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

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 6 ·
Replies
6
Views
5K
Replies
12
Views
3K
Replies
7
Views
5K
  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 7 ·
Replies
7
Views
3K