How Much Does It Cost to Run Household Appliances?

  • Thread starter Thread starter Dx
  • Start date Start date
  • Tags Tags
    Electric
AI Thread Summary
The discussion focuses on calculating the cost of energy for household appliances. For a 9V battery powering a CD player, the cost of energy is determined to be $8.98 per kWh, based on the battery's current and usage time. In the second question, the cost contribution of an AC unit and lightbulbs to a monthly electric bill is calculated, resulting in a total monthly cost of $11,760. The calculations involve determining power usage and applying the cost per kWh for accurate billing. Overall, the thread emphasizes the importance of correct unit conversions and careful mathematical operations in energy cost calculations.
Dx
I have two questions that i need help with, please?

1) A 9V battery costs $1.49 and will run a portable CD player for 6 hours. Suppose the battery supplies a current of 25mA to the player. What is the cost of energy in dollars per kWh?

I started out using ohms law but am lost for there. How do i set up this for a formula i can solve for dollars per kWh?




2) The monthly(30days) electric bill included the cost of running a AC unit for 2hr/day at 5000W, and a series connection of ten 60W lightbulbs for 5hr/day. How much did these items contribute to the cost of the monthly electric bill if electricity costs 8.0 per kWh?

1st i added the ten 60W = 600W. Next, i took (2hr/d)*(30d) = 60h. Also took (5hr/d)*(30d)=150h, i added 5KW + 600W = 5.6KW for total wattage. I plugged in a formula like so (5.6k)*(2.1k)* ?

How do i find the daily cost of electricity, my answer is 2.86x10^-1, i got that from (8/kWh) / 4weeks = ans. Then ans / 7 days = 2.86 x 10^-1

But i am not right, what am i doing wrong, can you help me solve this please?
 
Last edited by a moderator:
Physics news on Phys.org
Originally posted by Dx
I have two questions that i need help with, please?

1) A 9V battery costs $1.49 and will run a portable CD player for 6 hours. Suppose the battery supplies a current of 25mA to the player. What is the cost of energy in dollars per kWh?

I started out using ohms law but am lost for there. How do i set up this for a formula i can solve for dollars per kWh?


This problem is pretty easy, so I don't want to give too much away. Here are the overall steps:

1. Calculate the power delivered to the CD player. You have all the info for that.
2. Calculate the total energy delivered to the CD player. You were given one piece of info for that, and you will calculate the other piece in Step 1.
3. Calculate the cost for that energy. You were given one piece of info for that, and you will calculate the other piece in Step 2.

Give it a shot.

2) The monthly(30days) electric bill included the cost of running a AC unit for 2hr/day at 5000W, and a series connection of ten 60W lightbulbs for 5hr/day. How much did these items contribute to the cost of the monthly electric bill if electricity costs 8.0 per kWh?

1st i added the ten 60W=6KW.

I assume you meant to say 60W+[/color]6kW. That's fine, just don't forget to convert kW-->103W.

Next, i took (2hr/d)*(30d) = 60h. Also took (5hr/d)*(30d)=1.5Kh,

You made a mistake. 1.5Kh=1500h, which is wrong.

I am having trouble following you from here. Please correct the above mistake, check your math carefully, and come back if you need more help.
 



1) To calculate the cost of energy in dollars per kWh, we can use the formula: Cost = Power x Time x Cost per unit. In this case, the power is 25mA (0.025A), the time is 6 hours, and the cost per unit is $1.49. So, the cost of energy for the battery is: Cost = 0.025A x 6h x $1.49 = $0.2245.

To find the cost per kWh, we need to convert the units to kW and hours. 1 kWh is equal to 1000 watts for 1 hour. So, we can divide the cost by the number of watts (25mA or 0.025A) and multiply it by 1000 to get the cost per kWh.

Cost per kWh = ($0.2245 / 0.025A) x 1000 = $8.98.

Therefore, the cost of energy for the battery is $8.98 per kWh.

2) To find the daily cost of electricity, we can use the formula: Cost = Power x Time x Cost per unit. In this case, the power is 5.6kW, the time is 7 hours (2 hours for AC and 5 hours for lightbulbs), and the cost per unit is $8.0 per kWh. So, the cost of electricity for one day is: Cost = 5.6kW x 7h x $8.0/kWh = $392.

To find the monthly cost, we can multiply the daily cost by the number of days in a month (30 days).

Monthly cost = $392 x 30 days = $11,760.

Therefore, the items contributed $11,760 to the monthly electric bill.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top