Engineering Power, Circuits & Electrical Outlets: Calculate Cost & Usage

[sabak22]

while studying. There are three 190 watts light bulbs and a radio with an internal resistance of 46.0 ohms plugged into the living room electrical outlets. You decide to have a break and have a snack, turning on the pop-corn machine that draws 6.00 A.

a)How much electrical power are you using? Data: Assume the electric company provide you with 110 V potential difference.

b)All the electrical outlets are part of the same 110-V circuit which has a 10-A circuit breaker. How much current in amperes (A) is required to feed all the devices described above? This tells you if you should or should not turn on the pop-corn machine.

c) You are so tired after studying that you leave all of these household devices turned on for 11.0 hours. Calculate the cost (in dollars) of the electric bill, if the commercial rate is $.09/kWh? (Do not enter units).

Sadly I am unable to answer to any of these questions, I have been stuck on these ones for so long!

 

[issacnewton]

first thing you should know is that all the appliances in the home are connected in parallel...
in the above problem, conceptualize each appliance as a resistor...there are 3 light bulbs, a radio and a pop-corn machine...so think of them as 5 resistors connected in parallel

 

[vk6kro]

You also need to know a few formulas.

First:

Power (in watts) = voltage (in volts) * current (in amps)

eg 190 watts = 110 volts times the current.

so current (in amps) = 190 watts / 110 volts.

Second:

the currents drawn by devices in parallel add up. So the 3 lamps draw 3 times as much current as one lamp does.

Third:

energy used (in watt hours) = power (in watts) * time (in hours)

Fourth:

a kilowatt hour is the energy used if one kilowatt is supplied for one hour. So, it is the product of the power times the time.

Fifth:
Ohms's Law.
Current (in amps) = voltage (in volts) divided by resistance (in Ohms)

 

[Naty1]

Here is an example:

A kilowatt is 1,000 watts, so a 100 watt light bulb is 100/1,000 or 0.1 kw.
A tenth of a kilowatt.

If a kilowatt hour costs $.09, 9 cents/hr, then 1,000 watts (a kilowatt) running for one hour costs $.09...then a 100 watt light bulb would cost 1/10 that or $.009...

Here in NJ,USA, we pay about $.18/kwh..double that of your example!

 

[sabak22]

Okay great so I got the first one right, Ptot= 1.49*10^(-3) W. But for the second part, what do they mean by "All the electrical outlets are part of the same 110-V circuit which has a 10-A circuit breaker"? And what am I expected to do?

 

[Redbelly98]

Moderator's note: thread moved to Homework & Coursework Questions. The usual rules for giving homework help are in effect.

 

[vk6kro]

Okay great so I got the first one right, Ptot= 1.49*10^(-3) W. But for the second part, what do they mean by "All the electrical outlets are part of the same 110-V circuit which has a 10-A circuit breaker"? And what am I expected to do?

That would be 1.49 milliWatts. Are you sure?
Try working in watts and drop the Scientific Notation if it confuses you.

A circuit breaker is like a fuse, except you can reset it. So they are saying the maximum current you can use is 10 amps.

 

[sabak22]

Thank vk6kro.
So what I did was, I took the Ptot and divided it by 110V which gave me the total current running through the circuit. :)

Now what about the last part?
"You are so tired after studying that you leave all of these household devices turned on for 11.0 hours. Calculate the cost (in dollars) of the electric bill, if the commercial rate is $.09/kWh?". I don't remember doing an example like this in class

 

[sabak22]

Nvm got it! i was quite simple.

 

[Redbelly98]

sabak22, please do the following:

1. Post your other problem (which I have deleted) in a new and separate thread, and do so here:

https://www.physicsforums.com/forumdisplay.php?f=158

2. Be sure to fill in all sections of the homework template that appears.

3. Please read the Private Message (PM) I sent you earlier, if you have not already seen it.

Thanks for your cooperation.