Objective: applying physics to real life by solving the cost to run machines running on volts

Concept: use equations to solve for kilowatt per hour and cost to run machines

Real Life Applications:   -electricians
                                            -power companies
 

Equations:

    power(watts) = volts x current(amps)

    watts/1000 = kilowatts

    kilowatts x hours = kw/hrs

    kw/hrs x days = total kw/hrs

    total kw/hrs x amount of money per kw/hr = total cost
 

Examples:

        1. A freezer is on approximatly for 12 hrs. The amount of volts is 110 and its current is 10 amps.  How much will the
        cost be in 6 months (about 180 days) if the cost is $0.15 per kw/hr?

        2. A computer monitor is on for 2 hrs a day.  Its current is 1.4 amps and runs on 115.7 volts.  There are about
        300 computers in the school.  How much does it cost to run for a month if the cost is $0.11 per kw/hr.

        3. A furnace is on for 11 hrs a day in the winter.  Its current is 4 amps and runs on 110 volts.  The cost is $0.12
        per kw/hr.  How much does it cost to run in a yr.
 
 
 

Solutions:

1.) 110 v x 10 amps = 1100 watts
     1100 w/1000 = 1.1 kw
     1.1 kw x 12 hrs = 13.5 kw/hrs
     13.2 kw/hrs x 180 days = 2376 kw/hrs
     2376 kw/hrs x $0.15 = $356.40
     * So, for the cost to be less on your bill get a freezer with less amps or move you freezer to a colder area like your garage.
2.)  115.7 v x 1.4 amps = 161.98 watts
     161.98 w/1000 = .16198 kw
     .16198 kw x 2 hrs = .32396 kw/hrs
     .32396 kw/hrs x 180 days = 58.3128 kw/hrs
     58.3128 kw/hrs x $0.11 = $6.414408 about $6.41

3)   110 v x .4 amps = 44 watts
     44/1000 = .044 kw
     .044 kw x 11 = .484 kw/hrs
     .484 kw/hrs x  365 days = 176.66 kw/hrs
     176.66 kw/hrs x $0.12 = $21.1992 about $21.20
 

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