Electric Circuits Legacy Problem #14 Guided Solution
Problem*
A central air conditioner in a typical American home operates on a 220-V circuit and draws about 15 A of current.
- Determine the power rating of such an air conditioner.
- Determine the energy consumed (in kW•hr) if operated for 8 hours per day.
- Determine the monthly cost (31 days) if the utility company charges 13 cents per kW•hr.
Audio Guided Solution
Electrical devices consume electrical energy, and the rate at which they consume this energy is referred to as the power. And the devices in the home that consume the most energy, or the energy at the greatest rate, tend to be those which are heating and cooling devices, such as the central air conditioner. Here in Part A, we're asked to calculate the power rating of an air conditioner that operates on a 220 volt circuit and draws about 15 amps of current. We use the equation that the power is equal to the current multiplied by the electric potential difference and expressed across the device. That would be the current of 15 amps multiplied by the electric potential difference of 220 volts. When we do the math, we end up getting a power in units of watts, that is 3300 watts. Now in Part B of the problem, we're asked if we use this device for 8 hours a day, how much energy will be consumed? Now we need to use the definition of power as the rate at which energy is used. In other words, P is equal to energy used divided by the time. We can rearrange the equation so that we can solve for the amount of energy used or consumed, and it becomes energy consumed equal power times time. So we're going to take our answer to A, the 3300 watts, and multiply it by 8 hours. Then we're going to move the decimal place three places to the left in order to convert to kilowatts times hour. That ends up being about 26.400 kilowatt times hours. Finally in Part C, we wish to find out how much this would cost if we were to use the air conditioner at this 8 hours per day for a total of 31 days in an area where households are charged 13 cents per kilowatt hours. So we have the amount of kilowatt hours used in a single day. It's 26.4 kilowatt hours. So we need to multiply by 31 days in order to get the energy used in a month. It turns out to be about 818.4 kilowatt hours, and we're charged 13 cents or .13 dollars for every one of these kilowatt hours. So if I take the 818.400 kilowatt hours and multiply by .13 dollars per kilowatt hours, I'll get about 106 dollars and 39 cents. That's for an entire month. That could brown to two significant digits as 110 dollars per month.
Solution
- 3300 W
- 26 kW•hr
- $110 per month (rounded from $106)
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities and record them in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(\descriptive{\text{δV}}{δV,change in voltage} = 9.0\unit{\volt}\); \(\descriptive{R}{R,resistance} = 0.025\unit{\ohm}\); \(\descriptive{I}{I,current} = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read About It!
Get more information on the topic of Electric Circuits at The Physics Classroom Tutorial.