Electric Circuits Legacy Problem #12 Guided Solution
Problem*
Having recently lost her job, Penny Penching is looking for every possible means of cutting costs. She decides that her 4.0-Watt clock radio alarm does not need to be on for 24 hours every day since she only needs it for waking up after her average 8-hour sleep. So, she decides to plug it in before going to sleep and to unplug it when waking. Penny pays 12 cents per kiloWatt•hour for her electricity. How much money is Penny able to save over the course of a month (31 days) with her new alarm clock usage pattern?
Audio Guided Solution
Electrical devices consume energy and that's what the consumer pays for. And the amount of energy that is used depends upon the time over which that device is used and the power rating of the device. Here, Penny Pinching is looking to save a few pennies and so she decides that she's not going to have her radio alarm clock on for the entire 24 hours every day. She's going to save 16 hours of energy usage by plugging it in before she goes to bed at night and then unplugging it when she wakes up. Thus saving 16 hours of usage of a 4 watt clock radio alarm. And so if we wish to calculate the amount of money that she saves by doing this, we need to begin by calculating the amount of energy used or energy she's not going to use. And so we're going to take the 16 hours of not using the alarm clock and multiply it by the 4.0 watts in order to get the amount of energy that that alarm clock is not being used. And so we're going to go 4 times 16 and that will give us energy in units of watt times hours. And then we're going to move the decimal place three places to the left to get it in units of kilowatt hours. For a single evening, that comes out to be 0.064 kilowatt times hours. And for a 31 day month, we can simply multiply this figure by 31 and we would find it would come out to be just short of 2 kilowatt hours. 1.98 kilowatt hours. Now Penny pays about 12 cents per every kilowatt hour of electrical energy that's used. So if we take this figure of 1.98 kilowatt hours and multiply it by 12 cents per kilowatt hours, we could get the money saved by this practice. And it comes out to be just a little short of 24 cents. We can round that to 24 cents. And that's the money saved for the hassle of having to unplug and replug and then perhaps even reset our alarm every evening for a month.
Solution
24 cents
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.
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