Electric Circuits Legacy Problem #15 Guided Solution

Sal Erbate celebrates the Christmas season by hanging 100 mini bulbs, light strings, out on his, outside of his home as well as his inside of his home, and he has an equivalent of 45 such strings. In part A of this multi-part problem, we're asked to determine the resistance of a single string of lights. Each one is powered by a 110-volt outlet. So to do so, what we need to do is find an equation that relates the given information – power, time, volts – to the unknown information – resistance. I can find a collection of equations on the overview page for this problem set if I use the link found below on this web page. One of the equations that strikes me as being useful is the equation that says power is equal to delta V squared over R, where R is the resistance and delta V is the electric potential difference impressed across the device. So I'm going to use that equation and rearrange it in order to solve for resistance and the equation becomes R equal delta V squared over P, where the delta V is 110 volts and the P is 40 watts. Plugging and chugging into the equation, I end up getting 302.5 and the unit would be ohms. I can round that to 3 times 10 to the second ohms. Now in part B of the problem, I'm asked to determine the amount of energy that is consumed by the lights over the course of 40 days. So I'm going to break my calculations down into several parts, first calculating the energy consumed in a single day by one strand of light, then multiplying by 45, and then multiplying by 40 days to get all 45 strings for 40 days. So I need the equation that relates power and time. That equation is the equation that says power is equal to the rate of energy consumption or energy consumed divided by time. I can rearrange the equation to become energy consumed equal power times time. So for just a single string of lights, 40 watts, I can multiply 40 watts times the seven hours that is being used in a single day. I would get 280 watt-hours. If I move the decimal place three places and left, it becomes .280 kilowatt-hours. That's for one string for seven hours. Now if I multiply by 45 strings, I'll get 12.6 kilowatt-hours. That takes care of all 45 strings for seven hours. That's a single day. Now if I multiply by 40 days, I'll get the energy consumed over the 40 days of the Christmas season. That comes out to be 504 kilowatt-hours. Now in part C in this problem, I wish to calculate the cost of this declaration. And so in doing so, I use the 12 cents per kilowatt-hour rate of charging that the utility company charges the urbane household. So I take the 12 cents or .12 dollars and I multiply it by the 504 kilowatt-hours and I would get 60 dollars and 48 cents, which I round to 60 dollars.