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Unformatted text preview: Fa09-McKee-MT2-sol Fall 2009 - McKee - Midterm 2 solutions Problem 1 The only meaningful way to compare the cost of the battery and the wall socket is cost per unit energy. In fact, cost per unit energy is the only information we are given about the expense of the wall socket. Cost per unit power doesnt make much sense, because power can be delivered over different lengths of time. The only way to compare cost per unit power is to specify that the power is delivered over a fixed length of time, but in that case, you are really comparing cost per unit energy anyway. Many students wanted to assume that the wall socket was delivering the same current as the battery. There was no reason to make this assumption. Students who used this approach found that the wall socket was delivering 3 Watts, but then to complete the problem correctly they needed to relate this power to the cost per kilowatt hour, which would mean relating the 3 Watts to a quantity of energy. Points were awarded according to the following rubric: At most 3 points were given for finding the power delivered by the battery. P = IV = (25 mA )(1 . 5 V ) = 37 . 5 mW = 0 . 0375 W At most 3 points were awarded for converting the power from the battery into the energy used over 820 hours: Energy = Power time = (0 . 0375 W )(820 hours ) = 30 . 75 W h = . 03075 kW h ( = 110700 J ) ( = 110 . 7 kJ ) The conversion to Joules or kilo Joules was not necessary, but many students chose to make this conver- sion. At most 2 points were awarded for finding the cost per unit energy for the battery, and then the final 2 points were awarded for finding the ratio of cost per unit energy. I did take off 2 points if students inverted the ratio they were asked to find (cost of battery to cost of wall socket ). $1 . 70 . 0375 kW h = $55 . 37 kW h $1 . 70 110700 kJ = $1 . 54 10- 5 J $55 . 37 kW h $0 . 10 kW h = $1 . 54 10- 5 J $2 . 78 10- 8 J = 550 times more expensive to use battery (2 sig figs) Problem 2 To simplify the problem, assume there is no charge on the wire. When the system reaches equilibrium,To simplify the problem, assume there is no charge on the wire....
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- Fall '08