EE40fa09_HW01sol

# EE40fa09_HW01sol - Fall 2009 EE40 Homework#1 Solutions...

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Unformatted text preview: Fall 2009 EE40 Homework #1 Solutions Problem 1 a) Electrons are attracted to the most positive potential, so: g G Â¡ g Â¢ Â£ g Â¢G Â¤ Â¥ Â¦ Â§ Â¥ Â¨ Â© Âª b) First, find the charge that the battery must supply: Â« Â¤ Â¬ Â­ g Â£ Â¬ Â¤ Â« g Â¤ Â®Âª C Convert this to the number of electrons N using the charge on a single electron e : Â¯ Â¤ Â¬ Â° Â¤ Â± Â­ Â²Âª Â³Âª c) The energy cost equals the price of the battery divided by the energy delivered (in kilowatt- hours). Â´ Â¤ Â¬ Â­ g Â¤ Âµ Â­ Â¶ Â­ g Â£ Â´ Â¤ 0.1 A Â­ 75 h Â­ 1.2 V Â¤ 9 WÂ­h Â£ EnerÂ·y cost Â¤ \$0.50 9 WÂ­h Â¤ \$Â±Â¸/kWÂ­h kWÂ­h kWÂ­h kWÂ­h Problem 2 a) First, find the power in the voltage source. Its voltage is given (10 V), and the current is fixed by the current source (2 A into the positive terminal), so we see that it is dissipating 20 W . The resistor has the same current, and we see from Ohmâ€™s law that it has a voltage of 10 V....
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EE40fa09_HW01sol - Fall 2009 EE40 Homework#1 Solutions...

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