hw1sol - 1(a 0.10 $/kWhr 1 hr 3600 s 1 kW 1000 J/s = 2.77...

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1. (a) 0 . 10 $ / kWhr × 1 hr 3600 s × 1 kW 1000 J / s = 2 . 77 × 10 - 8 $ / J (b) 3 . 20 $ / gal × 264 gal / m 3 × 1 m 3 720 kg × 1 kg 4 . 5 × 10 7 J = 2 . 61 × 10 - 8 $ / J (c) 0 . 1 A × 20 hr × 3600 s [ 1 ] hr = 7200C; 7200C × 1 . 5 V = 10800 J; $0 . 75 10800 J = 6 . 94 × 10 - 5 $ / J Gasoline and residential electric power have almost the same cost per joule; AA bat- teries are thousands of times more expensive than the other two sources. 2. Cross-sectional area of wire: 3 . 30 × 10 - 6 m 2 10A = 10 C / s Use one second as the “measurement interval” for easy calculations One second’s worth of electrons fit inside 7 . 35 × 10 - 10 m 3 of copper; given the cross- section of the wire, that’s 2 . 23 × 10 - 4 m long The drift velocity of the electrons is 2 . 23 × 10 - 4 m / s 25 C = 298.15K Mean thermal energy = 6 . 17 × 10 - 21 J Kinetic energy = 1 2 mv 2 , so v = q 2 E m Thermal velocity = 1 . 16 × 10 5 m / s Working backwards: 28 m / s drift velocity means all the electrons in 9.24m 3 of copper cross the boundary in 1 second; that’s 1 . 26 × 10 6 C 1.26MA to match a cheetah with drift velocity Working backwards: An electron moving at 28 m / s
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This note was uploaded on 10/31/2010 for the course EE 100 taught by Professor Boser during the Fall '07 term at Berkeley.

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hw1sol - 1(a 0.10 $/kWhr 1 hr 3600 s 1 kW 1000 J/s = 2.77...

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