# Solution1 - UNIVERSITY OF CALIFORNIA AT BERKELEY EECS...

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UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 1 of 4 EECS 40/42/100, Spring 2007 Prof. Chang-Hasnain Homework #1 Solution Due at 6 pm in 240 Cory on Wednesday, 1/24/07 Total Points: 100 1. Hambley, P1.8 [11 points] (a) The graph of i(t) should look something like this: (The axes labels are t(ms) and i(A)) The important points to note: - The graph is a sine wave. [1 point] - The amplitude is 10A, meaning the peaks are at 10A and -10A. [1 point] - One period ends when the argument to sin() is 2 π . This occurs when ! ! 2 200 = " t , or when (solving for t ) ms s t 10 01 . 0 = = [1 points] (b) C t dt t dt t i q dt dq i ms 0318 . 0 ) 200 cos( 200 10 ) 200 sin( 10 ) ( 005 . 0 0 005 . 0 0 5 0 = ! = ! = = " = # # \$ \$ \$ [4 points] (c) We can repeat the same calculation replacing 5ms with 10ms, or we can notice that integrating sin() over one full period gives us 0 coulombs. [4 points] 2. Hambley, P1.17 [10 points] t e t i t v t p ! = " = 20 ) ( ) ( ) ( W [4 points] 20 20 ) ( 0 0 = ! = = " ! " # t e dt t p w J [4 points] w > 0, so the element absorbs energy. [2 points] 3. Hambley, P1.22 [9 points] The power the cell can deliver is p = vi = 0.12W. [3 points] In 75 hours, the energy is W = pT = 9Whr = 0.009kWhr [3 points] So the cost per kWhr is \$0.50/0.009kWhr = \$55.56/kWhr.

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