Solution2 - UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department EECS 40 Spring 2007 Prof Chang-Hasnain Homework#2 Solutions Due at 6 pm in 240

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UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 1 of 3 EECS 40, Spring 2007 Prof. Chang-Hasnain Homework #2 Solutions Due at 6 pm in 240 Cory on Wednesday, 1/31/06 Total Points: 100 Put (1) your name and (2) discussion section number on your homework. You need to put down all the derivation steps to obtain full credits of the problems. Numerical answers alone will at best receive low percentage partial credits. No late submission will be accepted expect those with prior approval from Prof. Chang-Hasnain. 1. Hambley, P2.6 (10pt) Combining the resistances shown in Figure P2.6b, we have eq eq eq eq R R R R + + = + + + = 1 2 1 1 1 1 1 eq eq eq eq R R R R + + = + ) 1 ( 2 ) 1 ( 0 2 ) ( 2 ) ( 2 = ! ! eq eq R R ! = 732 . 2 eq R ( ! " = 732 . 0 eq R is dropped) 2. Hambley, P2.11 (10pt) For operation at the lowest power, we have 2 1 2 120 300 R R P + = = At the high power setting, we have 2 2 1 2 120 120 1200 R R P + = = Solving these equations we find ! = = 24 2 1 R R The intermediate power setting is obtained by operating one of the elements from 120V
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This homework help was uploaded on 04/02/2008 for the course EE 40 taught by Professor Chang-hasnain during the Spring '07 term at University of California, Berkeley.

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Solution2 - UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department EECS 40 Spring 2007 Prof Chang-Hasnain Homework#2 Solutions Due at 6 pm in 240

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