Solution10

Solution10 - UNIVERSITY OF CALIFORNIA AT BERKELEY EECS...

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UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 1 of 10 EECS 40, Fall 2006 Prof. Chang-Hasnain Homework #10 Solutions Due at 6 pm in 240 Cory on Wednesday, 04/18/07 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, P10.46 (a) The integral of ) sin( t V m ! = over one cycle is zero, so the dc voltmeter reads zero. [2pts] (b) " m T t t m T T T m avg V t V T dt dt t V T V = # $ % ( ) = # $ % ( + = = = * * 2 / 0 2 / 2 / 0 ) cos( 1 0 ) sin( 1 [4pts] (c) m T T m T m avg V dt t V dt t V T V 2 ) sin( ) sin( 1 2 / 2 / 0 = # $ % ( ) + = * * [4pts] 2. Hambley, P10.48 As in Problem P10.47, the peak voltage must be 10 V. For a full-wave rectifier, the capacitance is given by Equation (10.12) in the text: F V T I C r L μ 417 ) 2 ( 2 ) 60 / 1 ( 1 . 0 2 = = = [4pts] The circuit diagram is: [6pts]
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UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 2 of 10 3. Hambley, P10.52 (a) The current pulse starts and ends at the times for which [5pts] 12 ) 200 sin( 20 ) ( = = t V t v B s ! Solving we find that ms 976 . 3 2 and ms 024 . 1 200 ) 6 . 0 ( sin 1 = ! = = = ! start end start t T t t " Between these two times the current is 80 12 ) 200 sin( 20 ) ( ! = t t i A sketch of the current to scale versus time is
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UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 3 of 10 (b) The charge following through the battery in one period is [5pts] end start end start end start t t t t t t t t dt t dt t i Q ! " # $ % = = = ( ( 80
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Solution10 - UNIVERSITY OF CALIFORNIA AT BERKELEY EECS...

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