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Chapter 03

# Chapter 03 - CHAPTER 3 Solutions for Exercises E3.1 v(t =...

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1 CHAPTER 3 Solutions for Exercises E3.1 V ) 10 sin( 5 . 0 ) 10 2 /( ) 10 sin( 10 / ) ( ) ( 5 6 5 6 t t C t q t v = × = = A ) 10 cos( 1 . 0 ) 10 cos( ) 10 5 . 0 )( 10 2 ( ) ( 5 5 5 6 t t dt dv C t i = × × = = E3.2 Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = 0. ms 4 ms 2 for 10 10 4 10 10 ms 2 0 for 10 10 0 ) ( ) ( 3 -6 3 2E 3 3 2E 0 3 3 0 3 0 × = + = = = + = t t dx dx t t dx dx x i t q t t t ms 4 ms 2 for 10 40 ms 2 0 for 10 / ) ( ) ( 4 4 = = = t t t t C t q t v ms 4 ms 2 for 10 10 40 ms 2 0 for 10 ) ( ) ( ) ( 3 + × = = = t t t t t v t i t p ms 4 ms 2 for ) 10 40 ( 10 5 . 0 ms 2 0 for 5 2 / ) ( ) ( 2 4 7 2 2 × = = = t t t t t Cv t w in which the units of charge, electrical potential, power, and energy are coulombs, volts, watts and joules, respectively. Plots of these quantities are shown in Figure 3.8 in the book. E3.3 Refer to Figure 3.10 in the book. Applying KVL, we have 3 2 1 v v v v + + = Then using Equation 3.8 to substitute for the voltages we have

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2 ) 0 ( ) ( 1 ) 0 ( ) ( 1 ) 0 ( ) ( 1 ) ( 3 0 3 2 0 2 1 0 1 v dt t i C v dt t i C v dt t i C t v t t t + + + + + = This can be written as ) 0 ( ) 0 ( ) 0 ( ) ( 1 1 1 ) ( 3 2 1 0 3 2 1 v v v dt t i C C C t v t + + + + + = (1) Now if we define ) 0 ( ) 0 ( ) 0 ( ) 0 ( and 1 1 1 1 3 2 1 3 2 1 eq v v v v C
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