ICE_Chapter_03 - CHAPTER 3 Solutions for Exercises E3.1 v...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
) 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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

ICE_Chapter_03 - CHAPTER 3 Solutions for Exercises E3.1 v...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online