Chapter 03

# Chapter 03 - CHAPTER 3 Exercises E3.1 v(t = q(t C = 10 6...

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1 CHAPTER 3 Exercises E3.1 V ) 10 sin( 5 . 0 ) 10 2 /( ) 10 sin( 10 / ) ( ) ( 5 6 5 6 t C q v = × = = A ) 10 cos( 1 . 0 ) 10 cos( ) 10 5 . 0 )( 10 2 ( ) ( 5 5 5 6 dt dv i = × × = = E3.2 Because the capacitor voltage is zero at = 0, the charge on the capacitor is zero at = 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 × = + = = = + = dx x ms 4 ms 2 for 10 40 ms 2 0 for 10 / ) ( ) ( 4 4 = = = ms 4 ms 2 for 10 10 40 ms 2 0 for 10 ) ( ) ( ) ( 3 + × = = = p ms 4 ms 2 for ) 10 40 ( 10 5 . 0 ms 2 0 for 5 2 / ) ( ) ( 2 4 7 2 2 × = = = Cv 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 + + = 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 + + + + + = This can be written as ) 0 ( ) 0 ( ) 0 ( ) ( 1 1 1 ) ( 3 2 1 0 3 2 1 + + + + + = (1) Now if we define ) 0 ( ) 0 ( ) 0 ( ) 0 ( and 1 1 1 1 3 2 1 3 2 1 eq + + = + + = we can write Equation (1) as ) 0 ( ) ( 1 ) ( 0 eq + = Thus the three capacitances in series have an equivalent capacitance given by Equation 3.25 in the book. E3.4 (a) For series capacitances: F 3 / 2 1 / 1 2 / 1 1 / 1 / 1 1 2 1 eq µ = + = + = (b) For parallel capacitances: F 3 2 1 2 1 eq = + = + = E3.5 From Table 3.1 we find that the relative dielectric constant of polyester is 3.4. We solve Equation 3.26 for the area of each sheet: 2 12 6 6 0 m 4985 . 0 10 85 . 8 4 . 3 10 15 10 = × × × × = = = ε r Cd A Then the length of the strip is m 93 . 24 ) 10 2 /( 4985 . 0 / 2 = × = = W L E3.6 () [] () V 10 sin 10 10 cos 1 . 0 ) 10 10 ( ) ( ) ( 4 4 dt t v = × = = () [] () J 10 cos 10 50 10 cos 1 . 0 10 5 ) ( ) ( 4 2 6 2 4 3 2 t Li t w × = × ×
3 E3.7 s 2 0 for V 10 25 10 5 . 7 6667 ) ( 10 150 1 ) 0 ( ) ( 1 ) ( 2 9 0 6 0 6 0 µ × = × = × = + = t xdx dx x v i L s 4 s 2 for V 1 . 0 10 5 . 7 6667 -6 E 2 0 6 = × = () s 5 s 4 for V 10 5 . 0 15 10 5 . 7 6667 5 t -6 E 4 -6 E 2 0 6 = + × = A plot of ( ) versus is shown in Figure 3.19b in the book. E3.8

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Chapter 03 - CHAPTER 3 Exercises E3.1 v(t = q(t C = 10 6...

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