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60287-Chapter_03

# 60287-Chapter_03 - CHAPTER 3 Exercises E3.1 v(t = q(t C =...

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CHAPTER 3 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 72

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) 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 C C C + + = + + = we can write Equation (1) as ) 0 ( ) ( 1 ) ( 0 v dt t i C t v t 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 = + = + = C C C (b) For parallel capacitances: F 3 2 1 2 1 eq µ = + = + = C C C 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 ε ε r Cd Cd A Then the length of the strip is m 93 . 24 ) 10 2 /( 4985 . 0 / 2 = × = = W A L E3.6 ( ) [ ] ( ) V 10 sin 10 10 cos 1 . 0 ) 10 10 ( ) ( ) ( 4 4 3 t t dt d dt t di L t v = × = = ( ) [ ] ( ) J 10 cos 10 50 10 cos 1 . 0 10 5 ) ( ) ( 4 2 6 2 4 3 2 2 1 t t t Li t w × = × × = = 73
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 t xdx dx x v i dx x v L t i t t t s 4 s 2 for V 1 . 0 10 5 . 7 6667 -6 E 2 0 6 µ µ = × = t xdx ( ) s 5 s 4 for V 10 5 . 0 15 10 5 . 7 6667 5 t -6 E 4 -6 E 2 0 6 µ µ = + × = t t dx xdx A plot of i ( t ) versus t is shown in Figure 3.19b in the book. E3.8 Refer to Figure 3.20a in the book. Using KVL we can write: ) ( ) ( ) ( ) ( 3 2 1 t v t v t v t v + + = Using Equation 3.28 to substitute, this becomes dt t di L dt t di L dt t di L t v ) ( ) ( ) ( ) ( 3 2 1 + + = (1) Then if we define 3 2 1 eq L L L L + + = , Equation (1) becomes: dt t di L t v ) ( ) ( eq = which shows that the series combination of the three inductances has the same terminal equation as the equivalent inductance. E3.9 Refer to Figure 3.20b in the book. Using KCL we can write: ) ( ) ( ) ( ) ( 3 2 1 t i t i t i t i + + = Using Equation 3.32 to substitute, this becomes ) 0 ( ) ( 1 ) 0 ( ) ( 1 ) 0 ( ) ( 1 ) ( 3 0 3 2 0 2 1 0 1 i dt t v L i dt t v L i dt t v L t i t t t + + + + + = This can be written as ) 0 ( ) 0 ( ) 0 ( ) ( 1 1 1 ) ( 3 2 1 0 3 2 1 i i i dt t v L L L t v t + + + + + = (1) Now if we define ) 0 ( ) 0 ( ) 0 ( ) 0 ( and 1 1 1 1 3 2 1 3 2 1 eq i i i i L L L L + + = + + =

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60287-Chapter_03 - CHAPTER 3 Exercises E3.1 v(t = q(t C =...

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