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HW37solution

# HW37solution - t-36 87 ◦ V(b P inst t = i in t v L t =...

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ECE 201 Spring 2010 Homework 37 Solutions Problem 5 (a) For Figure (a), I 2 eff = 1 T T 0 i 2 ( t ) dt = 1 9 3 0 9 dt + 6 3 16 dt = 25 3 I eff = 5 / 3 For Figure (b), I 2 eff = 1 4 1 0 9 dt + 3 2 16 dt = 25 / 4 I eff = 2 . 5 (b) Current through R L is given by i ( t ) × 60 / (30 + 60) = (2 / 3) i ( t ). Power absorbed by R L is then given by the square of the e ff ective current through it times the resistance. Thus P = 4 9 × 25 3 × 30 = 1000 9 W 1

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(c) P = 4 9 × 25 4 × 30 = 250 3 W Problem 7 In all parts of the problem, the following identities will be used, 2 π /m 0 cos( mnx ) dx = 0 2 π /m 0 sin( mnx ) dx = 0 Here m and n are integers. (a) V 2 1 eff = 1 T T 0 v 2 1 ( t ) dt = 1 T T 0 [102 + 40 cos(20 t ) + 2 cos(40 t )] dt = 102 V 1 eff = 10 . 0995 (b) V 2 2 eff = 1 T T 0 v 2 2 ( t ) dt = 1 T T 0 [50 { 1 + cos(4 t ) } + 12 . 5 { 1 + cos(8 t ) } + 50 { cos(6 t ) + cos(2 t ) } ] dt = 62 . 5 V 2 eff = 7 . 9057 (c) V 2 3 eff = 1 T T 0 v 2 3 ( t ) dt = 1 T T 0 [50 { 1 + cos(4 t ) } + 36 . 4277 { 1 + cos(8 t ) } + 6 . 2499 { 1 - cos(8 t ) } + . . . ] dt = 92 . 6776 V 3 eff = 9 . 6269 2
Problem 9 (a) Z L = 1 5 + j (30 × 5 × 10 - 3 ) - 1 = 0 . 8(4 - j 3) V L = I in × Z L = 4 2 (4 - j 3) v L ( t ) = 20 cos(30 t - 36 .

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Unformatted text preview: t-36 . 87 ◦ ) V (b) P inst ( t ) = i in ( t ) v L ( t ) = 100 cos(30 t ) cos(30 t-36 . 87 ◦ ) W P average = V m I m 2 cos( θ v-θ i ) = 50 cos 36 . 87 ◦ = 40 W Problem 10 (a) I s = 50 ±-90 ◦ 6 + j 12-j 4 = 5 ±-143 . 13 ◦ ( magnitude = 5) (b) P average = V eff I eff cos( θ v-θ i ) = 5 × 50 cos(-53 . 13 ◦ ) = 150 W 3 (c) P average = R | I s | 2 = 6 × 25 = 150 W ( same as part ( b )) (d) I s = 50 ±-90 ◦ 30 + j 50-j 10 = 1 ±-143 . 13 ◦ ( magnitude = 1) P average = V eff I eff cos( θ v-θ i ) = 1 × 50 cos(-53 . 13 ◦ ) = 30 W 4...
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