HW37solution

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

This preview shows pages 1–4. Sign up to view the full content.

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 ² ± 3 0 9 dt + ± 6 3 16 dt ³ = 25 3 I eff =5 / 3 For Figure (b), I 2 eff = 1 4 ² ± 1 0 9 dt + ± 3 2 16 dt ³ = 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±ective current through it times the resistance. Thus P = 4 9 × 25 3 × 30 = 1000 9 W 1

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

View Full Document
(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 ) ² - 1 =0 . 8(4 - j 3) V L = I in × Z L = 4 2 (4 - j 3) v L ( t ) = 20 cos(30

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 01/22/2012 for the course IE 370 taught by Professor Chunghorng,r during the Spring '08 term at Purdue University.

### Page1 / 4

HW37solution - t-36 . 87 ◦ ) V (b) P inst ( t ) = i in (...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online