20095ee10_1_HW1_Edition8

20095ee10_1_HW1_Edition8 - 4) 2.21 a) i 1 = 2 A, b) P ( 5 )...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
UCLA E LECTRICAL  E NGINEERING  D EPARTMENT :  EE 10: C IRCUIT  A NALYSIS  1   HOMEWORK 1: Problems from Textbook: with Selected Numerical Answers 1) 1.21  a)  P MAX  (delivered) = 450 W , b) P MAX  (extracted) = 450 W , c) P AVG  ( 0  to  5ms) =   0 , d) P AVG  (0 to  6.25ms) =  57.3 W 2) 1.24  3) 1.26 Note: compute total power of all sources that deliver power (or equivalently  show a negative absorbed power).  Note further that the sum of power delivered must  equal sum of power absorbed
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4) 2.21 a) i 1 = 2 A, b) P ( 5 ) = 320 W, P(25 ) = 400 W, P(70 ) = 280 W, P(10 ) = 360 W, P(8 ) = 800 W, c) P(180V) = 2160 W 5) 2.26 a) P(130) = -130 * 15 = -1950 W, P(460) =-460 * 30 = -13800 W, b) P(2 ) = 450 W P(10 ) = 2250 W P(2 ) = 1800 W P(25 ) = 2500 W P(10 ) = 6250 W P(100 ) = 2500W 1...
View Full Document

This note was uploaded on 10/15/2009 for the course EE 10 taught by Professor Chang during the Spring '07 term at UCLA.

Ask a homework question - tutors are online