Soln7 - I + V ab 20 I 2 I 1 I I I I j 12 2 j 8 5 j 60 10 j...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
EE411 Spring 2008 HW#7 Solution (Q2) 9.36 Let Z be the input impedance at the source. 20 10 100 200 mH 100 3 j x x j L j = = ⎯→ ω 500 200 10 10 1 1 F 10 6 j x x j C j = = ⎯→ μ 1000//-j500 = 200 –j400 1000//(j20 + 200 –j400) = 242.62 –j239.84 o j Z 104 . 6 2255 84 . 239 62 . 2242 = = mA 896 . 3 61 . 26 104 . 6 2255 10 60 o o o I = = ) 896 . 3 200 cos( 1 . 266 o t i = mA (Q6) 9.66 ) j 12 ( 145 170 5 j 60 ) 10 j 40 )( 5 j 20 ( ) 10 j 40 ( || ) 5 j 20 ( T = + + = + = Z = T Z 14.069 – j1.172 Ω = 14.118 -4.76 ° ° = ° ° = = 76 . 94 25 . 4 76 . 4 - 118 . 14 90 60 T Z V
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: I + V ab 20 I 2 I 1 I I I I j 12 2 j 8 5 j 60 10 j 40 1 + + = + + = I I I j 12 j 4 5 j 60 5 j 20 2 + = + = 2 1 ab 10 j 20-I I V + = I I V j 12 40 j 10 j 12 ) 40 j (160-ab + + + + + = I I V 145 j)(150)-12 ( j 12 150-ab + = + = ) 76 . 97 25 . 4 )( 24 . 175 457 . 12 ( ab = V = ab V 52.94 273 V...
View Full Document

This note was uploaded on 10/26/2009 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas at Austin.

Page1 / 2

Soln7 - I + V ab 20 I 2 I 1 I I I I j 12 2 j 8 5 j 60 10 j...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online